NCERT Solutions For Class 8 Maths Chapter 10 Exponents And Powers

Exponents And Powers

Exponents And Powers Introduction

If a is any non-zero integer and n is a positive integer, then

x a (n times) is written as a”,

i.e., a” is the continued product of a multiplied by itself n times.

Here, ‘o’ is called the base, and V is called the ‘exponent’ or ‘index’.

The number a” is read as a raised to the power of or simply as ‘nth power of o’.

The notation a” is called the exponential or power notation.

We can write large numbers more conveniently using exponents.

For example :

10000 = 104; 243 = 35; 128 = 27, etc.

Now, we shall learn about negative exponents

Powers With Negative Exponents

If a is any non-zero integer and m is a positive integer, then

⇒ \(a^{-m}=\frac{1}{a^m}\)

Note: a m is called the multiplicative inverse of am as a-m x am = 1

Am and a ~m are multiplicative inverses of each other.

Note 2: \(a^m=\frac{1}{a^{-m}}\)

Question: What is 10 ~10 equal to?

Solution: \(10^{-10}=\frac{1}{10^{10}}\)

Question 1. Find the multiplicative inverse of the following:

  1. 2-4
  2. 10-5
  3. 7-2
  4. 5-3
  5. 10-100

Solution:

The multiplicative inverse is \(2^{-4}=\left(\frac{1}{2^{-4}}\right) \text { is } 2^4\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

The multiplicative inverse of \(10^{-5}=\left(\frac{1}{10^{-5}}\right) \text { is } 10^5\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

The multiplicative inverse of \(7^{-2}=\left(\frac{1}{7^{-2}}\right) \text { is } 7^2\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

The multiplicative inverse \(5^{-3}=\left(\frac{1}{5^{-3}}\right) \text { is } 5^3\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

The multiplicative inverse\(10^{-100}=\left(\frac{1}{10^{-100}}\right) \text { is } 10^{100}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

Question 2. Expand the following numbers using exponents :

  1. 1025.63
  2. 1256.249

Solution:

1025.63

1 x 1000 + 0 x 100 + 2 x 10 + 5 x 1 + 6 x\(\frac{1}{10}+3 \times \frac{1}{100}\)

= 1 x 103 + 0 x 102 + 2 x 101 + 5 x 10° + 6 x 10-1 + 3 x 10-2

1256.249

= 1 x 1000 + 2 x 100 + 5 x 10

+ 6x 1+ 2 x \( \frac{1}{10}+4 \times \frac{1}{100}+9 \times \frac{1}{1000}\)

= 1 X 103 + 2 X 102 + 5 X 101 + 6 x 10° + 2 x 10-1 + 4 x 10-2 + 9 x lO-3

Laws Of Exponents

If a, b are non-zero integers and m, n are any integers, then

  1. am x an = am+ n
  2. \(\frac{a^m}{a^n}=a^{m-n}\)
  3. (am)n = amn
  4. am x bm = (ab)m
  5. \(\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^m\)
  6. a° = 1
  7. \(\left(\frac{a^{-m}}{b^{-n}}\right)=\frac{b^n}{a^m}\)
  8. \(\left(\frac{a}{b}\right)^{-m}=\left(\frac{b}{a}\right)^m\)

Remember

an” = 1 = n = 0

1n = 1 where n is any integer.

(- 1)n = 1 where n is any even integer.

(-1)n =-1 where n is any odd integer.

Q. Simplify and write in exponential form:

(-2)-3X (-2)-4

p3 x p -I0

32 x 3-5 x 36

Solution:

(-2)~3 x (-2)-4 = (- 2)(“3) + (-4>

am x an = am +n

= (-2)-7 = {(-1) x 2} -7

⇒ \(\frac{1}{\{(-1) \times 2\}^7}= \frac{1}{(-1)^7 \times(2)^7}\)

(ab)m = am bm

⇒ \(\frac{1}{(-1) \times 2^7}=-\frac{1}{2^7}\)

⇒ \(\mid(-1)^{\text {odd integer }}=-1\)

⇒ \(p^3 \times p^{-10}=p^{3+(-10)}=p^{-7}=\frac{1}{p^7}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

32 x 3-5 x 36 = 32 + (-5) + 6 = 33

Exponents And Powers Exercise 10.1

Question 1. Evaluate:

  1. 3-2
  2. (-4)-2
  3. \(\left(\frac{1}{2}\right)^{-5}\)

Solution:

⇒ \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\left(\frac{1}{2}\right)^{-5}=\left(\frac{2}{1}\right)^5\)

⇒ \( a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{2^5}{1^5}\)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \(\frac{32}{1}=32\)

1n = 1 where n is an integer

Question 2. Simplify and express the result in power notation with a positive exponent

  1. \((-4)^5 \div(-4)^8\)
  2. \(\left(\frac{1}{2^3}\right)^2\)
  3. \((-3)^4 \times\left(\frac{5}{3}\right)^4\)
  4. (3-7 – 3-10) x 3-5
  5. 2-3 x (-7) -3

Solution:

1. (-4)6 ÷ (-4)8

⇒ \(\frac{(-4)^5}{(-4)^8}\)

⇒ \((-4)^{5-8}\)

⇒ \((-4)^{-3}\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(=\frac{1}{(-4)^3}  a^{-m}=\frac{1}{a^m}\)

which is the required form.

2. \(\left(\frac{1}{2^3}\right)^2 =\frac{1^2}{\left(2^3\right)^2}\)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \(\frac{1}{2^{3 \times 2}} \)

⇒ \( \mid\left(a^m\right)^n=a^m\)

⇒ \(\frac{1}{2^6}\)

which is the required form

3. \((-3)^4 \times\left(\frac{5}{3}\right)^4=\{(-1) \times 3\}^4 \times \frac{5^4}{3^4} \)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \((-1)^4(3)^4 \times \frac{5^4}{3^4}\)

I (ab)m = am bm

= (- 1)4 x 54

= 1 x 54

(-1) even integer = 1

= 54

which is the required form.

4. (3-7 / 3-10) x 3-5

⇒ \(=\frac{3^{-7}}{3^{-10}} \times \frac{1}{3^5}\)

⇒ \( a^{-m}=\frac{1}{a^m}\)

⇒ \(3^{-7-(-10)} \times \frac{1}{3^5}\)

⇒ \(\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(3^3 \times \frac{1}{3^5} \)

⇒ \(\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(3^{3-5} \)

⇒ \(3^{-2} \)

⇒ \(a^{-m}=\frac{1}{a^m}\)

which is the required form.

5. \(2^{-3} \times(-7)^{-3}\)

⇒ \(\frac{1}{2^3} \times \frac{1}{(-7)^3}\)

⇒ \(\frac{1}{[2 \times(-7)]^3}=\frac{1}{a^m}\)

(ab)m = amb

⇒ \(\frac{1}{(-14)^3}\)

which is the required form.

Question 3. Find the value of:

(3° + 4 -1) x 22

(2-1 x 4-1)/2 -2

⇒ \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)

⇒ \(\left(3^{-1}+4^{-1}+5^{-1}\right)^0\)

⇒ \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2.\)

Solution:

1. (3° + 4-1) x 22

⇒ \(\left(1+\frac{1}{4}\right) \times 4\)

⇒ \(a^{-m}=\frac{1}{a^m}, \quad a^0=1\)

⇒ \(\frac{5}{4} \times 4=5\)

(2-1 x 4-1) + 2″2

= {2 -l x (22)-1} + 2 -2

= {2-1 x 22x(-1>} -f2-2

(am)n = amn

= (2-’X2-2)T2-2 = 2(-1)+(-2) 2-2

am x a” = am+”

= 2-3 4- 2″2

⇒ \(\frac{2^{-3}}{2^{-2}}=2^{-3-(-2)} \)

⇒ \(\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(2^{-1}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{1}{2}\)

3.\(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)

⇒ \(\frac{1^{-2}}{2^{-2}}+\frac{1^{-2}}{3^{-2}}+\frac{1^{-2}}{4^{-2}}\)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \(\frac{2^2}{1^2}+\frac{3^2}{1^2}+\frac{4^2}{1^2}\)

⇒ \(\frac{4}{1}+\frac{9}{1}+\frac{16}{1}\)

= 4 + 9 + 16 = 29

4. [3-1 + 4-1 + 5-1]0

⇒ \({\left[\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right]^0 }\)

⇒ \(\left(\frac{20+15+12}{60}\right)^0\)

⇒ \(\quad \text { | LCM }(3,4,5)=60\)

⇒ \(\left(\frac{47}{60}\right)^0=1\)

\(a^0=1\)

Aliter

(3-1 + 4-1 +5-1)° = 1

a°= 1

5. \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2\)

⇒ \(\left(\frac{-2}{3}\right)^{(-2) \times 2}\)

⇒ \(\mid\left(a^m\right)^n=a^{m n}\)

⇒ \(\left(\frac{-2}{3}\right)^{-4}\)

⇒ \(\left(\frac{3}{-2}\right)^4 \)

⇒ \( \left(\frac{a}{b}\right)^{-m}=\left(\frac{b}{a}\right)^m\)

⇒ \(\frac{3^4}{(-2)^4}=\frac{3^4}{(-1 \times 2)^4}\)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \(\frac{3^4}{(-1)^4(2)^4}\)

⇒ \((a b)^m=a^m b^m\)

⇒ \(\frac{3^4}{1 \times 2^4} \)

⇒ \((-1)^{\text {even integer }}=1\)

⇒ \(\frac{3^4}{2^4}\)

⇒ \(\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}=\frac{81}{16}\)

Question 4. Evaluate:

  1.  \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
  2.  \(\left(5^{-1} \times 2^{-I}\right) \times 6^{-1}\)

Solution:

1. \(\frac{8^{-1} \times 5^3}{2^{-4}}\)

⇒ \(\frac{2^4 \times 5^3}{8^1}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{16 \times 125}{8}=250\)

2. \(\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}=\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}\)

Question 5. Find the value of m for which 5m + 5-3 = 55
Solution:

5m + 5-3 = 55

⇒ \(\frac{5^m}{5^{-3}} =5^5\)

⇒ \(5^{m-(-3)} =5^5\)

⇒ \(\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(5^{m+3} =5^5\)

bases are equal exponents are equal

m + 3 = 5

m = 5-3

m = 2

Question 6. Evaluate:

  1. \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
  2. \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}.\)

Solution:

⇒ \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)

⇒ \(\left(\frac{1^{-1}}{3^{-1}}-\frac{1^{-1}}{4^{-1}}\right)^{-1}\)

⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

⇒ \(\left\{\frac{\frac{1}{1^1}}{\frac{1}{3^1}}-\frac{\frac{1}{1^1}}{\frac{1}{4^1}}\right\}^{-1}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\left(\frac{3^1}{1^1}-\frac{4^1}{1^1}\right)^{-1}\)

⇒ \(a^1=a\)

⇒ \(\left(\frac{3}{1}-\frac{4}{1}\right)^{-1}\)

⇒ \( (3-4)^{-1}\)

⇒ \((-1)^{-1}=\frac{1}{(-1)^1}\)

⇒ \(| a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{1}{(-1)}\)

\((-1)^{\text {odd integer }}=-1\)= -1

⇒ \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}=\frac{5^{-7}}{8^{-7}} \times \frac{8^{-4}}{5^{-4}}\)

⇒ \(\frac{5^{-7}}{5^{-4}} \times \frac{8^{-4}}{8^{-7}}\)

⇒ \(=5^{(-7)-(-4)} \times 8^m=\frac{a^m}{b^m}\)

⇒ \(\frac{a^m}{a^n}=a^{m-n}\)

⇒ \(5^{-7+4} \times 8^{-4}+7 \)

⇒ \(\frac{1}{5^3} \times 8^3=\frac{8^3}{5^3}\)

⇒ \(\frac{8 \times 8 \times 8}{5 \times 5 \times 5}=\frac{512}{125}\)

Question 7. Simplify:

  1. \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \quad(t \neq 0)\)
  2. \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)

Solution:

1. \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\)

⇒ \(\frac{25 \times 5^3}{10} \times \frac{t^8}{t^4}\)

⇒ \(\frac{625}{2} \times \frac{t^8}{t^4}\)

⇒ \(\frac{625 t^4}{2}\)

⇒ \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)

⇒ \(\frac{3^{-5} \times(2 \times 5)^{-5} \times(5 \times 5 \times 5)}{5^{-7} \times(2 \times 3)^{-5}}\)

⇒ \(\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}} \)

⇒ \(\quad(a b)^m=a^m b^m\)

⇒ \(\frac{5^{-5} \times 5^3}{5^{-7}}=\frac{5^{(-5)+3}}{5^{-7}}\)

⇒ \(\mid a^m \times a^n=a^{m+n}\)

⇒ \(\frac{5^{-2}}{5^{-7}}=5^{(-2)-(-7)}\)

⇒ \(5^{-2+7}=5^5\)

Use Of Exponents To Express Small Numbers In Standard Form

A number is said to be in standard form if expressed in the form K x 10″ where 1 < K < 10 and n is an integer. A number written in standard form is said to be expressed in scientific notation.

Tiny numbers can be expressed in standard form using negative exponents.

1. To express a large number, we move the decimal point to the left such that only one digit is left to the left side of the decimal point and multiply the resulting number by 10n where n is the number of places to which the decimal point has been moved to the left.

For example : 270,000,000,000 = 2.7 x 1011

(Decimalpointhas have been moved to the left for11 places)

2. To express a number (< 1), we move the decimal point to the right such that only one digit is left to the left side of the decimal point and multiply the resulting number by 10~’1 where n is the number of places to which the decimal point has been shifted to the right.

For example : 0.000 0009 = 9 x 10-7

(Decimalpointhas have been shifted to the right for 7 places.)

Question 1. Identify huge and very small numbers from the above facts and write them in the following table:

NCERT Solutions For Class 8 Maths Chapter 10 Large And very Small

Solution:

NCERT Solutions For Class 8 Maths Chapter 10 Very Large Numbers

Question 2. Write the following numbers in standard form:

  1. 0.000000564
  2. 0.0000021
  3. 21600000
  4. 15240000

Solution:

1.  0.000000564

= 5.64 x 10-7

Moving decimal 7 places to the right

2.  0.0000021

0.0000021 = 2.1 x 10-6

Moving decimal 6 places to the right

3.  21600000

21600000 = 2.16 x 107

Moving decimal 7 places to the left

4.  15240000

15240000 = 1.524 x 107

Moving decimal 7 places to the left

Question 2. Write all the facts given in the standard form.
Solution:

(1) The distance from the Earth to the Sun is 1.496 x 1011 m

149, 600,000,000 = 1.496 x 1011.

Moving the decimal 11 places to the left

(2) The speed of light is 3 x 108 m/sec.

300, 000, 000 = 3 x108.

Moving decimal 8 places to the left

The thickness of the Class VII Mathematics book is 2 x 101 mm.

20 = 2 x 10 = 2 x 101

The average diameter of a Red Blood Cell is 7 x 10 6 mm.

0.00 0007 = 7 x 10-6

Moving decimal 6 places to the right

The thickness of human hair is in the range of 5 x 10 -3 cm to 1 x I0″2cm

⇒ \(0.005=\frac{5}{1000}=\frac{5}{10^3}=5 \times 10^{-3}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(0.01=\frac{1}{100}=\frac{1}{10^2}=1 \times 10^{-2}\)

The distance of the moon from the Earth is 3.84467 x 108m

384,467,000 = 3.84467 x 108

I Moved the decimal 8 places to the left

(7) The size of a plant cell is 1.275 x 10-5m

0.00001275 = 1.275 x 105

Moving decimal 5 places to the right

The average radius of the Sun is 6.95 x105km

695000 = 695 x 1000 = 695 x 103 = 6.95 x 105

(9) Mass of fuel in a space shuttle

solid rocket booster is 5.036 x 105 kg

503600

= 5036 x 100 = 5036 x 102

= 5.036 x 103 x 102

= 5.036 x 103+2

am x an = am+n

= 5.036 x 105

(10) Thickness of a piece of paper is 1.6 x10-3 cm

0.0016 = 1.6 x 10-3

Moving decimal 3 places to the right

The diameter of wire on a computer chip is 3 x 10-6 m

is 3 x 10-6

0.000003 = 3 x 013

Moving decimal 6 places to the right

(12) The height of Mount Everest is 8.848 x 103  m.

8848 = 8.848 x 1000 = 8.848 x 103

Exponents And Powers Exercise 10.2

Question 1. Express the following numbers in standard form:

  1. 0.0000000000085
  2. 0.00000000000942
  3. 6020000000000000
  4. 0.00000000837
  5. 31860000000.

Solution:

1. 0.0000000000085

0.0000000000085 = 8.5 x 10-12

Moving the decimal 12 places to the right

2. 0.00000000000942

0.00000000000942 = 9.42 x 10-15

Moving the decimal 12 places to the right

3. 6020000000000000

6020000000000000 = 6.02 x 1015

Moving the decimal 15 places to the left

4. 0.00000000837

0.00000000837 = 8.37 x 10-9

Moving decimal 9 places to the right

5. 31860000000

31860000000 = 3.186 x 1010

Moving decimal 10 places to the left

Question 2. Express the following numbers in the usual form:

  1. 3.02 x 10-6
  2. 4.5 x 104
  3. 3 x 10-8
  4. 1.0001 x 109
  5. 5.8 x 1012
  6. 3.61492 x 106

Solution:

⇒ \(3 \times 10^{-8}=\frac{3}{10^8}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(=\frac{3.02}{1000000}\)

⇒ 0.00000302

2. 4.5 x 104 = 4.5 x 10000 = 45000

3.  \(3.02 \times 10^{-6}=\frac{3.02}{10^6}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(\frac{3}{100000000}\)

=0.00000003

4. 1.0001 X 109

= 1.0001 X 1000,000,000

= 1,000,100,000

5. 5.8 X 1012

= 5.8 X 1,000,000,000,000

= 5,800,000,000,000

6. 3.61492 x 106

= 3.61492 x 1,000,000

= 3,614,920

Question 3. Express the number appearing in the following statements in standard form:

  1. 1 micron is equal to \(\frac{1}{1000000} m\)
  2. The charge of an electron is 0. 000, 000,000,000,000,000,1 6 coulomb.
  3. The size of bacteria is 0. 0000005 m
  4. The size of a plant cell is 0. 00001 275 m
  5. The thickness of thick paper is 0.07 mm.

Solution:

1.  \(\frac{1}{1000000} \mathrm{~m}= \frac{1}{10^6}\)

⇒ \(1 \times 10^{-6} \mathrm{~m}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

which is the required standard form.

2.  0.000,000,000,000,000,000,16 coulomb

⇒ \(\frac{16}{100,000,000,000,000,000,000} \text { coulomb }\)

⇒ \(frac{16}{10^{20}} \text { coulomb }\)

⇒ \(\frac{1.6 \times 10}{10^{20}} \text { coulomb }\)

⇒ \(\frac{1.6 \times 10^1}{10^{20}} \text { coulomb } \quad \mid a^1=a\)

⇒ \(1.6 \times 10^{1-20} \text { coulomb } \frac{a^m}{a^n}=a^{m-n}\)

⇒ \(1.6 \times 10^{-19} \text { coulomb }\)

which is the required standard form.

3. 0.0000005 m

⇒ \(\frac{5}{10000000} \mathrm{~m} \)

⇒ \(\frac{5}{10^7} \mathrm{~m}\)

⇒ \(5 \times 10^{-7} \mathrm{~m}\)

which is the required standard form.

4.  \(0.00001275 \mathrm{~m} =\frac{1275}{100,000,000} \mathrm{~m}\)

⇒ \(\frac{1275}{10^8} \mathrm{~m}\)

⇒ \(\frac{1275}{10^3 \times 10^5} \mathrm{~m}\)

| am x a” = am + n

⇒ \(\frac{1275}{10^3} \times 10^{-5} \mathrm{~m}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

⇒ \(1.275 \times 10^{-5} \mathrm{~m}\)

which is the required standard form.

5. \( 0.07 \mathrm{~mm} =\frac{7}{100}\)

⇒ \(\frac{7}{10^2}=7 \times 10^{-2} \mathrm{~mm}\)

⇒ \(a^{-m}=\frac{1}{a^m}\)

which is the required standard form

Question 4. In a stack, there are 5 books each of thickness 20 mm, and 5 paper sheets each of thickness 0. 016mm. What is the total thickness of the stack?
Solution:

Total thickness of books = 5 x 20 mm = 100 mm

Total thickness of paper sheets = 5 x 0.016 mm = 0.080 mm

Total thickness of the stack

= Total thickness of books + Total thickness of paper sheets

= 100 mm + 0.080 mm

= (100 + 0.080) mm

= 100.080 mm

= 1.0008 x 102 mm.

Moving decimal 2 places to the left

Hence, the total thickness of the stack is 1.0008 x IQ2 mm.

Exponents And Powers Multiple-Choice Question And Solutions

Question 1. am x an is equal to

  1. am+n
  2. am-n
  3. amn
  4. an-m

Solution: 1. am+n

Question 2. am ÷ am is equal to

  1. am+n
  2. am-n
  3. amn
  4. an-m

Solution: 1. am+n

Question 3. (am)is equal to

  1. am+n
  2. am-n
  3. amn
  4. an-m

Solution: 3. amn

Question 4. Am x is equal to

  1. (ab)m
  2. (ab)-m
  3. amb
  4. abm

Solution: 1. (ab)m

Question 5. a0 is equal to

  1. 0
  2. 1
  3. -1
  4. a

Solution: 2. 1

Question 6. \(\frac{a^m}{b^m}\) is equal to

  1. \(\left(\frac{a}{b}\right)^m\)
  2. \(\left(\frac{b}{a}\right)^m\)
  3. \(\frac{a^m}{b}\)
  4. \(\frac{a}{b^m}\)

Solution: 1. \(\left(\frac{a}{b}\right)^m\)

Question 7. 2 x 2 x 2 x 2 x 2 is equal to

  1. 24
  2. 23
  3. 22
  4. 25

Solution: 4. 25

Question 8. In 102, the exponents

  1. l
  2. 2
  3. 10
  4. 1

Solution: 2. 2

Question 9. In 102, the base is

  1. 1
  2. 0
  3. 10
  4. 100

Solution: 3. \(\frac{1}{10}\)

Question 10. 10-1 is equal to

  1. 10
  2. -1
  3. \(\frac{1}{10}\)
  4. \(-\frac{1}{10} \text {. }\)

Solution: 3. \(\frac{1}{10}\)

Question 11. The multiplicative inverse of 2-3 is

  1. 2
  2. 3
  3. -3
  4. 23

Solution: 4. 23

Question 12. The multiplicative inverse of 105 is

  1. 5
  2. 10
  3. 10-5
  4. 105

Solution: 3. 10-5

Question 13. The multiplicative inverse of \( \frac{1}{2^2}\)

  1. 2-2
  2. 22
  3. 2
  4. 1

Solution: 2. 22

Question 14. The multiplicative inverse of 10 “10 is

  1. 10
  2. \(\frac{1}{10}\)
  3. 10-10
  4. 1010

Solution: 4. 1010

Question 15. The multiplicative inverse of am is

  1. a
  2. m
  3. am
  4. a-m

Solution: 5. a-m

Question 16. 53 x 5-1 is equal to

  1. 5
  2. 53
  3. 5-1
  4. 52

Solution: 4. 52

Question 17. (-2)5 x (- 2)6 is equal to

  1. 2
  2. -2
  3. -5
  4. 6

Solution: 2. -2

Question 18. 32 x 3-4 x 35 is equal to

  1. 3
  2. 32
  3. 33
  4. 35

Solution: 3. 33

19. (- 2) 2 is equal to

  1. \(\frac{1}{4}\)
  2. \(\frac{1}{2}\)
  3. \(-\frac{1}{2}\)
  4. \(-\frac{1}{4}\)

Solution: 1. \(\frac{1}{4}\)

Question 20. \(\left(\frac{1}{2}\right)^{-4}\) is equal to

  1. 2
  2. 24
  3. 1
  4. 2-4

Solution: 2. 24

Question 21. (20 + 4-1) x 22 is equal to

  1. 2
  2. 3
  3. 4
  4. 5

Solution: 4. 5

Question 22. (2-1 + 3-1 + 5-1)0 is equal to

  1. 2
  2. 3
  3. 5
  4. 1

Solution: 4. 1

Question 23. 3m÷ 3-3 = 35 ⇒ m is equal to

  1. 1
  2. 2
  3. 3
  4. 4

Solution: 2. 2

Question 24. (-2)m+1 x (-2)4 = (- 2)6 ⇒ m =

  1. 0
  2. 1
  3. -1
  4. none of these

Solution: 2. 1

Question 25. (-1)60 is equal to

  1. -1
  2. 1
  3. 50
  4. -50

Solution: 2. 1

Question 26. (-1)51 is equal to

  1. -1
  2. 1
  3. 51
  4. -51

Solution: 1. -1

Question 27. 149600000000 is equal to

  1. 1.496 x 1011
  2. 1.496 x lO10
  3. 1.496 x 1012
  4. 1.496 x 105

Solution: 1. 1.496 x 1011

Question 28. 300000000 is equal to

  1. 3 x 108
  2. 3 x 107
  3. 3 x 106
  4. 3 x 109

Solution: 1. 3 x 108

Question 29. 0.000007 is equal to

  1. 7 x 10-6
  2. 7 x 10-6
  3. 7 x 10-4
  4. 7 x 10-3

Solution: 1. 7 x 10-6

Question 30. 384467000 is equal to

  1. 3.84467 x 1o8
  2. 3.84467 x 103
  3. 3.84467 x 107
  4. 3.84467 x 106

Solution: 1. 3.84467 x 108

Question 31. 0.00001275 is equal to

  1. 1.275 x 10-6
  2. 1.275 x 10-3
  3. 1.275 x 104
  4. 1.275 x 103

Solution: 1. 1.275 x 10-6

Question 32. 695000 is equal to

  1. 6.95 x 105
  2. 6.95 x 103
  3. 6.95 x 106
  4. 6.95 x 104

Solution: 1. 6.95 x 105

Question 33. 503600 is equal to

  1. 5.036 x 105
  2. 5.036 x 106
  3. 5.036 x 104
  4. 5.036 x 107

Solution: 1. 5.036 x 105

Question 34. 0.0016is equal to

  1. 1.6 x 10 -3
  2. 1.6 x 10-2
  3. 1.6 x 10 -4
  4. 1.6 x lO-5

Solution: 1. 1.6 x 10-3

Question 35. 0.000003 is equal to

  1. 3 x 10-6
  2. 3 x 106
  3. 3 x 105
  4. 3 x 10-5

Solution: 1. 3 x 10-6

Question 36. 8848 is equal to

  1. 8.848 x 103
  2. 8.848 x 102
  3. 8.848 x 10
  4. 8.848 x 104

Solution: 1. 8.848 x 103

Question 37. 1.5 x 1011 is equal to

  1. 150000000000
  2. 15000000000
  3. 1500000000
  4. 1500000000000

Solution: 1. 150000000000

Question 38. 2.1 x 10-6 is equal to

  1. 0.0000021
  2. 0.000021
  3. 0.00021
  4. 0.0021.

Solution: 1. 0.0000021

Question 39. 2.5 x 104 is equal to

  1. 25
  2. 250
  3. 2500
  4. 25000

Solution: 4. 25000

Question 40. 0.07 x 1O10is equal to

  1. 700000000
  2. 7000000
  3. 7000
  4. 7

Solution: 1. 700000000

Exponents And Powers True-False

Write whether the following statements are True or False:

1. The value of \(\left\{(-1)^{-1}\right\}^{-1}\) is 1: False

2. The reciprocal of \(\left(\frac{4}{3}\right)^0\) is 1: True

3. The standard form of \(\frac{1}{1000000}\) is 1.0 x 10 -6: True

4. If 6m + 6 “3 = 66, then the value of m is 3: True

5. 2345.6 = 2 x 1000 + 3 x 100 + 4 x 10 + 5 x 1 + 6 x 10 – 1: True

Exponents And Powers Fill in the Blanks

1. (1000)° = 1

2. The standard form of 1,234,500,000,000 is: 1.2345 x 1012

3. The multiplicative inverse of(-3) ~2 is: (-3)2

4. (- 9)4 -5- (- 9)10 is equal to (-9)-6

5. The value of (2 -1 + 3 -1 + 4 -1)° is : 1

6. Write 1.0002 x 109 in the usual form: 1000200000

7. Write the reciprocal of 10 _1: 10

8. Find the value of*if* “3 = (100)1-4 + (100)°: 100

9. By what number should (-8) -1 be divided, so that the quotient may be equal to (-8) -1: 1

10. If = \(\frac{5^m \times 5^2 \times 5^{-3}}{5^{-5}}=5^4\) then find the value of m: 0

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration

Mensuration

Mensuration Introduction

We know that the perimeter of a closed figure is the distance around its boundary. Also, the area of a closed figure is the measurement of the region covered by it.

  • Moreover, the volume of a solid is the amount of space occupied by it. We know how to find the areas and perimeters of various plane figures such as triangles, parallelograms, rectangles, rhombuses, squares, circles, pathways, and borders in rectangular shapes, etc.
  • Here, we shall learn to solve the problems related to perimeters and areas of general quadrilaterals and trapeziums.
  • We shall also learn to solve the problems related to areas of polygons (regular and irregular) by using the formula for the area of a triangle and that for the area of a trapezium.
  • Moreover, we shall also learn to find out the surface areas, and volumes of cubes, cuboids, and cylinders.

Area Of A Polygon

We use the method of triangulation which means splitting into triangles.

Question 1.  Divide the following polygon (Figures) into parts (triangles and trapezium) to find out its area.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Polygons

Solution:

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Polygons Triangles And Trapezium

2. Polygon ABODE is divided into parts as shown below. Find its area ifAD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm, and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Solution:

Area of polygon ABCDE = Area of A AFB + Area of trapezium FBCH + Area of A CHD + Area of A ADE …..(1)

Area of Δ AFB = (I) =\(\frac{1}{2}\) x AF x BF = \(\frac{1}{2}\) x 3 x 2 = 3 cml2

Area of trapezium FBCH = (2) = FH x \(\frac{(\mathrm{BF}+\mathrm{CH})}{2}=3 \times \frac{(2+3)}{2}=7.5 \mathrm{~cm}^2\)

FH = AH-AF = 6- 3 = 3

Area of Δ CHD = (3) = \(\frac{1}{2} \times \mathrm{HD} \times \mathrm{CH}=\frac{2 \times 3}{2}=3 \mathrm{~cm}^2\)

HD = AD -AH = 8 – 6 = 2cm

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Triangles and Trapezium

Area of Δ ADE = (4) \(\frac{1}{2} \times \mathrm{AD} \times \mathrm{GE}=\frac{1}{2}\) x 8 x 2.5 = 10 cm2

From (1),

Area of polygon ABCDE = Area [(1) + (2) + (3) + (4)]

= 3 + 7.5 + 3 + 10 = 23.5 cml2

3. Find the area of polygon MNOPQR (Figure), if

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Area Of Polygon MNOPQR

MP = 9 cm, MD = 7 cm, MC = 6 cm, MB – 4 cm, MA = 2 cm, NA, OC, QD, and RB are perpendicular to diagonal MP.

Solution:

Area of polygon MNOPQR

= Area of A MAN + Area of trapezium ACON + Area of A OOP + AreaofAPDQ + Area of trapezium DBRQ + AreaofARBM …(1)

Given:

MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm MA = 2 cm

Area of A MAN = (1) = \(\frac{\mathrm{MA} \times \mathrm{AN}}{2}=\frac{2 \times 2.5}{2}=2.5 \mathrm{~cm}^2\)

Area of trapezium ACON = (2) = \(\frac{(\mathrm{AN}+\mathrm{CO}) \times \mathrm{AC}}{2}=\frac{(\mathrm{AN}+\mathrm{CO}) \times(\mathrm{MC}-\mathrm{MA})}{2}\)

\(=\frac{(2.5+3) \times(6-2)}{2}=11 \mathrm{~cm}^2\)

Area of A 0CP = (3) = \(\frac{\mathrm{CP} \times \mathrm{OC}}{2}=\frac{(\mathrm{MP}-\mathrm{MC}) \times \mathrm{OC}}{2}=\frac{(9-6) \times 3}{2}\) = 4.5 cml2

AreaofAPDQ = (4) = \(\frac{\mathrm{PD} \times \mathrm{DQ}}{2}=\frac{(\mathrm{MP}-\mathrm{MD}) \times \mathrm{DQ}}{2}=\frac{(9-7) \times 2}{2}\) = 2cm2

Area of trapezium DBRQ = (5) \(\frac{(\mathrm{DQ}+\mathrm{BR}) \times \mathrm{BD}}{2}=\frac{(2+2.5) \times(\mathrm{MD}-\mathrm{MB})}{2}\)

⇒  \(\frac{(2+2.5) \times(7-4)}{2}=6.75 \mathrm{~cm}^2\)

Area of ARBM= (6) =\(\frac{\mathrm{MB} \times \mathrm{RB}}{2}=\frac{4 \times 2.5}{2}=5 \mathrm{~cm}^2\)

From (1),

Area of polygon MNOPQR

= Area [(1) + (2) + (3)+ (4)+ (5) + (6)]

= 2.5 cml2 + 11cml2 + 4.5 cml2 + 2 cml2 + 6.75 cml2 + 5 cml2

= 31.75 cml2

Mensuration Exercise 9.1

Question 1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and the perpendicular distance between them is 0.8 m.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Perpendicular

Solution:

Area of the top surface of the table

⇒  \(\frac{1}{2} h(a+b)\)

⇒  \(\frac{1}{2} \times 0.8 \times(1.2+1)\)

⇒  \(0.88 \mathrm{~m}^2\)

Question 2. The area of trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Area Of The Trapezium

Solution:

Let the length of the other parallel side be b cm

Area of trapezium = \(\frac{1}{2} h(a+b)\)

⇒  \(34 =\frac{1}{2} \times 4 \times(10+b)\)

⇒  \(34 =2 \times(10+b)\)

⇒  \(10+b =\frac{34}{2}\)

10 + b = 17

b = 17 – 10

b = 7cm

Hence, the length of another parallel side is 7cm

Question 3. The length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m, and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Fence Of A Trapezium

Solution:

Fence of the trapezium-shaped field

ABCD = 120 m

AB + BC + CD + DA = 120

AB + 48 + 17 + 40 = 120

AB + 105 = 120

AB = 120 – 105

AB = 15 m

Area of the field

⇒  \(\frac{(B C+A D) \times A B}{2}\)

⇒  \(\frac{(48+40) \times 15}{2}=660 \mathrm{~m}^2 .\)

Hence, the area of the trapezium-shaped field is 660 ml2

Question 4. The diagonal of a quadrilateral-shaped field is 24 m and the perpendiculars dropped on it. from the remaining opposite vertices are 8 m and 13 m. Find the area, of the field

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Area, Of The Field

Solution: Area of the field

⇒  \(\frac{1}{2} d\left(h_1+h_2\right)\)

⇒  \(\frac{24 \times(8+13)}{2}=\frac{24 \times 21}{2}\)

= 12 x 21 = 252 ml2.

Hence, the area of the quadrilateral-shaped field is 252 ml2.

Question 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:

Area of the rhombus

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Diagonals of a Rhombus

⇒  \(\frac{1}{2} \times d_1 \times d_2=\frac{1}{2} \times 7.5 \times 12\)

= 45cml2

Hence, the area of the rhombus is 45cml2

Question 6. Find the area of a rhombus whose side is 5 cm. and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:

Area of the rhombus

= base (b) x altitude (h)

= 5 x 4.8 = 24 cml2

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Area of the rhombus

Area of the rhombus

⇒  \(\frac{1}{2} \times d_1 \times d_2\)

⇒  \(24 =\frac{1}{2} \times 8 \times d_2\)

⇒  \(24 =4 d_2\)

⇒  \(d_2 =\frac{24}{4}=6 \mathrm{~cm}\)

Hence, the length of the other diagonal of a rhombus is 6 cm.

Question 7. The floor of a building consists of 3,000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is 4.
Solution:

Area of a tile = \(=\frac{1}{2} \times d_1 \times d_2=\frac{1}{2} \times 45 \times 30\)

= 675 cm2

Area of the floor = 675 x 3,000 cm2 = 20,25,000 cm2

⇒  \(\frac{20,25,000}{100 \times 100} \mathrm{~m}^2\)

lm2 = 100 X 100 cml2

202.50 ml2

The cost of polishing per ml2 = 4

Total cost of polishing the floor

= 202.50 x 4

= $ 810.

Hence, the total cost of polishing the floor is? 810

Question 8. Mohan wants to buy a trapezium field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10,500 m2 and the perpendicular distance between the two parallel sides is 100 m., find the length of the side along the river.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Mohan wants To Buy A Trapezium

Solution:

Let the length of the side along the road be m. Then, the length of the side along the river is 2x m.

Area of the field = 10,500square meters

⇒ \(\frac{1}{2} h(a+b)=10500\)

⇒ \(\frac{100 \times(2 x+x)}{2}=10,500\)

⇒ \(150 x =10,500\)

⇒ \(x =\frac{10,500}{150}\)

x= 70

2x = 2 x 70 = 140m

Hence, the length of the side along the river is 140 m.

Question 9. The top surface of the raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Solution:

Area of the octagonal surface

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Area of the octagonal surface

= Area of rectangular surface + 2(Area of trapezoidal surface)

Since the octagon is regular, therefore, the two trapeziums will be congruent.

Hence, their surface area of 111 will be equal.

⇒ \(11 \times 5+2 \times\left[\frac{(5+11) \times 4}{2}\right] \mathrm{m}^2\)

55 + 64 m2 = 119 ml2

Question 10. There isapentagonalshaped’park as shown in the figure. ForfindingitsareaJyoti and Kavita divided it into two different ways

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Pentagonal shaped park

Find the area of this park using both ways. Can you suggest some other way offind¬ ing its area?

Solution: Jyoti’s diagram

Here, the pentagonal shape is split into two congruent trapeziums.

Area of the park

⇒ \(2 \times\left[\frac{(15+30)}{2} \times \frac{15}{2}\right]\)

⇒ \(\frac{675}{2}\)

= 337.5 m2

Kavita’s diagram:

Here, the pentagonal shape is split into a square and a triangle.

Area of the park

= Area of square + Area of triangle

⇒ \(=15 \times 15 \mathrm{~m}^2+\frac{15 \times(30-15)}{2} \mathrm{~m}^2\)

⇒ \(=225 \mathrm{~m}^2+\frac{225}{2} \mathrm{~m}^2\)

= 225 m2 + 112.5 ml2

= 337.5 ml2

Another way of finding the area

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Another Way Of Finding The Area

Here, the pentagonal shape is split into three triangles out of which two triangles are congruent.

Area of the park \(=\frac{15 \times(30-15)}{2} \mathrm{~m}^2 +2 \times\left[\frac{15 \times 15}{2}\right] \mathrm{m}^2 \)

Question 11. The diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is the same

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Adjacent Picture

Solution:

Area of the right section of the frame

⇒ \(\frac{28+20}{2} \times \frac{24-16}{2} \mathrm{~cm}^2\)

⇒ \(\frac{48}{2} \times \frac{8}{2}=24 \times 4=96 \mathrm{~cm}^2\)

The section is a trapezium

Similarly, the area of the left section of the frame = 96 cml2

The area of the upper section of the frame

⇒ \(\frac{24+16}{2} \times \frac{28-20}{2} \mathrm{~cm}^2\)

⇒ \(\frac{40}{2} \times \frac{8}{2}=20 \times 4 \mathrm{~cm}^2\)

= 80 cml2

Similarly, the area of the lower section of the frame

= 80 cml2

Solid Shapes

Two-dimensional figures are the faces of three-dimensional shapes. If two faces of a shape are identical, then they are called congruent faces.

Surface Area Of Cube, Cuboid, And Cylinder

The surface area of a solid is the sum of the areas of its faces.

Mensuration Cuboid

Total surface area of a cuboid = 2 (lb + bh + hl)

where l, b, and h are the length, width, and height of the cuboid respectively

Question Find the total surface area of the following cuboids (Figure):

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Total surface Area Of The Cubiod

Solution:

The total surface area of the first cuboid

= 2(lb + bh + hi)

= 2(6 x 4 + 4 x 2 + 2 x 6)

= 2(24 + 8 + 12) = 88 cml2

The total surface area of the second cuboid

= 2(lb+ bh + hi)

= 2(4 x 4 + 4 x 10 + 10 x 4)

= 2(16 + 40 + 40) = 192 cml2

Question 1 . Cover the lateral surface of a cuboidal duster (which your teacher uses in the classroom) using a strip of brown sheet of paper, such that it just fits around the surface. Remove the paper. Measure the area of the paper. Is it the lateral surface area of the duster?
Solution:

(i) Yes, we can say that the area of this strip of brown sheet paper is equal to the lateral surface area of the duster.

Question 2. Measure the length, width, and height of your classroom and find

  1. The total surface area of the room, ignoring the area of windows and doors.
  2. The lateral surface area of this room.
  3. The total area of the room which is to be white-washed.

Solution:

2.  Please find yourself

Question 3. Can we say that the total surface area of a cuboid

= lateral surface area + 2 x area of base?

Question 4. If we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change?
Solution:

1. Yes, we can say that the total surface area of a cuboid

= lateral surface area + 2 x area of base.

2. Lateral surface area of cuboid (i) = 2(1 + b) h

The lateral surface area of the cuboid (ii)

= 2(h + b) l. These results are different.

Hence, yes; the lateral surface area will change if we interchange the length of the base and the height of a cuboid

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Height Of A cuboid

Mensuration Cube

The total surface area of a cube = 6l2, where l is the side of the cube. The lateral surface area of a cube = 4l2, where l is the side of the cube.

Question: Draw the pattern shown on a squared paper and cut it out. (You know that this pattern is a net of a cube). Fold it along the lines and tape the edges to form a cube.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Squared Paper Cube

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Width And Height Of The Cube

  1. What is the length, width, and height of the cube? Observe that all the faces of a cube are square. This makes the length, height, and width of a cube equal.
  2. Write the area of each of the face. Are they equal?
  3. Write the total surface area of this cube
  4. If each side of the cube is l, what will be the area of each face?
  5. Can we say that the total surface area of a cube of side l is 6l2?

Solution:

(a) Length of the cube

= Width of the cube

= Length of the cube of side s

Area of each face = l x l = l2.

Yes, they are equal.

(c) Total surface area of this cube = 6l2.

Area of each face = l2.

Yes, we can say that the total surface area of a cube of side l is 6l2

10. Find the surface area of Cube A and the lateral surface area of Cube B.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Surface Area of cube

Solution:

The surface area of the cube (A)

= 6l2

= 6(10)2cm2

= 600 cm2

The lateral surface area of the cube (B)

= 4l2

= 4 x 8 x 8cm2

= 256 cm2

Question .1 Two cubes each with side b are joined to form a cuboid (Figure). What is the surface area, of this cuboid?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Surface Area Of This Cuboid

How will you arrange 12 cubes of equal length to form a cuboid of the smallest surface area. ?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Cuboid Of Smallest Surface Area

After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of the same dimensions (Figure).

How many have no face painted? 1 face painted? 2faces painted? 3faces painted?
Solution:

By joining two cubes on each side b,

we get a cuboid whose

length (L) = b + b = 2b units

breadth (B) = b units

and height (H) = b units

The surface area of this cuboid

= 2(L x B + B x H + H x L)

= 2[2b x b + b x b + b x 2b] sq. units

= 2[2b2 + b2 + 2b2 sq. units

= 10b2 sq. units

≠ 12b2. (No, it is not equal to 1262)

Again, by joining three cubes each with ® side b, we get a cuboid whose

length (L) = 6 + 6 + 6 = 36 units

breadth (B) = 6 units

height (H) = 6 units

The surface area of the cuboid formed by joining the three cubes

= 2 (L x B + B x H + H x L)

= 2 x [3b x b + b x b + b x 3b] sq. units

= 2 x [3b2 + b2 + 3b2] sq. units

= 14b2 sq. units

≠1862(No, it is not equal to 18b2)

(ii) In the first arrangement,

L = 66, B = 26, H = 26

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Surface Area Of The Cuboid Formed Cubes

Surface area

= 2 x (6b x 2b + 2b x 2b + 2b x 6b)

= 2 x (12b2 + 4b2 + 12b2)

= 56b2

In the second arrangement,

L = 4b

B = 6

H = 3b

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Surface Area Of The Cuboid Formed Cubes

Surface area

= 2 x (4b x b + b x 3b + 3b x 4b)

= 2 x (4b2 + 3b2 + 12b2)

= 38b2

Hence, for the smallest surface area, the second arrangement must be made.

(Hi) 16 cubes have no faces painted.

24 cubes have 1 face painted.

16 cubes have 2 faces painted.

8 cubes have 3 faces painted.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration 8 Cubes Have 3 Faces Painted.

Mensuration Cylinders

Formulae :

The lateral (curved) surface area of a cylinder = 2kvh

Total surface area of a cylinder = 2nr (h + r)

where r is the radius of the base and h is the height of the cylinder.

We take n to be \(\frac{22}{7}\) unless otherwise stated

Question .1  Find the total surface area of the following cylinders.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Total Surface area

Solution:

For First Cylinder

r = 14 cm

h = 8 cm

Total surface area of the cylinder = 2nr(r + h)

\(=2 \times \frac{22}{7} \times 14 \times(14+8) \mathrm{cm}^2\)

= 2 x 22 x 2 x 22 = 88 x 22

= 1936 cm2

For Second Cylinder

\(r=\frac{2}{2} \mathrm{~m}=1 \mathrm{~m}\)

h = 2 m

The total surface area of the cylinder

= 2πr (r + h)

⇒ \(2 \times \frac{22}{7} \times 1 \times(1+2) \mathrm{m}^2\)

⇒ \(\frac{132}{7} \mathrm{~m}^2=18 \frac{6}{7} \mathrm{~m}^2\)

Question 2.  Note that the lateral surface area of a cylinder is the circumference of the base x height of the cylinder. Can. we write the lateral surface area of the cuboid as the perimeter of the base x height of the cuboid.

Solution:

Lateral surface area

= 2 x [(l x h) + (b x h)]

= 2 (l + b) x h

= Perimeter of base b x height of the cuboid

Yes, we can write the surface area of a cuboid as

perimeter of base x height of the cuboid

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Perimeter Of Base Height Of The Cuboid

Mensuration Exercise 9.2

Question 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires a lesser amount of material to make?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Two Cuboidal boxes

Solution:

First Cuboidal Box

l = 60 cm, b = 40 cm, h = 50 cm

Total surface area

= 2(lb + bh + hl)

= 2(60 x 40 + 40 x 50 + 50 x 60) cm2

= 2(2400 + 2000 + 3000) cm2

= 2(7400) cm2

= 14800 cm2

Second Cuboidal Box

l = 50 cm, b = 50 cm, h = 50 cm

Total surface area

= 2(lb + bh + hl)

= 2(50 x 50 + 50 x 50 + 50×50) cm2

= 2(2500 + 2500 + 2500) cm2

= 2(7500) cm2

= 15000 cm2

Since the total surface area of the first cuboidal box is less than the total surface area of the second cuboidal box, therefore, box (a) requires a lesser amount of material to make.

Question 2. A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:

Here,

Length of the suitcase (l) = 80 cm

The breadth of the suitcase (b) = 48 cm

Height of the suitcase (h) = 24 cm

The total surface area of the suitcase

= 2(lb + bh + hl)

= 2(80 x 48 + 48 x 24 + 24 x 80) cm2

= 2(3840 + 1152 + 1920) cm2

= 2(6912) cm2

= 13824 cm2

Width of tarpaulin = 96 cm

Length of tarpaulin required to cover 1 suitcase

⇒ \(\frac{\text { Total surface area of the suitcase }}{\text { Width of tarpaulin }}\)

⇒ \(=\frac{13824}{96}=144 \mathrm{~cm}\)

Length of tarpaulin required to cover 100 such suitcases

= 144 x 100 cm = 14400 cm

⇒ \(\frac{14400}{100} \mathrm{~m}=144 \mathrm{~m}\)

Hence, 144 m of tarpaulin is required.

Question 3. Find the side of a cube whose surface area is 600 cm2.
Solution:

Let the side of the cube be a cm. Then, the total surface area of the cube

= 6a2 cm2

According to the question,6a2 = 600

⇒ \(a^2=\frac{600}{6}\)

a2= 100

a = √1oo

a = 10 cm

Hence, the side of the cube is 10 cm.

Question 4. Rukhsar painted the outside of the cabinet measuring 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Rukhsar painted The Outside Of The Cabinet

Solution:

I = 2m 7 cm

b = 1 m

h = 1.5 m

Required area

= 2[2 x 1 + 1 x 1.6 + 1.5 x 2] -2xl

= 2[2 +1.5 + 3] – 2

= 13 m2-2 m2

= 11 m2

Hence, she covered 11 m2 of surface area.

Question 5. Daniel is painting the walls and ceiling of the cuboidal hall with length, breadth, and height of 15 m, 10 m, and 1 m respectively. From each can ofpaint100 m2 of area is painted, How many cans of paint will she need, to paint the room?
Solution:

Z = 15 m

b = 10 m

h = 7 m

Surface area to be painted

= 2(1 x 6 + b x h + h x l) – l x b

= 2 (15 x 10 + 10 x 7 + 7 x 15) m2 – (15 x 10) m2

= 2(150 + 70 + 105) m2 – 150 m2

= 2(325) m2 – 150 m2

= 650 m2 – 150 m2

= 500 m2

Number of cans needed \(=\frac{\text { Surface area to be painted }}{\text { Area painted by } 1 \text { can }}\)

⇒ \(=\frac{500}{100}=5\)

Hence, she will need 5 cans of paint to paint the room.

Question 6. Describe how the two figures (below) are alike and how they are different. Which box has a larger lateral surface area?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Larger Lateral Surface Area

Solution:

Likeness→ Both have the same heights.

Difference → One is a cylinder, the other is a cube;

The cylinder is a solid obtained by revolving a rectangular area about one side.

A cube is a solid enclosed by six square faces.

A cylinder has two circular faces whereas a cube has six square faces

For First Box

Diameter = 7 cm

Radius (r) =\(\frac{\text { Diameter }}{2}=\frac{7}{2} \mathrm{~cm}\)

Height (h) = 7 cm

Lateral surface area = 2nrh

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times 7\) = 154 cm2

For Second Box

l = 7 cm

b = 7 cm

h = 7 cm

Lateral surface area

= 4 = 4 x (7)2

= 196cm2

Hence, the second box has a larger lateral surface area

Question 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Sheet Of Metal Is Required

Solution:

r = 7 m

h = 3 m

Total surface area

= 2πr(r + h)

⇒ \(2 \times \frac{22}{7} \times 7 \times(7+3)\) = 440 m2

Hence, 440 m2 of metal sheet is required.

Question  8. The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and forms a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.
Solution:

The lateral surface area of the hollow

cylinder = 4224 cm2

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The perimeter of rectangular sheet

The area of the rectangular sheet = 4224 cm2

Length x width = 4224 cm2

Length x 33 = 4224

Length = \(\frac{4224}{33}\)

Length = 128 cm

The perimeter of the rectangular sheet

= 2(Length + Breadtn)

= 2(128 + 33) cm

= 2(161) cm

= 322 cm

Hence, the perimeter of the rectangular sheet is 322 cm.

Question 9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and the length is 1 m.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Diameter Of The Road Roller

Solution:

The diameter of the road roller = 84 cm

Radius (r) of the road roller \(\frac{84}{2} \mathrm{~cm}=42 \mathrm{~cm}\)

Length (h) of the road roller = 1 m = 100 cm

The lateral surface area of the road roller = 2πrh

⇒ \(2 \times \frac{22}{7} \times 42 \times 100\)

= 26,400 cm2

The area of the road is covered in 1 complete revolution

= 26,400 cm2

The area of the road covered 750 complete revolutions

= 26,400 x 750 cm2

= 1,98,00,000 cm2

⇒ \(\frac{1,98,00,000}{100 \times 100} \mathrm{~m}^2\)

= 1,980 m2

Question 10. A company packages its milk ponder in a cylindrical container whose base has a diameter of 14 cm and a height of 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area, of the label

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration For a cylindrical container

Solution:

For a cylindrical container

The diameter of the base = 14 cm

Radius of the base (r) =\(\frac{14}{2} \mathrm{~cm}\)

= 7 cm

Height (h) = 20 cm

The curved surface area of the container = 2πrh

⇒ \(=2 \times \frac{22}{7} \times 7 \times 20\)

= 880 cm2

Surface area of the label

= 880 cm2

⇒ \(-2\left(2 \times \frac{22}{7} \times 7 \times 2\right) \mathrm{cm}^2\)

= 880 cm2 – 176 cm2

= 704 cm2

Hence, the surface area of the label is 704 cm2.

Aliter:

The label is in the form of a cylinder for which

radius (R) =\(\frac{14}{2}=7 \mathrm{~cm}\)

height (H) = 20 – (2 + 2) = 16 cm

Area of the label = 2πRH

⇒ \(=2 \cdot \frac{22}{7} \cdot 7 \cdot 16=704 \mathrm{~cm}^2\)

Hence, the surface area of the label is 1704 cm2.

Volume Of Cube, Cuboid And Cylinder

The volume of a three-dimensional object is the amount of space occupied by it. Volume is measured in cubic units.

Question 1 cubic cm

= 1 cm x 1 cm x 1 cm

= 1 cm3

= 10 mm x 10 mm x 10 mm

= mm3

1 cubic m

= lm x lm x lm = lm3

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm3

= 0.1 cm x 0.1 cm x 0.1 cm

cm3

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm3

= 0.1 cm x 0.1 cm x 0.1 cm

=100cm3

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm3

= 0.1 cm x 0.1 cm x 0.1 cm

cm3

Cuboid Formula

Volume of cuboid = Ibh

where l, b, and h are the length, width, and height of the cuboid respectively.
OR

The volume of the cuboid = area of the base x height cuboid.

YouTakecan36arrangecubes of the qualm size many(ie., ways.lengthObserveofeachthecubefollowingis same).tableArrangeandfillthemin the to blank torn a

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration 36 Cubes Of Them Equal Size

We have the following observations:

Since then, we have used 36 cubes of equal size to form these 36 cubic units. Also, the volume of each cuboid is the height of the cuboid. From the above example ese cuboids, the volume of each cuboid = l x b x h

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Volume Of cuboid

Since l x b is the area of its b

The volume of cuboid = Area of base x Therefore we can also say that,

Question 1.  Can you think of such objects whose

Solution: The volume of godow reservoirs, etc., can be found by this method.

Question Find the volume of the following cuboids (Figure).

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Cuboids

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Volume Of The Cuboid

Solution:

(i) Volume of the cuboid

= l x b x h

= (8x3x2) cm3

= 48 cm3

(ii) Volume of the cuboid

= Area of Base x Height

⇒ \(24 \times \frac{3}{100}\)

⇒ \(\frac{72}{100}=0.72 \mathrm{~m}^3\)

Cube

The volume of the cube = 13, where l is the side of the cube.

Question 1.  Find the volume of the following cubes.

  1. with a side of 4 cm
  2. with a side of 1.5 m

Solution:

Volume of the cube = (Side)3

= 4 x 4 x 4 cm3

= 64 cm3

(b) Volume of the cube = (Side)3

= 1.5 x 1.5 x 1.5 m3

= 3.375 m3

Question 2.  Arrange 64 cubes of equal size in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid shapes of the same volume have the same surface area?
Solution:

Some arrangements are as follows :

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Shapes Of Same Volume Have Same Surface Area

The surface area in the arrangement

(1) = 2 x (64 x l + i x l + l x 64)

258 square units

2 x (32 x 2 + 2 x l + i x 32)

= 196 square units

= 2 x (16 x 2 + 2 x 2 + 2 x 16)

= 136 square units

= 2 x (16 x 4 + 4 x 1 + 1x 16)

= 168 square units

=2 x (8 x 4 + 4 x 2 + 2 x 8) = 112 square units

= 2 x (4 x 4 + 4 x 4 + 4 x 4)

= 96 square units, etc.

Also, the volume of the cuboid obtained in each case is 64 cubic units. So, No, we cannot say that solid shapes of the same volume need to have the same surface area.

Question 3. The company sells biscuits. For packing purposes, they are using cuboidal boxes :

box A → 3 cm x 8 cm x 20 cm,

box B → 4 cm x 12 cm x 10 cm.

  1. What size of the box will be economical for the company? Why?
  2. Can you suggest any other size (dimensions) that has the same volume but is more economical than these?

Solution:

Box A: Total surface area

= 2 x (3 x 8 + 8 x 20 + 20 x 3) cm2

= 2 x (24 +160 + 60) cm2

= 2 x 244 cm2 = 488 cm2

Box B: Total surface area

= 2 x (4 x 12 + 12 x 10 + 10 x 4) cm2

= 2 x (48 + 120 + 40) cm2

= 2 x208 cm2 = 416 cm2.

-Since the total surface area of box B is lesser than the total surface area of box A, therefore, the size of 4 cm x 12 cm x 10 cm of the

The box will be economical for the company to use for packing purposes.

(The volume of both the boxes are same as given below)

Again, the volume of the cuboidal box A

= 3 x 8 x 20

= 480 cm3 and, volume of the cuboidal box B

= 4 x 12 x 10

= 480 cm3

Yes, we can suggest another size (dimensions) which has the same volume but is more economical than these. It is 10 m x 6 m x 8 m.

Volume = 10 x 6 x 8 = 480 cm3

Total surface area = 2 x (10 x 6 + 6 x 8 + 8 x 10) cm2 = 376 cm2.

This size (dimensions) has the same volume but it is more economical than box A and box B.

Cylinder Formula:

Volume of cylinder = πr2h

where r is the radius of the base and h is the height of the cylinder

Question 1.   Find the volume of the following cylinders :

Solution:

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration cylinders

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration cylinder

r = 7 cm, h = 10 cm

The volume of the cylinder = πr2h

⇒ \(=\frac{22}{7} \times 7^2 \times 10\)

⇒ \(=\frac{22}{7} \times 7 \times 7 \times 10\)

= 1540 cm3

The Volume of the cylinder

= Area of base x height

= (250 x 2) m3

= 500 m3

Volume And Capacity

There is not much difference between these two words.

Volume refers to the amount of space occupied by an object.

Capacity refers to the quantity that a container holds.

Note: If a water tin holds 100 cm3 of water, then the capacity of the water tin is 100 cm3.

Capacity is also measured in terms of liters.

The relation between liter and cm3 is,

1 mL = 1 cm3, 1 L = 1000 cm3.

lm3 = 10,00,000 cm3= 1000 L.

Mensuration Exercise 9.3

Question 1. Given a cylindrical tank, in which situation will you find surface area, and in which situation is volume?

  1. To find how much it can hold,
  2. Number of cement bags required to plaster it.
  3. To find the number of smaller tanks that can be filled with water from it.

Solution:

Volume

Surface area

Volume.

Question 2. The diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of cylinder B is 14 cm and the height is 7cm. Without doing any calculations canyon suggest whose volume is greater? Verify it by finding the volume of both cylinders. Check whether the cylinder with greater volume also has a greater surface area.

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Volume Of Cylinder B is greater

Solution:

The volume of cylinder B is greater.

For Cylinder A

⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

h = 14 cm

Volume = nr2h

⇒ \(=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14\)

= 539 cm3

For Cylinder B

⇒ \(r=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}\)

h = 7 cm

Volume = nr2h

⇒ \(\frac{22}{7} \times 7 \times 7 \times 7\)

= 1078 cm3.

By actual calculation of volumes of both the cylinders, it is verified that the volume of cylinder B is greater

Surface Area

For Cylinder A

Surface area = 2jir (r + h)

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times\left(\frac{7}{2}+14\right)\)

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{35}{2}\)

= 385 cm2

For Cylinder B

Surface area

=27tr(r + h)

⇒ \(2 \times \frac{22}{7} \times 7 \times(7+7)\)

⇒ \(2 \times \frac{22}{7} \times 7 \times 14\)

By actual calculation of the surface area of both the cylinders, we observe that the cylinder with greater volume (Cylinder B) has a greater surface area.

Question 3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3.
Solution:

Height of the cuboid = \(=\frac{\text { Volume of the cuboid }}{\text { Base area of the cuboid }}\)

⇒ \(=\frac{900}{180}\) = 5 cm

Question 4. A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with sides 6 cm can be placed in the given cuboid?
Solution:

Volume of the cuboid = 60 x 54 x 30 cm3

= 97200 cm3

The volume of a small cube

= 6 x 6 x 6 cm3

= 216 cm3

Number of small cubes that can be placed in the given cuboid

⇒ \(\frac{\text { Volume of the cuboid }}{\text { Volume of a small cube }}\)

⇒ \(\frac{97200}{216}=450\)

Hence, 450 small cubes can be placed in the given cuboid.

Question 5. Find the height of the cylinder whose volume is 1.54 m2 and the diameter of the base is 140 cm.
Solution:

The diameter of the base = 140 cm

Radius of the base (r)= \(=\frac{140}{2} \mathrm{~cm}=70 \mathrm{~cm}\)

Area of the base = nr2

⇒ \(=\frac{22}{7} \times 70 \times 70\)

= 15400 cm2

The volume of the cylinder

= 1.54 m3

= 1.54 x 100 x 100 x 100 cm3

= 15,40,000 cm3

Height of the cylinder

⇒ \(\frac{\text { Volume of the cylinder }}{\text { Area of the base of the cylinder }}\)

⇒ \(\frac{(2)}{(1)}\)

⇒ \(=\frac{15,40,000}{15400}\)

= 100 cm

= 1 m

Hence, the height of the cylinder is 1 m.

Question 6. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration For milk tank

solution:

For milk tank

r = 1.5 m

h = 7 m

Capacity = nFh

⇒ \(=\frac{22}{7} \times 1.5 \times 1.5 \times 7\)

⇒ \(=\frac{22}{7} \times \frac{15}{10} \times \frac{15}{10} \times 7\)

= 49.5 m3 I

= 49.5 x 1000 L I

[lm3 = 1000 L]

= 49500 L.

Hence, the quantity of milk that can be storedin the tank is 49500 litres.

Question 7. If each edge of a cube is doubled,

  1. how many times will its surface area increase?
  2. how many times will its volume increase?

Solution:

Let the original edge of the cube be a cm.

Then, its new edge = 2a cm

The original surface area of the cube = 6a2 cm2

The new surface area of the cube

= 6(2a)2 cm2

= 24a2 cm2

= 4 (6a2 cm2)

= 4 original surface area,

Hence, its surface area will increase 4 times.

Originalvolume ofthe cube = a3 cm3

A new volume of the cube

= (2a)3 cm3

= 8a3 cm3

= 8 x the original volume of the cube.

Hence, its volume will increase 8 times

Question 8. Water is poured into a cuboidal g reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the BL number of hours it will take to fill the reservoir

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration Volume Of Reservoir

Solution:

Volume of reservoir

= 108 m3

= 108 x 1000 L

[1 m3 = 1000 L]

= 108000 L

Water poured per minute = 60 L

Time taken to fill the reservoir

⇒ \(\frac{\text { Volume of the reservoir }}{\text { Water poured per minute }}\)

⇒ \(\frac{108000}{60} \mathrm{~m}\)

⇒ \(\frac{108000}{60 \times 60} \text { hours }\)

= 30 hours

Hence, the number of hours it will take to fill the reservoir is 30

Mensuration Multiple-Choice Question And Solutions

Question 1. 1 cm3 =

  1. 1000 mm3
  2. 100 mm3
  3. 10 mm3
  4. \(\frac{1}{1000} \mathrm{~mm}^3 .\)

Solution: 1. 1000 mm3

Question 2. 1 m3 =

  1. 1000000 cm3
  2. 100 cm3
  3. 10 cm3
  4. \(\frac{1}{1000} \mathrm{~cm}^3 .\)

Solution: 1. 1000000 cm3

Question 3. 1 mm3 =

  1. 0.001 cm3
  2. 0.01cm3
  3. 0.1 cm3
  4. 100 cm3

Solution: 1. 0.001 cm3

Question 4. 1 cm3 =

  1. 0.000001m3
  2. 0.01m3
  3. 0.1m3
  4. 1000m3

Solution: 1. 0.000001m3

Question 5. The surface area of a cuboid of length l, breadth b, and height h is

  1. lbs
  2. lb + bh + hl
  3. 2 (lb + bh + hi)
  4. 2 (l+ b)h.

Solution: 3. 2 (lb + bh + hi)

Question 6. The surface area of a cube of edge a is

  1. 4a2
  2. 6a2
  3. 3a2
  4. a2

Solution: 2. 6a2

Question 7. The total surface area of a cylinder of base radius r and height h is

  1. 2πr (r + h)
  2. πr (r + h)
  3. 2πrh
  4. 2πr2

Solution: 1. 2πr (r + h)

Question 8. The volume of a cuboid of length l, breadth b, and height h is

  1. lbh
  2. lb + bh + hl
  3. 2 (lb + bh + hl)
  4. 2(l + b)h.

Solution: 1. lb

Question 9. The volume of a cube of edge a is

  1. a2
  2. a3
  3. a4
  4. 6a2

Solution: 2. a3

Question 10. The volume of a cylinder of base radius and height his

  1. 2πrh
  2. πr2h
  3. 2πr (r + h)
  4. \(\frac{1}{3} \pi r^2 h .\)

Solution: 2. nr2h

Question 11. 1 L =

  1. 10 cm3
  2. 100 cm3
  3. 1000 cm3
  4. 10000 cm3

Solution: 3. 1000 cm3

Question 12. 1 m3 =

  1. 1L
  2. 10 L
  3. 100L
  4. 1000 L

Solution: 4. 1000 L

Question 13. The area ofthe trapezium

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Area Of The Trapezium

  1. 9 cm2
  2. 6 cm2
  3. 7 cm2
  4. 24 cm2

Solution: 1. 9 cm2

Area \(=\frac{(4+2) 3}{2}=9 \mathrm{~cm}^2\)

Question 14. The area of the trapezium

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Area Of The Trapezium

  1. 6 cm2
  2. 4 cm2
  3. 3 cm2
  4. 9 cm2

Solution: 1. 6 cm2

Area \(=\frac{(3+1) 3}{2}=6 \mathrm{~cm}^2\)

Question 15. The perimeter of the trapezium

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The perimeter Of The Trapezium

  1. 12 cm
  2. 24 cm
  3. 6 cm
  4. 18 cm.

Solution: 1. 12 cm

Perimeter = 3+ 3 + 2 + 4 = 12 cm

Question 16. The area of a rhombus is 60 cm2. One diagonal is 10 cm. The other diagonal is

  1. 6 cm
  2. 12 cm
  3. 3 cm
  4. 24 cm.

Solution: 2. 12 cm

⇒ \(\frac{1}{2}\)x 10 X = d260 => d2 = 12cm

Question 17. The area of a trapezium is 40 cm2. Its parallel sides are 12 cm and 8 cm. The distance between the parallel sides is

  1. 1cm
  2. 2cm
  3. 3cm
  4. 4cm

Solution: 4. 4cm

⇒ \(\frac{(12+8) d}{2}=\) – 40 =» d = 4cm

Question 18. 8 persons can stay in a cubical room. Each person requires 27 m3 of air. The side of the cube is

  1. 6m
  2. 4m
  3. 3m
  4. 2m

Solution: 1. 6m

Volume = 8 x 27 = 216 m3.

Side = 216 = 6 m

Question 19. If the height of the cuboid becomes zero, it will take the shape of a

  1. cube
  2. parallelogram
  3. circle
  4. rectangle

Solution: 4. rectangle

Height = \(\frac{80}{20}=4 \mathrm{~m} .\)

Question 20. The volume of a room is 80 m3. The area of the floor is 20 m2. The height of the 1 room is

  1. 1m
  2. 2m
  3. 3m
  4. 4m

Solution: 4. 4m

Question 21. The floor of a room is a square of side 6 m. Its height is 4 m. The volume of the room is

  1. 140 m3
  2. 142 m3
  3. 144 m3
  4. 145 m3

Solution: 3. 144 m3

Volume = 6x6x4 = 144 m3

Question 22. The base radius and height of a right circular cylinder are 14 cm and 5 cm respectively. Its curved surface is

  1. 220 cm2
  2. 440 cm2
  3. 1232 cm2
  4. 2π x 14 x (14 + 5) cm2

Solution: 2. 440 cm2

Curved surface =\(2 \times \frac{22}{7} \times 14 \times 5\)

= 440 cm2.

Question 23. The heights of the two right circular cylinders are the same. Their volumes are respectively 16πm3  and 8lπm3. The ratio of their base radius

  1. 16:81
  2. 4:9
  3. 2: 3
  4. 9: 4.

Solution: 2. 4:9

⇒ \(\frac{\pi r_1^2 h}{\pi \pi_2^2 h}=\frac{16 \pi}{81 \pi} \Rightarrow \frac{r_1}{r_2}=\frac{4}{9}\)

Question 24. The ratio of the radii of two right circular o cylinders is 1: 2 and the ratio of their heights is 4: 1. The ratio of their volumes is

  1. 1:1
  2. 1:2
  3. 2:1
  4. 4:1

Solution: 1. 1:1

⇒ \(\frac{r_1}{r_2}=\frac{\pi(1)^2 4}{\pi(2)^2 1}=1: 1 \text {. }\)

Question 25. A glass in the form of a right circular cylinder is half full of water. Its base radius is 3 cm and its height is 8 cm. The volume of water is

  1. 18π cm3
  2. 36π cm3
  3. 9π cm3
  4. 36π cm3

Solution: 2. 36πcm3

Volume = \(\frac{1}{2} \pi \times 3 \times 3 \times 8=36 \pi \mathrm{cm}^3 .\)

Question 26. The base area of the right circular cylinder is 16ft cm3. Its height is 5 cm. Its curved surface area is

  1. 40π cm2
  2. 30π cm2
  3. 20π cm2
  4. 10π cm2

Solution: 1. 40π cm2

πr2 = 16π => r = 4 cm

Curved surface area = 2xπX4x5 = 40π cm2

Question 27. The base radius and height of a right circular cylinder are 5 cm and 10 Its total surface area is

  1. 150ft cm2
  2. 150 cm2
  3. 300ft cm2
  4. 300n cm2

Solution: 1.150π cm2

Total surface area = 2πr (h + r)

= 2π5 (10 + 5) = 150π cm2.

Mensuration True-False

Write whether the following statements are True or False:

1. The volume of a solid is the measurement of the space occupied by it: True

2. The areas of any two faces of a cuboid are equal:  False

3. Two cuboids with equal volumes essentially have equal surface areas: False

4. The radii of two cylinders having the same volume are in the ratio 1 : 3. Then, the ratio of their heights is 9: 1: True

5. The volume of a cube is 216 cm3. Its surface area is 216 cm3: True

Mensuration Fill In The Blanks

1. The areas of any two faces of a cube are_______: Equal

2. The total surface area of a cube, whose volume is 1 cm3, is_____:   1cm2

3. 1 litre = _______cm31000

4. 1 mm =_______ m : \(\frac{1}{1000} \mathrm{~m}\)

5. The ratio of radii of two cylinders is 1: 2 and heights are in the ratio 4: 5. Find the ratio of their volumes:  1:5

6. A metal sheet 64 cm long, 27 cm broad, and 8 cm thick is melted into a cube. Find the side of the cube:  24 cm

7. A cube of side 2 cm is cut into 1 cm cubes. What is the percentage increase in the volume after such a cutting:   0%

8. Find the area of the following figure: 10 cm2

NCERT Solutions For Class 8 Maths Chapter 9 Mensuration The Area

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities

Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expressions

While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. Note that the sum of x number of like terms is another like term whose coefficient is the sum of the coefficients of the like terms being added.

Addition And Subtraction Of Algebraic Expressions Exercise 8.1

Question 1. Add the following.

xb – bc, bc – cx, cx – xb

x – b + xb, b – c + be, c – x + xc

2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

l2 + m2, m2 + n2 n2 + l2, 2lm + 2mn + 2nl

Solution: 1. xb – bc, bc – cx, cx – xb

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expressions

Thus, the sum of the given expressions is 0.

2. x-b + xb, b-c + bc, c-x+xc

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expression

Thus, the sum of the given expressions is ab + bc + cx

3. 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Adding The Three Expressions

4. l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expression

Thus, the sum of the given expressions is 2(l2 + m2 + n2 + Im + mn + nl).

Question 2.

  1. Subtract 4x – 7xb + 3b + 12 from 12x – 9xb + 5b – 3
  2. Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + lOxyz
  3. Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from IS – 3p – 11q + 5pq – 2pq2 + 5p2q

Solution:

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Algebraic Expressions

Thus, the required answer is 8a – 2ab + 26- 15

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraices Expressions

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraices Expression

Thus, the required answer is 28 + bp- 18q + 8pq- Ipq2 + p2q

8.2 Multiplication Of Algebraic Expressions: Introduction

There exist x number of situations when we need to multiply algebraic expressions. For example, in finding the xerox of x rectangle whose sides are given xs expressions

Question 1. Can you think of two more such situations, where we may need to multiply algebraic expressions?

[Hint: Think of speed and time; Think of interest to be paid, the principal and the rate of simple interest; etc.].

Solution:

Distance = Speed x Time

Simple Interest

Principal x Rate of simple interest

⇒ \(=\frac{\text { per annum } \times \text { Time in years }}{100}\)

Multiplying A Monomial By A Monomial

A monomial multiplied by an x monomial always gives an x monomial.

Multiplying Two Monomials

In the product of two monomials

Coefficient = coefficient of first monomial x coefficient of second monomial

Algebraic factor – algebraic factor of first monomial x algebraic factor of second monomial

Multiplying Three Or More Monomixls

We first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.

Rules Of Signs

(1) The product of two factors is positive or negative accordingly xs the two factors have like signs or unlike signs. Note that

  1. (+) x (+) = +
  2. (+) X (-) = –
  3. (-) x (+) = –
  4. (-) x (-) = +

(2) If x is x vxrixble xnd p, q xre positive inteqers, then

xp x xq = xp+q

Question 1. Find 4x x 5y x 7z. First, find 4x x 5y and. multiply it by 7z; or first, find 5y x 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution:

4x x 5y x 7z

4x x 5y = (4 x 5) x (x x y)

= 20 x (xy) = 20 xy

(4x x 5y) x 7z = 20xy x lz

= (20 x 7) x (xy x z)

= 140 x (XYZ)

= 140xyz …(1)

OR

5y x 7z = (5 x 7) x (y x z)

– 35 X (yz) = 35yz

4x X (5y X 7z) = 4x X 35yz

= (4 x 35) x (x x yz)

= 140 x (xyz)

= I40xyz…..(2)

We observe from Eqns. (1) and (2) that the result is the same. It shows that the product of monomials is associative, i.e., the order in which we multiply monomials does not matter.

Addition And Subtraction Of Algebraic Expressions Exercise 8.2

Question 1. Find the product of the following pairs of monomials:

  1. 4, 7p
  2. -4p, 7p
  3. -4p, 7pq
  4. 4p3, -3p
  5. 4p, 0.

Solution:

(1)  4, 7p

4 x 7p = (4 x 7) x p

= 28 x p

= 28p

(2)  – 4p, Ip

(-4p)x(7p) = {(-4) x 7} x (P x P)

= (-28) x p2

= -28p2

(3)  -4p, 7pq

(- 4p) x (Ipq) = {(- 4) x 7} x (p x (pq)}

= (-28) x (p x p x q)

= (-28) x p2

= -28p2q

(4)  4p2, – 3p

(4p3) x ( -3p) = {4 x (_ 3)} x (p3 Xp)

= (-12) X p4

= – 12p4

(5)  4p, 0

(4p) x 0 = (4×0) x p

0 x p = 0

Question 2. Find the press of rectangles with the following pairs of monomials xs their lengths and breadths respectively:

(p, q) ; (10m, 5n); (20x2 5y2); (4x, 3.x2); (3mn, 4np).

Solution:

1.  (p, q)

Length = p

Brexdth = q

Area of the rectangle

= Length x Brexdth

= P x q

= pq

2.  (10m, 5n)

Length = 10 m

Brexdth = 5 n

Area of the rectangle

= Length x Brexdth

= (10m) x (5n)

= (10 x 5) x (m x n)

= 50 x (mn)

= 50 mn

3. (2Ox2 , 5y2)

Length = 20X2

Brexdth = 5y2

Area of the rectangle

= Length x Brexdth

= (20x2) x (5y2)

= (20 x 5) x (x2 x y2)

= 100 x (x2/)

= 100x2y2

4.  (4x,3x2)

Length = 4x

Brexdth = 3X2

Area of the rectangle

= Length x Brexdth

= (4x) x (3x2)

= (4 x 3) x (x x x2)

= 12 x x3 = 12X3

5.  (3mn, 4np)

Length = 3mn

Brexdth = 4np

Area of the rectangle

= Length x Brexdth

= (Smn) x (4np)

= (3×4) x (mn) x (np)

= 12 x m x (n x n)xp

= 12 mn2p.

Question 3. Complete the table of products.

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Algebraic Expression

Solution:

We have.

1. 2x x -5y = {2 x (- 5)} x (x x y) = – lOxy

2x x 3X2 = (2 x 3) x (x x x2) = 6X3

2x x (- 4xy) = {2 x (-4)} x x x (xy)

= -8 X (x X x) X y

= – 8 x2y

2x X 7x2y = (2 X 7) X (x X x2) x y = 14x2y

2x x (-9X2y2) = {2 x (-9)> x (x x x2) Xy2

= – 18x3y2

2. (-5y)x(2x)

= {(-5) x 2} X (y X x)

= – 10 30

(_5y) x (- 5p)

= {(-5) x (-5)} x y xy

= 25 y2

(~5y) x 3X2

= {(-5) x 3} x yx2

= — 15.x2y

(-5y) x (— 430’)

= {(-5) X (- 4)} x XX (y X y)

= 20 xy2

(-5y) x (7x2y)

= {(-5) X 7} X X2 X (yXy)

= -35 x2y2

(~5y) x (-qx2ÿ)

= {(-5) x (- 9)} x x2x(yxy2)

= 45 x2y3

3. 3x2 x 2x = (3 x 2) x (x2 x x) = 6X2

3X2 x 3x2 = (3 x 3) x (x2 x x2) = 9 x4

3x4 x (- 4 xy) = 3 x (- 4)} x (x2 x x) x y = – 12x3y

3x2 x 7x2y

= (3 x 7) x (x2 x x1) xy

= 21×4;

3x2 x (- 9X2 y2)

= {3 x (- 9)} x (x2 x x2)xy2

= – 27xy

4. (-4xy) x 2x

= {(_4) x 2} x {x x x) xy

= -8x2y

(-4xy) x (-5y)

= {(- 4) x (- 5)} X X x (y x y)

= 20xy2

(-4x2y) x 3X2

= {(- 4) x 3} x (ix x2) x y

= – 12x2y

(-430) x (- 4ry)

= {(-4) X (_4)}x (ixi)x(y Xy)

= 16x2y2

(-4xy) X Ixÿy

= {(-4) X 7} X (x X X2) X (y X y)

= -28x3y2

(- 4xy) X (- 9X2/)

={(-4) x (-9)} X (x X x2) X (y X y2)

= 36x3y3

5. 7x2y x 2x

= (7 x 2) x (x2 x x) x y

= 14x2y

7x2y x (- 5y)

= {7 x (-5)} x x2 X (y Xy)

= – 35x2y2

7x2y X(3x2)

– (7 x 3) X (*2 x x2) Xy = 21x4y

7x2y x (- 4 xy)

= {7 X (-4)} X (x2 x x) X (y X y)

= -28x3y2

7x2y x 7x2p

= (7 x 7) X (x2 X X2) x (y X y)

= 49x4y2

7-X2y X (_ 9x2ÿ2)

= {7X(-9)} X (x2X X2) X (y x y2)

= -63x4y3

(- 9x2,y2) x 2x

= {(-9) x 2} x x( x x) x y2

= -18x3y2

(-9xV) x (-5y)

= {(-9) x (-5)} x x2 x x2 (y2x y)

– 45x2y3

(- 9x2y2) x (3X2)

= {(- 9) x 3} x (x2 x x2) x y2

= – 27x4 y2

(- 9x2y) x (- 4xy)

= {(-9) x (-4)} x (x2 x X) x (y2 x y)

= 36 x4

(~9x2y2) x (7x2y)

= {(-9)x7}x(x2Xx2)X(y2xy)

= – 63 xy

(-9x2y2) x (- 9xY)

= {(-9) x (-9)}x ft? xÿ x (y2 x y2)

= 81 x4y4

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Algebraic Expression

Question 4. Obtain the volume of rectangular boxes with the following length, breadth, and height respectively:

  1. 5x, 3x2, 7x.4
  2. 2p, 4q, 8r
  3. xy, 2x2y, 2xy2
  4. x, 2b, 3c.

Solution:

1.  5x, 3x2, 7×4

Length = 5x

Brexdth = 3x2

Height = 7×4

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (5x) x (3x2) x (7×4)

= (5 x 3 x 7) x (x x x2 x x4)

= 105×7

2.   2p, xq, 8r

Lenqth = 2p

Brexdth = xq

Heiqht = 8r

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (2p) x (xq) x (8r)

= (2 x 4 x 8) x (p x q x r)

= 64 pqr

3.  xy, 2x2y, 2xy2

Length = xy

Brexdth = 2x2y

Height = 2xy2

The volume of the rectangular box

= Length x Breadth x Height

= (xy) x (2x2y) x (2xy2)

= (2 x 2) x (x x x2 x x) x (yxyxy2)

= 4x4y4

4.  x, 2b, 3c

Length = x

Brexdth = 2b

Height = 3c

The volume of the rectangular box

= Length x Breadth x Height

= (x) x (2b) x (3c)

= (2 x 3) x (x x b x c)

= 6xbc

Question 5. Obtain the product of

  1. xy, yz, zx
  2. x, -x2, x3
  3. 2, 4y, 8y2, 16y3
  4. x, 2b, 3c,6xbc
  5. m, -mn, mn

Solution:

1.  xy, yz, zx

Required product

= (xy) x (yz) x (zx)

= (x x X) X (y X y) X (Z X z)

= x2 x x 2z2

= x22y2z2

2.  x, -x2, x3

Required product

= (x) x (-x2) x (x2)

= – (x x x2 x x3)

= -x6

3.  2, 4y, 8y2 16y3

Required product

= (2) x (4y) x (8y2) x (16y3)

= (2 x 4 x 8 x 16) x (y x y2 x y3)

= 1024y6

4.  x, 2b, 3c, 6xbc

Required product

= (x) x (26) x (3c) x (6xbc)

= (2 x 3 x 6) x (x x x) x (6 x 6) x(c x c)

= 36 x2b2c2

5.  m, – mn, mn

Required product

= (m) x (-mn) x (mnp)

= (- 1) x (m x m x m) x (n xn) xp

= -m2n2p

Multiplying A Monomial By A Polynomial

While multiplying x polynomial by x monomial, we multiply every term in the polynomial by the monomial.

Using the distributive law, we carry out the multiplication term by term.
It states that if P, Q, and R are three monomials, then

  1. P x (Q + R) = (P x Q) + (p x R)
  2.  (Q + R) x p = (Q x P) + (R x P)

Question. Find the product:

1.  2x(3x + 5xy) (H)x,2 (2xb – 5c).

Solution:

1.  2x (3x + 5ry)

2x (3x + 5xy) = (2x) x (3x) + (2x) x (5xy)

= (2 x 3) x(xxx) + (2 x 5) x (x x x) x y

= 6x2 + 10 x2y

2.  a2 (2xb-5c)

x2(2xb – 5c) = (x2) x (2xb) – (x2) x (5C)

= (1 x 2) x (x2 x x) X b -(1 X 5) X x2 X c

= 2a3b – 5a2c

8.4.2 Multiplying A Monomixl By A Trinomial

By using the distributive law, we carry out the multiplication term by term

Question. Find the product: (4p2 + 5p + 7) x 3p.
Solution:

(4p2 + 5p + 7) x 3p.

= (4p2x 3p) + (5p x 3p)+ (7 x 3p)

= (4 x 3) x (p2 x P) + (5 x 3) x (p x p) + (7 x 3) x p

= 12p3+ 15p2 + 21p

Addition And Subtraction Of Algebraic Expressions Exercise 8.3

Question 1. Carry out the. multiplication of the expressions in exch for the following pairs:

  1. 4p, q + r
  2. ab, a – b
  3. a + b, 7a2b2
  4. a2-9, 4a
  5. PQ + qr + rp, 0

Solution:

1.   4p, q + r

(4p) x (q+ r) = (dp) x (q) + (4 p) x (r)

By distributive law

= (4 x 1) x p x q + (4 x 1) x p x r

= 4pq + 4pr

(2) xb, x-b

(ab) x (a-b) = (ab) x (a) – (ab) x (b)

By distributive law

= (1 x 1) x (a x a) xb

– (1 x 1) a x (b x b)

= a2b – ab2

3.  a+b, 7a2b2

(x +b) x (7a2b2)

= x x 7a2 b2 + 5 x 7×2 b2

By distributive law

= (1 x 7) x (a x x2) x b2 + (1 x 7) x a2 x (a x b2)

= 7a3b2 + 7a2b3

a2– 9, 4a

(a2– 9) x (4a)

= a2x 4a – 9 x 4a

By distributive law

= ( 1 x 4) x (a2 x a) – (9 x 4) x a

= 4a3-36a

PQ + qr + rp, 0

(PQ + qr+ rp) x (0)

= (pq) x 0 + (qr) x 0 + (rp) x 0.

By distributive law

= 0 + 0 + 0 = 0.

Question 2. Complete the table:

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expression Complete the Table

Solution:

We have,

(1) a x (b + c + d)

= a x b + a x c + a x d

= ab + ac + ad

(2) (X + y – 5) x 5xy

= pC x 5pc_y + y x 5*y + (-5) x 5xy

= (1 X 5) X (x x X) X y

+ (1 x 5) X X X (y X y)

+ {(-5) x 5}x xy

= 5x2y + 5xy2 – 25xy

(3) p x (6p2 – 7p + 5)

= p x 6p2 + p x (- 7p) + p x 5

= (1 x 6) x (p xp2)

+ {1 x (- 7)} x (p x p) + 5 x p

= 6p3 – 7p2 + 5p

(4) 4p2q2 x (p2 – q2)

= 4p2q2 x p2 + 4p2q2 x (-q2)

= (4 x 1) x (p2 x p2) x q2

+ {4 X (- 1)} X p2 X (q2 X q2)

= 4p4 q2 – 4p2 q4.

(5)(a + b + c) x abc

= a x abc + b x abc + c x abc

= (ax a) x b x c

+ a x (b x b) x c

+ a x b x (c x c)

= a2 bc + ab2 c + abc2

NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expression Complete the Tables

Question 3. Find the product:

(a2) x(2a22) x (4a26)

⇒ \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)\)

⇒ \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

x x x2 x x3x x4

Sol. (i) (x2) x (2×22) x (4×26)

= (1 x 2 x 4) x (x2 x x22 x x26)

= 8 X a2+22+26

Bylaws of exponents

= 8 x a50 = 8×50

⇒ \(\left(\frac{2}{3} x y\right) \times\left(-\frac{9}{10} x^2 y^2\right)\)

⇒ \(\left\{\frac{2}{3} \times\left(-\frac{9}{10}\right)\right\} \times\left(x \times x^2\right) \times\left(y \times y^2\right)\)

⇒ \(-\frac{3}{5} x^3 y^3\) I by laws of exponents

3.   \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

⇒ \(=\left\{\left(-\frac{10}{3}\right) \times \frac{6}{5}\right\} \times\left(p \times p^3\right) \times\left(q^3 \times q\right)\)

= -4p4q4.    Bylaws of exponents

(4) X x x2 X x3 X x4

X x x2 x x3 X x4 = (1 X 1 X 1 X 1) X X1 X x2 X x3 X x4

= (1) X x 1+2+3+4

= X10.

Bylaws of exponents

Question 4. (x) Simplify: 3x (4x -5) +3 and find Us values for (i) x = 3, (ii) x =\(\frac{1}{2}\)

b) Simplify: x(x2 + x + 1) + 5 xnd find its vxlue for

(i) x = 0, (ii) x = l xnd (Hi) x=-l.

Solution :

1.  3x (4x-5) + 3

= (3x) (xx) – (3x) (5) + 3

= (3 x 4) x (x x x) – (3 x 5) x x + 3

= 12x2 – 15x + 3

(i) Whenx = 3,

12x2 – 15.x + 3

= 12(3)2 – 15(3) + 3

= 12 x 9 – 45 + 3

= 108 – 45 + 3 = 66

2.  When x =\(\frac{1}{2}\)

12X2 – 15.x + 3

⇒ \(12\left(\frac{1}{2}\right)^2-15\left(\frac{1}{2}\right)+3\)

⇒ \(12\left(\frac{1}{4}\right)-\frac{15}{2}+3\)

⇒ \(3-\frac{15}{2}+3\)

⇒ \(6-\frac{15}{2}\)

⇒ \(\frac{12-15}{2}\)

⇒ \(-\frac{3}{2}\)

(b) a(a2 + x + 1) + 5

a x a2 + a x a + (a x 1 + 5 )

= a3 + a2 + a+ 5

1.  When a = 0

a3 + a2+ a + 5 = (0)3 + (0)2 + (0) + 5

=0+0+0+5=5

2.  When a =1

a3 + a2 + a+ 5 = (l)3 + (l)2 + (1) + 5

=l+l+l+5=8

3.  When a=-l

a3 +a2 + a+ 5

= (- 1)3 + (- l)2 + (- 1) + 5

= —1 + 1 — 1 + 5 = 4

Question  5.

  1. add : p(p – q), q(q – r) and r(r-p)
  2. add: 2x(z – x – y)and 2y (z -y- x)
  3. Subtract: 31 (l – 4m. + 5n) from 41 (10n-3m + 21)
  4. Subtract: 3x(x +b+c)-2b(x-b+c) from 4c(- x + b+ c).

Solution:

(1) First expression

= p(p-q)=pxp-pxq

= p2 – pq

Second expression

= q(q-r)

= q x q – q x r

= q2 – qr

Third expression

= r(r – p)

=r x r – r x p

= r2 – rp

adding the three expressions,

image

First expression

= 2x (z – x – y)

= (2x) X (2)-(2x) X (x) – (2x) X (y)

= 2xz -2x2 – 2xy

Second expression

= 2y(z-y-x)

= (2y) x (z) – (2y) x (y) – (2y)x(x)

= 2yz – 2y2 – 2yx

adding the two expressions,

image

First expression

= 31(1 – 4m + 5n)

= (31) x (l) – (31) x (4m) + (31) x (bn)

= 312 – 12lm + 15In

Second expression

= 41 (10n – 3m + 2l)

= (4l) x (lOn) – (4l) x (3m) + (4l) x (2l)

= 40ln – 12lm + 8l2

Subtrading,

image

First expression

= 3x (x + 6 + c) – 26 (x – 6 + c)

= [(3x) x (x) + (3x) x (6) + (3x) x (c)]

-[(26) x (x) + (26) x (6) – (26) x (c)]

= [3a2 + 3ab + 3ac]- [2ab- 2b2 + 2bc]

= 3a2 + 2b2 + 3ab – 2ab- 2bc + 3ac

– 3a2 + 2b2 + ab- 2bc + 3ac

Second expression

= 4c (- x + 6 + c)

= 4c x (-x) + 4c x 6 + 4c x c

= -4xc + 46c + 4c2

Subtracting

image

8.5 Multiplying A Polynomial By A Polynomial

We multiply the term of one polynomial by each term of the other polynomial. Also, we combine the like terms in the product.

Multiplying A Binomial By A Binomial

We use distributive law and multiply each of the two terms of one binomial by exchanging the two terms of the other binomial and combining like terms in the product.

Thus, if P, Q, R, and S are four monomials, then

(P + Q) x (R + S) = P x (R + S) +Qx(R + S)

= (P x R + P x S) + (QXR + QXS)

= PR + PS + QR + QS.

8.5.2 Multiplying A Binomial By A Trinomial

We use distributive law and multiply each of the three terms in the trinomial by exchanging the two terms in the binomial and combining like terms in the product.

Addition And Subtraction Of Algebraic Expressions Exercise 8.4

Question 1. Multiply the binomial:

  1. (2x + 5) and (4x – 3)
  2. (y – 8) and (3y – 4)
  3. (2.51 – 0.5 m) and (2.51 + 0.5m)
  4. (x + 3b) and (x + 5)
  5. (2pq + 3q2) and (3pq – 2q2)
  6. \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

Solution:

(i) (2x + 5) xnd (4x – 3)

(2x + 5) x (4x – 3)

= (2x) x (4x – 3) + 5 x (4x – 3) by distributive law

= (2x) x (4x) – (2x) x (3) + (5) x (4x) – (5) x (3)

= 8x2 – 6x + 20x – 15

= 8x2 + (20x – 6x) – 15 Combining like terms

= 8x2 + 14x – 15

2. (y – 8) and (3y – 4) by distributive law

(y – 8) x (3y – 4)

= y x (3y – 4) – 8 x (3y – 4) – (8) x (3y) + 8 x 4

= (y) x (3y) – (y) x (4)

= 3y2 – 4y -24y + 32 Combining like terms

3. (2.5 l – 0.5 m) xnd (2.5 l + 0.5 m)

(2.5 l – 0.5 m) x (2.5 l + 0.5 m)

= (2.5/) x (2.5/ + 0.5 m) – (0.5 m) x (2.5/ + 0.5 m) by distributive law

= (2.5 Z) x (2.5 Z) + (2.5 /) x (0.5 m) – (0.5 m) x (2.5 /) – (0.5 m) x (0.5 m)

= 6.25 12 + 1.25 Zm – 1.25 m/- 0.25 m2 Combining like terms

= 6.25 l2 + (1.25 lm – 1.25 lm) – 0.25 m2

= 6.25 l2 – 0.25 m2

4. (x + 3 b) and (x + 5)

(a + 35) x (x + 5)

= a x (c +5) + (3b) x (x + 5) by distributive law

= (a) x (x) + (a) x (5) + (3b) x (x) + (3b) x (5)

= ax + 5a + 3bx + 15b

5.  (2pq + 3q2) xnd. (3pq – 2q2)

(2pq + 3q2) x (3pq – 2q2)

= (2pq) x (3pq – 2q2) + (3q2) x (3pq – 2q2) by distributive law

= (2pq) x (3pq) – (2pq) x (2q2) + (3q2) x (3pq) – (3q2) x (2q2)

= 6p2q2 – 4pq3 + 9pq3 – Qqi

= 6p2q2 + (9pq3 – 4pq3) – 6q4 Combining like terms

= 6p2q2 + 5pqz – 6q4

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \times 4\left(x^2-\frac{2}{3} b^2\right)=\left(\frac{3}{4} x^2+3 b^2\right) \times\left(4 x^2-\frac{8}{3} b^2\right)\)

⇒ \(\frac{3}{4} x^2 \times\left(4 x^2-\frac{8}{3} b^2\right)+3 b^2 \times\left(4 x^2-\frac{8}{3} b^2\right)\) by distributive law

⇒ \(\left(\frac{3}{4} x^2\right) \times\left(4 x^2\right)-\left(\frac{3}{4} x^2\right) \times\left(\frxc{8}{3} b^2\right)+\left(3 b^2\right) \times\left(4 x^2\right)-\left(3 b^2\right) \times\left(\frac{8}{3} b^2\right)\)

= 3a4 – 2a2b2 + 12b2a2 – 8b4

= 3a4 + (12a2b2 -2a2b2) -8b4

= 3a4 + 10a2b2 – 8b4Combining like terms

Question 2. Find the product:

  1. (5 – 2x) (3 + x)
  2. (x + 7y) (7x-y)
  3. (x2 + b) (x + b2)
  4. (p2 – q2) (2p + q)

Solution:

(i) (5 – 2x) (3 + x)

= (5) x (3 + x) – (2x) x (3 + x)

= (5) x (3) + (5) x (x) – (2x) X (3) – (2x) X (X)

= 15 + 5x – 6x – 2x2

= 15 – x – 2x2

2.  (x + 7y) (7x – y)

= (x) x (7x – y) + (7y) x (7x – y)

= (x) x (7x) – (x) x (y) + (7y) x (7x) – (7y) x (y)

= 7x2 – xy + 49yx – 7y2

= 7x2 + 48xy – 7y2

3.  (a2 + b) (a + b2)

= a2 x (a+ b2) + b x (a + b2)

= (a2) x (a) + (a2) x (b2) + (b) x (a) + (b) x (b2)

= a3 + a2b2 + ab + b3

= a3 + a2b2 + ab + b3

4.  (p2 – q2) (2p + q)

= p2 x (2p + q) – q2 x (2p + q)

= (p2) x (2p) + (p2) x (q) – (q2) x (2p)- (q2) x (q)

= 2p3+ p2q – 2q2p – q3

= 2p3 + p2q – 2pq2 – q3

Question 3. Simplify :

  1. (x2 – 5) (x+ 5) + 25
  2. (x2 + 5) (b3 + 3) + 5
  3. (l + s2) (l2 – s)
  4. (x + b)(c – d) + (x – 6) (c + d)+ 2 (xc + bd)
  5. (x + y) (2x + y) + (x + 2y)(x – y)
  6.  (x + y) (x2 – xy + y2)
  7. (1.5x – 4y)(1.5x + 4y + 3)- 4.5x + I2y
  8. (x + b + c) (x. + b – c).

Solution:

1.   (x2 – 5) (x + 5) + 25

= x2(x + 5) – 5(x + 5) + 25

= (x2 x X) + (x2 x 5) – (5 x X) – (5 x 5) + 25

= x3 + 5X2 – 5x – 25 + 25

= x3 + 5×2 – 5x

2.  (a2 + 5) (b3 + 3) + 5

= a2(b3 + 3) + 5(b3 + 3) + 5

= (a2 x b3) + (a2 x 3) + (5 x b3) +(5a3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 3a2 + 5b3 + 20

3.  (t + s2) (t2 – s)

= t(t2 – s) + s2(t2 – s)

= (tx t2) – (t X s) + (s2 X t2) – (s2 X s)

= t2 -ts + s2t2 – s3

4.  (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2(oc + bd)

= ac – ad + be – bd + ac + ad – be – bd + 2ac + 2bd by distributive law

= (ac + ac + 2ac) + (ad- ad) + (be- be) + (2bd – bd – bd)

= 4ac combines the like terms

5.  (x + y) (2x + y) + (x + 2y) (x – y)

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)

= (X X 2x) + (x X y) + (y X 2x) + (y x y) + (x x x) – (x X y) (2y x x)- (2y x y

By distributive law

– 2×2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2

= (2x2 + x2) + (xy + 2xy – xy + 2xy) + (y2 – 2y2)

= 3x2 + 4xy – y2

6.  (x + y) (x2 – xy + y2)

x(x2 – xy + y2)= y(x2 – xy + y2)

(x x x2) – (x x xy) + (x x y2) + (y x x2) – (y x xy) + (y x y2 ) | by distributive law

x3 – x2y + xy2 + x2y – xy2 + y3

= x-3 + x2y – x2y) + (xy2 – xy2) + y3   combining the like terms

= x3 + y2

7.  (1.5x – 43-) (1.5x + 4y + 3) – 4.5x + 12y

= 1.5x (1.5x + 4y + 3)- 4y (1.5x + 4y + 3) – 4.5x + 12y | by distributive law

= (1.5x x 1.5x) + (1.5x x 4y) + (1.5x X 3) – (4y x 1.5x) – (4y x 4y) – (4y x 3) – 4.5x + 12y | by distributive law

2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y

= 2.25x2 + (6xy – 6xy) – 16y2 + (4.5x – 4.5x) + (12y – 12y) I combining the like terms

= 2.25x2 – 16y2

8.  (a + b + c) (a + b – c)

= a(a + b-c) + b(a + b-c) + c(a + 6-c) I by distributive law

= a2 + ab – ac + ab + b2 – be + ae + be – c2    I by distributive law

= a2+ (ab + ab) + (ac – ac) + b2 + (be – be) – c2 combining the like terms

= a2 + 2ab + b2 – c2

Addition And Subtraction Of Algebraic Expressions Multiple-Choice Questions And Solutions

Question 1. The expression x + 3 is in

  1. one variable
  2. two variables
  3. no variable
  4. none of these.

Solution: 1. one variable

Question 2. The expression 4xy + 7 is in

  1. one variable
  2. two variables
  3. no variable
  4. none of these.

Solution: 2. two variables

Question 3. The expression x + y + z is in

  1. one variable
  2. no variable
  3. three variables
  4. two variables.

Solution: 3. three variables

Question 4. The value of 5x when x = 5 is

  1. 5
  2. 10
  3. 25
  4. -5.

Solution: 3. 25

Question 5. The value of x2 – 2x + 1 when x = 1 is

  1. l
  2. 2
  3. -2
  4. 0.

Solution: 4. 0

Question 6. The value of x2 + y2 when x = 1, y = 2 is

  1. l
  2. 2
  3. 4
  4. 5.

Solution: 4. 5

Question 7. The value of x2 – 2yx + y2 when x = 1, y = 2 is

  1. l
  2. -l
  3. 2
  4. -2.

Solution: 1. 1

Question 8. The value of x2 – xy + y2 when x = 0, y = 1 is

  1. 0
  2. -l
  3. l
  4. none of these.

Solution: 3. 1

Question 9. The sum of lx, lOx, and 12x is

  1. 17x
  2. 22x
  3. 19x
  4. 29x.

Solution: 4. 29x.

Question 10. The sum of 8pq and -17pq is

  1. pg
  2. 9pqr
  3. -9pq
  4. -pg.

Solution: 3.-9pq

Question 11. The sum of 5x2, -7X2, 8x2, 11x2 and -9x2 is

  1. 2x2
  2. 4x2
  3. 6x2
  4. 8x2

Solution: 4. 8x2.

Question 12. The sum ofx2 y2-z2 and z2-x2 is

  1. 0
  2. 3x2
  3. 3y2
  4. 3z2

Solution: 1. 0

Question 13. What do you get when you subtract – 3xy from 5xy?

  1. 3xy
  2. 5xy
  3. 8xy
  4. xy.

Solution: 3. 8xy

Question 14. The result of the subtraction of law from 0 is 4

  1. 0
  2. 7x
  3. -7x
  4. x

Solution: 3. -7x

Question 15. The result of subtraction of 3x from – 4x is

  1. -7x
  2. 7x
  3. x
  4. -x.

Solution: 1. -7x

Question 16. The product of 4mn and 0 is

  1. 0
  2. 1
  3. mn
  4. 4mn.

Solution: 1. 0

Question 17. The product of 5x and 2y is

  1. xy
  2. 2×7
  3. 5xy
  4. 10xy.

Solution: 4. 10xy.

Question 18. The product of lx and -12x is

  1. 84x2
  2. -84x2
  3. x2
  4. -x2.

Solution: 2. -84x2

Question 19. The area of a rectangle whose length and breadth are 9 and 4y respectively is

  1. 4y3
  2. 9y3
  3. 36y3
  4. 13y3

Solution: 3. 36y3

Question 20. The area of a rectangle with length 2l2m and breadth 3Im? is

  1. 6l3m3
  2. l3m3
  3. 2l3m3
  4. 4l3m3

Solution: 1. 6l3m3

Question 21. The volume of a cube of side 2a is

  1. 4a3
  2. 2a
  3. 8a3
  4. 8.

Solution: 3. 8a3

Question 22. The volume of a cuboid of dimensions a, 6, c is

  1. abc
  2. a2b2c2
  3. a3b3c3
  4. none of these.

Solution: 1. abc

Question 23. The product of x2, – x3, – x4 is

  1. x9
  2. x5
  3. x7
  4. x6

Solution: 1. x9

Addition And Subtraction Of Algebraic Expressions True-False

Write whether the following statements are True or False:

1. In a polynomial, the exponents of the variables are always non-negative integers:   True

2. a (b + c) = ab + ae is called the distributive property : True

3. a2 – b2 is the product of (a – b) and (a + b): True

Addition And Subtraction Of Algebraic Expressions Fill in the Blanks

1. The value of, when 92 -72 = 8p, is___________ : 4

2. On subtracting -a2b2 from 2ab2, we get___________:  2a2b2

3. The difference between the squares of two consecutive natural numbers is their_______: Sum

4. Area of a rectangle with length 3 ab2 and breadth 4 ac2 is_______:  12a2b2c2

5. Write the product of the monomials -12a, -15a2, a2b:  180 a3b3

6. Add a2 + b2 – c2  b2  + c2 – a2  and c2  + a2 – b2a2 + b2 + c2

7. Find the value of, if lOOOy = (981)2 – (19)2962

NCERT Exemplar Solutions For Class 6 Maths Chapter 12 Ratio And Proportion

Class 6 Maths Chapter 12 Ratio And Proportion

Question 1. The ratio of books to 20 books is

  1. 2:5
  2. 5:2
  3. 4:5
  4. 5:4

Solution: (1): The ratio of 8 books to 20 books

= 8 books: 20 books

⇒ \(=\frac{8}{20}=\frac{2}{5}=2: 5\)

Question 2. The ratio of the number of sides of a square to the number of edges of a cube is

  1. 1:2
  2. 3:2
  3. 4:1
  4. 1 :3

Solution: (4) : Required ratio = \(\frac{\text { Number of sides of a square }}{\text { Number of edges of a cube }}=\frac{4}{12}\)

⇒ \(=\frac{1}{3}=1: 3\)

Read and Learn More Class 6 Maths Exemplar Solutions

Question 3. A picture is 60 cm wide and 1.8 m long. The ratio of its width to its perimeter in the lowest form is

  1. 1:2
  2. 1:3
  3. 1:4
  4. 1:8

Solution:

(4) : Width of the picture = 60 cm and length of the picture = 1.8 m = 1.8 x 100 cm = 180 cm

Primetor of the picture = 2* (180 + 60) cm

= 2 x 240 cm = 480 cm

The required ratio \(=\frac{\text { Width of the picture }}{\text { Perimeter of the picture }}\)

⇒ \(\frac{60 \mathrm{~cm}}{480 \mathrm{~cm}}=\frac{1}{8}=1: 8\)

Question 4. Neelam’s annual income is ₹  288000. Her annual savings amount to ₹  36000. The ratio of her savings to her expenditure is
Solution:

  1. 1:8
  2. 1:7
  3. 1:6
  4. 1:5

Solution: (2):

Given

Neelam’s annual income =₹ 288000

Her savings =₹  36000

Her expenditure = Annual income – Savings

= ₹  288000- ₹  36000 =₹  252000

The required ratio = \(=\frac{\text { Savings }}{\text { Expenditure }}\)

⇒ \(=\frac{₹ 36000}{₹ 252000}=\frac{1}{7}=1: 7\)

Question 5. The mathematics textbook for Class VI has 320 pages. The chapter’symmetry’runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is

  1. 11: 320
  2. 3: 40
  3. 3: 80
  4. 272: 320

Solution: (3):

Given

The mathematics textbook for Class VI has 320 pages. The chapter’symmetry’runs from page 261 to page 272.

Total number of pages = 320

The number of pages of the chapter ‘symmetry’ = 272- 261 +1 = 12

The required ratio \(=\frac{\text { Number of pages of chapter ‘symmetry’ }}{\text { Total number of pages of the book }}\)

⇒ \(=\frac{12}{320}=\frac{3}{80}=3: 80\)

NCERT Exemplar Solutions For Class 6 Maths Chapter 12 Ratio And Proportion

Question 6. In a box, the ratio of red marbles to blue marbles is 7: 4. Which of the following could be the total number of marbles in the box?

  1. 18
  2. 19
  3. 21
  4. 22

Solution: (4):

Given

In a box, the ratio of red marbles to blue marbles is 7: 4.

Since the sum of the given ratio is 7 + 4 = 11 and from the given options, only 22 is Divisible by 11.

Option (D) is correct.

Question 7. On a shelf, books with green covers and those with brown covers are in the ratio 2 : 3. If there are 18 books with green covers, then the number of books with brown covers is

  1. 12
  2. 24
  3. 27
  4. 36

Solution:

Given

On a shelf, books with green covers and those with brown covers are in the ratio 2 : 3. If there are 18 books with green covers

(3): Let the number of green cover and brown cover books be 2x and 3x respectively. Since, the number of green cover books = 18

⇒  2x = 18

⇒ \( x=\frac{18}{2}=9\)

The number of brown cover books = 3×9 = 27

Question 8. The greatest ratio among the ratios 2 : 3, 5: 8, 75:121 and 40: 25 is

  1. 2 : 3
  2. 5: 8
  3. 75: 121
  4. 40: 25

Solution: (4): we have \(2: 3=\frac{2}{3}, 5: 8=\frac{5}{8}, 75: 121=\frac{75}{121}\) and \(40: 25=\frac{40}{25}\)

Ratio and Proportion The greatest ratio among the ratios

The LCM of 3, 8, 121 and 25 is 2 x 2 x 2 x 3 x 5 x 5x11x11 = 72600

Making the denominator of each ratio equal to 72600, we get

⇒ \(\frac{2}{3}=\frac{2 \times 24200}{3 \times 24200}=\frac{48400}{72600}\)

⇒ \(\frac{5}{8}=\frac{5 \times 9075}{8 \times 9075}=\frac{45375}{72600}\)

⇒ \(\frac{75}{121}=\frac{75 \times 600}{121 \times 600}=\frac{45000}{72600}\)

⇒ \(\frac{40}{25}=\frac{40 \times 2904}{25 \times 2904}=\frac{116160}{72600}\)

On comparing the numerator of the above ratios, we get \(\frac{40}{25}\) is the greatest ratio, i.e., 40: 25 is the greatest ratio among the given ratios.

Question 9. There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is: 20

  1. \(\frac{b}{b+g}\)
  2. \(\frac{g}{b+g}\)
  3. \(\frac{b}{g}\)
  4. \(\frac{b+g}{b}\)

Solution: (1) :

Given

There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is: 20

Total number of students = number of boys + number of girls = b+g

The required ratio \(=\frac{\text { Number of boys }}{\text { Total number of students }}\)

⇒ \(\frac{b}{b+g}\).

Question 10. If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then the ratio of the distance travelled by them in one hour is
Solution:

  1. 1: 2
  2. 4: 5
  3. 5: 8
  4. 8: 5

Solution: (3):

Given

If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds

Distance travelled by bus in 4 hours = 160 km

Distance travelled by bus in 1 hour \(=\frac{160}{4} \mathrm{~km}=40 \mathrm{~km}\)

Distance travelled by train in 5 hours = 320 km

Distance travelled by train in1 hour \(=\frac{320}{5} \mathrm{~km}=64 \mathrm{~km}\)

The required ratio = \(\frac{40 \mathrm{~km}}{64 \mathrm{~km}}=\frac{5}{8}=5: 8\)

Question 11. Algebra 11
Solution:

To get the missing number, we consider the fact that 4 * 5 = 20, i.e., we get 20 when we multiply 5 by 4. This indicates that to get the missing number, 3 must also be multiplied by 4.

When we multiply, we have, 3 x 4 = 12

Hence, the ratio is \(\frac{12}{20}.\)

Ratio and Proportion In order to get the missing number

Question 12. Ratio and Proportion In order 12
Solution:

To get the missing number, we consider the fact that 9 * 2 = 18, i.e., when we multiply 9 by 2 we get 18. This indicates that to get the missing number, 2 must also be multiplied by 2.

When we multiply, we have, 2×2 = 4

Hence, the ratio is \(\frac{4}{18}.\)

Ratio and Proportion In order to get the missing

Question 13. Ratio and Proportion In order 13
Solution:

To get the missing number, we consider the fact that 3.2 x 2.5 = 8, i.e., when we multiply 3.2 by 2.5 we get 8. This indicates that to get the missing number, 4 must also be multiplied by 2.5.

When we multiply, we have, 4 x 2.5 = 10

Hence, the ratio is \(\frac{8}{10}.\)

Ratio and Proportion In This indicates missing number

Question 14. Ratio and Proportion In order 14
Solution:

To get the missing number, we consider the fact that 40 x 1.125 = 45, i.e., when we multiply 40 by 1.125 we get 45. This indicates that to get the missing number of the first ratio, 16 must also be multiplied by 1.125.

When we multiply, we have, 16 * 1.125 = 18.

Hence, the first ratio is \(\frac{18}{45}\)

Similarly, to get the third ratio we multiply both terms of the second ratio by 1.5.

Hence, tine third ratio is \(\frac{24}{60}\)

Ratio and Proportion 14

Question 15.Ratio and Proportion In order 15
Solution:

To get the missing number, we consider the fact that 36 x 1.75 = 63, i.e., we get 63 when we multiply 36 by 1.75. This indicates that to get the missing number of the second ratio, 16 must also be multiplied by 1.75.

When we multiply, we have, 16 * 1.75 = 28

Hence, the second ratio is \(\frac{28}{63} \text {. }\)

Similarly, to get the third ratio we multiply both terms of the first ratio by 2.25

Hence, the third ratio is \(\frac{36}{81} \text {. }\)

And to get the fourth ratio we multiply both terms of the first ratio by 3.25.

Hence, the fourth ratio is \(\frac{52}{117}\)

Ratio and Proportion 15

Question 16. \(\frac{3}{8}=\frac{15}{40}\)
Solution: True

We have, \(\frac{15}{40}=\frac{3}{8}\)

Question 17. \(4: 7=20: 35\)
Solution: True

We have, \(20: 35=\frac{20}{35}=\frac{4}{7}=4: 7\)

Question 18. 0.2:5 = 2:0.5
Solution: False

⇒ \(0.2: 5=\frac{0.2}{5}=\frac{2}{50}=\frac{1}{25}\)

⇒ \(2: 0.5=\frac{2}{0.5}=\frac{20}{5}=\frac{4}{1}\)

Since, \(\frac{1}{25} \neq \frac{4}{1}\)

0.2: 5* 2: 0.5

Question 19. 3:33 = 33:333
Solution: False

We have \(3: 33=\frac{3}{33}=\frac{1}{11}\)

⇒ \(\text { and } 33: 333=\frac{33}{333}=\frac{11}{111}\)

Since, \(\frac{1}{11} \neq \frac{11}{111}\)

3: 33 x 33: 333

Question 20. 1 5 m : 40 m = 35 m: 65 m
Solution: False

We have \(15 \mathrm{~m}: 40 \mathrm{~m}=\frac{15}{40}=\frac{3}{8}\)

⇒ \(\text { and } 35 \mathrm{~m}: 65 \mathrm{~m}=\frac{35}{65}=\frac{7}{13}\)

Since,\(\frac{3}{8} \neq \frac{7}{13}\)

15 m : 40 m* 35 m : 65 m

Question 21. 27 cm2: 57 cm2 = 18 cm: 38 cm
Solution: True

We have, 27 cm2: 57 cm2

⇒ \(=\frac{27}{57}=\frac{9}{19}\)

and 18 cm : 38 cm \(=\frac{18}{38}=\frac{9}{19}\)

27 cm2: 57 cm2 = 18 cm : 38 cm

Question 22. 5 kg: 7.5 kg = 7.50:? 5
Solution: False

We have, 5 kg: 7.5 kg

⇒ \(=\frac{5}{7.5}=\frac{50}{75}=\frac{2}{3}\)

and ? 7.50:? 5 =\(\frac{7.50}{5}=\frac{750}{500}=\frac{3}{2}\)

5 kg: 7.5 kg*? 7.50:? 5

Question 23. 20 g : 100 g = 1 metre: 500 cm
Solution: True

We have, \(20 \mathrm{~g}: 100 \mathrm{~g}=\frac{20}{100}=\frac{1}{5}=1: 5\) and 1 metre : 500 cm = 100 cm : 500 cm

⇒ \(=\frac{100}{500}=\frac{1}{5}=1: 5\)

20 g : 100 g =1 metre : 500 cm

Question 24. 1 2 hours: 30 hours = 8 km: 20 km
Solution: True

We have 12 hours : 30 hours = \(=\frac{12}{30}\)

⇒ \(=\frac{2}{5}=2: 5\)

⇒ \(\text { and } 8 \mathrm{~km}: 20 \mathrm{~km}=\frac{8}{20}=\frac{2}{5}=2: 5\)

12 hours: 30 hours- 8 km: 20 km

Question 25. The ratio of 0 kg to 1 00 kg is 1: 1 0
Solution: True

The ratio of 10 kg to 100 kg \(=\frac{10}{100}=\frac{1}{10}=1: 10\)

Question 26. The ratio of 1 50 cm to 1 metre is 1: 1 .5
Solution: False

The ratio of 150 cm to 1 metre = 150 cm : 1 metre = 150 cm : 100 cm

⇒ \(=\frac{150}{100}=\frac{3}{2}=3: 2\)

⇒ \(\text { and } 1: 1.5=\frac{1}{1.5}=\frac{10}{15}=\frac{2}{3}=2: 3\)

Since, 3:2*2:3

150 cm : 1 metre*1 : 1.5

Question 27. 25 kg: 20 g = 50 kg : 40 g
Solution: True

We have, 25 kg : 20 g = 25 * 1000 g : 20 g

⇒ \(=\frac{25000}{20}=\frac{1250}{1}=1250: 1\)

and 50 kg : 40 g = 50 x 1000 g : 40 g

⇒ \(=\frac{50000}{40}=\frac{1250}{1}=1250: 1\)

25 kg : 20 g = 50 kg : 40 g

Question 28. The ratio of 1 hour to one day is 1: 1.
Solution: False

The ratio of1 hour to 1 day =1 hour : 1 day

= 1 hour : 24 hours = \(=\frac{1}{24}=1: 24 \neq 1: 1\)

Question 29. The ratio of 4: 16 is in its lowest form
Solution: False

We have, 4:16 =\(\frac{4}{16}=\frac{1}{4}=1: 4\)

4: 16 is not in its lowest form.

Question 30. The ratio 5: 4 is different from the ratio 4: 5.
Solution: True

We have, 5 : 4 = \(\frac{5}{4} \text { and } 4: 5=\frac{4}{5}\)

5: 4 x 4: 5

Question 31. A ratio will always be more than 1.
Solution: False

A ratio may be less than 1.

Question 32. A ratio can be equal to 1.
Solution: True

Question 33. If b: a = c: d, then a, b, c, d are in proportion.
Solution: False

If b: a = c : d, then b, a, c, d are in proportion.

34. The two terms of a ratio can be in two different units_____.
Solution: False

Question 35. A ratio is a form of comparison by________.
Solution: Division

Question 36. 20 m : 70 m = ?8: ?________.
Solution:

28: We have, 20 m:70 m \(=\frac{20}{70}=\frac{2}{7}=2: 7\)

Ratio and Proportion In order 36

Ratio and Proportion In order 36

Ratio and Proportion In order 36.1

20 m : 70 m = ? 8 : ? 28

Question 37. There is a number in the box such thatRatio and Proportion Rectangle, 24, 9, and 12 are in proportion. The number in the box is _______.
Solution:

18: Since, 24, 9, and 12 are in proportion.

Ratio and Proportion 37

Ratio and Proportion 37..

Ratio and Proportion 37.1

Question 38. If two ratios are equal, then they are in______.
Solution: Proportion

Ratio and Proportion Proportion

Question 39. The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is______.
Solution:

3:7: Since each square is of unit length. The perimeter of the shaded portion is 6 units and the perimeter of the whole figure is 14 units.

The required ratio = \(\frac{6}{14}=\frac{3}{7}=3: 7\)

Question 40. The ratio of the area of the shaded portion to that of the whole figure is________.
Solution:

1 : 6 : Area of shaded portion = (1 unit) x (2 units) = 2 sq units

and area of whole figure = (3 units) x (4 units) = 12 sq units

The required ratio = \(\frac{2}{12}=\frac{1}{6}=1: 6\)

Question 41. The sleeping time of a python in a 24-hour clock is represented by the shaded portion in the given figure.

Ratio and Proportion Sleeping time of a python

The ratio of sleeping time to awaking time is______.

Solution: 3:1: Sleeping time = 18 hours and awaking time = (24- 18) hours = 6 hours

The required ratio = \(\frac{18}{6}=\frac{3}{1}=3: 1\)

Question 42. A ratio expressed in the lowest form has no common factor other than _____ in its terms.
Solution: One

43. To find the ratio of two quantities, they must be expressed in______ units.
Solution: Same

Question 44. The ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to_____.
Solution: 100 paise or 1 rupee : We have, 5 paise : 25 paise = 20 paise : Ratio and Proportion Rectangle

Ratio and Proportion Rectangles

Ratio and Proportion Rectangles

Ratio and Proportion Rectangle Paise

5 paise : 25 paise = 20 paise : 100 paise

Question 45. Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in the lowest form is_______.
Solution:

Given

Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes.

149 : 160: Time taken by Saturn

= 9 hours 56 minutes

= (9 x 60 + 56) minutes

= (540 + 56) minutes

= 596 minutes

Time taken by Jupiter = 10 hours 40 minutes

= (10 x 60 + 40) minutes

= 640 minutes

The required ratio = \(=\frac{\text { Time taken by Saturn }}{\text { Time taken by Jupiter }}\)

⇒ \(=\frac{596 \text { minutes }}{640 \text { minutes }}=\frac{149}{160}=149: 160\)

Question 46. 10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda. The amount of caustic soda needed for 1 litre of water to make the same type of solution is______.
Solution:

Given

10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda.

100 g: Let the required amount of caustic soda be x g.

According to the question,

10 g: x g = 100 mL : 1 L

=> 10 g : .v g = 100 mL : 1000 mL

⇒ \(\frac{10}{x}=\frac{100}{1000} \Rightarrow \frac{10}{x}=\frac{1}{10}\)

X 1-10×10 ⇒ x — 100

The required amount of caustic soda is 100 g.

Question 47. The marked price of a table is ₹ 625 and its sale price is ₹ 500. What is the ratio of the sale price to the marked price?
Solution:

Given

The marked price of a table is ₹625 and its sale price is ₹500.

4:5: We have given the marked price = ₹625 and selling price = ₹500

The required ratio \(=\frac{\text { Selling price }}{\text { Marked price }}\)

⇒ \(=\frac{₹ 500}{₹ 625}=\frac{4}{5}=4: 5\)

Question 48. Which pair of ratios are equal? And why?

  1. \(\frac{2}{3}, \frac{4}{6}\)
  2. \(\frac{8}{4}, \frac{2}{1}\)
  3. \(\frac{4}{5}, \frac{12}{20}\)

Solution:

1. We Have, \(\frac{2}{3} \text { and } \frac{4}{6}\)

The simplest form of \(\frac{4}{6} \text { is } \frac{2}{3}\)

⇒ \(\frac{2}{3}=\frac{4}{6}\)

2. We have, \(\frac{8}{4} \text { and } \frac{2}{1}\)

The simplest form of\(\)

⇒ \(\frac{8}{4}=\frac{2}{1}\)

3. We have, \(\frac{4}{5} \text { and } \frac{12}{20}\)

The simplest form of \(\frac{12}{20} \text { is } \frac{3}{5}\)

⇒ \(\frac{4}{5} \neq \frac{12}{20}\)

Question 49. Which ratio is larger 1 0: 21 or 21: 93?
Solution:

We have, 10: 21 \(=\frac{10}{21} \text { and } 21: 93=\frac{21}{93}\)

Ratio and Proportion In order ratio is larger

Now, the I, CM of 21; and 93 is 3 * 7 x 31 – 651 Mnking the denominator of each ratio equal to 651, we have

⇒ \(\frac{10}{21}=\frac{10 \times 31}{21 \times 31}=\frac{310}{651}\)

⇒ \(\text { and } \frac{21}{93}=\frac{21 \times 7}{93 \times 7}=\frac{147}{651}\)

⇒ \(\frac{310}{651}>\frac{147}{651}\)

⇒ \(\frac{10}{21}>\frac{21}{93}\)

Thus, the ratio 10: 21 is larger.

Question 50. Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio of 7: 2. How much Khoya did she use?
Solution:

Given

Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio of 7: 2.

The ratio of Khoya to sugar = 7:2

Let the quantity of Khoya be 7x kg and sugar be 2x kg.

7x + 2x =18 => 9x =18 => x = \(\frac{18}{9}=2\)

The quantity of Khoya be (7 x 2) kg = 14 kg

Question 51. A line segment 56 cm long is to be divided into two parts in the ratio of 2: 5. Find the length of each part.
Solution:

Given

A line segment 56 cm long is to be divided into two parts in the ratio of 2: 5.

The length of the line segment = 56 cm

Let the lengths of the two parts be 2 cm and 5 cm

2x + 5x = 56

7x = 56 ⇒ x \(=\frac{56}{7}=8\)

The length of one part is (2 x 8) cm =16 cm and length of other part is (5 x 8) cm = 40 cm

Question 52. The number of milk teeth in human beings is 20 and the number of permanent teeth is 32. Find the ratio of the number of milk teeth to the number of permanent teeth.
Solution:

Given

The number of milk teeth = 20

and the number of permanent teeth = 32

The required ratio \(=\frac{\text { Number of milk teeth }}{\text { Number of permanent teeth }}\)

⇒ \(\frac{20}{32}=\frac{5}{8}=5: 8\)

Question 53. The sex ratio is defined as the number of females per 1,000 males in the population. Find the sex ratio if there are 3732 females per 4000 males in a town.
Solution:

Given

The sex ratio is defined as the number of females per 1,000 males in the population.

Number of females = 3732

Number of males = 4000

The required sex ratio = \(\frac{3732}{4000}=\frac{933}{1000}\)

Thus, there are 933 females per 1000 males.

Question 54. In a year, Ravi earns? 360000 and paid? 24000 as income tax. Find the ratio of his

  1. Income to income tax.
  2. Income tax to income after paying income tax.

Solution: Ravi’s annual income =? 360000

Income tax paid by him = ? 24000

1. The required ratio \(=\frac{\text { The amount of income }}{\text { The amount of income tax }}\)

⇒ \(=\frac{₹ 360000}{₹ 24000}=\frac{15}{1}=15: 1\)

After paying income tax, the remaining amount of income =? (360000- 24000) = ? 336000

The required ratio \(\begin{aligned}
& =\frac{\text { The amount of income tax }}{\text { The remaining amount of income }} \\
& \text { after paying the income tax }
\end{aligned}\)

⇒ \(=\frac{₹ 24000}{₹ 336000}=\frac{1}{14}=1: 14\)

Question 55. Ramesh earns₹ 28000 per month. His wife Rama earns₹ 36000 per month. Find the ratio of

Ramesh’s earnings to their total earnings

Rama’s earnings to their total earnings.

Ramesh’s monthly earnings =₹ 28000

Rama’s monthly earnings =₹ 36000

Their total earnings =₹ (28000 + 36000) = ₹ 64000

1. Required ratio = \(=\frac{\text { Ramesh’s monthly earnings }}{\text { Their total monthly earnings }}\)

⇒ \(=\frac{₹ 28000}{₹ 64000}=\frac{7}{16}=7: 16\)

Question 56. Required ratio

⇒ \(=\frac{\text { Rama’s monthly earnings }}{\text { Their total monthly earnings }}\)

⇒ \(=\frac{₹ 36000}{₹ 64000}=\frac{9}{16}=9: 16\)

Question 56. Of the 288 persons working in a company, 11 2 are men and the remaining are women. Find the ratio of the number of

  1. men to that of women.
  2. men to the total number of persons.
  3. women to the total number of persons

Solution:

Total number of persons working in a company = 288

Number of men = 112

Number of women = 288 -112 = 176

Required ratio=

⇒ \(=\frac{\text { Number of men }}{\text { Number of women }}=\frac{112}{176}\)

⇒ \(=\frac{7}{11}=7: 11\)

Required ratio \(=\frac{\text { Number of men }}{\text { Total number of persons }}\)

⇒ \(=\frac{112}{288}=\frac{7}{18}=7: 18\)

Required ratio \(=\frac{\text { Number of women }}{\text { Total number of persons }}\)

⇒ \(=\frac{176}{288}=\frac{11}{18}=11: 18\)

Question 57. A rectangular sheet of paper is of length 1.2 m and width 21 cm. Find the ratio of the width of the paper to its length.
Solution:

Given

A rectangular sheet of paper is of length 1.2 m and width 21 cm.

Length of paper = 1.2 m = 1.2 x 100 cm = 120 cm

And the width of the paper = is 21 cm

The required ratio = \(=\frac{\text { width of paper }}{\text { length of paper }}\)

⇒ \(=\frac{21 \mathrm{~cm}}{120 \mathrm{~cm}}=\frac{7}{40}=7: 40\)

Question 58. A scooter travels 1 20 km in 3 hours and a train travels 1 20 km in 2 hours. Find the ratio of their speeds.

\(\left(\text { Hint: Speed }=\frac{\text { distance travelled }}{\text { time taken }}\right)\)

Solution:

Given

A scooter travels 1 20 km in 3 hours and a train travels 1 20 km in 2 hours.

We know that Speed = \(\frac{\text { Distance }}{\text { Time }}\)

The speed of the scooter = \(\frac{120 \mathrm{~km}}{3 \text { hours }}\)

= 40 km/hour

And the speed of the train = \(\frac{120 \mathrm{~km}}{2 \text { hours }}\)

Thus, the required ratio \(=\frac{\text { Speed of the scooter }}{\text { Speed of the train }}\)

⇒ \(=\frac{40 \mathrm{~km} / \text { hour }}{60 \mathrm{~km} / \text { hour }}=\frac{2}{3}=2: 3\)

Question 59. An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes. What is the ratio of lunch breaks to the total period in the office?
Solution:

Given

An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes.

Opening time of office = 9 a.m.

Closing time of office = 5.30 p.m.

Total period in office = 8 hours 30 minutes

= (8 x 60 + 30) minutes = 510 minutes

The duration of lunch break = 30 minutes

The required ratio \(=\frac{\text { Duration of lunch break }}{\text { Total time period in office }}\)

⇒ \(=\frac{30 \text { minutes }}{510 \text { minutes }}=\frac{1}{17}=1: 17\)

Question 60. The shadow of a 3 m long stick is 4 m long. At the same time of the day, if the shadow of a flagstaff is 24 m long, how tall is the flagstaff?
Solution:

Given

The shadow of a 3 m long stick is 4 m long. At the same time of the day, if the shadow of a flagstaff is 24 m long

Let the length of the flagstaff be x m.

According to the given question,

3m : xm:: 4m: 24m

⇒ \(\frac{3}{x}=\frac{4}{24} \Rightarrow 4 \times x=3 \times 24\)

⇒ \(x=\frac{3 \times 24}{4}=18\)

Thus, the length of the flagstaff is 18 m.

Question 61. A recipe calls for 1 cup of milk for every 2– cups of flour to make a cake that would feed 6 persons. How many cups of both flour and milk will be needed to make a similar cake for 8 people?
Solution:

Given

A recipe calls for 1 cup of milk for every 2– cups of flour to make a cake that would feed 6 persons.

Quantity of both milk and flour to make a cake for 6 persons

⇒ \(=\left(1+2 \frac{1}{2}\right) \mathrm{cups}\)

⇒ \(=\left(1+\frac{5}{2}\right) \text { cups }=\left(\frac{2+5}{2}\right) \mathrm{cups}=\frac{7}{2} \text { cups }\)

Let the quantity of both milk and flour to make a cake for 8 persons be x cups.

According to the given question,

⇒ \(\frac{7}{2} \text { cups : } 6 \text { persons }:: x \text { cups }: 8 \text { persons }\)

⇒ \(\frac{\frac{7}{2}}{6}=\frac{x}{8} \Rightarrow 6 \times x=8 \times \frac{7}{2}\)

⇒ \(x=\frac{28}{6} \Rightarrow x=\frac{14}{3}\)

Thus, \(\frac{14}{3}=4 \frac{2}{3} \text { cups }\) of both flour and milk will be needed to make a cake for 8 people.

Question 62. In a school, the ratio of the number of large classrooms to small classrooms is 3: 4. If the number of small rooms is 20, then find the number of large rooms.
Solution:

Given

In a school, the ratio of the number of large classrooms to small classrooms is 3: 4. If the number of small rooms is 20,

Let the number of large classrooms is 3x and the number of small classrooms be 4x.

We have given, the number of small rooms = 20

⇒ \(4 x=20 \Rightarrow x=\frac{20}{4} \Rightarrow x=5\)

The number of large classrooms is 3 x 5 = 15

Question 63. Samira sells newspapers at Janpath CCrossing daily. On a particular day, she had 312 newspapers out of which 216 were in English and the remaining in Hindi. Find the ratio of

  1. the number of English newspapers to the number of Hindi newspapers.
  2. the number of Hindi newspapers to the total number of newspapers.

Solution:

Given

Samira sells newspapers at Janpath CCrossing daily. On a particular day, she had 312 newspapers out of which 216 were in English and the remaining in Hindi.

Total number of newspapers = 312.

Number of English newspapers = 216

Number of Hindi newspapers- 31 2- 216

= 96

The required ratio

⇒ \(=\frac{\text { Number of English newspapers }}{\text { Number of Hindi newspapers }}\)

⇒ \(=\frac{216}{96}=\frac{9}{4}=9: 4\)

The required ratio

⇒ \(=\frac{\text { Number of Hindi newspapers }}{\text { Total number of newspapers }}\)

⇒ \(=\frac{96}{312}=\frac{4}{13}=4: 13\)

Question 64. The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 1 44 and the number of Christian students is 72. Find the ratio of

Solution: Given

The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 1 44 and the number of Christian students is 72.

The number of Hindu students to the number of Christian students.

The number of Muslim students to the total number of students.

Number of Hindu students = 288

Number of Muslim students = 252

Number of Sikh students = 144

Number of Christian students = 72

Total number of students in the school = 288 + 252 + 144 + 72 = 756

The required ratio

⇒ \(=\frac{\text { Number of Hindu students }}{\text { Number of Christian students }}\)

⇒ \(=\frac{288}{72}=\frac{4}{1}=4: 1\)

The required ratio \(=\frac{\text { Number of Muslim students }}{\text { Total number of students }}\)

⇒ \(=\frac{252}{756}=\frac{1}{3}=1: 3\)

Question 65. When Chinmay visited Chowpati in Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5: 4. If the total number of food stalls is 117, find the number of each type of food stalls.
Solution:

Given

When Chinmay visited Chowpati in Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5: 4. If the total number of food stalls is 117,

Let the number of North Indian food stalls be 5x and the number of South Indian food stalls be 4x.

The total number of food stalls = 117

5x + 4x = 117 => 9x = 117

⇒ \(x=\frac{117}{9} \Rightarrow x=13\)

Thus, the number of North Indian food stalls is 5 x 13 = 65

And the number of South Indian food stalls is 4×13 = 52

Question 66. At the parking stand of Ramleela ground, Kartik counted that there were 115 cycles, 75 scooters and 45 bikes. Find the ratio of the number of cycles to the total number of vehicles.
Solution:

Given

At the parking stand of Ramleela ground, Kartik counted that there were 115 cycles, 75 scooters and 45 bikes.

Number of cycles = 115

Number of scooters = 75

Number of bikes = 45

Total number of vehicles = 115 + 75 + 45 = 235

The required ratio \(=\frac{\text { Number of cycles }}{\text { Total number of vehicles }}\)

⇒ \(=\frac{115}{235}=\frac{23}{47}=23: 47\)

Question 67. A train takes 2 hours to travel from Ajmer to Jaipur, which are 130 km apart. How much time will it take to travel from Delhi to Bhopal which are 780 km apart if the train is travelling at a uniform speed?
Solution:

Given

A train takes 2 hours to travel from Ajmer to Jaipur, which are 130 km apart.

A train covers a distance of 130 km in 2 hours and it covers a distance of 780 km in x hours.

According to the question, 130 km: 2 hours:: 780 km: x hours

⇒ \(\frac{130}{2}=\frac{780}{x} \Rightarrow 130 \times x=2 \times 780\)

⇒ \(x=\frac{2 \times 780}{130}=12\)

Thus, the required time is 12 hours.

Question 68. The length and breadth of a school ground are 150 m and 90 m respectively, while the length and breadth of a mela ground are 210 m and 126 m, respectively. Are these measurements in proportion?
Solution:

Given

The length and breadth of a school ground are 150 m and 90 m respectively, while the length and breadth of a mela ground are 210 m and 126 m, respectively.

The ratio of length to breadth of school ground = 150 m : 90 m \(=\frac{150}{90}=\frac{5}{3}=5: 3\)

The ratio of length to breadth of mela ground = 210 m: 126 m

⇒ \(=\frac{210}{126}=\frac{5}{3}=5: 3\)

Hence, both ratios are equal.

The given measurements are in proportion.

Question 69. In the given figure, the comparative areas of the continents are given :

  1. What is the ratio of the areas of
  2. Africa to Europe
  3. Australia to Asia
  4. Antarctica to a Combined area of North America and South America.

Ratio and Proportion (Comparative areas of the continents)

Solution:

Let each square be of unit length.

The required ratio = \(=\frac{26 \text { sq units }}{10 \text { sq units }}\)

⇒ \(=\frac{13}{5}=13: 5\)

The required ratio = \(=\frac{\text { Area of Australia }}{\text { Area of Asia }}\)

⇒ \(=\frac{8 \mathrm{sq} \text { units }}{44 \text { sq units }}\)

⇒ \(=\frac{2}{11}:=2: 11\)

The required ratio = \(=\frac{\text { Area of Antarctica }}{\text { Area of (North America }+ \text { South America) })}\)

⇒ \(=\frac{13 \mathrm{sq} \text { units }}{(17+18) \mathrm{sq} \text { units }}=\frac{13 \mathrm{sq} \text { units }}{35 \mathrm{sq} \text { units }}\)

⇒ \(=\frac{13}{35}=13: 35\)

Question 70. A tea merchant blends two varieties of tea costing her? 234 and 1 30 per kg in the ratio of their costs. If the weight of the mixture is 84 kg, then find the weight of each variety of tea.
Solution:

Given

A tea merchant blends two varieties of tea costing her? 234 and 1 30 per kg in the ratio of their costs. If the weight of the mixture is 84 kg,

The ratio of the costs of two varieties of tea is? 234 per kg? 130 per kg.

Sum of ratios = ?(234 + 130) per kg = ? 364 per kg.

Total weight of the mixture = 84 kg

The weights of each variety of tea are given by

⇒ \(\frac{234 \text { per } \mathrm{kg}}{364 \text { per } \mathrm{kg}} \times 84 \mathrm{~kg}=54 \mathrm{~kg}\)

⇒ \(\text { and } \frac{130 \text { per } \mathrm{kg}}{364 \text { per } \mathrm{kg}} \times 84 \mathrm{~kg}=30 \mathrm{~kg}\)

Question 71. An alloy contains only zinc and copper and they are in the ratio of 7: 9. If the weight of the alloy is 8 kg, then find the weight of copper in the alloy.
Solution:

Given

An alloy contains only zinc and copper and they are in the ratio of 7: 9. If the weight of the alloy is 8 kg

The ratio of the weight of zinc and copper is 7: 9.

Let the weight of zinc be lx kg and the weight of copper be 9x kg.

The weight of the alloy = 8 kg

7x + 9x = S => 16x = 8 => x \(=\frac{8}{16}=\frac{1}{2}\)

So, the weight of copper = \(9 \times \frac{1}{2} \mathrm{~kg}=\frac{9}{2} \mathrm{~kg}\)

⇒ \(=4 \frac{1}{2} \mathrm{~kg}\)

Question 72. In the following figure, each division represents 1 cm:

Question 72In the following figure, each division represents 1 cmExpress numerically the ratios of the following distancesi AC : AFii AG: ADiii BF : AIiv CE : DI

Express numerically the ratios of the following distances:

  1. AC: AF
  2. AG: AD
  3. BF: AI
  4. CE: DI

Solution:

1. AC:AF \(=\frac{A C}{A F}=\frac{2 \mathrm{~cm}}{5 \mathrm{~cm}}=\frac{2}{5}=2: 5\)

2. \(A G: A D=\frac{A G}{A D}=\frac{6 \mathrm{~cm}}{3 \mathrm{~cm}}=\frac{2}{1}=2: 1\)

3. \(B F: A I=\frac{B F}{A I}=\frac{4 \mathrm{~cm}}{8 \mathrm{~cm}}=\frac{1}{2}=1: 2\)

4. \(C E: D I=\frac{C E}{D I}=\frac{2 \mathrm{~cm}}{5 \mathrm{~cm}}=\frac{2}{5}=2: 5\)

Question 73. Find two numbers whose sum is 100 and whose ratio is 9: 16.
Solution:

The ratio of the two numbers is 9: 16.

Let the numbers be 9x and 16. v.

The sum of numbers = 100

⇒ 9.v + 16.v = 100 => 25.v = 100

\(x=\frac{100}{25} \Rightarrow x=4\)

Thus, the required numbers are 9 x 4 = 36 and 16×4 = 64

Question 74. In Figure (1) and Figure (2), find the ratio of the area of the shaded portion to that of the whole figure:

Ratio and Proportion In orderS haded portion

Solution:

Required ratio \(=\frac{\text { Area of the shaded portion }}{\text { Area of the whole figure }}\)

⇒ \(=\frac{8 \text { sq units }}{16 \text { sq units }}=\frac{1}{2}=1: 2\)

Lei the area of each small triangle is 1 sq unit.

Required ratio

⇒ \(=\frac{\text { Area of shaded portion }}{\text { Area of the whole figure }}\)

⇒ \(=\frac{16 \mathrm{sq} \text { units }}{32 \mathrm{sq} \text { units }}=\frac{1}{2}=1: 2\)

Question 75. A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript. What is the ratio of the number of pages typed to the number of pages left?
Solution:

Given

A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript.

Total number of pages = 40

Number of pages typed = 30

The number of pages left = 40- 30 = 10

The required ratio \(=\frac{\text { Number of pages typed }}{\text { Number of pages left }}\)

⇒ \(=\frac{30}{10}=\frac{3}{1}=3: 1\)

Question 76. In a floral design made from tiles each of dimensions 40 cm by 60 cm (see figure), find the ratios of:

  1. The perimeter of the shaded portion to the perimeter of the whole design.
  2. The area of the shaded portion to the area of the unshaded portion.

Ratio and Proportion In order shaded portion

Solution:

Length of one tile = 60 cm

And the breadth of one tile = 40 cm

Required ratio \(=\frac{\text { Perimeter of shaded portion }}{\text { Perimeter of whole design }}\)

⇒ \(=\frac{(4 \times 60+6 \times 40) \mathrm{cm}}{(8 \times 60+10 \times 40) \mathrm{cm}}=\frac{480 \mathrm{~m}}{880 \mathrm{~m}}\)

⇒ \(=\frac{6}{11}=6: 11\)

Area ofone tile- 60 an x 40 an = 2400 cm2 Required ratio

⇒ \(=\frac{\text { Area of the shaded portion }}{\text { Area of the unshaded portion }}\)

⇒ \(=\frac{6 \times \text { area of } 1 \text { tile }}{14 \times \text { area of } 1 \text { tile }}\)

⇒ \(=\frac{6 \times 2400 \mathrm{~cm}^2}{14 \times 2400 \mathrm{~cm}^2}=\frac{6}{14}=\frac{3}{7}=3: 7\)

Question 77. In the given figure, what is the ratio of the areas of

1. Shaded portion 1 to shaded portion 2?

Ratio and Proportion (Comparative areas of the continents)

2. shaded portion 2 to shaded portion 3?

3. shaded portions 1 and 2 taken together and shaded portion 3?

Solution:

Area of shaded portion I = 5 x 5 sq units = 25 sq units

Area of shaded portion HI = 7 x 5 sq units = 35 sq units

Area of shaded portion II = Area of the whole figure- Area of a shaded portion (I + III)

= [10 x 10- (25 + 35)] sq units

= [100- 60] sq units = 40 sq units

1.  Required ratio

⇒ \(=\frac{\text { Area of shaded portion I }}{\text { Area of shaded portion II }}\)

⇒ \(=\frac{25 \mathrm{sq} \text { units }}{40 \mathrm{sq} \text { units }}=\frac{5}{8}=5: 8\)

Required ratio \(=\frac{\text { Area of shaded portion II }}{\text { Area of shaded portion III }}\)

\(=\frac{40 \text { sq units }}{35 \text { sq units }}=\frac{8}{7}=8: 7\)

Required ratio \(=\frac{\text { Area of shaded portion (I + II) }}{\text { Area of shaded portion III }}\)

\(=\frac{(25+40) \text { sq units }}{35 \text { sq units }}=\frac{65 \text { sq units }}{35 \text { sq units }}\) \(=\frac{13}{7}=13: 7\)

Question 78. A car can travel 240 km in 15 litres of petrol. How much distance will it travel in 25 litres of petrol?
Solution:

Given

A car can travel 240 km in 15 litres of petrol.

The distance can be travelled in 15 litres of petrol = 240 km

The distance can be travelled in 1 litre of petrol \(=\frac{240}{15} \mathrm{~km}=16 \mathrm{~km}\)

Distance can be travelled in 25 litres of petrol = (25 x 16) km = 400 km

Question 79. Bachhu Manjhi earns ₹ 24000 in 8 months. At this rate,

How much does he earn in one year?

In how many months does he earn₹ 42000?

Solution:

Given

Bachhu Manjhi earns ₹ 24000 in 8 months.

In 8 months, money earned by Bachhu Manjhi = ₹ 24000

In 1 month, money earned by him

\(=₹ \frac{24000}{8}=₹ 3000\)

In 1 year, he earns =? (12 x 3000)

= ₹ 36000

Number of months for earning₹ 3000 = 1

Number of months for earning₹ 42000

\(=\frac{4200}{300} = 14 \)

Question 80. The yield of wheat from 8 hectares of land is 360 quintals. Find the number of hectares of land required for a yield of 540 quintals.
Solution:

Given

The yield of wheat from 8 hectares of land is 360 quintals.

Land required for a yield of 360 quintals of wheat \(=\frac{8}{360} \text { hectares }=\frac{1}{45} \text { hectares }\)

Land required for yield of 540 quintals of wheat = \(=\frac{1}{45} \times 540 \text { hectares }=12 \text { hectares }\)

Question 81. The earth rotates 360° about its axis in about 24 hours. By how many degrees will it rotate in 2 hours?
Solution:

Given

The earth rotates 360° about its axis in about 24 hours.

Rotation of earth in 24 hours = 360°

Rotation of earth in1 hour \(=\frac{360^{\circ}}{24}\) = 15°

Rotation of earth in 2 hours = 2 x 15° = 30°

Question 82. Shivangi is suffering from anaemia as the haemoglobin level in her blood is lower than the normal range. The doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets₹ 17, then what amount will she be required to pay for her medical bill for 15 days?
Solution:

Given

Shivangi is suffering from anaemia as the haemoglobin level in her blood is lower than the normal range. The doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets ₹ 17

Shivangi will consume 2 x 15 tablets i.e., 30 tablets in 15 days.

Cost of 10 tablets = ₹17

Thus, the cost of 30 tablets = \(₹ \frac{17}{10} \times 30\)

= ₹ 51

Question 83. The quarterly school fee in Kendriya Vidyalaya for Class VI is₹ 540. What will be the fee for seven months?
Solution:

Given

The quarterly (3 months) fee =₹ 540

The fee for 1 month = \(₹ \frac{540}{3}=₹ 180\)

Thus, the fee for 7 months =? (180 x 7)

= ₹ 1260

The fee for seven months= ₹ 1260

Question 84. In an electron, the votes cast for two of the candidates were in the ratio 5::7. If the successful candidate received 20734 votes, how many votes did his opponent receive?
Solution:

Given

In an electron, the votes cast for two of the candidates were in the ratio 5::7. If the successful candidate received 20734 votes

Let the number of votes cast for the candidates be 5x and 7x.

According to the question,

7x = 20734 => \(\)

⇒ \(x=\frac{20734}{7}=2962\)

⇒ \(x=\frac{20734}{7}=2962\)

Number of votes received by opponent candidate = 5×2962 = 14810

5. A metal pipe 3 metres long was found to weigh 7.6 kg. What would be the weight of the same kind of 7.8 m long pipe?
Solution:

Given

A metal pipe 3 metres long was found to weigh 7.6 kg.

The weight of 3 metre long pipe = 7.6 kg

The weight of1 metre long pipe = \(=\frac{7.6}{3} \mathrm{~kg}\)

Thus, the weight o7.8-metreetre long pipe \(=\frac{7.6}{3} \times 7.8 \mathrm{~kg}=19.76 \mathrm{~kg}\)

Question 86. A recipe for raspberry jelly calls for 5 cups of raspberry juice and \(2 \frac{1}{2}\) cups of sugar. Find the amount of sugar needed for 6 cups of the juice?
Solution:

Given

A recipe for raspberry jelly calls for 5 cups of raspberry juice and \(2 \frac{1}{2}\) cups of sugar.

The requirement of sugar for 5 cups of raspberry juice \(=2 \frac{1}{2} \text { cups }=\frac{5}{2} \text { cups }\)

The requirement of sugar for 1 cup of raspberry juice \(=\left(\frac{5}{2} \div 5\right) \text { cups }\)

⇒ \(\left(\frac{5}{2} \times \frac{1}{5}\right) \text { cups }=\frac{1}{2} \text { cups }\)

Thus, the requirement of sugar for 6 cups of the raspberry juice \(=6 \times \frac{1}{2} \mathrm{cups}=3 \mathrm{cups}\)

Thus, the requirement of sugar for 6 cups of raspberry juice = \(=6 \times \frac{1}{2} \text { cups }=3 \text { cups }\)

Question 87. A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row. How many plants did the worm destroy in the whole field?
Solution:

Given

A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row.

Total number of plants = number of rows x number of plants in each row

1890 = number of rows x 63

⇒m\(\frac{1890}{63}=\text { number of rows }\)

number of rows = 30

Now, the number of plants destroyed worms in 1 row = IS

Total number of plants destroyed by worm in 30 rows = 18 x 30 = 540

Question 88. The length and breadth of the floor of a room are 5 m and 3 m, respectively. Forty tiles, each witan h area are used to cover the floor partially. Find the ratio of the tiled and the non tiled portion of the floor.
Solution:

Given

The length and breadth of the floor of a room are 5 m and 3 m, respectively. Forty tiles, each witan h area are used to cover the floor partially.

Tire total area of the floor = 5 m x 3 m = 15 m2

Tire area covered with tiles \(=40 \times \frac{1}{16} \mathrm{~m}^2\)

⇒ \(=\frac{5}{2} \mathrm{~m}^2\)

The non tiled area = \(\left[15-\frac{5}{2}\right] \mathrm{m}^2\)

⇒ \(=\left[\frac{30-5}{2}\right] \mathrm{m}^2=\frac{25}{2} \mathrm{~m}^2\)

The required ratio \(=\frac{\text { Tiled area }}{\text { Non tiled area }}\)

⇒ \(=\frac{\frac{5}{2} \mathrm{~m}^2}{\frac{25}{2} \mathrm{~m}^2}=\frac{5}{2} \times \frac{2}{25}=\frac{1}{5}=1: 5\)

Question 89. A carpenter had a board which measured 3 m x 2m. She cut out a rectangular piece of 250 cm x 90 cm. What is the ratio of the area of the cutout piece and the remaining piece?
Solution:

Given

A carpenter had a board which measured 3 m x 2m. She cut out a rectangular piece of 250 cm x 90 cm.

The area of the board = 3 m x 2 m = 6 m2 = 6 x 10000 cm2 = 60000 cm2

And the area of cut out piece = 250 cm x 90 cm = 22500 cm2

The area of remaining piece = (60000-22500) cm2

= 37500 cm2

Thus, the required ratio \(=\frac{\text { The area of cut out piece }}{\text { The area of remaining piece }}\)

⇒ \(=\frac{22500 \mathrm{~cm}^2}{37500 \mathrm{~cm}^2}=\frac{3}{5}=3: 5\)

NCERT Exemplar Solutions For Class 6 Maths Chapter 11 Algebra

Class 6 Maths Chapter 11 Algebra

Question 1. If each matchbox contains 50 matchsticks, the number of matchsticks required to fill n such boxes is

  1. 50 + n
  2. 50n
  3. 50 ÷ n
  4. 50 – n

Solution: (2): Number of matchsticks in one box = 50

The number of matchsticks required to fill n boxes = 50 x n = 50n

Question 2. Amulya is x years of age now. 5 years ago her age was

  1. (5 -x) years
  2. (5 +x) years
  3. (x- 5) years
  4. (5 ÷ X) years

Solution: (3): The present age of Amulya is x years.

Question 3. Which of the following represents 6 x x

  1. 6x
  2. \(\frac{x}{6}\)
  3. 6+x
  4. 6-x

Solution: (1): 6 x x can be represented as 6x.

Read and Learn More Class 6 Maths Exemplar Solutions

Question 4.  Which of the following is an equation?

  1. x + 1
  2. x – 1
  3. x – 1=0
  4. x+1 >0

Solution:(3): Equation means an expression with a variable, constants and the sign of equality (=).

x- 1 = 0 is an equation.

Question 5. If x takes the value 2, then the value of + 1 0 is

  1. 20
  2. 12
  3. 5
  4. 8

Solution:(2) : When we put x = 2 in the given expression x + 10, we get

x + 10 = 2 + 10 = 12.

NCERT Exemplar Solutions For Class 6 Maths Chapter 11 Algebra

Question 6. If the perimeter of a regular hexagon is x metres, then the length of each of its sides is

  1. (x + 6) metres
  2. (x ÷ 6 ) meters
  3. (x- 6) metres
  4. (6 ÷ x) metres

Solution: (2): Perimeter of a regular hexagon

= 6 x Length of each side

⇒ x m = 6 x Length of each side

⇒ Length of each side = (x + 6) m

Question 7. Which of the following equations has = 2 as a Solution?

  1. x + 2 = 5
  2. x – 2 = 0
  3. 2x + 1 =0
  4. x + 3 = 6

Solution: (2) : (A) Putting x = 2 in x + 2, we get 2+2=4*5

Putting x = 2 in x- 2, we get 2- 2 = 0

Putting x = 2 in 2x + 1, we get 2×2+l=5*0

Putting x = 2 in x + 3, we get 2 + 3 = 5* 6

Thus, the above conditions shows that x = 2 is the solution of x- 2 = 0

Question 8. For any two integers x and y, which of the following suggests that the operation of addition is commutative?

  1. x + y = y + x
  2. x + y > x
  3. x – y = y – x
  4. x x y = y x x

Solution: (A): x + y- y + x shows that addition is commutative.

Question 9. Which of the following equations does not have a solution in integers?

  1. x + 1 =1
  2. x – 1 =3
  3. 2x+ 1 =6
  4. 1 -x = 5

Solution: (3) : (A) x +1 =1

x + 1- 1 = 1 – 1

[Subtracting1 from both sides]

⇒ x = 0, which is an integer

x-1 =3

⇒ x-l + l= 3 + 1 [Adding1 to both sides]

⇒  x = 4, which is an integer.

2x +1 = 6

⇒ 2x + 1- 1 =6- 1

[Subtracting1 from both sides]

2x = 5

⇒ \(\frac{2 x}{2}=\frac{5}{2}\) [Dividing both sides by 2]

⇒ \(x=\frac{5}{2} \text {, }\) which is not an integer.

1 -x = 5

1-x-l =5-1

[Subtracting1 from both sides]

⇒ -x = 4

⇒ -(-x) =-4

[Multiplying both sides by (-1)]

⇒ x = -4, which is an integer

Thus, the above conditions show that equation 2x + 1 = 6 does not have a solution in integers.

Question 10. In algebra, an x b means ab, but in arithmetic 3×5 is

  1. 35
  2. 53
  3. 15
  4. 8

Solution: (3) : 3 x 5 means 15.

Question 11. In algebra, letters may stand for

  1. known quantities
  2. Unknown quantities
  3. Fixed numbers
  4. None of these

Solution: (2): In algebra, letters may stand for unknown quantities.

Question 12. “Variable” means that it

  1. Can take different values
  2. Has a fixed value
  3. Can take only 2 values
  4. Can take only three values

Solution: (1): “Variable” means that it can take different values.

Question 13. 10 -x means

  1. 10 is subtracted x times
  2. x is subtracted 10 times
  3. x is subtracted from 1 0
  4. 10 is subtracted from x

Solution: (3): 10- x means x is subtracted from 10.

Question 14. Savitri has a sum of ₹ x. She spent ₹ 1000 on groceries,₹ 500 on clothes and ₹ 400 on education, and received ₹ 200 as a gift. How much money (in ₹) is left with her?

  1. x – 1700
  2. x – 1900
  3. x + 200
  4. x – 2100

Solution: (1):

Given

Savitri has a sum of ₹ x. She spent ₹ 1000 on groceries,₹ 500 on clothes and ₹ 400 on education, and received ₹ 200 as a gift.

Amount of money Savitri has = ₹ x.

Amount of money spent by her =₹ (1000 + 500 + 400) = ₹ 1900

Amount of money received by her as a gift = ₹ 200

Amount of money left with her

= ₹(x- 1900 + 200)

= ₹ (x- 1700)

Question 15. The perimeter of the triangle shown in the given figure is

Algebra Perimeter of the triangle

  1. 2x+y
  2. x + 2y
  3. x + y
  4. 2x-y

Solution: (1): The perimeter of the triangle = Sum of all sides

=x+x+y

= 2x + y

Question 16. The area of a square with each side x is

  1. xxx
  2. 4x
  3. x + x
  4. 4 = x + x

Solution: (1) : The area of a square = (side) x (side) =x*x

Question 17. The expression obtained when x is multiplied by 2 and then subtracted from 3 is

  1. 2x- 3
  2. 2x + 3
  3. 2x- 3
  4. 3x- 2

Solution : x is multiplied by 2 and then subtracted from 3 =3-2 x * x = 3-2x

Question 18. \(\frac{q}{2}=3\) has a solution

  1. 6
  2. 8
  3. 3
  4. 2
  5. Solution: (1): We have, \(\frac{q}{2}=3\) Multiplying both sides by 2, we get

⇒ \(\Rightarrow \quad \frac{q}{2} \times 2=3 \times 2\)

q = 6

Question 19. x-4 =- 2 has a solution

  1. 6
  2. 2
  3. -6
  4. -2

Solution:(2) : We have, x- 4 =-2

⇒  x -4 + 4 = -2 + 4

[Adding 4 to both sides]

⇒  x = 2

Question 20. \(\frac{4}{2}=2\) denotes a V

  1. Numerical equation
  2. Algebraic expression
  3. Equation with a variable
  4. False statement

Solution: (1): \(\frac{4}{2}=2\) denotes a numerical equation.

Question 21. Kanta has p pencils in her box. She puts q more pencils in the box. The total number of pencils with her is.

  1. p+q
  2. PQ
  3. p-q
  4. \(\frac{p}{q}\)

Solution: (1): Kanta has p pencils in her box. After putting q more pencils in the box, she has a total number of pencils = p + q.

Question 22. The equation 4x = 16 is satisfied by the following value of x

  1. 4
  2. 2
  3. 12
  4. -12

Solution:(1) : We have, 4x = 16

⇒ \(\frac{4 x}{4}=\frac{16}{4}\)

x = 4

Question 23. I think of a number and on adding 13 to it, I get 27. The equation for this is

  1. x – 27 = 13
  2. x – 13 = 27
  3. x + 27 = 13
  4. x + 13 = 27

Solution: (D): Let the number be x.

According to the question, x + 13 = 27

Question 24. The distance (in km) travelled in h hours at a constant speed of 40 km per hour is_______.
Solution: 40 h : Distance = Speed * Time = (40 x h) km = 40 h km

Question 25. p kg of potatoes are bought for ₹70. The cost of 1 kg of potatoes (in ₹) is_______.
Solution:

⇒ \(\frac{70}{p}\) : The cost of p kg of potatoes = ₹ 70

The cost of1 kg of potatoes = \(₹ \frac{70}{p}\)

Question 26. An auto rickshaw charges ₹ 10 for the first kilometre then ₹ 8 for each such subsequent kilometre. The total charge (in ₹) for d kilometres is_______.
Solution:

Given

An auto rickshaw charges ₹ 10 for the first kilometre then ₹ 8 for each such subsequent kilometre.

2 + 8d: For the first kilometre, rickshaw charges = ₹ 10.

And the charges for subsequent kilometres = ₹ 8.

The charges for d kilometres

= 10 + (d- 1)8

= 10 +8d-8 = 2 + 8d

Question 27. lf 7x + 4 = 25, then the value of x is_______.
Solution: 3:

We have given, 7x + 4 = 25

⇒  7x + 4- 4 = 25- 4

[Subtracting 4 from both sides]

⇒ \(\frac{7 x}{7}=\frac{21}{7}\)

⇒  x = 21

⇒  x = 3

Question 28. The solution of equation 3x + 7 = -20 is _______.
Solution:

-9 : We have, 3x + 7 =- 20

3x + 7- 7 =- 20- 7

[Subtracting 7 from both sides]

⇒  3x = -27

[Dividing both sides by 3]

⇒  x = -9

Question 29. ‘exceeds by 7’ca can be expressed as_______.
Solution: x = y + 7: x exceeds y by 7 can be expressed as = y + 7 or x-y = 7.

Question 30. ‘8 more than three times the number x7 can be written as_______.
Solution: 3x + 8 : 8 more than three times the number x can be expressed as 3x + 8

Question 31. Many pencils bought for? x at the rate of ₹ 2 per pencil is_______.
Solution: \(\frac{x}{2}\) Number of pencils bought for ₹ 2 =1

Number of pencils bought for ₹ x = \(\frac{1}{2} \times x=\frac{x}{2}\)

Question 32. The number of days in w weeks is_______.
Solution: 7W : Number of days in a week = 7

Number of days in w weeks = 7 * w = 7w

Question 33. Annual salary at r rupees per month along with a festival bonus of? 2000 is_______.
Solution: ₹ (12r + 2000) : We have given, monthly salary = ₹ r

Annual salary = ₹ (12 * r) = ₹ 12r

Now, total annual salary with a festive bonus = ₹ (12r + 2000)

Question 34. The two-digit number whose ten digit is Y and unit’s digit is’ is_______.
Solution: + u: The digit at the unit’s place is V.

And the digit at ten’s place is Y.

The number can be expressed as 10 x t + u = lOf + u

Question 35. The variable used in the equation 2p +8 = 18 is _______.
Solution: p : p is the variable in the equation Ip + 8 = 18

Question 36. x metres = _______centimeters.
Solution:

x : 1 metre = 100 cm

⇒  x metres = lOOx cm

Question 37. p litres = _______.millilitres.
Solution:

lOOOp : 1 litre = 1000 millilitres

⇒ p litres = lOOOp millilitres

Question 38. r rupees = _______ paise.
Solution:

Floor : 1 rupee = 100 paise

⇒  r rupees = lOOr paise

Question 39. If their present age of Ramandeep is n years, then her age after 7 years will be_______.
Solution:

(n + 7) years: Present age of Ramandeep = n years

Her age after 7 years = (n + 7) years

Question 40. If I spend f rupees from 100 rupees, the money left with me is_______rupees.
Solution:

100 – f: I have 100 rupees and spend f rupees.

Money left with me = (100 -f) rupees 40.

Question 41. 0 is a solution of the equation x+ 1 = 0
Solution: False

We have given, x +1 = 0

⇒  — 1 – 0- 1

[Subtracting 1 from both sides]

⇒  x –1, which is the solution of the given equation.

Question 42. The equations x + 1 = 0 and 2x + 2 = 0 have the same solution.
Solution: True

We have given, x + 1 = 0

⇒  x +1 -1 =0-1

[Subtracting 1 from both sides]

⇒  x = —1 …(1)

and 2x + 2 = 0

2x + 2-2 = 0-2

[Subtracting 2 from both sides]

⇒  2x = -2

⇒ \(\frac{2 x}{2}=\frac{-2}{2}\)

[Dividing both sides by 2]

⇒  x =-1 …(2)

Thus, (1) and (2) imply that both equations have the same solution.

Question 43. If m is a whole number, then 2m denotes a multiple of 2.
Solution: True

Question 44. The additive inverse of an integer x is 2x.
Solution: False

Since the additive inverse of x is -x.

Question 45. If x is a negative integer, – x is a positive integer.
Solution: True

The negative of a negative integer is always a positive integer.

Question 46. 2x- 5 > 1 1 is an equation.
Solution: False

Since an equation includes a sign of equality(=)

Question 47. In an equation, the LHS is equal to the RHS.
Solution: True

Question 48. In the equation 7k- 7 = 7, the variable is 7.
Solution: False

Since, in the equation 7k-7 = 7, the variable is k.

Question 49. a = 3 Is a solution of the equation 2a- 1 = 5
Solution: True

We have, 2a- 1 = 5

⇒  2n- 1 +1 = 5 + 1

[Adding 1 to both sides]

⇒  2a = 6

⇒  \(\frac{2 a}{2}=\frac{6}{2}\) [Dividing both sides by 2]

⇒  a = 3, which is the solution of the given equation

Question 50. The distance between New Delhi and Bhopal is not a variable.
Solution: True

Question 51. t minutes are equal to 60f seconds.
Solution: True

Since, 1 minute = 60 seconds

t minutes = 60 x f seconds = 60f seconds

Question 52. x = 5 is the solution of the equation 3x + 2 = 20
Solution: False

⇒  3x + 2- 2 = 20- 2

[Subtracting 2 from both sides]

⇒  3x = 18

⇒  \(\quad \frac{3 x}{3}=\frac{18}{3}\) [Dividing both sides by 3]

⇒  x = 6, which is the solution of the given equation.

Question 53. ‘One-third of a number added to itself gives 8’, can be expressed as\(\frac{x}{3}+8=x\)
Solution: False

Let the number be x.

One third of the number = \(\frac{x}{3} \text {. }\)

According to the given question, \(\frac{x}{3}+x=8\)

Question 54. The difference between the ages of the two sisters Leela and Yamini is a variable.
Solution: False

The difference between the ages of the two sisters will be a fixed number.

Question 55. The number of lines that can be drawn through a point is a variable. ,
Solution: False

Question 56. One more than twice the number.
Solution:

Let the number be x.

Twice the number = 2x.

Now, 1 more than 2x can be expressed as 2x + l.

Question 57. 20°C less than the present temperature.
Solution:

Let the present temperature be t°C.

Now, 20°C less than t = (t- 20)°C

Question 58. The successor of an integer.
Solution:

Let the integer be n.

Now, the successor of the integer n = n +1.

Question 59. The perimeter of an equilateral triangle, if the side of the triangle is m.
Solution: We have given, the side of the equilateral triangle = m.

The perimeter of an equilateral triangle

= 3 x side

= 3 m

Question 60. Area of the rectangle with length k units and breadth n units.
Solution:

We have given, a rectangle of length = k units and breadth = n units.

Area of the rectangle = length * breadth = (k x n) sq units = kn sq units

Question 61. Omar helps his mother 1 hour more than his sister does.
Solution:

Let Omar’s sister help her mother for x hours.

Omar helps his mother for (x +1) hours.

Question 62. Two consecutive odd integers.
Solution: Two consecutive odd integers are (2n + 1) and (2n + 3), where n is any integer.

Question 63. Two consecutive even integers.
Solution: Two consecutive even integers are 2m and 2m + 2, where m is any integer.

Question 64. Multiple of 5.
Solution:

A multiple of 5 can be expressed as 5n, where n is an integer.

Question 65. The denominator of a fraction is 1 more than its numerator.
Solution:

Let the numerator of the fraction be x.

The denominator of the fraction can be expressed as x + 1.

The fraction \(=\frac{x}{x+1}\)

Question 66. The height of Mount Everest is 20 times the height of the Empire State Building.
Solution:

Let the height of the Empire State Building be y.

Height of Mount Everest = 20y.

Question 67. If a notebook costs ₹ p and a pencil costs ₹ 3, then the total cost (in ₹) of two notebooks and one pencil.
Solution:

The cost of a notebook = ₹ p

Cost of 2 notebooks = ₹ 2p.

The cost of a pencil =₹ 3.

Cost of two notebooks and one pencil = ₹(2p + 3)

Question 68. Z is multiplied by 3 and the result is subtracted from 1 3.
Solution:

z multiplied by -3 = -3z

– 3z subtracted from 13 =13- (-3z) = 13 + 3z

Question 69. p is divided by 11 and the result is added to 1 0.
Solution:

p divided by 11 = \(\frac{p}{11} .\)

\(\frac{p}{11}\) added to 10 = \(\frac{p}{11}+10 .\)

Question 70. x times 3 is added to the smallest natural number.
Solution: x times of 3 = 3x

And the smallest natural number =1.

So, 3x added to1 = 3x + 1.

Question 71. 6 times q is subtracted from the smallest two-digit number.
Solution:

6 times q = 6q.

The smallest two-digit number = 10

6q subtracted from 10 = 10- 6q.

Question 72. Write two equations for which 2 Is the solution.
Solution: The required equations are 3i/ + 4 = 10 and 2. V- 3 = 1, i.e., for both equations, 2 is the solution.

Question 73. Write an equation for which 0 is a solution.
Solution: The required equation is 2f + 3 = 3, which has solution t = 0.

Question 74. Write an equation whose solution is not a whole number.
Solution: The required equation is x + 1 = 0, and its solution is x = -1, which is not a whole number.

Question 75. A pencil costs? p and a pen cost 5p.
Solution: A pen costs 5 times the cost of a pencil.

Question 76. Leela contributed ₹ y towards the Prime Minister’s Relief Fund. Leela is now left with ₹ (y+ 10000).
Solution: The amount left with Leela is? 10,000 more than the amount she contributed towards the Prime Minister’s Relief Fund.

Question 77. Kartik is n years old. His father is In years old.
Solution: The age of Kartik’s father is seven times the age of Kartik.

Question 78. The maximum temperature on a day in Delhi was p°C. The minimum temperature was (P-10)°C.
Solution: The minimum temperature on a day in Delhi was 10°C less than the maximum temperature.

Question 79. John planted two plants last year. His friend Jay planted 2f + 1 0 plants that year.
Solution: Last year Jay planted 10 more plants than twice the plants planted by his friend John.

Question 80. Sharad used to take p cups of tea a day. After having some health problems, he takes p – 5 cups of tea a day.
Solution: Sharad reduced the consumption of tea per day by 5 cups after having some health problems.

Question 81. The number of students dropping out of school last year was m. Number of students dropping out of school this year is m- 30.
Solution: The number of students dropping out this year is 30 less than the number of students who dropped out last year.

Question 82. The price of petrol was ₹ p per litre last month. The price of petrol now is ₹ (p- 5) per litre.
Solution: The price of petrol per litre decreased this month by 5 than its price last month.

Question 83. Khader’s monthly salary was P in the year 2005. His salary in 2006 was ₹ (P + 1 000).
Solution: Khader’s monthly salary increased by? 1000 in the year 2006 than that of 2005.

Question 84. The number of girls enrolled in a school last year was g. The number of girls enrolled this year in the school is 3g- 1 0.
Solution: The number of girls enrolled this year is 10 less than 3 times the girls enrolled last year.

Question 85. Translate each of the following statements into an equation, using x as the variable:

  1. 13 subtracted from twice a number gives 3.
  2. One-fifth of a number is 5 less than that number.
  3. Two-thirds of the number is 12.
  4. 9 added to twice a number gives 1 3.
  5. 1 subtracted from one-third of a number gives 1.

Solution:

Let the number be x.

Twice of the number x = 2x

According to question, 2x- 13 = 3

Let the number be x.

One-fifth of the number x = \(\frac{x}{5}.\)

5 less than the number x = x- 5

According to question, \(\frac{x}{5}=x-5\)

Let the number be x.

Two-thin! of the number x \(=\frac{2}{3} x\)

According to question, \(\frac{2}{3} x=12\)

Let the number be x.

Twice of the number x = 2x.

Now, 9 is added to 2x = 9 + 2x

According to the question, 9 + 2x = 13.

Let the number be x.

One-third of the number x = \(\frac{x}{3}\)

Now, 1 is subtracted from \(\frac{x}{3}=\frac{x}{3}-1\)

According to question, “\(\frac{x}{3}-1=1 \text {. }\)

Question 86. Translate each of the following statements into an equation :

  • The perimeter (p) of an equilateral triangle is three times its side (a).
  • The diameter (d) of a circle is twice its radius (r).
  • The selling price (s) of an item is equal to the sum of the cost price (c) of an item and the profit (p) earned.
  • Amount (a) is equal to the sum of principal (p) and interest (/).

Solution:

We have given the perimeter of an equilateral triangle

= 3 (the side of an equilateral triangle)

⇒  p = 3a

We have given the diameter of a circle

= 2 (the radius of the circle)

⇒  d = 2r

We have given, Selling price = cost price + profit

⇒  s = c + p

We have given, Amount = principal + interest

⇒  a = p + i

Question 87. Let Kanika’s present age be x years. Complete the following table, showing the ages of her relatives:

Algebra Situation Described In Ordinary Expressions

Solution:

We have given the present age of Kanika = x years.

Her brother’s age = (x- 2) years

Her father’s age = (x + 35) years

Her mother’s age = (x + 35- 3) years = (x + 32) years

Her grandfather’s age = 8x years

Question 88. If m is a whole number less than 5, complete the table and by inspection of the table, find the solution of the equation 2m- 5 = -1 :

Algebra whole number less than 5

Solution:

Given

If m is a whole number less than 5, complete the table and by inspection of the table

For m = 0, 2m-5=2×0-5=0-5=-5

For m = 1, 2m-5=2xl-5=2-5=-3

For m = 3, 2m-5=2×3-5=6-5=l

For m = 4, 2m-5=2×4-5=8-5=3

Algebra whole number less than

Thus, the solution of the equation 2m- 5 =-1 is m = 2.

Question 89. A class with p students has planned a picnic. 50 per student is collected, out of which ₹ 1800 is paid in advance for transport. How much money is left with them to spend on other items?
Solution:

Given

A class with p students has planned a picnic. 50 per student is collected, out of which ₹ 1800 is paid in advance for transport.

Number of students in the class = p

The total amount collected from p students =₹ 50p

Amount paid in advance for transport =₹ 1800

The amount left with them = ₹ (50p- 1800)

Question 90. In a village, there are 8 water tanks to collect rainwater. On a particular day, x litres of rainwater is collected per tank. If 100 litres of water was already there in one of the tanks, what was the total amount of water in the tanks on that day?
Solution:

Given

In a village, there are 8 water tanks to collect rainwater. On a particular day, x litres of rainwater is collected per tank. If 100 litres of water was already there in one of the tanks

Number of water tanks = 8

Rainwater collected by each tank = x litres

Rainwater collected by 8 tanks = 8x litres

But 100 litres of water was already there in one of the tanks.

Total amount of water in the tanks = (8x + 100) litres.

Question 91. What is the area of a square whose side is m cm?
Solution:

We have given, the side of a square = m cm

Area of the square = side x side = m x m sq cm

Question 92. The perimeter of a triangle is found by using the formula P = a + b + c, where a, b and c are the sides of the triangle. Write the rule that is expressed by this formula in words.
Solution:

Given

The perimeter of a triangle is found by using the formula P = a + b + c, where a, b and c are the sides of the triangle.

The perimeter of a triangle is the sum of all its sides.

Question 93. The perimeter of a rectangle is found by using the formula P = 2 (I + w), where l and w are respectively the length and breadth of the rectangle. Write the rule that is expressed by this formula in words.
Solution:

Gien

The perimeter of a rectangle is found by using the formula P = 2 (I + w), where l and w are respectively the length and breadth of the rectangle.

The perimeter of a rectangle is twice the sum of its length and breadth.

Question 94. On my last birthday, I weighed 40 kg. If I put on m kg of weight after a year, what is my present weight?
Solution:

Given

On my last birthday, I weighed 40 kg. If I put on m kg of weight after a year

Present weight = weight on last birthday + weight put on after a year.

D 40 kg + tn kg

= (40 + m) kg

Question 95. Length and breadth of rcm and tern, respectively.

  1. What will be the length (in cm) of the aluminium strip required to frame the board, if 1 0 cm extra strip is required to fix it properly?
  2. Ifx nails are used to repair one board, how many nails will be required to repair 15 such boards?
  3. If 500 sq cm extra cloth per board is required to cover the edges, what will be the total area of the cloth required to cover 8 such boards?
  4. What will be the expenditure for making 23 boards if the carpenter charges? x per board.

Solution:

We have given, the length of the board = r cm and breadth = t cm.

The length of the aluminium strip to frame the board = Perimeter of the board

= 2(length + breadth)

= 2(r + t) cm

But a 10 cm extra strip is required to fix it properly.

The total length of the aluminium strip = 2(r + t) cm + 10 cm.

Number of nails required to repair 1 board = x.

Number of nails required to repair 15 boards = 15 x  x = 15 x.

The area of the cloth required for the board

= Area of the board

= length x breadth

= r cm x t cm

= (rt) sq cm

The area of the cloth required for 8 boards

= 8 x (rt) sq cm

= 8rt sq cm

But 500 sq cm extra cloth per board is required to cover the edges.

For 8 boards we need 8 x 500 sq cm = 4000 sq cm extra cloth.

Thus, the total area of the cloth required = (8rt + 4000) sq cm

The carpenter charges for 1 board = %x.

The carpenter charges for 23 boards = ?(23*x) = ?23x

Question 96. Sunita is half the age of her mother Geeta. Find their ages

  1. After 4 years.
  2. Before 3 years.

Solution:

Let the present age of Sunita be x years.

The present age of her mother Geeta = 2(Sunita’s present age) = 2x years.

1.  After 4 years,

Sunita’s age = (x + 4) years

Geeta’s age = (2x + 4) years

2.  Before 3 years,

Sunita’s age = (x- 3) years

Geeta’s age = (2x- 3) years.

Question 97. Match the items of Column I with that of Column 2:
Solution:

(1)⇒ (B), (2) ⇒(E), (3) ⇒ (C), (4) ⇒ (C), (5) ⇒ (A)

The number of comers of a quadrilateral is 4, i.e., a constant.

The variable in the equation 2p + 3 =5 is p.

We have given, x + 2 = 3

⇒ x+2-2=3-2

[Subtracting 2 from both sides]

⇒ x = 1, which is the solution of the given equation.

We have given, 2p + 3 = 5

⇒ 2p + 3 — 3 =5- 3

[Subtracting 3 from both sides]

⇒ 2p = 2

⇒ \(\frac{2 p}{2}=\frac{2}{2}\)

[Dividing both sides by 2]

p = 1, which is the solution of the given equation.

The equality sign (=) is used in an equation

NCERT Exemplar Solutions For Class 6 Maths Chapter 7 Fractions

Class 6 Maths Chapter 7 Fractions

Question 1. The fraction which is not equal to\(\frac{4}{5}\)

  1. \(\frac{40}{50}\)
  2. \(\frac{12}{15}\)
  3. \(\frac{16}{20}\)
  4. \(\frac{9}{15}\)

Solution:

4. Since, \(\frac{9}{15}=\frac{9 \div 3}{15 \div 3}=\frac{3}{5} \text {, }\) which is not equal to \(\frac{4}{5}\)

\(\frac{9}{15}\) is the required fraction.

NCERT Exemplar Solutions For Class 6 Maths Chapter 7 Fractions

Question 2. The two consecutive integers between which the fraction \(\frac{5}{7}\) lies are

  1. 5 and 6
  2. 0 and 1
  3. 5 and 7
  4. 6 and 7

Solution:

Read and Learn More Class 6 Maths Exemplar Solutions

2. Since we know that a proper fraction whose numerator is less than its denominator always lies between 0 and 1.

∴ \(\frac{5}{7},\) which is also a proper fraction, must
lie between 0 and 1.

Question 3. when \(\frac{1}{4}\) is written with denominator as 12, its numerator is

  1. 3
  2. 8
  3. 24
  4. 12

Solution:

1. Since, \(\frac{1}{4}=\frac{1}{4} \times \frac{3}{3}\) [By multiplying the numerator and denominator by 3 to get 12 as the denominator] \(=\frac{3}{12},\) which shows that the required numerator is 3.

Question 4. Which of the following is not in the lowest form?

  1. \(\frac{7}{5}\)
  2. \(\frac{15}{20}\)
  3. \(\frac{13}{33}\)
  4. \(\frac{27}{28}\)

Solution:

In the case of \(\frac{7}{5}\) the common factor of 7 and 5 is

i.e. already given in the simplest form.

Similarly,

In \(\frac{15}{20}\) the common factor of 15 and 20 is 4

\(\frac{13}{33}\) the common factor of 13 and 33 is 1

i.e., already given in the simplest form

\(\frac{27}{28}\) the common factor of 27 and 28 is 1

i.e., already given in the simplest form.

So, the above cases show that \(\frac{15}{20}\) is not in the lowest form.

Question 5. If \(\frac{5}{8}=\frac{20}{p}\) then value of P is

  1. 23
  2. 2
  3. 32
  4. 16

Solution:

3. We have given, \(\frac{5}{8}=\frac{20}{p}\)

To find the value of p, we must multiply the numerator and denominator of \(\frac{5}{8} \text { by } 4.\)

∴ \(\frac{5}{8}=\frac{5 \times 4}{8 \times 4}=\frac{20}{32}\)

∴ \(\frac{20}{32}=\frac{20}{p}\)

∴ The value of p is 32.

Question 6. Which of the following is not equal to the others?

  1. \(\frac{6}{8}\)
  2. \(\frac{12}{16}\)
  3. \(\frac{15}{25}\)
  4. \(\frac{18}{24}\)

Solution:

3. Firstly, we will simplify all the fractions into their lowest form, we get,

\(\frac{6}{8}=\frac{6 \div 2}{8 \div 2}=\frac{3}{4}, \frac{12}{16}=\frac{12 \div 4}{16 \div 4}=\frac{3}{4}, \frac{15}{25}=\frac{15 \div 5}{25 \div 5}=\frac{3}{5}\) and \(\frac{18}{24}=\frac{18 \div 6}{24 \div 6}=\frac{3}{4}\)

Thus, on comparing the above cases, observe that \(\frac{3}{5}=\frac{15}{25}\) is not equal to the above-given fractions.

Question 7. Which of the following fractions is the greatest?

  1. \(\frac{5}{7}\)
  2. \(\frac{5}{6}\)
  3. \(\frac{5}{9}\)
  4. \(\frac{5}{8}\)

Solution:

2. We have given, \(\frac{5}{7}, \frac{5}{6}, \frac{5}{9} \text { and } \frac{5}{8}\) Since we know that, among all the fractions with the same numerator, the one with a smaller denominator will be the greatest

\(\frac{5}{6}\) is the required fraction.

Question 8. Which of the following fractions is the smallest?

  1. \(\frac{7}{8}\)
  2. \(\frac{9}{8}\)
  3. \(\frac{3}{8}\)
  4. \(\frac{5}{8}\)

Solution:

3. We have given \(\frac{7}{8}, \frac{9}{8}, \frac{3}{8} \text { and } \frac{5}{8}\)

Since we know that, among all the fractions with the same denominator, the one with a smaller numerator will be the smallest.

∴ \(\frac{3}{8}\) is the required fraction.

QueSomeon 9. Sum of \(\frac{4}{17} \text { and } \frac{15}{17} \text { is }\)

  1. \(\frac{19}{17}\)
  2. \(\frac{11}{17}\)
  3. \(\frac{19}{34}\)
  4. \(\frac{2}{17}\)

Solution:

1.  We have, \(\frac{4}{17}+\frac{15}{17}\)

⇒ \(=\frac{4+15}{17}\)

⇒ \(\left[\begin{array}{l}
\text { Sum of like fractions } \\
=\frac{\text { Sum of their numerators }}{\text { common denominator }}
\end{array}\right]\)

⇒ \(=\frac{19}{17}\)

Question 10. On subtracting \(\frac{5}{9} \text { from } \frac{19}{9} \text {, }\) the result is

  1. \(\frac{24}{9}\)
  2. \(\frac{14}{9}\)
  3. \(\frac{14}{18}\)
  4. \(\frac{14}{0}\)

Solution:

According to the question, \(\frac{19}{9}-\frac{5}{9}\)

⇒ \(\left[\begin{array}{l}\text { Difference of like fraction } \\
=\frac{\text { Difference of their numerator }}{\text { Common denominator }}
\end{array}\right]\)

∴ \(=\frac{14}{9}\)

Question 11. \(\frac{11}{7}\) can be expressed in the form

  1. \(\frac{33}{7}\)
  2. \(\frac{39}{7}\)
  3. \(\frac{33}{4}\)
  4. \(\frac{39}{4}\)

Solution:

We have, \(5 \frac{4}{7}=\frac{(5 \times 7)+4}{7}=\frac{35+4}{7}=\frac{39}{7}\)

Question 13. A number representing a part of a ______ is called a fraction.
Solution: whole

Question 14. A fraction with a denominator greater than the numerator is called a______fraction.
Solution: 
Proper

Question 15. Fractions with the same denominator are called _____ fractions.
Solution: Like

Question 16. \(13 \frac{5}{18}\) is a _____ fraction.
Solution:
Mixed

Question 17. \(\frac{18}{5}\) is an _____ fraction.
Solution:
Improper

Question 18. \(\frac{7}{19}\) is an _____ fraction.
Solution: Proper

Question 19. \(\frac{5}{8} \text { and } \frac{3}{8} \text { are }\) proper _____ fraction.
Solution: Like

Question 20. \(\frac{6}{11} \text { and } \frac{6}{13}\) proper _____ fraction.
Solution: Unlike

Question 21. The fraction \(\frac{6}{15}\) in simplest form is_____.
Solution:

\(\frac{2}{5}: \text { We have, } \frac{6}{15}=\frac{6 \div 3}{15 \div 3}=\frac{2}{5}\)

Question 22. The fraction \(\frac{17}{34}\)  in simplest form is _____.
Solution:

\(\frac{1}{2} \text { : We have, } \frac{17}{34}=\frac{17 \div 17}{34 \div 17}=\frac{1}{2}\)

Question 23. \(\frac{18}{135} \text { and } \frac{90}{675}\)  are proper, unlike and _____ fractions.
Solution: 
Equivalent

Question 24. \(8 \frac{2}{7}\)  is equal to the improper fraction______.
Solution:

\(\frac{58}{7}: 8 \frac{2}{7}=\frac{(8 \times 7)+2}{7}=\frac{56+2}{7}=\frac{58}{7}\)

Question 25. \(\frac{87}{7}\)  is equal to the mixed fraction______.
Solution:

\(12 \frac{3}{7}:\)

Fraction

∴ \(\frac{87}{7}=12 \frac{3}{7}\)

Question 26. \(\frac{17}{9}+\frac{41}{9}=\)______.
Solution:

⇒ \(\frac{58}{9}: \frac{17}{9}+\frac{41}{9}=\frac{17+41}{9}=\frac{58}{9}\)

Question 27. \(\frac{67}{14}-\frac{24}{14}=\)_______.
Solution:

\(\frac{43}{14}: \frac{67}{14}-\frac{24}{14}=\frac{67-24}{14}=\frac{43}{14}\)

Question 28. \(\frac{17}{2}+3 \frac{1}{2}=\)______.
Solution:

\(\text { 12: } \frac{17}{2}+3 \frac{1}{2}=\frac{17}{2}+\frac{7}{2}=\frac{17+7}{2}=\frac{24}{2}=12\)

Question 29. \(9 \frac{1}{4}-\frac{5}{4}=\)______.
Solution:

\(8: 9 \frac{1}{4}-\frac{5}{4}=\frac{37}{4}-\frac{5}{4}=\frac{37-5}{4}=\frac{32}{4}=8\)

Question 30. Fractions with the same numerator are called fractions.
Solution:
False

Because fractions with the same denominators are called fractions.

Question 31. Fraction \(\frac{18}{39}\) is in its lowest form
Solution: 
False

Since the common factor of 18 and 39 is 3 and its simplest form is

⇒ \(\frac{18 \div 3}{39 \div 3}=\frac{6}{13}\)

Question 32. Fractions  \(\frac{15}{39} \text { and } \frac{45}{117}\) are equivalent fractions.
Solution: 
True

⇒\(\text { Since, } \frac{45 \div 3}{117 \div 3}=\frac{15}{39}\)

∴ Both are equivalent

Question 33. The sum of two fractions is always a fraction.
Solution: 
True

Let, \(\frac{2}{3} \text { and } \frac{4}{3}\) are two fractions.

When we add them, we get \(\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}\)

\(=\frac{6}{3}=\frac{2}{1} \text {, }\) which is a fraction.

Question 34. The result obtained by subtracting a fraction from another fraction is necessarily a fraction.
Solution:  False

Let \(\frac{1}{2} \text { and } \frac{2}{3}\) are two fractions.

When we subtract \(\frac{2}{3} \text { from } \frac{1}{2} \text {, }\)

⇒ \(\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=\frac{-1}{6}\)

Which is not a fraction.

Question 35. If a whole or an object is divided into several equal parts, then each part represents a fraction.
Solution: True

Question 36. The fraction represented by the shaded portion in the adjoining figure is \(\frac{3}{8}.\)

shaded portion

Solution: True

The total number of parts in a given figure is 8 and the shaded parts are 3.

∴ The required fraction is \(\frac{3}{8}\)

Question 37. The fraction represented by the unshaded portion in the adjoining figure is\(\frac{5}{9}.\)

unshaded Portion

Solution: False

The total number of parts in a given figure is 9 of which unshaded parts are 4.

∴ The required fraction is \(\frac{4}{9}.\)

Question 38. \(\text { 38. } \frac{25}{19}+\frac{6}{19}=\frac{31}{38}\)
Solution: 
False

⇒ \(\frac{25}{19}+\frac{6}{19}=\frac{25+6}{19}=\frac{31}{19} \neq \frac{31}{38}\)

∴ The above condition is false.

Question 39. \(\frac{8}{18}-\frac{8}{15}=\frac{8}{3}\)
Solution: False

Fraction condition

The L.C.M. of 18 and 15 is 2 * 3 x 3 x 5- 90

∴ \(\frac{8}{18}-\frac{8}{15}=\frac{8 \times 5}{90}-\frac{8 \times 6}{90}\)

⇒\(=\frac{40}{90}-\frac{48}{90}=-\frac{8}{90} \neq \frac{8}{3}\)

∴ The above condition is false.

Question 40. \(\frac{7}{12}+\frac{11}{12}=\frac{3}{2}\)
Solution: True

⇒ \(\frac{7}{12}+\frac{11}{12}=\frac{7+11}{12}=\frac{18}{12}=\frac{18 \div 6}{12 \div 6}=\frac{3}{2}\)

Question 41. \(\frac{16}{25}>\frac{13}{25}\)
Solution: True

The given fractions are like fractions. On comparing the numerators, we get

⇒ \(\frac{16}{25}>\frac{13}{25}\)

Question 42. \(\frac{11}{16} \ldots \frac{14}{15}\)
Solution:

< : The L.C.M. of 16 and 15 is 2x2x2x2x3x5 = 240

⇒ \(\text { Thus, } \frac{11}{16}=\frac{11 \times 15}{16 \times 15}=\frac{165}{240}\)

⇒ \(\text { and } \frac{14}{15}=\frac{14 \times 16}{15 \times 16}=\frac{224}{240}\)

Comparing Fraction

On comparing, we observe that

∴ \(\frac{224}{240}>\frac{165}{240} \text {, i.e., } \frac{11}{16}<\frac{14}{15}\)

Question 43. \(\frac{8}{15} \ldots \frac{95}{14}\)
Solution:

< : The L.C.M. of 15 and 14 is 2 x 3 x 5 x 7 = 210

⇒ \(\text { Thus, } \frac{8}{15}=\frac{8 \times 14}{15 \times 14}=\frac{112}{210}\)

⇒ \(\text { and } \frac{95}{14}=\frac{95 \times 15}{14 \times 15}=\frac{1425}{210}\)

Fraction Of Comparing

On comparing, we observed that

⇒ \(\frac{1425}{210}>\frac{112}{210} \text { i.e., } \frac{8}{15}<\frac{95}{14}\)

Question 44. \(\frac{12}{75} \ldots \frac{32}{200}\)
Solution:

= : The L.C.M. of 75 and 200 is 2x2x2x3x5x5 = 600

⇒ \(\text { Thus, } \frac{12}{75}=\frac{12 \times 8}{75 \times 8}=\frac{96}{600}\)

⇒ \(\text { and } \frac{32}{200}=\frac{32 \times 3}{200 \times 3}=\frac{96}{600}\)

Observe Fraction

On comparing, we observe that

⇒ \(\frac{12}{75}=\frac{32}{200}\)

Question 45. \(\frac{18}{15} \ldots 1.3\)
Solution:

< : \(1.3=\frac{13}{10}\)

The L.C.M. of 15 and 10 is 2 x 3 x 5 = 30

⇒ \(\text { Now, } \frac{18}{15}=\frac{18 \times 2}{15 \times 2}=\frac{36}{30} \text { and } \frac{13}{10}=\frac{13 \times 3}{10 \times 3}=\frac{39}{30}\)

Comparing

⇒ \(\frac{36}{30}<\frac{39}{30}\)

∴ \(\text { Thus, } \frac{18}{15}<1.3\)

Question 46. Write the fraction represented by the shaded portion of the adjoining figure:

Fraction represented by the shaded

Solution:

In the given figure, the total parts in which the figure has been divided is 8 out of which 7 parts are shaded.

∴ The required fraction is \(\frac{7}{8}.\)

Question 47. Write the fraction represented by the unshaded portion of the adjoining figure:

unshaded portion of the adjoining

Solution:

In the given figure, the total number of parts in which the 1444 figure has been divided is 15, out of which 4 are unshaded.

∴ The required fraction is \(\frac{4}{15}.\)

Question 48. Ali divided one fruit cake equally among six persons. What part of the cake he gave to each person?
Solution:  Since Ali has to divide one fruit cake equally among 6 persons

Each person will get \(\frac{1}{6}\) part.

Question 49. Express \(6 \frac{2}{3}\) as an improper fraction.
Solution:

We have \(6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Question 50. Arrange the fractions \(\frac{2}{3}, \frac{3}{4}, \frac{1}{2} \text { and } \frac{5}{6} \text { in }\) in ascending order
Solution:

⇒ \(\text { We have, } 6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Firstly find the L.C.M. of 3, 4, 2, and 6.

Ascending Order

∴ The L.C.M. of 3, 4, 2 and 6 is 2 x 2 x 3 = 12

⇒ \(\text { Now, } \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}, \frac{3}{4}=\frac{3 \times 3}{4 \times 3}=\frac{9}{12} \text {, }\)

⇒ \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12} \text { and } \frac{5}{6}=\frac{5 \times 2}{6 \times 2}=\frac{10}{12}\)

On comparing the above, we get

⇒ \(\frac{6}{12}<\frac{8}{12}<\frac{9}{12}<\frac{10}{12} \text { i.e., } \frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{5}{6}\)

Question 51. Arrange the fractions \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text { in }\) descending order.
Solution:

⇒ We have given, \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text {. }\)

The L.C.M. of 7, 8, 5 and 4 is 2x2x2x5x7 = 280

⇒ \(\text { Now, } \frac{6}{7}=\frac{6 \times 40}{7 \times 40}=\frac{240}{280}\)

⇒  \(\frac{7}{8}=\frac{7 \times 35}{8 \times 35}=\frac{245}{280}\)

⇒  \(\frac{4}{5}=\frac{4 \times 56}{5 \times 56}=\frac{224}{280} \text { and } \frac{3}{4}=\frac{3 \times 70}{4 \times 70}=\frac{210}{280}\)

Decesding order

On comparing the above, we get;

⇒  \(\frac{245}{280}>\frac{240}{280}>\frac{224}{280}>\frac{210}{280}\)

⇒ \(\text { i.e., } \frac{7}{8}>\frac{6}{7}>\frac{4}{5}>\frac{3}{4}\)

Question 52. Write \(\frac{3}{4}\) as a fraction with denominator 44.
Solution:

⇒ \(\text { Let } \frac{3}{4}=\frac{?}{44}\)

Then, we have to find the missing numeral. To get 44 in the denominator, we multiply by 4 by 11.

So, we multiply the numerator and denominator by 11.

∴ \(\frac{3}{4}=\frac{3 \times 11}{4 \times 11}=\frac{33}{44}\)

\(\text { Hence, } \frac{3}{4} \text { and } \frac{33}{44}\) are equivalent fractions.

Question 53. Write \(\frac{5}{6}\) as a fraction with the numerator 60.
Solution:

Let, \(\frac{5}{6}=\frac{60}{?}\)

Then, we have to find the missing numeral. To get 60 in the numerator, we multiply 5 by

So, we multiply the numerator and denominator by 12.

∴ \(\frac{5 \times 12}{6 \times 12}=\frac{60}{72}\)

⇒ \(\text { Hence, } \frac{5}{6} \text { and } \frac{60}{72}\) are equivalent fractions.

Question 54. Write \(\frac{129}{8}\) as a mixed fraction
Solution:

We have, \(\frac{129}{8}=16 \frac{1}{8}\)

Mixed Fraction

Question 55. Add the fractions \(\frac{3}{8} \text { and } \frac{2}{3} \text {. }\)
Solution:

L.C.M. 0f 8 and 3 is 2x2x2x3 = 24

Now, \(\frac{3}{8}+\frac{2}{3}=\frac{3 \times 3}{8 \times 3}+\frac{2 \times 8}{3 \times 8}\)

Add the Fraction

∴  \(=\frac{9}{24}+\frac{16}{24}=\frac{9+16}{24}=\frac{25}{24}\)

Question 56. Add the fractions \(\frac{3}{8} \text { and } 6 \frac{3}{4} \text {. }\)
Solution:

L.C.M. of 8 and 4 is 2 x 2 x 2 = 8

Now, \(\frac{3}{8}+6 \frac{3}{4}=\frac{3}{8}+\frac{27}{4}\)

Add the Fractions

⇒  \(\frac{3}{8}+\frac{27 \times 2}{4 \times 2}\)

⇒  \(\frac{3}{8}+\frac{54}{8}=\frac{3+54}{8}=\frac{57}{8}=7 \frac{1}{8}\)

Question 57. Subtract \(\frac{1}{6} \text { from } \frac{1}{2} \text {. }\)
Solution:

The L.C.M. of 6 and 2 = 6

Subtract Fraction

Now, \(\frac{1}{2}-\frac{1}{6}=\frac{1 \times 3}{2 \times 3}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}\)

⇒ \(\frac{3-1}{6}=\frac{2}{6}=\frac{2 \div 2}{6 \div 2}\)

[H.C.F. of 2 and 6 is 2]

⇒ \(\frac{1}{3}\)

Question 58. Subtract \(8 \frac{1}{3} \text { from } \frac{100}{9}\).
Solution:

The L.C.M. of 3 and 9 = 9

Subtract Fraction

⇒ \(\frac{100}{9}-8 \frac{1}{3}=\frac{100}{9}-\frac{25}{3}\)

⇒ \(=\frac{100}{9}-\frac{25 \times 3}{3 \times 3}=\frac{100}{9}-\frac{75}{9}=\frac{100-75}{9}=\frac{25}{9}\)

⇒ \(2 \frac{7}{9}\)

Question 59. Subtract \(1 \frac{1}{4} \text { from } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

Subtract Fractions

Now, \(6 \frac{1}{2}-1 \frac{1}{4}=\frac{13}{2}-\frac{5}{4}\)

⇒ \(\frac{13 \times 2}{2 \times 2}-\frac{5}{4}=\frac{26}{4}-\frac{5}{4}=\frac{26-5}{4}\)

⇒ \(\frac{21}{4}=5 \frac{1}{4}\)

Question 60. Add \(1 \frac{1}{4} \text { and } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

⇒ \(\text { Now, } 1 \frac{1}{4}+6 \frac{1}{2}=\frac{5}{4}+\frac{13}{2}\)

⇒ \(\frac{5}{4}+\frac{13 \times 2}{2 \times 2}=\frac{5}{4}+\frac{26}{4}=\frac{5+26}{4}=\frac{31}{4}\)

⇒ \(7 \frac{3}{4}\)

⇒ \(\begin{array}{l|l}
2 & 2,4 \\
\hline 2 & 1,2 \\
\hline & 1,1
\end{array}\)

Question 61. Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening. Find the distance traveled by her altogether on that day.
Solution:

Given

Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening.

The distance travelled by Katrina in the morning  \(=6 \frac{1}{2} \mathrm{~km}=\frac{13}{2} \mathrm{~km}\)

The distance traveled by Katrina in the evening \(=8 \frac{3}{4} \mathrm{~km}=\frac{35}{4} \mathrm{~km}\)

Total =distance traveled by her

⇒ \(\frac{13}{2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

⇒ \(\frac{13 \times 2}{2 \times 2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

∴ [ L.C.M. of 2 and 4 = 4]

⇒ \(\frac{26}{4} \mathrm{~km}+\frac{35}{4} \mathrm{~km}=\frac{(26+35) \mathrm{km}}{4}\)

⇒ \(\frac{61}{4} \mathrm{~km}=15 \frac{1}{4} \mathrm{~km}\)

Question 62. A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\) find the number of parts in which the rectangle has been divided.
Solution:

Given

A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\)

Let the number of parts in which the rectangle has been divided be x.

According to the question, \(\frac{1}{4}=\frac{16}{x}\)

By cross-multiplication, x = 16 Χ 4 = 64

∴ The required number of parts is 64.

Question 63. The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\). Express the size as an improper fraction.
Solution:

Given

The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\).

We have given, you the grip size of a tennis racquet \( =11 \frac{9}{80} \mathrm{~cm}=\frac{889}{80} \mathrm{~cm}\), which is the required improper fraction.

Question 64. On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain. What fraction of the food eaten is not available for the next level?
Solution:

Given

On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain.

We have given, \(\frac{1}{10}\) of the food eaten is turned into the organism’s own body.

The required fraction of the food eaten not available for the next level is

⇒ \(1-\frac{1}{10}=\frac{10-1}{10}\)

[ L.C.M. of1 and 10 is 10]

⇒ \(\frac{9}{10}\)

Question 65. Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years. What fraction of his age till retirement was he in the job?
Solution:

Given

Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years.

Mr. Rajan got a job at the age of 24 years.

He retired at the age of 60 years.

He worked for (60- 24) years = 36 years

∴ The required fraction \(=\frac{36}{60}=\frac{36 \div 12}{60 \div 12}=\frac{3}{5}\)

[ H.C.F. of 36 and 60 is 12]

Question 66. The food we eat remains in the stomach for a maximum of 4 hours. For what fraction of a day, does it remain there? 
Solution:

Given

The food we eat remains in the stomach for a maximum of 4 hours.

Total number of hours in a day 24 hours

The required fraction \(=\frac{4}{24}= \div {4+4}{24+4}= \div {1}{6}\)

[ H.C.F. of 4 and 24 is 4]

Question 67. It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007. Find the fraction of:

  1. The quantity of diesel saved to the quantity of petrol saved.
  2. The quantity of diesel saved to the quantity of CNG saved.

Solution:

Given

It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007.

By the end of the year 2007,

The quantity of CNG saved 33000 tonnes

The quantity of diesel saved 3300 tonnes and

The quantity of petrol saved 21000 tonnes

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of petrol saved }}=\frac{3300}{21000}\)

⇒ \(\frac{3300 \div 300}{21000 \div 300}\)

[H.C.F. of 3300 and 21000 is 300]

\(=\frac{11}{70}\)

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of CNG saved }}=\frac{3300}{33000}\)

⇒ \(\frac{3300 \div 3300}{33000 \div 3300}\)

[H.C.F. of 3300 and 33000 is 3300]

∴ \(=\frac{1}{10}\)

Question 68. A cup is \(\frac{1}{3}\) full of milk. What part of the cup is still to be filled with milk to make it full?
Solution:

Given

A cup is \(\frac{1}{3}\) full of milk.

The remaining part of the cup which is still to be filled with milk  \(=1-\frac{1}{3}\)

⇒ \(\frac{3-1}{3}\)

⇒  \(\frac{2}{3}\)

Question 69. Mary bought \(3 \frac{1}{2} \mathrm{~m}\) of lace. She used \(1 \frac{3}{4} \mathrm{~m}\) of Hires lace for her new dress. How much lace is left with her?
Solution:

Given

Mary bought the lace = \(3 \frac{1}{2} \mathrm{~m}\) = \(\frac{7}{2} \mathrm{~m}\)

Lace used by Mary \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\)

She is left with \(\frac{7}{2} m-\frac{7}{4} m\)

⇒\(\frac{7 \times 2}{2 \times 2} m-\frac{7}{4} m=\frac{14 m-7 m}{4}\)

[L.C.M. of 2 and 4 is 4]

⇒ \(\frac{7}{4} m=1 \frac{3}{4} m \text { of lace }\)

Question 70. When Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)  What was her weight on Monday?
Solution:

Given

Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)

Sunita had gained \(=1 \frac{1}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg}\)

Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}\)

Her total weight on Monday

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{5}{4} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}+\frac{5 \times 2}{4 \times 2} \mathrm{~kg}\)

[L.C.M. of 8 and 4 is 8]

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{10}{8} \mathrm{~kg}\)

⇒ \(\frac{(371+10)}{8} \mathrm{~kg}=\frac{381}{8} \mathrm{~kg}=47 \frac{5}{8} \mathrm{~kg}\)

Question 71. Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday. How many liters of juice did he purchase together in two days?
Solution:

Given

Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday.

Sunil purchased juice on Monday = \(12 \frac{1}{2}\)liters \(=\frac{25}{2} \text { litres }\)

and on Tuesday \(14 \frac{3}{4}\)liters  \(=\frac{59}{4} \text { litres }\)

The total quantity of juice Sunil purchased in two days  \(=\left(\frac{25}{2}+\frac{59}{4}\right) \text {liters}\)

⇒ \(=\left(\frac{25 \times 2}{2 \times 2}+\frac{59}{4}\right) \text { litres }=\left(\frac{50}{4}+\frac{59}{4}\right) \text { litres }\)

⇒ \(=\left(\frac{50+59}{4}\right) \text { litres }=\frac{109}{4} \text { litres }=27 \frac{1}{4} \text { litres }\)

Question 72. Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends. How many liters of juice is left with her?
Solution:

Given

Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends.

Total quantity of juice = \(5 \frac{1}{2}\)liters

= \(\frac{11}{2} \text { litres }\)

Nazima gave to her friends \(=2 \frac{3}{4} \text { litres }\)

⇒ \(\frac{11}{4} \text { litres }\)

The required quantity of juice she is left \(=\left(\frac{11}{2}-\frac{11}{4}\right) \text { litres }=\left(\frac{11 \times 2}{2 \times 2}-\frac{11}{4}\right) \text { litres }\)

⇒ \(\left(\frac{22-11}{4}\right) \text { litres }=\frac{11}{4} \text { litres }=2 \frac{3}{4} \text { litres }\)

Question 73. Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it. What is the length of the remaining piece?
Solution:

Given

Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it.

Total length of a wooden board \(=150 \frac{1}{4} \mathrm{~cm}=\frac{601}{4} \mathrm{~cm}\)

The carpenter sawed off a piece of length \(=40 \frac{1}{5} \mathrm{~cm}=\frac{201}{5} \mathrm{~cm}\)

The length of the remaining piece \(=\left(\frac{601}{4}-\frac{201}{5}\right) \mathrm{cm}=\left(\frac{601 \times 5}{4 \times 5}-\frac{201 \times 4}{5 \times 4}\right) \mathrm{cm}\)

⇒\(\left(\frac{3005}{20}-\frac{804}{20}\right) \mathrm{cm}=\frac{2201}{20} \mathrm{~cm} \text { or } 110 \frac{1}{20} \mathrm{~cm}\)

Question 74. Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town. How much did he travel to reach the town?
Answer:

Given

Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town.

Nasir traveled by a bus = \(3 \frac{1}{2} \mathrm{~km}\)

⇒ \(\frac{7}{2} \mathrm{~km}\)

Nasir walked \(=1 \frac{1}{8} \mathrm{~km}=\frac{9}{8} \mathrm{~km}\)

Total distance traveled by him latex]=\frac{7}{2} \mathrm{~km}+\frac{9}{8} \mathrm{~km}=\frac{7 \times 4}{2 \times 4} \mathrm{~km}+\frac{9}{8} \mathrm{~km}[/latex]

⇒ \(\left(\frac{28}{8}+\frac{9}{8}\right) \mathrm{km}\)

∴ \(\left(\frac{28+9}{8}\right) \mathrm{km}=\frac{37}{8} \mathrm{~km} \text { or } 4 \frac{5}{8} \mathrm{~km}\)

Question 75. The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg. How much more did Neetu’s fish weigh than that of Narendra?
Solution:

Given

The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg.

The weight of fish caught by Neetu \(=3 \frac{3}{4} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}\)

The weight of fish caught by Narendra \(=2 \frac{1}{2} \mathrm{~kg}=\frac{5}{2} \mathrm{~kg}\)

∴ Neetu’s fish weigh more than that of Narendra by \(\frac{15}{4} \mathrm{~kg}-\frac{5}{2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{5 \times 2}{2 \times 2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{10}{4} \mathrm{~kg}\)

⇒ \(\frac{(15-10)}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg} \text { or } 1 \frac{1}{4} \mathrm{~kg}\)

Question 76. Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf. How much cloth must he buy in all?
Solution:

Given

Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf.

Neelam’s father purchased the length of cloth for the skirt \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\) and for the scarf \(=\frac{1}{2} \mathrm{~m}\)

Total length he buys in all = \(=\frac{7}{4} m+\frac{1}{2} m\)

⇒ \(\frac{7}{4} m+\frac{1 \times 2}{2 \times 2} m=\frac{7}{4} m+\frac{2}{4} m\)

⇒ \(\frac{(7+2)}{4} m=\frac{9}{4} m \text { or } 2 \frac{1}{4} m\)

Question 77. What is wrong with the following additions?

Additions
Solution:

  1. Equal denominators are added.
  2. Numerators and denominators are added

Question 78. Match the fractions of Column I with the shaded or marked portion of figures of Column II:

The shaded portion

Solution:

1→(D); 2→(A); 3→(E); 4→(B);

Marked portion in (A) \(=\frac{6}{10}\)

Shaded fraction in (B) \(=\frac{6}{16}\)

Shaded fraction in (C) \(=\frac{6}{7}\)

Shaded fraction in (D) \(=\frac{4}{4}+\frac{2}{4}=\frac{6}{4}\)

Shaded fraction in (E) \(=\frac{6}{6}\)

Question 79. Find the fraction that represents the number of natural numbers to total numbers in the collection 0, 1, 2, 3, 4, 5. What fraction will it be for whole numbers?
Solution:

Out of 0, 1, 2, 3, 4, 5 → 1, 2, 3, 4 and 5 are the natural numbers.

The fraction that represents the number of natural numbers to the total numbers \(=\frac{5}{6}\) and the whole numbers are 0, 1, 2, 3, 4, and 5.

∴ The fraction that represents the number of whole numbers to the total numbers \(=\frac{6}{6}\)

Question 80. Write the fraction representing the total number of natural numbers in the collection of numbers -3, -2, -1, 0, 1, 2, 3. What fraction will it be for whole numbers? What fraction will it be for integers?
Solution:

Out of -3, -2, -1, 0, 1, 2, 3 → 1, 2, and 3 are the natural numbers, 0, 1, 2, and 3 are the whole numbers and -3, -2, -l, 0, 1, 2, 3 are integers.

∴ The fraction representing the natural numbers to the total numbers \(=\frac{3}{7}\)

The fraction representing the whole numbers to the total numbers \(=\frac{4}{7}\)

The required fraction representing the integers to the total numbers  \(=\frac{7}{7}\)

Question 81. Write a pair of fractions whose sum is \(\frac{7}{11}\) and the difference is \(\frac{2}{11}\)
Solution: Let one fraction be x

Another fraction be \(\frac{7}{11}-x\)

Now, according to the question,

⇒ \(x-\left(\frac{7}{11}-x\right)=\frac{2}{11}\)

⇒ \(\Rightarrow x-\frac{7}{11}+x=\frac{2}{11}\)

⇒ \(\frac{11 x}{11}-\frac{7}{11}+\frac{11 x}{11}=\frac{2}{11}\)

⇒ \(\frac{11 x-7+11 x}{11}=\frac{2}{11}\)

⇒ \(22 x-7=\frac{2}{11} \times 11\)

⇒ \(22 x-7=2 \Rightarrow 22 x=2+7=9\)

⇒ \(x=\frac{9}{22}\)

Thus, one fraction is \(\frac{9}{22}\) and another fraction is \(\frac{7}{11}-\frac{9}{22}=\frac{7 \times 2}{11 \times 2}-\frac{9}{22}=\frac{14-9}{22}=\frac{5}{22}\)

Question 82. What fraction of a straight angle is a right angle?
Solution: Since we know that the measurement of a straight angle is 180° and a right angle is 90°

∴ The required fraction is \(\frac{90^{\circ}}{180^{\circ}}=\frac{1}{2} .\)

Question 83. Put the right card in the right bag.

Right Card And Bag Card
Right Card And Right Bag

Solution:

We know that if the numerator is less than the denominator, then the fraction is less than 1.

If the numerator is equal to the denominator, then the fraction is equal to 1 and if the numerator is greater than the denominator, then the fraction is greater than 1.

The cards in Bag 1 are

1.  \(\frac{3}{7},\)

4. \(\frac{8}{9}\)

5.  \(\frac{5}{6},\)

6. \(\frac{6}{11},\)

8. \(\frac{19}{25}\)

9. \(\frac{2}{3}\) and

10. \(\frac{13}{17}\)

The cards in Bag 2 are

2.  \(\frac{4}{4}\)

7. \(\frac{18}{18}\)

And cards in B ag 3 are

3. \(\frac{9}{8} .\)

NCERT Exemplar Solutions For Class 6 Maths Chapter 10 Mensuration

Class 6 Maths Chapter 10 Mensuration

Question 1. The following figures are formed by joining six unit squares. Which figure has the smallest perimeter?

Mensuration The Six Unit Squares

  1. 2
  2. 3
  3. 4
  4. 1

Solution: (4): Perimeter of the figure (1) = 10 units

Perimeter of figure (2) = 12 units

Perimeter of figure (3) = 14 units

Perimeter of figure (4) = 14 units

So, figure (1) has the smallest perimeter

Read and Learn More Class 6 Maths Exemplar Solutions

Question 2. A square-shaped park ABCD of side 100 m has two rectangular flower beds each of size 10 m x 5 m (see figure). The length of the boundary of the remaining park is

Mensuration The Two Equal Rectangle

  1. 360 m
  2. 400 m
  3. 340 m
  4. 460 m

Solution: (2): Length of required boundary

= Perimeter of the remaining park

= 90m + 95m + 10m + 5m + 90m + 95m + 10m + 5m

= 400 m

NCERT Exemplar Solutions For Class 6 Maths Chapter 10 Mensuration

Question 3. The side of a square is 1 0 cm. How many times will the new perimeter become if the side of the square is doubled?

  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times

Solution: (1): Side of square = 10 cm

Perimeter of square = 4 x 10 cm = 40 cm

Now, new side of square = 2 x 10 cm = 20 cm

Perimeter of new square = 4 x 20 cm = 80 cm = 2 x 40 cm …(ii)

(1) and (2) show that the new perimeter will be 2 times the perimeter of a given square.

Question 4. The length and breadth of a rectangular sheet column II: of paper are 20 cm and 10 cm, respectively. A rectangular piece is cut from the sheet as shown in the figure. Which of the following statements is correct for the remaining sheet?

Mensuration The Rectangle Piece

  1. The perimeter remains the same but the area changes.
  2. The area remains the same but the perimeter changes.
  3. Both area and perimeter are changing.
  4. Both the area and perimeter remain the same.

Solution: (1): The given figure shows that the perimeter will remain the same but the area changes

Question 5. Two regular hexagons of perimeter 30 cm each are joined as shown in the figure. The perimeter of the new figure is

Mensuration The Same But Perimeter Change

  1. 65 cm
  2. 60 cm
  3. 55 cm
  4. 50 cm

Solution: (4): Perimeter of a regular hexagon = 6 x side

30 cm= 6 x side => side \(\frac{30}{6} \mathrm{~cm}=5 \mathrm{~cm}\)

Each side of a regular hexagon = 5 cm

The perimeter of the new figure = 10 x side- 10 x 5 cm = 50 cm

Question 6. In the given figure, which of the following is a regular polygon? All have equal sides except (i)

Mensuration The Regular Polygon

  1. 1
  2. 2
  3. 3
  4. 4

Solution: (2): Since we know that in a regular polygon, all angles and all sides are equal. Thus figure (2) satisfies the above condition.

Question 7. Match the shapes (each side measures 2 cm) in column 1 with the corresponding perimeters in column 2.

Mensuration The Shapes

Solution: (A) → (4); (B) → (1); (C) → (2); (D) → (3)

1.  The given figure has 14 sides.

Each side measures 2 cm.

∴ The perimeter of the given figure = 14 x 2 cm = 28 cm

2.  The given figure has 8 sides.

∴ The perimeter of the given figure = 8 x 2 cm = 16 cm

3. The given figure has 10 sides.

∴ The perimeter of the given figure = 10 x 2 cm = 20 cm

4.  The given figure has 12 sides.

∴  The perimeter of the given figure = 12 x 2 cm = 24 cm

Question 8. Match the following:

Mensuration The Match The Following

Solution: (A) → (3); (B) → (3); (C) → (2); (D) → (1)

1.  The perimeter of a rectangle

= 2 (length + breadth)

= 2(6 + 4) = 2(10) = 20

2. Perimeter of a square = 4 x side

= 4×5 = 20

3. The perimeter of an equilateral triangle

= 3 x side = 3×6 = 18

4. Perimeter of an isosceles triangle

= Sum of all sides = 4 + 2 + 4 = 10

Question 9. The perimeter of the shaded portion in the given figure is

Mensuration The Perimeter Of The Shaded Portion

AB+ _ + _ + _ + _ + _ + _ + HA

Solution:

EM, MD, DE, EK NG, GH: Perimeter of the shaded portion of the given figure

= AB + BM +MD + DE + EN + NG + GH +HA

Question 10. The amount of region enclosed by a plane closed figure is called its____.
Solution: Area

11. The area of a rectangle with a length of 5 cm and breadth of 3 cm is____.
Solution:

15 cm2: Area of a rectangle

= length x breadth = (5×3) cm2= 15 cm2

Question 12. A rectangle and a square have the same perimeter (see figure).

Mensuration The A rectangle and a square

  1. The area of the rectangle is____.
    The area of the square is____.

Solution:

12 sq units: Area of a rectangle

= length x breadth

= (6×2) sq units = 12 sq units

16 sq units: We have given,

Perimeter of a rectangle = Perimeter of a square

⇒  2(length + breadth) = 4 x side

⇒  2(6 + 2) = 4 x side ⇒  2(8) = 4 x side

⇒  \(\frac{16}{4}=\text { side }\)

Side = 4 units

∴ The area of square = side x side = (4×4) sq units = 16 sq units

Question 13.

  1. 1 m = ____ cm.
  2. 1 sq cm = ____ cm x 1 cm.
  3. 1 sq m = 1 m x ____ m = 100 cm x ____ cm.
  4. 1 sq m = ____ sq cm

Solution:

  1. 100: 1 ni = 100 cm
  2. 1 : 1 sq cm =1 cm x 1 cm
  3. 1 and 100: 1 sq m =1 m x1 m = 100 cm x 100 cm
  4. 10000: 1 sq m = 10000 sq cm

Question 14. If the length of a rectangle is halved and the breadth is doubled then the area of the rectangle obtained remains the same.
Solution: True

Let l and b are the dimensions of a given rectangle, whereas l’ and b’ are the dimensions of a new rectangle.

According to question, \(l^{\prime}=\frac{1}{2} l \text { and } b^{\prime}=2 b\)

Area of new rectangle = \(l^{\prime} b^{\prime}=\frac{1}{2} l \times 2 b=l b \text {, }\)

which is the area of the given rectangle

Question 15. The area of the square is doubled if the side of the square is doubled.
Solution: False

Let the side of a given square be a.

∴ Area of the square = a2

The side of a new square = 2a

∴ Area of the new square = (2a)2 = 4a2

which concludes that the area of the new square is four times the area of the given square.

Question 16. The perimeter of a regular octagon of side 6 cm is 36 cm.
Solution: False

The side of a regular octagon = 6 cm

The perimeter of a regular octagon = 8 x side

= 8x6cm = 48cm

Question 17. A farmer who wants to fence his field must find the perimeter of the field.
Solution: True

Question 18. An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
Solution: False

To build a compound wall on all sides of a house, the engineer must find the perimeter of the compound

Question 19. To find the cost of painting a wall we need to find the perimeter of the wall.
Solution: False

To find the cost of painting a wall we need to find the area of the wall.

Question 20. To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Solution: True

Question 21. Four regular hexagons are drawn to form the design as shown in the figure. If the perimeter of the design is 28 cm, find the length of each side of the hexagon.

Mensuration The Perimeter length of each side of the hexagon

Solution:

The perimeter of the design = 28 cm

⇒  14 x side = 28 cm

⇒  \(\text { side }=\frac{28}{14} \mathrm{~cm}=2 \mathrm{~cm}\)

Question 22. The perimeter of an isosceles triangle is 50 cm. If one of the two equal sides is 18 cm, find the third side.
Solution:

Mensuration The Triangle

Given

The perimeter of an isosceles triangle is 50 cm. If one of the two equal sides is 18 cm

Let AABC be the given isosceles triangle, where AB = CA = 18 cm.

Perimeter of the AABC = 50 cm

AB + BC + CA = 50 cm A

⇒  18 cm + BC + 18 cm = 50 cm

⇒  36 cm + BC = 50 cm

⇒  BC = 50 cm- 36 cm

⇒  BC = 14 cm

The third side BC = 14 cm

Question 23. The length of a rectangle is three times its breadth. The perimeter of the rectangle is 40 cm. Find its length and width.
Solution:

Given

The length of a rectangle is three times its breadth. The perimeter of the rectangle is 40 cm.

Let l and b be the length and breadth respectively of the given rectangle.

According to question,l = 3b

Perimeter of the rectangle = 2(1 + b)

⇒  40 cm = 2(3b + b)

⇒  40 cm = 2 x 4b => 40 cm = 8b

⇒  \(\frac{40}{8} \mathrm{~cm}=b \Rightarrow b=5 \mathrm{~cm}\)

The required length = 3x5cm = 15cm and breadth = 5 cm

Question 24. There is a rectangular lawn 10 m long and 4 m wide in front of Meena’s house (see figure). It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance. Find the length of the fencing.

Mensuration The length of fencing

Solution:

Given

There is a rectangular lawn 10 m long and 4 m wide in front of Meena’s house (see figure). It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance.

Length of rectangular lawn = 10 m

And the breadth of rectangular lawn = 4 m

The length of fencing along the two smaller

sides of lawn = 4m + 4m = 8 m

Andthelengthoffencingalongthelongerside of lawn leaving a gap of1 m for the entrance =10m-lm = 9m

Total length of fencing=8m +9m=17m

Question 25. The region given in the figure is measured by taking Mensuration The Area Of Of Triangleit as a unit. What is the area of the region?

Mensuration The perimeters

Solution:

Given

he region given in the figure is measured by taking Mensuration The Area Of Of Triangleit as a unit.

Since, in the given figure there are 13 rectangular shapes.

Area of the region = 13 sq units.

Question 26. Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140 cm. Find the side of this field in metres.
Solution:

Given

Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140 cm.

The perimeter of a square field = the distance around a square field

⇒ 4 x side = length of 200 rods

⇒ 4 x side = 200 x 140 cm

[ The length of the rod = 140 cm]

Side = \(\frac{200 \times 140}{4} \mathrm{~cm}\)

⇒ Side = 7000 cm \(=\frac{7000}{100} \mathrm{~m}=70 \mathrm{~m}\)

∴ The side of the field = 70 m

Question 27. The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6 km. What is the length of the field?
Solution:

Given

The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6 km.

We have given,

The length of a rectangular field = 2 (breadth)

The perimeter of field = 2(length + breadth)

Jamal jogged around it 4 times and covered a distance of 6 km.

4[2(length + breadth)] = 6 km

⇒  8(length + breadth) = 6 x 1000 m

(2 breadth + breadth) \(\frac{6 \times 1000}{8} \mathrm{~m}\)

⇒  3 breadth = 3 x 250 m

⇒  breadth = \(\frac{3 \times 250}{3} \mathrm{~m}=250 \mathrm{~m}\)

∴ Length of the field = 2 x 250 m = 500 m

Question 28. Three squares are joined together as shown in the figure. Their sides are 4 cm, 10 cm and 3 cm. Find the perimeter of the figure.

Mensuration The Three squares

Solution:

Given

Three squares are joined together as shown in the figure. Their sides are 4 cm, 10 cm and 3 cm.

We have

Mensuration The Three square

The perimeter of the given figure

= (10 + 3 + 3 + 3 + 7 + 10 + 6 + 4 + 4 + 4) cm

= 54 cm

Question 29. In the given figure all triangles are equilateral and AB = 8 units. Other triangles have been formed by taking the midpoints of the sides. What is the perimeter of the figure?

Mensuration The triangles are equilateral

Solution:

Given

In the given figure all triangles are equilateral and AB = 8 units. Other triangles have been formed by taking the midpoints of the sides.

We have given, all the triangles are equilateral and AB = 8 units

Triangles Are Equilateral

Now, the perimeter of the given figure

= AE + EF + FG + GH +HI+ID + DC + CK + KN + NO + OM +ML + LJ + JB + BU + US + SR +RT+TQ + QP + PA

=(4+2+1 +1+1+2+4+4+2+1+1 +1+2 + 4 + 4 + 2 +1 +1 +1 + 2 + 4) units

= 45 units

The perimeter of the figure = 45 units

Question 30. The length of a rectangular field is 250 m and the width is 150 m. Anuradha runs around this field 3 times. How far did she run? How many times she should run around the field to cover a distance of 4 km?
Solution:

Given

The length of a rectangular field is 250 m and the width is 150 m. Anuradha runs around this field 3 times.

Let ABCD be the given rectangular field where length = AB = 250 m and breadth = AD = 150 m

Mensuration The Length of a rectangular

The perimeter of the rectangle ABCD

= 2(AB+AD)

= 2(250 + 150) m

= 2(400) m = 800 m

Since Anuradha runs around the field 3 times.

∴ She covers a distance of 3 x 800 m = 2400 m = 2 km 400 m [ 1 km = 1000 m]

Now, the number of times she will run to cover the distance of 4 km or 4000 m \(=\frac{4000}{800}=5\)

Question 31. Bajinder runs ten times around a square track and covers 4 km. Find the length of the track.
Solution:

Given

Bajinder runs ten times around a square track and covers 4 km.

Let a be the side of the square track.

The perimeter of the square track = 4a

Bajinder runs ten times around the square track and covers 4 km.

10 x 4a = 4 km

40a = 4 x 1000 m

⇒ \(a=\frac{4 \times 1000}{40} \mathrm{~m}=100 \mathrm{~m}\)

∴ Length of the track = 4a = 4 x 100 m = 400 m

Question 32. The lawn in front of Molly’s house Is 12 m x 8 m, whereas the lawn in front of Dolly’s house is 1 5 m x 5 m. A bamboo fencing is built around both the lawns. How much fencing is required for both?
Solution:

Given

The lawn in front of Molly’s house Is 12 m x 8 m, whereas the lawn in front of Dolly’s house is 1 5 m x 5 m. A bamboo fencing is built around both the lawns.

The perimeter of the lawn in front of Molly’s house= 2(length + breadth) = 2(12 + 8) m = 2(20) m = 40 m

The perimeter of the lawn in front of Dolly’s house = 2(length + breadth) = 2(15 + 5) m = 2 x 20 m = 40 m

∴ Total length of fencing for both the lawns = (40 + 40) m = 80 m

Question 33. The perimeter of a regular pentagon is 1540 cm. How long is it on each side?
Solution:

The perimeter of the regular pentagon = 1540 cm

⇒ 5 x side = 1540 cm

⇒ Side = \(\frac{1540}{5} \mathrm{~cm}=308 \mathrm{~cm}\)

Each side of the regular pentagon is 308 cm.

Question 34. The perimeter of a triangle is 28 cm. One of its sides is 8 cm. Write all the sides of the possible isosceles triangles with these measurements.
Solution:

Given

The perimeter of a triangle is 28 cm. One of its sides is 8 cm.

Let AABC be an isosceles triangle,

where AB = 8 cm

Case 1: If AB = BC

AB = BC = 8 cm

Mensuration The perimeter of a triangle

The perimeter of the triangle ABC = 28 cm

⇒  AB + BC+ CA = 28 cm

⇒  8 cm + 8 cm + CA = 28 cm

⇒  16 cm + CA = 28 cm

⇒  CA = 28 cm- 16 cm = 12 cm

∴ Sides are 8 cm, 8 cm and 12 cm

Case 2: If BC = CA

The perimeter of the triangle ABC = 28 cm

⇒  AB + BC + CA = 28 cm

⇒  8 cm + 2BC = 28 cm

⇒  2BC = 28 cm- 8cm = 20 cm

⇒  BC = 10 cm

∴ Sides are 10 cm, 10 cm and 8 cm

Question 35. The length of an aluminium strip is 40 cm. If the lengths in cm are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with 1 5 cm length and 5 cm breadth can be made from this strip.)
Solution:

Given

The length of an aluminium strip is 40 cm. If the lengths in cm are measured in natural numbers,

Perimeter of rectangular frame = Length of the aluminium strip

2(length + breadth) = 40 cm

length + breadth = \(\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm}\)

The possible measurement of rectangular frames are 1 cm x 19 cm, 2 cm x 18 cm, 3 cm x 17 cm, 4 cm x 16 cm, 5 cm x 15 cm, 6 cm x 14 cm, 7 cm x 13 cm, 8 cm x 12 cm, 9 cm x 11 cm, 10 cm x 10 cm

Question 36. The base of a tent is a regular hexagon with a perimeter of 60 cm. What is the length of each side of the base?
Solution:

Given

The base of a tent is a regular hexagon with a perimeter of 60 cm.

The perimeter of the regular hexagon= 60 cm

⇒  6 x side = 60 cm

⇒  \(=\frac{60}{6} \mathrm{~cm}=10 \mathrm{~cm}\)

∴ Each side of the base is 10 cm.

Question 37. In an exhibition hall, there are 24 display boards each of length 1 m 50 cm and breadth 1 m. There is a 100 m long aluminium strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.
Solution:

Given

In an exhibition hall, there are 24 display boards each of length 1 m 50 cm and breadth 1 m. There is a 100 m long aluminium strip, which is used to frame these boards.

Length of display board =1 m 50 cm

= 1.50 m

Breadth of display board =1 m

Perimeter of one display board

= 2(length + breadth) = 2(1.50 + 1) m

= 2(2.50) m = 5 m

∴ Perimeter of 24 display boards

= (24 x 5) m = 120 m

The number of boards will be framed using the aluminium strip of 100 m long = \(\frac{100}{5}=20\)

The length of the aluminium strip required to frame the remaining boards = 120 m- 100 m = 20 m

Question 38. In the above question, how many square metres of cloth is required to cover all the display boards? What will be the length in metres of the cloth used, if its breadth is 120 cm?
Solution:

Length of display board =1 m 50 cm

= 1.5 m

Breadth of display board =1 m

∴ Area of 24 display boards = 24(/ x b)

= 24(1.5 x 1) m2= 24 x 1.5 m2= 36 m2

Therefore, 36 m2 of cloth is required to cover all the display boards.

Now, breadth of cloth = 120 cm = \(\frac{120}{100} \mathrm{~m}\)

∴ Length of cloth = \(\frac{\text { area }}{\text { breadth }}\)

⇒  \(\frac{36 \times 100}{120} \mathrm{~m}=30 \mathrm{~m}\)

Question 39. What is the length of the outer boundary of the park shown in the figure? What will be the total cost of fencing it at the rate of? 20 per metre? There is a rectangular flower bed in the centre of the park. Find the cost of manuring the flower bed at the rate of? 50 per square metre.

Mensuration The length of outer boundary of the Park

Solution:

The length of outer boundary of the park = (200 + 300 + 80 + 300 + 200 + 260) m = 1340 m

The cost of fencing the park at the rate of? 20 per metre = ₹ (20 x 1340) = ₹ 26800 Area of the flowerbed = (100 x 80) m2 = 8000 m2 Now, the cost of manuring the flower bed at the rate of? 50 per square metre

= ₹(50×8000)

= ₹ 400000

Question 40. The total cost of fencing the park shown in the figure is 55000. Find the cost of fencing per metre.

Mensuration The total cost of fencing the park

Solution:

Given

The total cost of fencing the park shown in the figure is 55000.

The perimeter of the park

= (150 + 100 + 120 + 180 + 270 + 280) m

= 1100 m

∴ Cost of fencing the park per metre \(=\frac{\text { Total cost of fencing }}{\text { Perimeter of the park }}\)

⇒ \(=₹\left(\frac{55000}{1100}\right)=₹ 50\)

Question 41. In the given figure each square is a length

Mensuration The square is of unit length

  1. What is the perimeter of the rectangle ABCD?
  2. What is the area of the rectangle ABCD?
  3. Divide this rectangle into ten parts of equal area by shading squares.
    (Two parts of equal area are shown here)
  4. Find the perimeter of each part which you have divided. Are they all equal?

Solution:

We have given each square is of unit length, where length = 10 units and breadth = 6 units

1.  Perimeter of ABCD = 2 (length +breadth)

= 2(10 + 6) units

= 2 x 16 units = 32 units

2.  Area of ABCD = length x breadth

= (10 x 6) sq units = 60 sq units

3.  Since, there are several ways to divide this rectangle into 10 parts, from which one of the ways is shown below :

Mensuration The square is of unit length

So, the above figure shows that there are 10 parts of equal area i.e., E, F, G, H, I, /, K, L, M and N.

The perimeter of each part is 12 units. Yes, the perimeter of each part is equal.

Question 42. The rectangular wall MNOP of the kitchen is covered with square tiles 15 cm in length (see figure). Find the area of the wall.

Mensuration The Rectangular wall MNOP Of A kitchen

Solution:

Given

The rectangular wall MNOP of the kitchen is covered with square tiles 15 cm in length (see figure).

Area of one square tile with side 15 cm = (15 x 15) cm2 = 225 cm2

Since the walls are covered with 28 square tiles.

∴Area of the wall = (225 x 28) cm2 = 6300 cm2

Question 43. The length of a rectangular field is 6 times its breadth. If the length of the field is 120 cm, find the breadth and perimeter of the field.
Solution:

Given

The length of a rectangular field is 6 times its breadth. If the length of the field is 120 cm

Length of rectangular field = 120 cm

According to the question,

⇒ 6(breadth) = 120 cm

⇒ breadth = \(\frac{120}{6} \mathrm{~cm}=20 \mathrm{~cm}\)

∴ Perimeter of the field = 2(length +breadth)

= 2(120 + 20) cm

= 2(140) cm = 280 cm

Perimeter of the field = 280 cm

Question 44. Anmol has a chart paper of measures 90 cm x 40 cm, whereas Abhishek has one which measures 50 cm x 70 cm. Which will cover more area on the table and by how much?
Solution:

Given

Anmol has a chart paper of measures 90 cm x 40 cm, whereas Abhishek has one which measures 50 cm x 70 cm.

The area of Anmol’s chart paper = (90×40) cm2 = 3600 cm2

And the area of Abhishek’s chart paper = (50 x 70) cm2 = 3500 cm2

The chart paper of Anmol will cover more area on the table than that of Abhishek by (3600- 3500) cm2 = 100 cm2

Question 45. A rectangular path of 60 m in length and 3 m in width is covered by square tiles of side 25 cm. How many tiles will there be in one row along its width? How many such rows will be there? Find the number of tiles used to make this path.
Solution:

Given

A rectangular path of 60 m in length and 3 m in width is covered by square tiles of side 25 cm.

Length of rectangular path = 60 m = 6000 cm

Width of rectangular path = 3 m = 300 cm

The side of a square tile = 25 cm

⇒ The number of tiles will be in one row along the width of the path = \(=\frac{\text { width of the path }}{\text { side of a tile }}=\frac{300}{25}=12\)

⇒ Number of rows = \(\frac{\text { length of the path }}{\text { side of a tile }}\)

⇒ \(=\frac{6000}{25}=240\)

∴ Total number of tiles required = 12 x 240

= 2880

Total number of tiles required = 2880

Question 46. How many square slabs each with side 90 cm are needed to cover a floor of area 81 sq m?
Solution:

Area of the floor = 81 m2 = 81 x 10000 cm2

= 810000 cm2

Area of one square slab = (90 x 90) cm2

= 8100 cm2

⇒ The required number of square slabs \(=\frac{\text { Area of the floor }}{\text { Area of a square slab }}=\frac{810000}{8100}=100\)

Question 47. The length of a rectangular field is 8 m and the breadth is 2 m. If a square field has the same perimeter as this rectangular field, find which field has the greater area.
Solution:

Given

The length of a rectangular field is 8 m and the breadth is 2 m. If a square field has the same perimeter as this rectangular field,

We have given, the perimeter of the rectangular field = perimeter of the square field

⇒ 2(length + breadth) – 4 * side

⇒ 2(8 + 2) m = 4 x side

⇒ 2(10) m = 4 x side

⇒ 20 m = 4 x side

⇒ \(\frac{20}{4} \mathrm{~m}=\text { side } \Rightarrow \text { side }=5 \mathrm{~m}\)

Now, area of the rectangular field = length x breadth = 8 m x 2 m = 16 m2

Area of the square field= side x side = 5mx5m = 25m2

∴ The square field has a greater area than that of the rectangular field.

Question 48. Parmindar walks around a square park once and covers 800 m. What will be the area of this park?
Solution:

Given

Parmindar walks around a square park once and covers 800 m.

The perimeter of the square park = 800m

⇒ 4 x side = 800 m

⇒ \(\text { side }=\frac{800}{4} m=200 \mathrm{~m}\)

∴ Area of the square park = side x side

= (200 x 200) m2

= 40000 m2

Question 49. The side of a square is 5 cm. How many times does the area increase, if the side of the square is doubled?
Solution:

Given

Side of the square = 5 cm

∴ Area of the square = side x side = 5 cm x 5 cm = 25 cm2…..(1)

Side of a new square

= 2 (side of the given square)

= 2 x 5 cm = 10 cm

∴ Area of the new square = side x side

= 10 cm x 10 cm

= 100 cm2 …(2)

Thus, (1) and (2) show that the area will be increased by 4 times.

Question 50. Amita wants to make rectangular cards measuring 8 cm x 5 cm. She has a square chart paper of side 60 cm. How many complete cards can she make from this chart? What area of the chart paper will be left?
Solution:

Given

Amita wants to make rectangular cards measuring 8 cm x 5 cm. She has a square chart paper of side 60 cm.

The square chart of length 60 cm can be divided into 7 parts each of length 8 cm along one side of chart paper and exactly in 12 parts each of length 5 cm along the other side of chart paper.

Mensuration Amita wants to make rectangular cards

The figure shows that the number of rectangular cards which can be made from the square chart paper is 7 x 12 = 84

Now, area of the square chart paper = (60 x 60) cm2

= 3600 cm2

Area of one rectangular card = (8 x 5) cm2 = 40 cm2

∴ Area of 84 rectangular cards = 84 x 40 cm2= 3360 cm2

∴ Area of the remaining chart paper = (3600- 3360) cm2= 240 cm2

Question 51. A magazine charges ₹ 300 per 10 sq cm area for advertising. A company decided to order a half-page advertisement. If each page of the magazine is 15 cm x 24 cm, what amount will the company have to pay for it?
Solution:

Given

A magazine charges ₹ 300 per 10 sq cm area for advertising. A company decided to order a half-page advertisement. If each page of the magazine is 15 cm x 24 cm

The area of one page of magazine = length x breadth

= (15 x 24) cm2 = 360 cm2

∴ Area of the half page of magazine \(=\frac{360}{2} \mathrm{~cm}^2=180 \mathrm{~cm}^2\)

Cost of advertising for 10 cm2 = ₹ 300

∴ Cost of advertising for 180 cm2 \(=₹ \frac{300}{10} \times 180=₹ 5400\)

Question 52. The perimeter of a square garden is 48 m. A small flower bed covers an 18 sq m area inside this garden. What is the area of the garden that is not covered by the flower bed? What fractional part of the garden is covered by a flower bed? Find the ratio of the area covered by the flower bed and the remaining area.
Solution:

Given

The perimeter of a square garden is 48 m. A small flower bed covers an 18 sq m area inside this garden.

The perimeter of the square garden= 48m

⇒ \(4 \times \text { side }=48 \mathrm{~m} \Rightarrow \text { side }=\frac{48}{4} \mathrm{~m}=12 \mathrm{~m}\)

Area of the square garden = side * side

= 12 m x 12 m

= 144 m2

The area of the small flower bed = is 18 m2

∴ The area of the garden that is not covered by the flower bed = 144 m2- 18 m2= 126 m2 The required fractional part of the garden which is covered by the flower bed

⇒ \(=\frac{\text { Area of the flower bed }}{\text { Area of the square garden }}\)

⇒ \(=\frac{18}{144}=\frac{1}{8}\)

The ratio of the area covered by the flower bed and the remaining area

⇒ \(\frac{18}{126}=\frac{1}{7} \text { i.e., } 1: 7\)

Question 53. The perimeter of a square and a rectangle is the same. If a side of the square is 15 cm and one side of the rectangle is 18 cm, find the area of the rectangle.
Solution:

Given

The perimeter of a square and a rectangle is the same. If a side of the square is 15 cm and one side of the rectangle is 18 cm

We have given,

Perimeter of the square= Perimeter of the rectangle

⇒ 4 x side = 2(length + breadth)

⇒ 4 x 15 cm = 2(18 cm + breadth)

⇒ \(\frac{4 \times 15}{2} \mathrm{~cm}=18 \mathrm{~cm}+\text { breadth }\)

30 cm = 18 cm + breadth

⇒ (30- 18) cm = breadth

=» breadth = 12 cm

Now, the area of the rectangle= length x breadth

= (18 x 12) cm2

= 216 cm2

Question 54. A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Solution:

Given

A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm

Length of one piece = Perimeter of the square

= 4 x 2 cm = 8 cm

Let the number of pieces be n.

We have given, the total area of n squares = 28 cm2

⇒ n (side)2 = 28

⇒ n (2)2 = 28

n x 4 = 28

⇒ \(n=\frac{28}{4}=7\)

The original length of the wire

=Number of pieces x length of one piece

=7×8 cm- 56 cm

Question 55. Divide the park shown in the figure of question 40 into two rectangles. Find the total area of this park. If one packet of fertilizer is used for 300 sq m, how many packets of fertilizer are required for the whole park?
Solution:

Area of rectangle ABCD=length x breadth

= (150 x 100) m2

= 15000 m2

Mensuration The perimeter Area of rectangle

And area of rectangle AGFE = length x breadth = (270×180) m2= 48600 m2

Total area of the park = 15000 m2+ 48600 m2 = 63600 m2

Number of packets of fertilizer used for 300 m2=1

Number of packets of fertilizer used for \(63600 \mathrm{~m}^2=\frac{63600}{300}=212\)

Question 56. The area of a rectangular field is 1600 sqm. If the length of the field is 80 m, find the perimeter of the field.
Solution:

Given

The area of a rectangular field is 1600 sqm. If the length of the field is 80 m,

Area of the rectangular field = 1600 m2

⇒ length x breadth= 1600 m2

⇒ 80 m x breadth = 1600 m2

⇒ breadth = \(\frac{1600 \mathrm{~m}^2}{80 \mathrm{~m}}=20 \mathrm{~m}\)

Now, the perimeter of the rectangular field

= 2(length + breadth)

= 2(80 + 20) m

= 2 x 100 m = 200 m

Question 57. The area of each square on a chessboard is cm. Find the area of the board.

  1. At the beginning of the game when all the chessmen are put on the board, write the area of the squares left unoccupied.
  2. Find the area of the squares occupied by chessmen.

Solution:

We have given, The area of one square on a chess board = 4 sq cm

Since there are 64 squares on a chessboard.

Area of the chess board = 64 x 4 sq cm = 256 sq cm

1.  At the beginning of the game when all the chessmen are put on the board, 32 squares are left unoccupied.

∴ The area of the squares left unoccupied

= 32 x area of one square

= 32 x 4 sq cm = 128 sq cm

2.  Since 32 squares are occupied by chessmen.

∴ Area of the squares occupied by chessmen = 32 x (area of one square)

= 32 x 4 sq cm = 128 sq cm

Question 58.

Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter of 36 cm and find their areas.

Find all the possible dimensions (in natural numbers) of a rectangle with an area of 36 sq cm, and find their perimeters.
Solution:

The perimeter of the rectangle = 36 cm

⇒ 2(length + breadth) = 36 cm

⇒ length + breadth = \(\frac{36}{2} \mathrm{~cm}=18 \mathrm{~cm}\)

The possible dimensions of the rectangles and their areas are as follows:

Mensuration The possible dimensions

The area of the rectangle = 36 sq cm

⇒ length x breadth = 36 sq cm

The possible dimensions of the rectangle and their perimeters are as follows:

Mensuration The possible dimensions rectangle and their perimeters

Question 59. Find the area and perimeter of each of the following figures, if the area of each small square is 1 sq cm

Mensuration The area and perimeter

Solution:

In the given figure (i), there are 11 small squares with an area of sq cm each.

Area of the given = (11 x 1) sq cm = 11 sq cm

Perimeter of the given  = (18 x 1) cm = 18 cm

In the given, there are 13 small squares with an area of 1 sq cm each.

Area of the given= (13 x 1) sq cm = 13 sq cm

Perimeter of the given  = (28 x 1) cm = 28 cm

In the given, there are 13 small squares with an area of 1 sq cm each.

Area of the given figure = (13 x 1) sq cm = 13 sq cm

Perimeter of the given figure = (28 x 1) cm = 28 cm

Question 60. What is the area of each small square in the given figure if the area of the entire figure is 96 sq cm? Find the perimeter of the figure.

Mensuration The area of each small square

Solution:

We have 24 small squares in the figure.

Area of the entire figure = 96 sq cm

The area of one small square

⇒ \(\frac{96}{24} \mathrm{sq} \mathrm{cm}\)

= 4 sqcm

Thus, the area of each small square = 4 sq cm

(side) x (side) = 4 sq cm

side = √4 cm = 2 cm

Now, the perimeter of the given figure

= 34 x 2 cm

= 68 cm

NCERT Exemplar Solutions For Class 6 Maths Chapter 9 Data Handling

Class 6 Maths Chapter 9 Data Handling

Question 1. Using tally marks, which one of the following represents the number eight:

Data Handling Using tally marks

Solution: (4): 8 can be represented as an image

Question 2. The marks (out of 0) obtained by 28 students in a Mathematics test are listed below:

8, 1, 2, 6, 5, 5, 5, 0, 1, 9, 7, 8, 0, 5, 8, 3, 0, 8, 1 0, 1 0, 3, 4, 8, 7, 8, 9, 2,0

The number of students who obtained marks more than or equal to 5 is

  1. 13
  2. 15
  3. 16
  4. 17

Solution: (4): The number of students who obtained marks more than or equal to 5 is 17

Question 3. In question 2 above, the number of students who scored marks less than 4 is

  1. 15
  2. 13
  3. 12
  4. 10

Solution: (4): The number of students who scored marks less than 4 is 10

Read and Learn More Class 6 Maths Exemplar Solutions

NCERT Exemplar Solutions For Class 6 Maths Chapter 9 Data Handling

Question 4. The choices of the fruits of 42 students in a class are as follows:

A,0,B,M,A,G,B,G,A,G, B,M,A,G,M,A,B,G,M,B, A,0,M,0,G,B,0,M,G,A, A,B,M,0,M,G,B,A,M,0,M,0,

where A, B, G, M, and O stand for the fruits Apple, Banana, Grapes, Mango, and Orange respectively.

Which two fruits are liked by an equal number of students?

  1. A and M
  2. M and B
  3. B and O
  4. B and G

Solution: (4): We have given two fruits Banana and Grapes, which are liked by an equal number of students i.e., 8.

Question 5. According to data from question 4, which fruit Is liked by most of the students?

  1. O
  2. G
  3. M
  4. A

Solution: (3): Mango is the only fruit liked by most students.

Question 6. In a bar graph, the width of bars may be unequal.
Solution: False

In a bar graph, bars have equal width.

Question 7. In a bar graph, bars of uniform width are drawn vertically only.
Solution: False

In a bar graph, bars of uniform width can be drawn vertically as well as horizontally.

Question 8. In a bar graph, the gap between two consecutive bars may not be the same.
Solution: False

The gap between two consecutive bars should be the same in a bar graph.

Question 9. In a bar graph, each bar (rectangle) represents only one value of the numerical data.
Solution: True

Question 10. To represent the population of different towns using a bar graph, it is convenient to take one unit length to represent one person.
Solution: False

Question 11. Pictographs and bar graphs are pictorial representations of the numerical data.
Solution: True

Question 12. An observation occurring five times in data is recorded as an image using tally marks.
Solution: False

Using tally marks, the occurrence of any observation five times can be represented as an image

Question 13. In a pictograph, if a symbol Data Handling Bookrepresents 50 books on a library shelf, then the symbol Data Handling Bookrepresents 25 books.

Solution: True

Question 14. A _______is a collection of numbers gathered to give some meaningful information.
Solution: Data

Question 15. The data can be arranged in a tabular form using_______marks.
Solution: Tally

Question 16. A_______ represents data through pictures of objects
Soluction: Pictograph

17. In a bar graph, _______ can be drawn horizontally or vertically.
Solution: Bars

Question 18. In a bar graph, bars of _______ width can be drawn horizontally or vertically with _______ spacing between them.
Solution: Uniform, Equal

Question 19. An observation occurring seven times in a data is represented as _______ using tally marks.
Solution: 

occurring seven times Data

Question 20. In a pictograph, if a symbol Data Handling Flowerrepresents 20 flowers in a basket thenData Handling Flowers stands for _______ flowers.
Solution: 60

Question 21. On the scale of 1 unit length = 10 crores, the bar of length 6 units will represent _______ crore, and of _______units will represent 75 crores.
Solution: 60, 7.5

Question 22. In an examination, the grades achieved by 30 students of a class are given below. Arrange these grades in a table using tally marks:

B, C, C, E, A, C, B, B, D, D, D, D, B, C, C, C, A, C, B, E, A, D, C, B, E, C, B, E, C, D

Solution:

Data Handling grades in a table using tally marks

Question 23. The number of two-wheelers owned individually by each of the 50 families is listed below. Make a table using tally marks.

1,1, 2, .1,1, 1,2, 1,2, 1,0, 1,1, 2, 3,1,2, 1,1,2, 1, 2, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 2, 1, 1, 4, 1, 3, 1, 1, 2, 1,1, 1,1, 2, 3, 2,1,1

Find the number of families having two or more, two-wheelers

Solution:

Data Handling Families having two or more, two-wheelers

So, 14 + 4 + 1 = 19 families have two or more, two-wheelers

Question 24. The lengths in centimeters (to the nearest centimeter) of 30 carrots are given as follows:

15, 22, 21, 20, 22,15, 15, 20, 20,15, 20, 18, 20, 22, 21, 20, 21, 18, 21, 18, 20, 18, 21, 18, 22, 20, 15,21,18, 20

Arrange the data given above in a table using tally marks and answer the following questions.

  1. What is the number of carrots which have a length of more than 20 cm?
  2. Which length of the carrots occur the maximum number of times? The minimum number of times?

Solution:

Data Handling Length of the carrots

  1. The number of carrots that have a length of more than 20 cm is 6 + 4 = 10.
  2. The carrots of length 20 cm occur a maximum number of times and the carrots of length 22 cm occur a minimum number of times

Question 25. Thirty students were interviewed to find out what they wanted to be in the future. Their responses are listed below:

doctor, engineer, doctor, pilot, officer, doctor, engineer, doctor, pilot, officer, pilot, engineer, officer, pilot, doctor, engineer, pilot, officer, doctor, officer, doctor, pilot, engineer, doctor, pilot, officer, doctor, pilot, doctor, engineer.

Arrange the data in a table using tally marks.

Solution:

data Handling Thirty students were interviewed

Question 26. Following are the choices of games for 40 students of Class 6:

football, cricket, football, kho-kho, hockey, cricket, hockey, kho-kho, tennis, tennis, cricket, football, football, hockey, kho-kho, football, cricket, tennis, football, hockey, khokho, football, cricket, cricket, football, hockey, kho-kho, tennis, football, hockey, cricket, football, hockey, cricket, football, kho-kho, football, cricket, hockey, football.

Arrange the choices of games in a table using tally marks.

  1. Which game is liked by most of the students?
  2. Which game is liked by the minimum number of students?

Solution: 1.

Data Handling Game is liked by minimum number of student

  1. Football is liked by most of the students.
  2. Tennis is liked by a minimum number of students.

Question 27. Fill in the blanks in the following table which represents the shirt size of 40 students of a school.

Data Handling Represents shirt size of 40 students of a school

Solution:

Data Handling Represents-shirt-size-of-40-students-of-a-school

Question 28. The following pictograph represents some surnames of people listed in the telephone directory of a city.

Data Handling surnames of people listed in the telephone

Observe the pictograph and answer the following questions:

  1. How many people have the surname ‘Roy’?
  2. Which surname appears the maximum number of times in the telephone directory?
  3. Which surname appears the least number of times in the directory?
  4. Which two surnames appear an equal number of times?

Solution:

Data Handling surnames of people listed in the telephones

  1. 400 people have the surname ‘Roy’.
  2. The surname ‘Patel’ appears the maximum number of times in the telephone directory.
  3. The surname ‘Saikia’ appears the least number of times in the telephone directory.
  4. The surnames ‘Rao’ and ‘Roy’ appear an equal number of times.

Question 29. Students of Class 6 in a school were given a task to count the number of articles made of different materials in the school. The information collected by them is represented as follows:

Students of Class VI in a school were given a task to count the number of articles made of different materials in school. The information collected by them is represented as follows.Observe

Observe the pictograph and answer the following questions:

  1. Which material is used in a maximum number of articles?
  2. Which material is used in a minimum number of articles?
  3. Which material is used in exactly half the number of articles as those made up of metal?
  4. What is the total number of articles counted by the students?

Solution:

Data Handling The Total Number Of Articles

  1. Metal is used in a maximum number of articles.
  2. Glass is used in the minimum number of articles.
  3. Rubber is used in exactly half the number of articles as those made up of metal.
  4. The total number of articles counted by the students is 160.

Question 30. The number of scouts in a school is depicted by the following pictograph:

Data Handling The number of scouts in a school is depicted

Observe the pictograph and answer the following questions:

  1. Which class has the minimum number of scouts?
  2. Which class has the maximum number of scouts?
  3. How many scouts are there in Class 6?
  4. Which class has exactly four times the scouts as that of Class X?
  5. What is the total number of scouts in the Classes 6 to 10?

Solution:

Data Handling Class X has minimum number of scouts

  1. Class 10 has a minimum number of scouts.
  2. Class 8 has a maximum number of scouts.
  3. There are 40 scouts in Class 6.
  4. Class 6 has exactly four times the scouts as that of Class 10.
  5. Total number of scouts in the Classes 6 to 10 is 160.

Question 31. A survey was carried out in a certain school to find out the popular school subjects among students of Classes VI to VIII. The data in this regard is displayed as a pictograph given below:

A survey was carried out in a certain school to find out the popular school subjects among students of Classes VI to VIII. - Sarthaks eConnect | Largest Online Education Community

  1. Which subject is most popular among the students?
  2. How many students like Mathematics?
  3. Find the number of students who like subjects other than Mathematics and Science.

Solution:

Data Handling Most popular subject among the Student

  1. Hindi is the most popular subject among the students.
  2. 175 students like Mathematics.
  3. A total of 200 + 150 + 75 = 425 students are there who like subjects other than Mathematics and Science.

Question 32. The following pictograph depicts the information about the areas in sq km (to the nearest hundred) of some districts of Chhattisgarh State:

The following pictograph depicts the information about the areas in sqkm (to nearest hundred) of some districts of Chhattisgarh State:(a) What is the area of Koria district?

  1. What is the area of the Koria district?
  2. Which two districts have the same area?
  3. How many districts have an area of more than 5000 square kilometers?

Solution:

Data Handling Some districts of Chhattisgarh State

  1. The area of Koria district is 6000 sq km.
  2. Raigarh and Jashpur districts have the same area of 6500 sq km.
  3. 4 districts have an area of more than 5000 square kilometers and they are Raigarh, Rajnandgaon, Koria, and Jashpur

Question 33. The number of bottles of cold drinks sold by a shopkeeper on six consecutive days is as follows:

Data Handling The number of bottles of cold drinks sold by a shopkeepers

Prepare a pictograph of the data using one symbol to represent 50 bottles.

Solution:

Data Handling The number of bottles of cold drinks sold by a shopkeepers

Question 34. The following table gives information about the circulation of newspapers (dailies) in a town in five languages.

Data Handling The circulation of newspaper

Prepare a pictograph of the above data, using a symbol of your choice, each representing 1000 newspapers.

Solution:

Data Handling The circulation of newspaper

Question 35. Annual expenditure of a company in the year 2007-2008 is given below:

Data Handling Company in the year

Prepare a pictograph of the above data using an appropriate symbol to represent ₹ 10 lakh.

Solution:

Data handling Data using an appropriate symbol to represent

Question 36. The houses following (out ofbar100) graphing a showdown using the number of different types of fuels for cooking.

Read the bar graph and answer the following questions:

The following bar graph shows the number of houses (out of 100) in a town using different types of fuels for cooking. - Sarthaks eConnect | Largest Online Education Community

  1. Which fuel is used in the maximum number of houses?
  2. How many houses are using coal as fuel?
  3. Suppose that the total number of houses in the town is ₹ 1 lakh. From the above graph estimate the number of houses using electricity.

Solution:

  1. LPG is used in a maximum number of houses.
  2. 10 houses are using coal as fuel.
  3. Since there are 5 houses using electricity out of 100. Therefore, out of ₹ 1 lakh, the electricity will be used by \(\frac{5}{100} \times 100000\) = 5000 houses.

Question 37. The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and answer the following questions.

The graph above shows the size of shoes worn by different students belonging to a school. What is the difference in the number of students wearing shoe number 7 and shoe number

  1. Find the number of students whose shoe sizes have been collected.
  2. What is the number of students wearing shoe size 6?
  3. What are the different sizes of the shoes worn by the students?
  4. Which shoe size is worn by the maximum number of students?
  5. Which shoe size is worn by the minimum number of students?
  6. State whether true or false: The total number of students wearing shoe sizes 5 and 8 is the same as the number of students wearing shoe size 6.

Solution:

  1. The total number of students whose shoe sizes have been collected = 250 + 200 + 300 + 400 + 150 = 1300
  2. The number of students wearing shoe size 6 is 300.
  3. The different sizes of the shoes worn by the students are 4, 5, 6, 7, and 8.
  4. The shoe size 7 is worn by the maximum number of students.
  5. The shoe size 8 is worn by the minimum number of students.
  6. False: Since the number of students wearing shoe sizes 5 and 8 is 350 and the number of students wearing shoe size 6 is 300.

Question 38. The following graph gives information about the number of railway tickets sold for different cities on a railway ticket counter between 6.00 am to 10.00 am. Read the bar graph and answer the following questions.

The following graph gives information about the number of railway tickets sold different cities on a railway ticket counter between 6.00 am to 10.00 am. Read the bar graph and answer the

  1. How many tickets were sold in all?
  2. For which city were the maximum number of tickets sold?
  3. For which city were the minimum number of tickets sold?
  4. Name the cities for which the number of tickets sold is more than 20.
  5. Fill in the blanks: The number of tickets sold for Delhi and Jaipur together exceeds the total number of tickets sold for Patna and Chennai by________.

Solution:

  1. Total number of tickets sold = 80 + 50 + 100 + 20 + 45 = 295
  2. The maximum number of tickets were sold for Delhi.
  3. The minimum number of tickets were sold for Chennai.
  4. The cities for which the number of tickets sold is more than 20 are as follows: Patna, Jaipur, Delhi, and Guwahati.
  5. 50: Number of tickets sold for Delhi and Jaipur together = 100 + 50 = 150.
  • Number of tickets sold for Patna and Chennai together = 80 + 20 = 100
  • Therefore, tickets sold for Delhi and Jaipur exceed the tickets sold for Patna and Chennai by 150- 100 = 50.

Question 39. The bar graph given below represents the approximate length (in kilometers) of some National Highways in India. Study the bar graph and answer the following questions:

The bar graph given below represents approximate length (in kilometres) of some National Highways in India. - Sarthaks eConnect | Largest Online Education Community

  1. Which National Highway (N.H.) is the longest among the above?
  2. Which National Highway is the shortest among the above?
  3. What is the length of National Highway 9?
  4. Length of which National Highway is about three times the National Highway l0?

Solution:

  1. The longest National Highway is N.H. 2.
  2. The shortest National Highway is N.H. 10.
  3. The length of National Highway 9 is 900 km.
  4. Length ofNational Highway 10 = 450km Length ofNational Highway 8 = 1400km

So, the length of National Highway 8 is about three times the of National Highway 10

Question 40. The bar graph given below represents the circulation of newspapers in different languages in a town. Study the bar graph and answer the following questions:

The bar graph shown above represents the circulation of newspapers in different languages in a town. Study the bar graph and answer the following question: What is the circulation of the English

What is the circulation of English newspapers?

Name the two languages in which the circulation of newspapers is the same.

By how much is the circulation of newspapers in Hindi more than the newspaper in Bengali?

Solution:

  1. The circulation of English newspapers is 1000.
  2. In Bengali and Marathi languages the circulation of newspapers is the same.
  3. The circulation of newspapers in Hindi is more than the newspaper in Bengali by 1400-600 = 800.

Question 41. Read the bar graph given below and answer the following questions:

Data Handling Academic years

  1. What information is given by the bar graph?
  2. In which year is the number of students maximum?
  3. In which year is the number of students twice that of 2001 -02?
  4. In which year did the number of students decrease as compared to the previous year?
  5. In which year is the increase in number of students maximum as compared to the previous year?

Solution:

  1. The given bar graph shows the number of students in different academic years.
  2. In the year 2005-06, the number of students is maximum.
  3. Since number of students in 2001-02 is150 and number of students in 2004-05 is 300. So, in the year 2004-05, the number of students is twice that of 2001-02.
  4. In the year 2003-04, the number of students decreased as compared to the previous year.
  5. The year is 2004-05in which the increase in number of students is maximum as compared to the previous year

Question 42. The lengths in km (rounded to the nearest hundred) of some major rivers of India are given below:

Data handling Nearst some major rivers of India

Draw a bar graph to represent the above information.

Solution:

Data Handling some major rivers of India

Question 43. The number of ATMs of different banks in a city is shown below:

Data Handling The number of ATMs of different banks

Draw a bar graph to represent the above information by choosing the scale of your choice.

Solution:

Data Handling The number of ATMs of different banks

Question 44. Number of mobile phone users in various age groups in a city is listed below:

Data Handling Number of mobile phone users in various age

Draw a bar graph to represent the above information.

Solution:

Data Handling Number of mobile phone users in various ages

Question 45. The following table gives the number of vehicles passing through a toll gate, every hour from 8.00 am. to 1.00 pm:

Data Handling Vehicles passing through a toll gate

Draw a bar graph representing the above data.

Solution:

Data Handling Vehicles passing through a toll gate

Question 46. The following table represents the income of a Gram Panchayat from different sources in a particular year:

Data Handling Gram Panchayat from different sources

Draw a bar graph to represent the above information.

Solution:

Data Handling Gram Panchayat from different sources

Question 47. The following table gives the data on several schools (stage-wise) in a country in the year 2002.

Data handling The data of number of schools (stage-wise) of a country in the year

Draw a bar graph to represent the above data:

Solution:

Data handling The data of number of schools (stage-wise) of a country in the year

Question 48. Home appliances sold by a shop in one month are given as below:

Data handling Home appliances sold by a shop

Draw a bar graph to represent the above information.

Solution:

Data handling Home appliances sold by a shop

Question 49. In a botanical garden, the number of different types of plants are found as follows:

Data Handling In a botanical garden, the number of different types of Plants

Draw a bar graph to represent the above information and answer the following questions:

  1. Which type of plant is the maximum in number in the garden?
  2. Which type of plant is the minimum in number in the garden?

Solution:

Data Handling In a botanical garden, the number of different types of Plants

  1. The tree is the maximum in number in the garden.
  2. Creeper is the minimum in number in the garden.

Question 50. Prepare a bar graph of the data given in question 28.

Solution:

Data Handling Number Of People Surnames

Question 51. Refer to question 39. Prepare a pictograph of the data by taking a suitable symbol to represent 200 kilometers.
Solution:

Data Handling Data by taking a suitable symbol

Question 52. Prepare a pictograph of the information given in question 38.
Solution:

Data Handling 10 Ticktes

Question 53. Refer to question 23. Prepare a bar graph of the data.
Solution:

Data Handling Number of families

Data handling Number Of Wheelers

Question 54. The following table shows the area of the land on which different crops were grown.

Data handling The area of the land

Prepare a pictograph by choosing a suitable symbol to represent 10 million hectares.

Solution:

Data Handling The area of the land suitable 50

Question 55. Refer to question 54. Prepare a bar graph of the data.
Solution:

Data Handling Area Of Land

NCERT Exemplar Solutions For Class 6 Maths Chapter 8 Decimals

Class 6 Maths Chapter 8 Decimals

Question 1. 0.7499 lies between

  1. 0.7 and 0.74
  2. 0.75 and 0.79
  3. 0.749 and 0.75
  4. 0.74992 and 0.75

Solution: (3) On observing the given options, we notice that 0.7499 lies between 0.749 and 0.75.

[As 0.7499 is the successor of 0.749 and the predecessor of 0.75]

Question 2. 0.023 lies between

  1. 0.2 and 0.3
  2. 0.02 and 0.03
  3. 0.03 and 0.029
  4. 0.026 and 0.024

Solution: (2): On observing the given options, we notice that 0.023 lies between 0.02 and 0.03.

Question 3. 0.07 + 0.008 is equal to

  1. 0.15
  2. 0.015
  3. 0.078
  4. 0.78

Solution: (3): We have to find the value of 0.07 + 0.008

Read and Learn More Class 6 Maths Exemplar Solutions

First converting the given decimals into like decimals, we get 0.070 and 0.008

By writing these decimals in column form and adding them, we get

⇒ \(\begin{array}{r}
0.070 \\
+0.008 \\
\hline 0.078 \\
\hline
\end{array}\)

Hence, 0.07 + 0.008 is 0.078

Question 4. Which of the following decimals is the greatest?

  1. 0.182
  2. 0.0925
  3. 0.29
  4. 0.038

Solution: (3): First converting the given decimals into like decimals, we get 0.1820, 0.0925, 0.2900 and 0.0380.

Thus, on comparing the above, 0.2900 is the greatest among the given decimals

Question 5. Which of the following decimals is the smallest?

  1. 0.27
  2. 1.5
  3. 0.082
  4. 0.103

Solution: (3): First converting the given decimals into like decimals, we get 0.270, 1.500, 0.082, and 0.103.

Thus, on comparing the above, 0.082 is the smallest among the given decimals.

NCERT Exemplar Solutions For Class 6 Maths Chapter 8 Decimals

Question 6. 1 3.572 correct to the tenth place is

  1. 10
  2. 13.57
  3. 14.5
  4. 13.6

Solution: (4): When 13.572 is estimated correctly to the tenth place, we get 13.6

Question 7. 15.8 -6.73 is equal to

  1. 8.07
  2. 9.07
  3. 9.13
  4. 9.25

Solution: (B): We have given, 15.8- 6.73. Firstly converting the given decimals into like fractions, then writing the decimals in column form and subtracting, we get

∴ \(\begin{array}{r}
15.80 \\
-6.73 \\
\hline 9.07 \\
\hline
\end{array}\)

Question 8. The decimal 0.238 is equal to the fraction

  1. \(\frac{119}{500}\)
  2. \(\frac{238}{25}\)
  3. \(\frac{119}{25}\)
  4. \(\frac{119}{50}\)

Solution:

(A): We have given 0.238, which can be written in the form of

⇒ \(\frac{238}{1000}=\frac{238 \div 2}{1000 \div 2}\left[\begin{array}{l}
\text { On dividing the numerator } \\
\text { and denominator by } 2
\end{array}\right]\)

∴ \(=\frac{119}{500}\)

Question 9. \(9+\frac{2}{10}+\frac{6}{100}\) is equal to the decimal number______.
Solution:

9.26: The L.C.M. of 10 and 100 is 100.

Decimal Number

⇒ \(9+\frac{2}{10}+\frac{6}{100}\)

⇒ \(=9 \times \frac{100}{100}+\frac{2}{10} \times \frac{10}{10}+\frac{6}{100}\)

⇒ \(\frac{900+20+6}{100}=\frac{926}{100}=9.26\)

Question 10. Decimal 1 6.25 is equal to the fraction______.
Solution:

⇒ \(\frac{65}{4}:\) We have

⇒ \(16.25=\frac{1625}{100}=\frac{1625 \div 25}{100 \div 25}\)

⇒ \(\frac{65}{4} \text { or } 16 \frac{1}{4}\)

Question 11. Fraction \(\frac{7}{25}\) is equal to the decimal number _______.
Solution:

0.28: We have, \(\frac{7}{25}=\frac{7}{25} \times \frac{4}{4}=\frac{28}{100}=0.28\)

Question 12. 4.55 + 9.73 = _______.
Solution: 14.28 : 4.55 + 9.73 = 14.28

Question 13. 8.76-2.68 = _______.
Solution: 6.08: 8.76- 2.68 = 6.08

Question 14. The value of 50 coins of 50 paisa = ₹ _______.
Solution:

25 : The value of 50 coins of 50 paisa = 50 x 50 paisa

= 2500 paisa \(=₹ \frac{2500}{100}\)

= ₹ 25

Question 15. 3 Hundredths + 3 tenths = _______.
Solution: 0.33 : 3 hundredths + 3 tenths = 0.33

Question 16. The place value of a digit at the tenth place is 10 times the same digit at the one’s place.
Solution: False

Since the place value of a digit at the tenth place is \(\frac{1}{10}\) times the same digit at the one’s place

∴ \(\frac{1}{10}=10x\)

Question 17. The place value of a digit at the hundredth place is \(\frac{1}{10}\) times the same digit at the tenth place.
Solution: True

Question 18. The decimal 3.725 is equal to 3.72 correct to two decimal places.
Solution: False

Since the digit at the thousandth place of 3.725 is 25

3.725 can be written as 3.73 (after rounding off at hundredths place)

Question 19. In the decimal form, fraction \(\frac{25}{8}=3.125\)
Solution: True

⇒ \(\frac{25}{8}=\frac{25 \times 125}{8 \times 125}\)

On multiplying the numerator and denominator by 125

∴ \(\frac{3125}{1000}=3.125\)

Question 20.The decimal 23.2 = \(23 \frac{2}{5}\)
Solution: False

Fraction Of decimal

We have \(23.2=\frac{232}{10}=\frac{116}{5}\)

⇒ \(23 \frac{1}{5}\)

Question 21. 3.03 + 0.016 = 3.019
Solution: False

Firstly convert 3.03 and 0.016 into like fractions, writing the decimals in column form and finally by adding we get,

⇒ \(\begin{array}{r}
3.030 \\
+0.016 \\
\hline 3.046
\end{array}\)

Question 22. 42.28-3.19 = 39.09
Solution: True

⇒ \(\begin{array}{r}
42.28 \\
-3.19 \\
\hline 39.09 \\
\hline
\end{array}\)

Question 23. 19.25 <19.053
Solution: False

Since, the digit at a tenth place of 19.25 is 2 and the digit at a tenth place of 19.053 is 0, where 2>0.

19.25 > 19.053

Question 24. 13.730 = 13.73
Solution: True

Since, after converting the given decimals into like decimals we get, 13.730 = 13.73.

Question 25. 3.25…3.4
Solution:

<: Converting the given decimals into like decimals, they become 3.25 and 3.40. The whole number part of these is the same. On comparing, we get their tenth digits 2 < 4

3.25 < 3.4

Question 26. 6.25…\( \frac{25}{4}\)
Solution:

=: \(6.25=\frac{625}{100}=\frac{625 \div 25}{100 \div 25}=\frac{25}{4}\)

∴ \(6.25\frac{25}{4}\)

Question 27. Arrange 12.142, 12.124, 12.104, 12.401 and 12.214 in ascending order.
Solution:

Given

12.142, 12.124, 12.104, 12.401 and 12.214

The digits are already given in the form of like decimals

Clearly,

12.104 < 12.124 < 12.142 < 12.214 < 12.401

Question 28. Write the largest four-digit decimal number less than 1 using the digits 1, 5, 3 and 8 once.
Solution:

The required number is 0.8531, which is the largest four-digit decimal number less than 1.

Question 29. Using the digits 2, 4, 5 and 3 once, write the smallest four-digit decimal number.
Solution:

The required number is 0.2345, which is the smallest four-digit decimal number

Question 30. Express \(\frac{11}{20}\) as a decimal.
Solution:

Given

latex]\frac{11}{20}[/latex]

We have, \(\frac{11}{20}=\frac{11 \times 5}{20 \times 5}=\frac{55}{100}=0.55\)

Question 31. Express \(3 \frac{2}{5}\) as a decimal.
Solution:

Given

\(3 \frac{2}{5}\)

We have \(3 \frac{2}{5}=\frac{(3 \times 5)+2}{5}=\frac{15+2}{5}=\frac{17}{5}\)

Now, \(\frac{17}{5}=\frac{17 \times 2}{5 \times 2}=\frac{34}{10}=3.4\)

Question 32. Express 0.041 as a fraction.
Solution:

Given 0.041

We have, \(0.041=\frac{41}{1000}\)

Question 33. Express 6.03 as a mixed fraction.
Solution:

Given 6.03

We have, \(6.03=\frac{603}{100}\)

⇒ \(6 \frac{\mathrm{3}}{100}\)

Decimal Mixed Fraction

Question 34. Convert 5201 g to kg.
Solution:

Given

5201

We have ,\(5201 \mathrm{~g}=\frac{5201}{1000} \mathrm{~kg}\)

[1kg = 1000 g]

= 5.201 kg

Question 35. Convert the 2009 paise to rupees and express the result as a mixed fraction.
Solution:

We have 2009 paise \(=₹ \frac{2009}{100}=₹ 20.09\)

Decimal Paise Mixed Fraction

∴ \(\text { or } ₹ 20 \frac{9}{100}\)

36. Convert 1 537 cm to m and express the result as an improper fraction.
Solution:

We have, 1537 cm \(=\frac{1537}{100} \mathrm{~m}\)

[1 m = 100 cm]

= 15.37 m

Question 37. Convert 2435 m to km and express the result as a mixed fraction
Solution:

Given 2435

we have, 2453 m \(=\frac{2435}{1000} \mathrm{~km}\)

[1 km = 1000 m]

= 2.435 km

Firstly, convert the fraction \(\frac{2435}{1000}\) into the simplest form, for dividing the numerator and denominator by 5, we get

Decimal Convert Mixed Fraction

⇒ \(\frac{2435 \div 5}{1000 \div 5} \mathrm{~km}=\frac{487}{200} \mathrm{~km}\)

∴ \(2 \frac{87}{200} \mathrm{~km}\)

Question 38. Round off 20.83 to the nearest tenths.
Solution: The estimated value of 20.83 to the nearest tenths is 20.8

Question 39. Round off 75.1 95 to the nearest hundredths.
Solution: The estimated value of 75.195 to the nearest hundredths is 75.20

Question 40. Round off 27.981 to the nearest tenths.
Solution: The estimated value of 27.981 to the nearest tenths is 28.0

Question 41. What should be added to 25.5 to get 50?
Solution: According to question, 25.5 +? = 50

⇒ \(\begin{array}{r}
50.0 \\
-25.5 \\
\hline 24.5 \\
\hline
\end{array}\)

? = 50 – 25.5

= 24.5

Thus, the required number is 24.5

Question  42. Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.
Solution:

Given

Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes.

Alok purchased,

Potatoes =1 kg 200 g = 1.200 kg

Dhania = 250 g = 0.250 kg

Onion = 5 kg 300 g = 5.300 kg

Palak = 500 g = 0.500 kg

Tomatoes = 2 kg 600 g = 2.600 kg

The total weight of the above purchases

⇒ \(\begin{array}{r}
1.200 \mathrm{~kg} \\
0.250 \mathrm{~kg} \\
5.300 \mathrm{~kg} \\
0.500 \mathrm{~kg} \\
+2.600 \mathrm{~kg} \\
\hline \underline{9.850 \mathrm{~kg}} \\
\hline
\end{array}\)

Question 43. Arrange in ascending order: 0.011, 1.001, 0.101, 0.110
Solution:

Given

0.011, 1.001, 0.101, 0.110

Since, all the decimals are already given in like fractions, i.e., 0.011, 1.001, 0.101, 0.110.

Arranging them in ascending order, we get 0.011, 0.101, 0.110, 1.001

Question 44. Add the following: 20.02 and 2.002
Solution:

We have, 20.02 and 2.002 To add the above decimals, we must convert them into like decimals first.

Writing 20.020 and 2.002 in a column

So,

⇒ \(\begin{array}{r}
20.020 \\
+2.002 \\
\hline 22.022 \\
\hline
\end{array}\)

which is the required sum.

Question 45. The energy content of different foods is as follows:

Decimal Food And Energy Content

Which food provides the least energy and which provides the maximum? Express the least energy as a fraction of the maximum energy.
Solution:

Milk provides the least energy and rice provides the maximum energy

The required fraction = \(\frac{3.0}{5.3}=\frac{3.0 \times 10}{5.3 \times 10}\)

∴ \(=\frac{30}{53}\)

Question 46. Which one is greater? 1 metre 40 centimetres + 60 centimetres or 2.6 metres.
Solution:

1 metre 40 centimetres + 60 centimetres = 1.40 metres + 0.60 metres

[ 100 centimetres =1 metre]

= 2.00 metres

Since, 2.6 >2.00

2.6 metres is greater.

Key Highlights of the Companies (Amendment) Act, 2020

The Companies Amendment Act 2020

BE it enacted by Parliament in the Seventy-first Year of the Republic of India as follows:

Short title and commencement:

This Act may be called the Companies (Amendment) Act, of 2020.

  1. It shall come into force on such date as the Central Government may, by notification in the Official Gazette, appoint:
  2. Provided that different dates may be appointed for different provisions of this Act and any reference in any such provision to the commencement of this Act shall be construed as a reference to the coming into force of that provision.

Amendment of section 2 18 of 2013: In the Companies Act, 2013 (hereinafter referred to as the principal Act), in section 2, in clause (52), the following proviso shall be inserted, namely:—

“Provided that such class of companies, which have listed or intend to list such class of securities, as may be prescribed in consultation with the Securities and Exchange Board, shall not be considered as listed companies.

Amendment of section: In section 8 of the principal Act, in sub-section (11) —

  • The words “with imprisonment for a term which may extend to three years or” shall be omitted;
  • For the words “twenty-five lakh rupees, or with both”, the words “twenty-five lakh rupees” shall be substituted.

Amendment of section 16: In section 16 of the principal Act;—

  • In subsection (1), in clause (2), for the words “six months”, the words “three months” shall be substituted;
  • For sub-section (3), the following sub-section shall be substituted, namely: “(3)if a company is in default in complying with any direction given under sub-section (1), the

Central Government shall allot a new name to the company in such manner as may be prescribed and the Registrar

  • shall enter the new name in the register of companies in place of the old name and issue a fresh certificate of incorporation with the new name, which the company shall use thereafter:
  • Provided that nothing in this sub-section shall prevent a company from subsequently changing its name by the provisions of section 13.”

Amendment of Section 26: In Section 26 Of The Principle Act, In Sub-Section (9)-

  • The words “with imprisonment for a term which may extend to three years or” shall be omitted;
  • For the words “three lakh rupees, or with both”, the words “three lakh rupees” shall be substituted.
  • Amendement Of Section 40. In section 40 of the*principal Act, in sub-section (5)-
  • The words “with imprisonment for a term which may extend to one year or” shall be omitted;
  • For the words “three lakh rupees, or with both”, the words “three lakh rupees” shall be substituted.

Amendment Of Section 8: In section 48 of the principal Act, sub-section (5) shall be omitted.

Amendment Of Section 56. In section 56 of the principal Act, for sub-section (6), the following sub-section shall be substituted, namely: “(6) Where any default is made in complying with the provisions of sub-sections (1) to (5), the company and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.

Amendment of section 59: In section 59 of the principal Act, sub-section (5) shall be omitted.

Amendment of section 62: In section 62 of the principal Act, in sub-section (1), in clause (a), in sub-clause, after the words “less than fifteen days”, the words “or such lesser number of days as may be prescribed” shall be inserted.

Amendment of section 64. In section 64 of the principal Act, in sub-section(2)-

  • For the words “one thousand rupees”, the words “five hundred rupees” shall be substituted;
  • For the words “or five lakh rupees whichever is less”, the words “subject to a maximum of five lakh rupees in case of a company and one lakh rupees in case of an officer who is in default” shall be substituted.

Amendment Of Section 66: In section 66 of the principal Act, sub-section (11) shall be omitted

Amendment of section 68: In section 68 of the principal Act, in sub-section (11)-

  • The words “with imprisonment for a term which may extend to three years or” shall be omitted;
  • For the words “three lakh rupees, or with both”, the words “three lakh rupees” shall be substituted.

Amendment of section 71: In section 71 of the principal Act, sub-section (11) shall be omitted.

Amendment of section 86: In section 86 of the principal Act, for sub-section (1), the following sub-section shall be substituted, namely:

“If any company is in default in complying with any of the provisions of this Chapter, the company shall be liable to a penalty of five lakh rupees and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.”.

Amendment of section 38: In section 80 of the principal Act, for sub-section (5), the following sub-section shall be substituted, namely—

(5)If a company does not maintain a register of members or debenture-holders or other security holders or fails to maintain them by the provisions of sub-section (1) or sub-section (2), the company shall be liable to a penalty of three lakh rupees and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.

Amendment of Section 89: In section 89 of the principal Act, (a) for sub-section (5), the following sub-section shall be substituted, namely—

  • “(5) If any person fails to make a declaration as required under sub-section (1) or sub-section (2) or sub-section (3), he shall be liable to a penalty of fifty thousand rupees and in case of continuing failure, with a further penalty of two hundred rupees for each day after the first during which such failure continues, subject to a maximum of five lakh rupees.”;
  • (2) for sub-section (7), the following sub-section shall be substituted, namely:— “(7) If a company, required to file a return under sub-section (6), fails to do so before the expiry of the time specified therein, the company and every officer of the company who is in default shall be liable to a penalty of one thousand rupees for each day during which such failure continues, subject to a maximum of five lakh rupees in the case of a company and two lakh rupees in case of an _officer who is in default.
  • After sub-section (10), the following sub-section shall be inserted, namely: ‘(11) The Central Government may, by notification, exempt any class or classes of persons from complying with any of the requirements of this section, except sub-section (10), if it is considered necessary to grant such exemption in the public interest and any such exemption may be granted either unconditionally or subject to such conditions as may be specified in the notification.

Amendment of section 90: In section 90 of the principal Act,—

  1. For sub-section (10), the following sub-section shall be substituted, namely:— ‘(10) If any person fails to make a declaration as required under subsection (1), he shall be liable to a penalty of fifty thousand rupees and in case of continuing failure, with a further penalty of one thousand rupees for each day after the first during which such failure continues, subject to a maximum of two lakh rupees.”
  2. For sub-section (11), the following sub-section shall be substituted, namely: “(11) If a company, required to maintain register under sub-section (2) and file the information under sub-section (4) or required to take necessary steps under sub-section (4A), fails to do so or denies inspection as provided therein, the company shall be liable to a penalty of one lakh rupees and in case of continuing failure, with a further penalty of five hundred rupees for each day,
  3. After the first during which such failure continues, subject to a maximum of five lakh rupees, every officer of the company who is in default shall be liable to a penalty of twenty-five thousand rupees and in case of continuing failure, with a further penalty of two hundred rupees for each day, after the first during which such failure continues, subject to a maximum of one lakh rupees.

Amendment of section 92: In section 92 of the principal Act,—

1. In sub-section (5),—

  • For the words “fifty thousand rupees”, the words “ten thousand rupees’1 shall be substituted;
  • For the words “five lakh rupees”, the words “two lakh rupees in case of a company and fifty thousand rupees in case of an officer who is in default11 shall be substituted;

2. In sub-section (6), for the words “punishable with fine which shall not be less than fifty thousand rupees but which may extend to five lakh rupees”, the words “liable to a penalty of two lakh rupees” shall be substituted.

Amendment of section 105: In section 105 of the principal Act, in sub-section (5),

  • 1. For the words “who knowingly issues the invitations as aforesaid or wilfully authorizes or permits their issue shall be punishable with a fine which may extend to one lakh rupees”, the words
  • “who invites as aforesaid or authorizes or permits their issue, shall be liable to a penalty of fifty thousand rupees11 shall be substituted;
  • In the proviso, for the word “punishable, the word liable shall be substituted.

Amendment of section 117: In section 117 of the Principal Act,—

  • For Sub-Section (2), The Following Sub-Section Shall Be Substituted, Namely:— “(2) If any company fails to file the resolution of the agreement under sub-section (1) before the expiry of the period specified therein, such company shall be liable to a penalty of ten thousand rupees and in case of continuing failure.
  • with a further penalty of one hundred rupees for each day after the first during which such failure continues, subject to a maximum of two lakh rupees, and every officer of the company who is in default including the liquidator of the company.
  • , if any, shall be liable to a penalty of ten thousand rupees and in case of continuing failure, with a further penalty of one hundred rupees for each day after the first during which such failure continues, subject to a maximum of fifty thousand rupees.11;

In subsection (3), in clause (g), for the second proviso, the following proviso shall be substituted, namely:—

  • “Provided further that nothing contained in this clause shall apply in respect of a resolution passed to grant loans, or give guarantee or provide security in respect of loans under clause (f)-of sub-section (3) of section 179 in the ordinary course of its business by,— (a) a banking company; -(b) company any class registeredof non-banking under.
  • 2 of 1934: Chapter 3B of the Reserve Bank of India Act, 1934, as may be prescribed in Consultation with the Reserve Bank of India;
  • 53 of 1987: Any class of housing finance company registered under the National Housing Bank Act, 1987, as may be prescribed in consultation with the National Housing Bank; and”.

Amendment of section 124: In section 124 of the principal Act, for sub-section (7), the following sub-section shall be substituted, namely:—

  • “(7) If a company fails to comply with any of the requirements of this section, such company shall be liable to a penalty of one lakh rupees and in case of continuing failure, with a further penalty of five hundred rupees for each day after the first during which such failure continues, subject to a maximum of ten lakh rupees and every officer of the company
  • who is in default shall be liable to a penalty of twenty-five thousand rupees and in case of continuing failure, with a further penalty of one hundred rupees for each day after the first during which such failure continues, subject to a maximum of two lakh rupees.”

Amendment of section 128: In section 128 of the principal Act, in sub-section (6)-

The words “with imprisonment for a term which may extend to one year or” shall be omitted;

The words “or with both” shall be omitted.

  • Insertion of new section 129A. Periodical financial results.
  • After section 129 of the principal Act, the following section shall be inserted, namely:—
  • “129A. The Central Government may require such class or classes of unlisted companies, as may be prescribed,—
  • To prepare the financial results of the company on such periodical basis in such form as may be prescribed;
  • To obtain approval of the Board of Directors and complete audit or limited review of such periodical financial results in such manner as may be prescribed; and

File a copy with the Registrar within thirty days of completion of the relevant period with such fees as may be prescribed.”

Amendment of section 134: In section 134 of the principal Act, for sub-section (8), the following sub-section shall be substituted, namely:

“(8) If a company is in default in complying with the provisions of this section, the company shall be liable to a penalty of three lakh rupees and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.”

Amendment of Section 135: In section 134 of the principal Act, for sub-section (8), the following sub-section shall be substituted, namely:

“(8) If a company is in default in complying with the provisions of this section, the company shall be liable to a penalty of three lakh rupees and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.”

Sub-section for such number of succeeding financial years and in such manner, as may be prescribed.’;

 For sub-section (7), the following sub-section shall be substituted, namely:—

  • ’(7) If a company is in default in complying with the provisions of sub-section (5) or sub-section (6), the company shall be liable to a penalty of twice the amount required to be transferred by the company to the Fund specified in Schedule VII or the Unspent Corporate Social Responsibility
  • Account, as the case may be, or one crore rupees, whichever is less, and every officer of the company who is in default shall be liable to a penalty of one-tenth of the amount required to be transferred by the company to such Fund specified in Schedule VII,
  • Unspent Corporate Social Responsibility Account, as the case may be, or two lakh rupees, whichever is less. (c) after sub-section (8),

The following sub-section shall be inserted, namely:—

“(9) Where the amount to be spent by a company under sub-section (5) does not exceed fifty lakh rupees, the requirement under sub-section (1) for the constitution of the Corporate Social Responsibility Committee shall not be applicable and the functions of such Committee provided under this section shall, in such cases, be discharged by the Board of Directors of such company.”

Amendment of section 135: In section 137 of the principal Act, in sub-section (3),— (a) for the words -one thousand rupees for every day during which the failure continues but which shall not be more than ten lakh rupees”, the words “ten thousand rupees and in case of continuing failure, with a further penalty of one hundred rupees for each day during which such failure continues, subject to a maximum of two lakh rupees,” shall be substituted; (b) for the words “one lakh rupees”, the words “ten thousand rupees” shall be substituted; (c) for the words “five lakh rupees”, the words “fifty thousand rupees” shall be substituted.

Amendment of section 140: In section 140 of the principal Act, in sub-section (3), for the words “five lakh rupees”, the words “two lakh rupees” shall be substituted.

Amendment of section 143: In section 143 of the principal Act, for sub-section (15), the following sub-section shall be substituted, namely:—

  • “(15) If any auditor, cost accountant, or company secretary in practice does not comply with the provisions of sub-section (12), he shall,—
  • (1) In case of a listed company, be liable to a penalty of five lakh rupees; and
  • (2) In case of any other company, be liable to a penalty of one lakh rupees.”.

Amendment of section 147: In section 147 of the principal Act, (a) in sub-section (1),—

  • The words “with imprisonment for a term which may extend to one year or” shall be omitted;
  • For the words “one lakh rupees, or with both”, the words ’one lakh rupees* shall be substituted;
  • In subsection (2), the word and figures, “section 143” shall be omitted.

Amendment of section 149: In section 149 of the principal Act, in sub-section (9), the following proviso shall be inserted, namely:—

“Provided that if a company has no profits or its profits are inadequate, an independent director may receive remuneration, exclusive of any fees payable under sub-section (5) of section 1 97, by the provisions of Schedule V.

Amendment of section 165: In section 165 of the principal Act, for sub-section (6), the following sub-section shall be substituted, namely:

“(6) If a person accepts an appointment as a director in violation of this section, he shall be liable to a penalty of two thousand rupees for each day after the first during which such violation continues, subject to a maximum of two lakh rupees.

Amendment of Section 167: In section 167 of the principal Act, in sub-section (2)-

  • The words ‘with imprisonment for a term which may extend to one year or” shall be omitted;
  • For the words “five lakh rupees, or with both”, the words “five lakh rupees” shall be substituted.

Amendment of section 172: For section 172 of the principal Act, the following section shall be substituted, namely:—

“172. If a company is in default in complying with any of the provisions of this Chapter and for which no specific penalty or punishment is provided therein, the company and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees, and in case of continuing failure, with a further penalty of five hundred rupees for each day during which such failure continues, subject to a maximum of three lakh rupees in case of a company and one lakh rupees in case of an officer who is in default.

Amendment of section 178: In section 178 of the principal Act, in sub-section (8), for the words ’punishable with fine which shall not be less than one lakh rupees but which may extend to five lakh rupees and every officer of the company who is in default shall be punishable with imprisonment for a term which may extend to one year or with fine which shall not be less than twenty-five thousand rupees but which may extend to one lakh rupees, or with both”, the words “liable to a penalty of five lakh rupees and every officer of the company who is in default shall be liable to a penalty of one lakh rupees shall be substituted.

Amendment of Section 184: In section 184 of the principal Act, in sub-section (4), for the words “punishable with imprisonment for a term which may extend to one year or with fine which may extend to one lakh rupees, or with both”, the words “liable to a penalty of one lakh rupees” shall be substituted.

Amendment of section 187: In section 1 87 of the principal Act, for sub-section (4), the following sub-section shall be substituted, namely:—

“(4) If a company is in default in complying with the provisions of this section, the company shall be liable to a penalty of five lakh rupees and every officer of the company who is in default shall be liable to a penalty of fifty thousand rupees.”.

Amendment of section 188: In section 188 of the principal Act, in sub-section (5) (a) in clause (i), for the words “punishable with imprisonment for a term which may extend to one year or with fine which shall not be less than twenty-five thousand rupees but which may extend to five lakh rupees, or with both”, the words “liable to a penalty of twenty-five lakh rupees” shall be substituted; (b) in clause

For the words “punishable with fine which shall not be less than twenty-five thousand rupees but which may extend to five lakh rupees”, the words “liable to a penalty of five lakh rupees” shall be substituted.

Amendment of section 197: In section 1 97 of the principal Act, in sub-section (3), after the words “whole-time director or manager,”, the words “or any other non-executive director, including an independent director” shall be inserted.

Amendment of section 204: 41. In section 204 of the principal Act, in sub-section (4), for the words “punishable with fine which shall not be less than one lakh rupees but which may extend to five lakh rupees”, the words “liable to a penalty of two lakh rupees” shall be substituted.

Amendment of section 232: In section 232 of the principal Act, for sub-section (8), the following sub-section shall be substituted, namely: “(8) If a company fails to comply with sub-section (5), the company and every officer of the company who is in default shall be liable to a penalty of twenty thousand rupees, and where the failure is a continuing one, with a further penalty of one thousand rupees for each day after the first during which such failure continues, subject to a maximum of three lakh rupees.”

Amendment of section 242: In section 242 of the principal Act, in sub-section (8) (a) the words “with imprisonment for a term which may extend to six months or” shall be omitted; (b) for the words “one lakh rupees, or with both”, the words “one lakh rupees” shall be substituted.

Amendment of section 243: In section 243 of the principal Act, in sub-section(2)-

1. The words “with imprisonment for a term which may extend to six months or” shall be omitted;

2. For the words “five lakh rupees, or with both”, the words “five lakh rupees” shall be substituted.

Amendment of section 247: In section 247 of the Principal Act, in sub-section (3), for the words “punishable with fine which shall not be less than twenty-five thousand rupees but which may extend to one lakh rupees”, the words “liable to a penalty of fifty thousand rupees” shall be substituted.

Amendment of section 284: In section 284 of the principal Act, for sub-section (2), the following sub-sections shall be substituted, namely:

“(2) If any person required to assist or cooperate with the Company Liquidatorundersub-section (1) does not assist or cooperate, the Company Liquidator may make an application to the Tribunal for necessary directions.

(3) On receiving an application under sub-section

(2), the Tribunal shall, by an order, direct the person required to assist or cooperate with the Company Liquidator to comply with the instructions of the Company Liquidator and to cooperate with him in discharging his functions and duties.”.

Amendment section 302: In section 302 of the principal Act,— (a) for sub-section (3), the following sub-section shall be substituted, namely:—

“(3) The Tribunal shall, within thirty days from the date of the order) forward a copy of the order to the Registrar who shall record in the register relating to the company a minute of the dissolution of the company; and

(2) direct the Company Liquidator to forward a copy of the order to the Registrar who shall record in the register relating to the company minute of the dissolution of the company.”;

(2) sub-section (4) shall be omitted.

Amendment of section 342: In section 342 of the principal Act, sub-section (6) shall be omitted.

Amendment of section 34: In section 347 of the principal Act, in sub-section (4),-

1. The words “with imprisonment for a term which may extend to six months or” shall be omitted;

For the words“fifty thousand rupees, or with both”, the words “fifty thousand rupees” shall be substituted.

Amendment of section 348: In section 348 of the principal Act,— (a) for sub-section (6), the following sub-section shall be substituted, namely “(6) Where a Company Liquidator, who is an insolvency professional registered under the Insolvency and Bankruptcy Code, 2016 is in default in complying with the provisions of this section, then such default shall be deemed to be a contravention of the provisions of the said Code, and the rules and regulations made thereunder for proceedings under Chapter VI of Part IV of that Code.”; (b) sub-section (7) shall be omitted.

Amendment of section 356: In section 356 of the Principal Act, for sub-section (2), the following sub-section shall be substituted, namely “(2) The Tribunal shall— (a) forward a copy of the order, within thirty days from the date thereof, to the Registrar who shall record the same; and (b) direct the Company Liquidator or the person on file who sea certified application copy ofthetheorderorder, was within made, thirty to days from the date thereof or such further period as allowed by the Tribunal, with the Registrar who shall record the same.”

Insertion of new chapter XXIA: After section 378 of the principal Act, the following
Chapter shall be inserted, namely

Chapter XXIA

Producer Companies

Part-1

Preliminary

Definitions: 378A. In this Chapter, unless the context otherwise requires

“Active Member” means a Member who fulfills the quantum and period of patronage of the Producer Company as may be required by the articles;

“Chief Executive” means an individual appointed as such under sub-section (1) of section 378W

  • 39 of 2002: “Inter-state co-operative society” means a multi-state co-operative society as defined in clause (p) of section 3 of the Multi-State Co-operative Societies Act, 2002, and includes any co-operative society registered under any other law for the time being in force, which has, after its formation, extended any of its objects to more than one State by enlisting
  • The participation of persons or by extending any of its activities outside the State, whether directly or indirectly or through an institution of which it is a constituent;
  • “limited return” means the maximum dividend as may be specified by the articles;
  • Member” means a person or Producer Institution (whether incorporated or not) admitted as a Member of a Producer Company and who retains the qualifications necessary for continuance as such;
  • “Mutual assistance principles” means the principles set out in subsection (2) of section 378G;
  • “Officer” includes any director or Chief Executive or Secretary or any person by whose directions or instructions part or whole of the business of the Producer Company is carried on;
  • “Patronage” means the use of services offered by the Producer Company to its Members by participation in its business activities;
  • “Patronage bonus” means payments made by a ProducerCompany out of its surplus income to the Members in proportion to their respective patronage;
  • “primary produce” means —
  • Produce of farmers, arising from agriculture including animal husbandry, horticulture, floriculture, pisciculture, viticulture, forestry, forest products, re-vegetation, bee raising, and farming plantation products), or from any other primary activity or service which promotes the interest of the farmers or consumers; or produce of persons engaged in handloom, handicraft, and other cottage industries; Or
  • Any product resulting from any of the above activities, including by-products of such products; or
  • Any product resulting from an ancillary activity that may assist or promote any of the aforesaid activities or anything ancillary thereto; or any activity that is intended to increase the production of anything referred to in sub-clauses

(1) to (4) or improve the quality thereof; means any person engaged in any activity connected with or relatable to any primary produce; “Producer Company” means a body corporate having objects or activities specified in section 378B and registered as Producer Company under this Act or under the Companies Act, 1956;

  • ‘Producer Institution” means a Producer Company or any other institution having only producer or producers or Producer Company or Producer Companies as its member whether incorporated or not having any of the objects referred to in section 378B and which agrees to make use of the services of the Producer Company or Producer Companies as provided in its articles;
  • withheld price means part of the price due and payable for goods supplied by any Member to the Producer Company; and as withheld by the Producer Company for payment on a subsequent date.

Objects of Producer Company: Part 2 Incorporation of Producer Companies and Other Flatters 378B. (1) The objects of the Producer Company shall relate to all or any of the following matters, namely:— “

  • Production, harvesting, procurement, grading, pooling, handling, marketing, selling, export of primary produce of the Members or import of goods or services for their benefit:
  • Provided that the Producer Company may carry on any of the activities specified in this clause either by itself or through another institution;
  • Processing including preserving, drying, distilling, brewing, venting, canning, and packaging of produce of its Members;
  • Manufacture, sale, or supply of machinery, equipment, or consumables mainly to its Members;
  • Providing education on the mutual assistance principles to its Members and others;
  • Rendering technical senses, consultancy services, training, research and development, and all other activities for the promotion of the interests of its Members;
  • Generation, transmission, and distribution of power, revitalization of land and water resources, their use, conservation, and communications relatable to primary produce;
  • Insurance of producers or their primary produce;
  • Promoting techniques of mutuality and mutual assistance;
  • Welfare measures or facilities for the benefit of Members as may be decided by the Board;
  • Any other activity, ancillary or incidental to any of the activities referred to in clauses (a) to (i) or other activities which may promote the principles of mutuality and mutual assistance amongst the Members in any other manner;
  • Financing of procurement, processing, marketing or other activities specified in clauses (a) to (j) which include extending of credit facilities or any other financial services to its Members.”.
  • Every Producer Company shall deal primarily with the produce of its active Members for carrying out any of its objects specified in this section.

Formation of Producer Company and its registration:

378C.

  1. Any ten or more individuals, each of them being a producer or any two or more Producer Institutions, or a combination of ten or more individuals and Producer Institutions, desirous of forming a Producer Company having its objects specified in section 378B and otherwise complying with the requirements of this Chapter and the provisions of this Act in respect of registration, may form an incorporated company as a Producer Company under this Act.
  2.  If the Registrar is satisfied that all the requirements of this Act have been complied with in respect of registration and matters precedent and incidental thereto, he shall, within thirty days of the receipt of the documents required for registration, register the memorandum, the articles, and other documents, if any, and issue a certificate of incorporation under this Act.
  3.  A Producer Company so formed shall have the liability of its Members limited by the memorandum to the amount, if any, unpaid on the shares respectively held by them and be termed a company limited by shares.
  4. The Producer Company may reimburse to its promoters all other direct costs associated with the promotion and registration of the company including registration, legal fees, printing of a memorandum and articles and the payment thereof shall be subject to the approval at its first general meeting of the Members.
  5. On registration under sub-section (2), the Producer Company shall become a body corporate as if it is a private limited company to which the provisions contained in this Chapter apply, without, however, any limit to the number of Members thereof, and the Producer Company shall not, under any circumstance, whatsoever, become or be deemed to become a public limited company under this Act.

Membership and voting rights of Members of Producer Company: 378D.

  1. In a case where the membership consists solely of individual Members, the voting rights shall be based on a single vote for every Member, irrespective of his shareholding or patronage of the Producer Company.
  2. In a case where the membership consists of Producer Institutions only, the voting rights of such ProducerInstitutions shall be determined based on their participation in the business of the Producer Company in the previous year, as may be
    specified by articles: Provided that during the first year of registration of a Producer Company, the voting rights shall be determined based on the shareholding by such Producer Institutions.
  3. In a case where the membership consists of individuals and Producer Institutions, the voting rights shall be computed based on a single vote for every Member.
  • The articles of any Producer Company may provide for the conditions, subject to which a Member may continue to retain his membership, and how voting rights shall be exercised by the Members.
  • Notwithstanding anything contained in sub-section (1) or sub-section (2), any Producer Company may, if so authorized by its articles, restrict the voting rights to active Members, in any special or general meeting.
  • No person, who has any business interest which conflicts with the business of the Producer Company, shall become a Member of that Company.
  • A Member, who acquires any business interest that conflicts with the business of the Producer Company, shall cease to be a Member of that Company and be removed as a Member by the articles.

Benefits to Members

378E.

(1) Subject to the provisions made in articles, every Member shall initially receive only such value for the produce or products pooled and supplied as the Board of Producer Company may determine, and the withheld price may be disbursed later in cash or kind or by allotment of equity shares, in proportion to the produce supplied to the Producer Company during the financial year Jo such extent and in such manner and subject to such conditions ‘as may be decided by the Board.

(2) Every Member shall, on the share capital contributed, receive only a limited return: Provided that every such Member may be allotted bonus shares by the provisions contained in section 378ZJ.

(3) The surplus if any, remaining after making provision for payment of limited return and reserves referred to in section 378ZI, may be disbursed as a patronage bonus, amongst the Members, in proportion to their participation in the business of the Producer Company, either in cash or by way of allotment of equity shares, or both, as may be decided by the Members at the general meeting.

Memorandum of Producer Company: 378F. The memorandum of association of every Producer Company shall state—

  • The name of the company with “Producer Company Limited” as the last words of the name of such Company;
  • The State in which the registered office of the Producer Company is to be situated;
  • The main objects of the Producer Company shall be one or more of the objects specified in section 378B;
  • The names and addresses of the persons who have subscribed to the memorandum;
  • The amount of share capital with which the ProducerCompany is to be registered and division thereof into shares of a fixed amount;
  • The names, addresses, and occupations of the subscribers being producers, who shall act as the first directors by sub-section (2) of section 378J;
  • That the liability of its members is limited;
  • Against the subscriber’s name, the number of shares each subscriber takes: Provided that no subscriber shall take less than one share;
  • That in case the objects of the Producer Company are not confined to one State, the States to whose territories the objects extend.’

Article Of Association:

378G

  • There shall be presented, for registration to the Registrar of the State to which the registered office of the Producer Company is, stated by the memorandum of association, to be situated—
  • Memorandum of the Producer Company;
  • Its articles are duly signed by the subscribers to the memorandum.
  • The articles shall contain the following mutual assistance principles, namely:
  • The membership shall be voluntary and available, to all eligible persons who can participate or avail of the facilities or services of the Producer Company, and are willing to accept the duties of membership;
  • Each Member shall save as otherwise provided in this Chapter, have only a single vote irrespective of the shareholding;
  • The Producer Company shall be administered by a Board consisting of persons elected or appointed as directors in a manner consistent with the provisions of this Chapter and the Board shall be accountable to the Members;
  • Particulars on limited return on share capital;
  • The surplus arising out of the operations of the Producer Company shall be distributed equitably by—
  • Providing for the development of the business of the Producer Company;
  • Providing for common facilities; and
  • Distributing amongst the Members, as may be admissible in proportion to their respective participation in the business. provision for the education of Members, employees, and others, on the principles of mutuality and techniques of mutual assistance;
  • The Producer Company shall actively cooperate with other Producer Companies (and other organizations following similar principles) at local, national, or international levels to best serve the interest of their Members and the communities it purports to serve.
  • Without prejudice to the generality of the foregoing provisions of sub-sections (1) and (2), the articles shall contain the following provisions, namely:
  • The qualifications for membership, the conditions for continuance or cancellation of membership, and the terms, conditions, and procedure for transfer of shares;
  • The manner of ascertaining the patronage and voting rights based on patronage;
  • Subject to the provisions contained in sub-section (1 ) of section 378N, the manner of constitution of the Board, its powers and duties, the minimum and maximum number of directors, manner of election and appointment of directors and retirement by rotation, qualifications for being elected or continuance as such, and the terms of office of the said directors,
  • Their powers and duties, conditions for election or co-option of directors, method of removal of directors and the filling up of vacancies on the Board, and the manner and the terms of appointment of the Chief Executive;

The election of the Chairman, term of office of directors and the Chairman, manner of voting at the general or special meetings of Members, procedure for voting, by directors at meetings of the Board, powers of the Chairman, and the circumstances under which the Chairman may exercise a casting vote;

  • The circumstances under which, and how, the withheld price is to be determined and distributed;
  • The manner of disbursement of patronage bonuses in cash by issue of equity shares, or both;
  • The contribution to be shared and related matters referred to in sub-section (2) of section 378ZI;
  • The matters relating to the issue of bonus shares out of general reserves as set out in section 378ZJ;
  • The basis and manner of allotment of equity shares of the Producer Company instead of the whole or part of the sale proceeds of produce or products supplied by the Members;
  • The amount of reserves, sources from which funds may be raised, limitation on raising of funds, restriction on the use of such funds, and the extent of debt that may be contracted and the conditions thereof;
  • The credit, loans, or advances which may be granted to a Member and the conditions for the grant of the same;
  • The right of any Member to obtain information relating to the general business of the company.
  • The basis and manner of distribution and disposal of funds available after meeting liabilities in the event of dissolution or liquidation of the Producer Company;
  • The authorization for division, amalgamation, merger, creation of subsidiaries and the entering into joint ventures and other matters connected therewith;
  • Laying of the memorandum and articles of the Producer Company before a special general meeting to be held within ninety days of its registration;
  • Any other provision, which the Members may, by special resolution recommend to be included in the articles.

Amendment Of Memorandum:

378H.

A Producer Company shall not alter the conditions contained in its memorandum except in the cases, by the mode and to the extent for which express provision is made in this Act. .

  • A Producer Company may, by special resolution, not inconsistent with section 378B, alter its objects specified in its memorandum.
  • A copy of the amended memorandum, together with a copy of the special resolution duly certified by two directors, shall be filed with the Registrar within thirty days from the date of adoption of any resolution referred to in subsection (2):
  • Provided that in the case of transfer of the registered office of a Producer Company from the jurisdiction of one Registrar to another, certified copies of the special resolution certified by two directors.
  • Shall be filed with both the Registrars within thirty days, and each Registrar shall record the same, and thereupon the Registrar from whose jurisdiction the office is transferred, shall forthwith forward to the other Registrar all documents relating to the Producer Company.
  • The alteration of the provisions of the memorandum relating to the change of the place of its registered office from one State to another shall not take effect unless it is approved by the Central Government on an application in such form and manner as may be prescribed.

Amendment Of Articles:

378-I

  • Any amendment of the articles shall be proposed by not less than two-thirds of the elected directors or by not less than one-third of the Members of the Producer Company, and adopted by the Members by a special resolution.
  • A copy of the amended articles together with the copy of the special resolution, both duly certified by two directors, shall be filed with the Registrar within fifteen days from the date of its adoption.

Option to inter-state co-operative societies to become Producer Companies:

378J:

Notwithstanding anything contained in sub-section (1) of section 378C, any inter-State co-operative society with objects not confined to one State may make an application to the Registrar for registration as a Producer Company under this Chapter.

  • Every application under sub-section (1) shall be accompanied by—
  • A copy of the special resolution, of not less than two-thirds of the total members of inter-state co-operative society, for its incorporation as a Producer Company under this Act;

A Statement Showing: Names and addresses or the occupation of the directors and the Chief Executive, if any, by whatever name called, of such co-operative; and

List of members of such inter-State co-operative society;

  • A statement indicating that the inter-state co-operative society is engaged in any one or more of the objects specified in section 378B;
  • A declaration by two or more directors of the inter-state co-operative society certifying that particulars given in clauses (a) to (c) are correct.
  • When an inter-state co-operative society is registered as a Producer Company, the words “Producer Company Limited” shall form part of its name with any word or expression to show its identity preceding it.
  • In compliance with the requirements of sub-sections (1) to (3), the Registrar shall, within thirty days of the receipt of the application, certify under his hand that the inter-State co-operative society applying for registration is registered and thereby incorporated as a Producer Company under this Chapter.

A co-operative society formed by producers, by federation or union of co-operative societies of producers or co-operatives of producers, registered under any law for the time being in force which has extended its objects outside the State, either directly or through a union or federation of co-operatives of which it is a constituent, as the case may be, and any

  • Federation or unions of such co-operatives, which has so extended any of its objects or activities outside the State, shall be eligible to apply under sub-section (1) and obtain registration as a Producer Company under this Chapter.
  • The inter-state co-operative society shall, upon registration under sub-section (1), stand transformed into a Producer Company, and thereafter shall be governed by the provisions of this Chapter to the exclusion of the law by which it was earlier governed, save in so far as anything done or omitted to be done before its registration as a Producer Company, and notwithstanding anything contained in any other law for the time being in force, no person shall have any claim against the co-operative institution or the company because of such conversion or transformation.
  • Upon registration as a Producer Company, the Registrar of Companies who registers the company shall intimate the Registrar with whom the erstwhile inter-State co-operative society was earlier registered for deletion of the society from its register.

Effect of incorporation of Producer Company:

378K:

Every shareholder of the inter-State co-operative society immediately before the date of registration of Producer Company (hereafter in this Chapter referred to as the date of transformation) shall be deemed to be registered on and from that date as a shareholder of the Producer Company to the extent of the face value of the shares held by such shareholder.

Vesting of undertaking in Producer Company:

378L:

All properties and assets, movable and immovable, of, or belonging to, the inter-State co-operative society as of the date of transformation, shall vest in the Producer Company. ,

  • All the rights, debts, liabilities, interests, privileges, and obligations of the inter-State co-operative society as on the date of transformation shall stand transferred to, and be the rights, debts, liabilities, interests, privileges, and obligations of, the Producer Company.
  • Without prejudice to the provisions contained in sub-section (2), all debts, liabilities, and obligations incurred, all contracts entered into, and all matters and things engaged to be done by, with or for, the society as on the date of transformation for or in connection with their purposes, shall be deemed to have been incurred, entered into, or engaged to be done by, with or for, the Producer Company.
  • All sums of money due to the inter-state co-operative society immediately before the date of transformation shall be deemed to be due to the Producer Company.
  • Every organization, which was being managed immediately before the date of transformation by the inter-State co-operative society shall be managed by the Producer Company for such period, to such extent and in such manner as the circumstances may require.
  • Every organization which was getting financial, managerial, or technical assistance from the inter-State co-operative society, immediately before the date of transformation, may continue to be given financial, managerial, or technical! assistance, as the case may be, by the Producer Company, for such period, to such extent and in such manner as that company may deem fit.
  • The amount representing the capital of the erstwhile inter-state co-operative society shall form part of the capital of the Producer Company.
  • Any reference to the inter-state co-operative society in any law other than this Act or in any contract or other instrument shall be deemed to be a reference to the Producer Company.
  • If, on the date of transformation, there is pending any suit, arbitration, appeal, or another legal proceeding of whatever nature by or against the inter-state co-operative society, the same shall not abate, be discontinued, or be in any way prejudicially affected because of the incorporation of the Producer
  • Company under section 378C or transformation of the inter-state co-operative society as a Producer Company under section 378J, as the case may be, but the suit, arbitration, appeal, or another proceeding, may be continued, prosecuted, and enforced by or against.
  • The Producer Company in the same manner and to the same extent as it would have, or may have been continued, prosecuted, and enforced by or against the inter-state co-operative society as if the provisions contained in this Chapter had not come into force.

Concession, etc., to be deemed to have been granted to Producer Company:

378M:

With effect from the date of transformation, all fiscal and other concessions, licenses, benefits, privileges, and exemptions granted to the inter-state co-operative society in connection with the affairs and business of the inter-state co-operative society under any law for the time being in force shall be deemed to have been granted to the Producer Company.

Provisions in respect of I officers and other employees of inter-State co-operative society:

378N:

Notwithstanding anything contained in section 378-0, all the directors in the inter-state co-operative society before the incorporation of the Producer Company shall continue in office for one year from the date of transformation and, by the provisions of this Act.

  • Every officer or other employee of the inter-State co-operative society (except a director of the Board, Chairman, or Managing Director) serving in its employment immediately before the date of transformation shall,
  • In so far as such officer or other employee is employed in connection with the inter-State co-operative society which has vested in the Producer Company by this Act, become, as from the date of transformation, an officer or, as the case may be, another employee
  • The Producer Company shall hold his office or service therein by the same tenure, at the same remuneration, upon the same terms and conditions, with the same obligations and with the same rights and privileges as to leave, leave travel concession, welfare scheme, medical benefits scheme, insurance, provident fund, other funds,
  • I retirement, voluntary retirement, gratuity, and other benefits as he would have held under the erstwhile inter-State co-operative society if its undertaking had not vested in the Producer Company and shall continue to do so as an officer or, as the case may be, other employees of the Producer Company.

14 Of 1947: Where an officer or other employee of the inter-State co-operative society opts under sub-section (2) not to be in employment or service of the Producer Company, such officer or other employee shall be deemed to have resigned.

  • Notwithstanding anything contained in the Industrial Disputes Act, 1947 or in any other law for the time being in force, the transfer of the services of any officer or other employee of the inter-state co-operative society to the Producer Company shall not entitle such officer or other employee to any compensation under this Act or any other law for the time being in force and no such claim shall be entertained by any court, tribunal or other authority.
  • The officers and other employees who have retired before the date of transformation from the service of the inter-State co-operative society and are entitled to any benefits, rights, or privileges, shall be entitled to receive the same benefits, rights, or privileges from the Producer Company.

The trusts of the provident fund or the gratuity fund of the inter-state co-operative society and any other bodies created for the welfare of officers or employees shall continue to discharge functions in the Producer Company as was being done hitherto in the inter-state co-operative society and any tax exemption granted to the provident fund or the gratuity fund would continue to be applied to the Producer Company.

  • Notwithstanding anything contained in this Act or any other law for the time being in force or the regulations of the inter-State co-operative society, no director of the Board, Chairman, Managing
  • Director or any other person entitled to manage the whole or substantial part of the business and affairs of the inter-State co-operative society shall be entitled to any compensation against the inter-State co-operative society or the Producer Company for the loss of office or the premature termination of any contract of management entered into by him with the inter-State co-operative society.

Number of Directors Part III

Management of Producer Company

378-0. Every Producer Company shall have at least five and not more than fifteen directors:

  • Provided that in the case of an inter-state co-operative society incorporated as a Producer Company, such company may have more than fifteen directors for one year from the date of its incorporation as a Producer Company.
    Appointment of directors 378P
  • Save as otherwise provided in section 378N, the Members who sign the memorandum and the articles may designate therein the Board of Directors, not less than five, who shall govern the affairs of the Producer Company until the directors are elected by the provisions of this section.
  • The election of directors shall be conducted within ninety days of the registration of the Producer Company:

Provided that in the case of an inter-state co-operative society that has been registered as a Producer Company under sub-section (4) of section 378J in which at least five directors [including the directors continuing in office under sub-section (1) of section 378N] hold office as

  • such on the date of registration of such company, the provisions of this sub-section shall have effect as if for the words “ninety days”, the words “three hundred and sixty-five days” had been substituted.
  • Every person shall hold office of a director for a period not less than one year but not exceeding five years as may be specified in the articles.
  • Every director, who retires by the articles, shall be eligible for re-appointment as a director.
  • Save as otherwise provided in subsection (2), the directors of the Board shall be elected or appointed by the Members in the annual general meeting.
  • The Board may co-opt one or more expert directors or an additional director not exceeding one-fifth of the total number of directors or appoint any other person as additional director for such period as the Board may deem fit:
  • Provided that the expert directors shall not have the right to vote in the election of the Chairman but shall be eligible to be elected as Chairman if so provided by its articles:
  • Provided further that the maximum period, for which the expert director or the additional director holds office, shall not exceed such period as may be specified in the articles.

Vacation of office by directors 378Q

The office of the director of a Producer Company shall become vacant if,—

  • He is convicted by a court of any offense involving moral turpitude and sentenced in respect thereof to imprisonment for not less than six months;
  • The Producer Company, in which he is a director, has made a default in repayment of any advances or loans taken from any company or institution or any other person, and such default continues for ninety days;
  • He has made a default in repayment of any advances or loans taken from the Producer Company in which he is a director;

The Producer Company, in which he is a director—

  • Has not filed the annual accounts and annual return for any continuous three financial years; or
  • Has failed to, repay its deposit or withheld price or patronage bonus or interest thereon on the due date, or pay a dividend and such failure continues for one year or more;
  • Default is made in holding an election for the office of director, in the Producer Company in which he is a director, by the provisions of this Act and articles;
  • The annual general meeting or extraordinary general meeting of the Producer Company, in which he is a director, is not called by the provisions of this Act except due to natural calamity or such other
    reason.
  • The provisions of sub-section (1) shall, as far as may be, apply to the director of a Producer! Institution which is a member of a Producer Company.

Power And Functions Of Board:

378R:

Subject to the provisions of this Act and articles, the Board of Directors of a Producer Company shall exercise all such powers and do all such acts and things, as that Company is authorized so to do.

  • In particular and without prejudice to the generality of the. foregoing powers, such powers may include all or any of the following matters, namely:—
  • determination of the dividend payable;
  • Determination of the quantum of withheld price and recommend patronage to be approved at a general meeting;
  • Admission of new Members;
  • Pursue and formulate the organizational policy, and objectives, establish specific long-term and annual objectives, and approve corporate strategies and financial plans;
  • Appointment of a Chief Executive and such other officers of the Producer Company, as may be specified in the articles;
  • Exercise superintendence, direction, and control over the Chief Executive and other officers appointed by it;
  • Cause proper books of account to be maintained; prepare annual accounts to be placed before the annual general meeting with the report of the auditor and the replies on qualifications, if any, made by the auditors;
  • Acquisition or disposal of property of the Producer Company in its ordinary course of business;
  • Investment of the funds of the Producer Company in the ordinary course of its business;
  • Sanction any loan or advance, in connection with the business activities of the Producer Company to any Member, not being a director or his relative;
  • Take such other measures or do such other – acts as may be required in the discharge of its functions or exercise of its powers.

All the powers specified in sub-sections (1) and (2) shall be exercised by the Board, through a resolution passed at its meeting on behalf of the Producer Company.

Explanation: For the removal of doubts, it is hereby declared that a director or a group of directors, do not constitute the Board, and shall not exercise any of the powers exercisable by it.

  • Matters are to be transacted at general meeting 378S. The Board of Directors of a Producer Company shall exercise the following powers on behalf of that Company, and it shall do so only through resolutions passed at the annual general meeting of its Members, namely:—
  • Approval of budget and adoption of annual accounts of the Producer Company;
  • Approval of patronage bonus;
  • Issue of bonus shares;
  • Declaration of limited return and decision on the distribution of patronage;
  • Specify the conditions and limits of loans that may be given by the Board to any director; and
  • Approval of any transaction of the nature as is to be reserved in the articles for approval by the Members.

Liability of directors:

378T:

When the directors vote for a resolution, or approve by any other means, anything done in contravention of the provisions of this Act or any other law for the time being in force or articles, they shall be jointly and severally liable to make good any loss or damage suffered by the Producer Company. ,

  • Without prejudice to the provisions contained in sub-section (1), the Producer Company shall have the right to recover from its director—
  • where such director has made any profit as a result of the contravention specified in sub-section (1), an amount equal to the profit so made;
  • where the Producer Company incurred a loss or damage as a result of the contravention specified in sub-section (1), an amount equal to that loss or damage.
  • The liability imposed under this section shall be in addition to and not in derogation of a liability imposed on a director under this. Act or any other law for the time being in force.

Committee of Directors:

378U:

The Board may constitute such number of committees as it may deem fit for assisting the Board in the efficient discharge of its functions:

  • Provided that the Board shall not delegate any of its powers or assign the powers of the Chief Executive, to any committee.
  • A committee constituted under sub-section (1) may, with the approval of the Board, co-opt such number of persons as it deems fit as members of the committee:
  • Provided that the Chief Executive appointed under section 378W or a director of the Producer Company shall be a member of such committee.
  • Every such committee shall function under the. general superintendence, direction, and control of the Board, for such duration, and in such manner as the Board may direct.
  • The fee and allowances to be paid to the members of the committee shall be such as may be determined by the Board.
  • The minutes of each meeting of the committee shall be placed before the Board at its next meeting.

Meetings of Board and Quorum:

378V:

A meeting of the Board shall be held not less than once every three months and at least four such meetings shall be held every year.

  • Notice of every meeting of the Board of Directors shall be given in writing to every director for the time being in India, and at his usual address in India to every other director.
  • The Chief Executive shall give notice as aforesaid not less than seven days before the date of the meeting of the Board and if he fails to do so, he shall be liable to a penalty of five thousand rupees:-
  • Provided that a meeting of the Board may be called at shorter notice and the reasons thereof shall be recorded in writing by the Board.
  • The quorum for a meeting of the Board shall be one-third of the total strength of directors, subject to a minimum of three.

Committee of Directors:

378U:

The Board may constitute such number of committees as it may deem fit for assisting the Board in the efficient discharge of its functions:

  • Provided that the Board shall not delegate any of its powers or assign the powers of the Chief Executive, to any committee.
  • A committee constituted under sub-section (1) may, with the approval of the Board, co-opt such number of persons as it deems fit as members of the committee:
  • Provided that the Chief Executive appointed under section 378W or a director of the Producer Company shall be a member of such committee.
  • Every such committee shall function under the general superintendence, direction, and control of the Board, for such duration, and in such manner as the Board may direct.
  • The fee and allowances to be paid to the members of the committee shall be such as may be determined by the Board.
  • The minutes of each meeting of the committee shall be placed before the Board at its next meeting.

Meetings of Board and Quorum:

378V:

A meeting of the Board shall be held not less than once every three months and at least four such meetings shall be held every year.

  • Notice of every meeting of the Board of Directors shall be given in writing to every director for the time being in India, and at his usual address in India to every other director.
  • The Chief Executive shall give notice as aforesaid not less than seven days before the date of the meeting of the Board and if he fails to do so, he shall be liable to a penalty of five thousand rupees:-
  • Provided that a meeting of the Board may be called at shorter notice and the reasons thereof shall be recorded in writing by the Board.
  • The quorum for a meeting of the Board shall be one-third of the total strength of directors, subject to a minimum of three ave as provided in the articles, directors including the co-opted director, may be paid such fees and allowances for attendance at the meetings of the Board, as may be decided by the Members in the general meeting.

Chief Executive and his functions 378W.

Every Producer Company shall have a full-time Chief Executive, by whatever name called, to be appointed by the Board from amongst persons other than Members.

  • The Chief Executive shall be ex officio director of the Board and such director shall not retire by rotation.
  • Save as otherwise provided in articles, the qualifications, experience, and the terms and conditions of service of the Chief Executive shall be such as may be determined by the Board.
  • The Chief Executive shall be entrusted with substantial powers of management as the Board may determine.
  • Without prejudice to the generality of subsection (4), the Chief Executive may exercise the powers and discharge the functions, namely:—
  • Do administrative acts of a routine nature including managing the day-to-day affairs of the Producer Company;
  • Operate bank accounts or authorize any person, subject to the general or special approval of the Board on this behalf, to operate the bank account;

Make arrangements for safe custody of cash and other assets of the Producer Company Sign such documents as may be authorized by the Board, for and on behalf of the company

  • Maintain proper books of account; prepare annual accounts and audit thereof; place the audited accounts before the Board and in the
    , the annual general meeting of the Members;
  • Furnish Members with periodic information to apprise them of the operation and functions of the Producer Company;
  • Make appointments to posts under the powers delegated to him by the Board;
  • Assist the Board in the formulation of goals, objectives, strategies, plans, and policies;
  • Advise the Board concerning legal and regulatory matters concerning the proposed, and ongoing activities and take necessary action in respect thereof;
  • Exercise the powers as may be necessary in the ordinary course of business;
  • Discharge such other functions, and exercise such other powers, as may be delegated by the Board.
  • The Chief Executive shall manage the affairs of the Producer Company under the general superintendence, direction, and control of the Board and be accountable for the performance of the Producer Company.

Secretary of Producer Company:

378X:

Every Producer Company having an average annual turnover exceeding five crore rupees or such other amount as may be prescribed in each of three consecutive financial years shall have a whole-time secretary.

15 Of 1980

No individual shall be appointed as v/hole-time secretary unless he possesses membership of the Institute of Company Secretaries of India constituted under the Company Secretaries Act, 1980.

  • If a Producer Company fails to comply with the provisions of sub-section (1), the Company and every officer of the Company who is in default shall be liable to a penalty of one hundred rupees for every day during which the default continues subject to a maximum of rupees one lakh:
  • Provided that in any proceedings against a person in respect of a default under this sub-section, no penalty shall be imposed if it is shown that all reasonable efforts to comply with the provisions of sub-section (1) were taken or that the financial position of the Company was such that it was beyond its capacity to engage a whole-time secretary.

Quorum:

378Y:

Unless the articles require a larger number, one-fourth of the total membership shall constitute the quorum at a general meeting, 378Z. Save as otherwise provided in sub-sections (1) and (3) of section 378D, every Member shall have one vote and in the case of equality of votes, the Chairman or the person presiding shall have a casting vote except in the case of election of the Chairman.

Annual General Meetings

Part IV

General Meetings

378ZA.

Every Producer Company shall in each year, hold, in addition to any other meetings, a general meeting, as its annual general meeting and shall specify the meeting as such in the notices calling

  • it, and not more than fifteen months shall elapse between the date of one annual general meeting of a Producer Company and that of the next: Provided that the Registrar may, for any special reason, permit extension of the time for holding any annual general meeting (not being the first annual general meeting) by a period not exceeding three months.
  • A Producer Company shall hold its first annual general meeting within ninety days from the date of its incorporation.
  • The Members shall adopt the articles of the Producer Company and appoint directors of its Board in the annual general meeting.

The notice calling the annual general meeting shall be accompanied by the following documents, namely:—

  • The agenda of the annual general meeting;
  • The minutes of the previous annual general meeting or the extraordinary general meeting;
  • The names of candidates for election, if any, to the office of director including a statement of qualifications in respect of each candidate;
  • The audited balance sheet and profit and loss accounts of the Producer Company and its subsidiary, if any, together with a report of the Board of Directors of such Company concerning—
    • The state of affairs of the Producer Company;
    • The amount proposed to be carried to reserve;
    • The amount to be paid as a limited return on share capital;
    • The amount proposed to be disbursed as a patronage bonus;

The material changes and commitments, if any, affecting the financial position of the Producer Company and its subsidiary, which have occurred between the date of the annual accounts of the Producer Company to which the balance sheet relates and the date of the report of the Board;

  • Any other matter of importance relating to energy conservation, environmental protection, expenditure, or earnings in foreign exchanges;
  • Any other matter which is required to be, or maybe, specified by the Board;
  • The text of the draft resolution for the appointment of auditors;
  • The text of any draft resolution proposing an amendment to the memorandum or articles is to be considered at the general meeting, along with the recommendations of the Board.
  • The Board of Directors shall, on the requisition made in writing, duly signed and setting out the matters for the consideration, made by one-third of the Members entitled to vote in any general meeting, proceed to call an extraordinary general meeting by the relevant provisions contained in Chapter VII.

Every annual general meeting shall be called, for a time during business hours, on a day that is not a public holiday and shall be held at the registered office of the Producer Company or at some other place within the city town or village in which the registered office of the Company is situated.

  • A general meeting of the Producer Company shall be called by giving not less than fourteen days prior notice in writing.
  • The notice of the general meeting indicating the date, time, and place of the meeting shall be sent to every Member and auditor of the Producer Company.
  • Unless the articles, of the Producer Company provide for a larger number, one-fourth of the total number of members of the Producer Company shall be the quorum for its annual general meeting.
  • The proceedings of every annual general meeting along with the report of the Board of Directors, the audited balance sheet, and the profit and loss account shall be filed with the Registrar within sixty days of the date on which the annual general meeting is held, with an annual return along with the filing fees as applicable under the Act.
  • ln the case where a Producer Company is formed by Producer Institutions, such Institutions shall be represented in the general body through the Chairman or the Chief Executive thereof who shall be competent to act on its behalf:
  • Provided that a Producer Institution shall not be represented if such Institution is in default or failure referred to in clauses (d) to (f) of sub-section (1) of section 378Q.

Share capital Part V

Share Capital and Member’s Rights

378ZB.

  • The share capital of a Producer Company shall consist of equity shares only.
  • The shares held by a Member in a Producer Company, shall as far as may be, be in proportion to the patronage of that company.

Special user rights 378ZC.

  • The producers, who are active Members may, if so provided in the articles, have special rights and the Producer Company may issue appropriate instruments to them in respect of such special rights.
  • The instruments of the Producer Company issued under sub-section (1) shall, after obtaining approval of the Board on that behalf, be transferable to any other active Member of that Producer Company.

Explanation: For this section, the expression “special right” means any right relating to the supply of additional produce by the active Member or any other right relating to his produce that may be conferred upon him by the Board.

Transferability of shares and attendant rights 378ZD.

Save as otherwise provided in sub-sections (2) to (4), the shares of a Member of a Producer Company shall not be transferable.

  • A Member of a Producer Company may, after obtaining the previous approval of the Board, transfer the whole or part of his shares along with any special rights, to an active Member at par value.
  • Every Member shall, within three months of his becoming a Member in the Producer Company, nominate, in the manner specified in articles, a person to whom his shares in the Producer Company shall vest in the event of his Death.
  • The nominee shall, on the death of the Member, become entitled to all the rights in the shares of the Producer Company, and the Board of that Company shall transfer the shares of the Deceased member to his nominee:
  • Provided that in a case where such nominee is not a producer, the Board shall direct the surrender
  • I share with special rights, if any, to the Producer Company at par value or such other value as may be determined by the Board.
  • Where the Board of a Producer Company is satisfied that
  • Any Member has ceased to be a primary producer;
  • Any Member has failed to retain his qualifications to be a Member as specified in articles, the Board shall direct the surrender of shares together with special rights, if any, to the Producer Company at par value or such! other value as may be determined by the Board:

I Provided that the Board shall not direct such! surrender of shares unless the Member has been served with a written notice and allowed to be heard.

Books of Account Part VI

Finance, Accounts and Audit

378ZE

Every Producer Company shall keep at its! registered office proper books of account concerning—

  • All sums of money received and expended by the Producer Company and the matters in respect of which the receipts and expenditure! take place;
  • All sales and purchase of goods by the Producer Company;
  • The instruments of liability executed by or on behalf of the Producer Company;
  • The assets and liabilities of the Producer Company;
  • In the case of a Producer Company engaged in production, processing, and manufacturing, the particulars relating to the utilization of materials or labor or other items of costs.
  • The balance sheet and profit and loss accounts of the Producer Company shall be prepared, as far as may be, under the provisions contained in section 129
  • Internal 2udrt 378ZF. Every Producer Company shall have an internal audit of its accounts carried out, at such interval and in such manner as may be specified in articles, by a chartered accountant as defined in clause (b) of sub-section (1) of section 2 of the Chartered

38 of 1949 Accountants Act, 1949.

Duties of auditor under this Chapter 378ZG:

Without prejudice to the provisions contained in section 143, the auditor shall report on the following additional matters relating to the Producer Company, namely:

  • The amount of debts due along with particulars of bad debts, if any;
  • The verification of cash balance and securities;
  • The details of assets and liabilities;
  • All transactions which appear to be contrary to the provisions of this Chapter;
  • The loans given by the Producer Company to the directors;
  • The donations or subscriptions given by the] Producer Company;
  • Any other matter as may be considered necessary) by the auditor.
  • Donation or subscription by Producer Company

1378ZH:

A Producer Company may, by special resolution, make a donation or subscription to any institution or individual for the purposes

  • Promoting the social and economic welfare of Producer Members or producers or the general public
  • Promoting the mutual assistance principles: Provided that the aggregate amount of all such] donations and subscriptions in any financial year shall not exceed three percent, of the net profit of the Producer Company in the financial year immediately preceding the financial year in which the donation or subscription was made:
  • Provided further that no Producer Company shall make directly or indirectly to any political party or for any political purpose to any person any contribution or subscription or make available any] facilities including personnel or material.

General and other reserves

Every Producer Company shall maintain a general serve in every financial year, in addition to any reserve maintained by it as may be specified in articles.

  • In a case where the Producer Company does not have sufficient funds in any financial year for transfer to maintain the reserves as may be specified in articles,
  • The contribution to the reserve shall be shared amongst the Members in proportion to their patronage in the business of that Company in that year.

Issue Of Bonus Shares:

378ZJ:

Any Producer Company may, upon ‘recommendation of the Board and passing of resolution in the general meeting, issue bonus shares by capitalization of amounts from general reserves referred to in section 378Z-I in proportion to the shares held by the Members on the date of the issue of such shares.

Loan, etc., to Members Part VII

Loans to Members and Investments

378ZK:

The Board may, subject to the provisions made in articles, provide financial assistance to the Members of the Producer Company by way of—

  • credit facility, to any Member, in connection with the business of the Producer Company, for a period not exceeding six months;
  • Loans and advances, against security specified in articles to any Member, repayable within a period exceeding three months but not exceeding seven years from the date of disbursement of such loan or advances:
  • Provided that any loan or advance to any director or his relative shall be granted. only after the approval by the Members in the general meeting.

Investment In Other Companies, Formation Subsidiaries, Etc.

378ZL:

The general reserves of any Producer Company shall be invested to secure the highest returns available from approved securities, fixed deposits, units, and bonds issued by the Government or cooperative or scheduled bank or in such other mode as may be prescribed.

  • Any Producer Company may, for the promotion of its objectives acquire the shares of another Producer Company.
  • Any Producer Company may subscribe to the share capital of, or enter into any agreement or other arrangement, whether by way of formation of its subsidiary company, joint venture or in any other manner with any body corporate, to promote the objects of the Producer) Company by special resolution on this behalf.
  • Any Producer Company, either by itself or together with its subsidiaries, may invest, by way of subscription, purchase, or otherwise, shares in any other company, other than a Producer Company, specified under sub-section (2), or subscription of capital under sub-section (3), for an amount not exceeding thirty percent, of the aggregate of its) paid-up capital and free reserves:
  • Provided that a Producer Company may, by) a special resolution passed in its general meeting and with prior approval of the Central Government, invest more than the limits specified in this| section.
  • All investments by a Producer Company may be) made if such investments are consistent with the| objectives of the Producer Company.
  • The Board of a Producer Company may, with the) previous approval of Members by a special resolution, dispose of any of its investments) referred to in sub-sections (3) and (4).
  • Every Producer Company shall maintain a register containing particulars of all the investments, showing the names of the companies in which shares have been acquired, the number and value of shares; the date of acquisition; and the manner) and the price at which any of the shares have been) subsequently disposed of.

The register referred to in sub-section (7) shall be kept at the registered office of the Producer Company and the same shall be open to inspection by any Member who may take extracts therefrom.

Penalty for contravention Part VIII Penalties:

378ZM:

If any person, other than a Producer Company registered under this Chapter, carries on business under any name which contains the words “Producer Company Limited”, he shall be punishable with a fine which may extend to ten thousand rupees for every day during which such name has been used by him.

If a director or an officer of a Producer Company, who wilfully fails to furnish any information relating to the affairs of the Producer Company required by a Member or a person duly authorized on this behalf, he shall be liable to imprisonment for a term which may extend to six months and with a fine equivalent to five percent, of the turnover of that Company during the preceding financial year.

If A Director Or Officer Of A Producer Company—

Fails to hand over the custody of books of account and other documents or property in his custody to the Producer Company of which he is a director or officer; or

Fails to convene the annual general meeting or other general meetings, he shall be punishable with a fine which may extend to one lakh rupees, and in the case of a continuing default or failure, with an additional fine which may extend to ten thousand rupees for every day during which such default or failure continues.

Part IX

Amalgamation, Merger or Division

378ZN:

A Producer Company may, by a resolution passed| at its general meeting,—

Decide to transfer its assets and liabilities, in whole or in part, to any other Producer Company, which agrees to such transfer by a resolution passed at its general meeting, for any of the objects specified in section 378B;

  • Divide itself into two or more new producer Companies.
  • Any two or more Producer Companies may, by a resolution passed at any general or special meetings of its Members, decide to—
  • Amalgamate and form a new Producer Company; or
  • Merge one Producer Company (hereafter in this Chapter referred to as “merging company”) with another Producer Company (hereafter in this Chapter referred to as “merged company”).
  • Every resolution of a Producer Company under this section shall be passed at its general meeting by a majority of total Members, with the right of vote not less than two-thirds of its Members present and voting and such resolution shall contain all particulars of the transfer of assets and liabilities, or division, amalgamation, or merger, as the case] may be.
  • Before passing a resolution under this section, the Producer Company shall give notice thereof in writing together with a copy of the proposed resolution to all the Members and creditors who may give their consent.
  • Notwithstanding anything contained in articles or any contract to the contrary, any Member, or any creditor not consenting to the resolution shall, during one month of the date of service of the notice on him, have the option,—
  • In the case of any such Member, to transfer his shares with the approval of the Board to any active Member thereby ceasing to continue as a Member of that Company; or
  • In the case of a creditor, to withdraw his deposit or loan or advance, as the case may be.
  • Any Member or creditor, who does not exercise his option within the period specified in sub-section (5), shall be deemed to have consented to the resolution.
  • A resolution passed by a Producer Company under this section shall not take effect until the expiry of one month or until the assent thereto of all the Members and creditors has been obtained, whichever is earlier.

The resolution referred to in this section shall provide for—

  • The regulation of the conduct of the affairs of the Producer Company in the future;
  • The purchase of shares or interest of any Members of the Producer Company by other Members or by the Producer Company;
  • The consequent reduction of its share capital, in case of the purchase of shares of one Producer Company by another Producer Company;
  • Termination, setting aside, or modification of any agreement, howsoever arrived between the company on the one hand and the directors,
  • secretaries, and manager on the other hand, apart from such terms and conditions as may, in the opinion of the majority of shareholders, be just and equitable in the circumstances of the case;
  • Termination, setting aside, or modification of any agreement between the Producer Company and any person not referred to in clause (d):
  • Provided that no such agreement shall be terminated, set aside, or modified except after giving due notice to the party concerned
  • Provided further that no such agreement shall be modified except after obtaining the consent of the party concerned;
  • The setting aside of any transfer, delivery of goods, payment, execution, or other act relating to property, made or done by or against
  • The Producer Company within three months before the date of passing of the resolution, which would if made or done against any individual, be deemed in his insolvency to be a fraudulent preference;
  • The transfer to the merged company of the whole or any part of the undertaking, property, or liability of the Producer Company;
  • The allotment or appropriation by the merged company of any shares, debentures, policies, or other like interests in the merged company;
  • The continuation by or against the merged company of any legal proceedings pending by or against any Producer Company;
  • The dissolution, without winding up, of any Producer Company;
  • The provision to be made for the Members or creditors who make dissent;
  • The taxes, if any, to be paid by the Producer Company;

Such incidental, consequential, and supplemental matters are necessary to ensure that the division, amalgamation or merger shall be fully and effectively carried out. When a resolution passed by a Producer Company under this section- takes effect, the resolution shall be a sufficient conveyance to vest the assets and liabilities in the transferee.

  • The Producer Company shall make arrangements for meeting in full or otherwise satisfying all claims of the Members and the creditors who exercise the option, within the period specified in sub-section (4), not to continue as the Member or creditor, as the case may be.
  • Where the whole of the assets and liabilities of a Producer Company is transferred to another Producer Company by the provisions of sub-section (9), or where there is a merger under sub-section (2), the registration of the first mentioned Company or the merging company, as the case may be, shall stand canceled and that Company shall be deemed to have been dissolved and shall cease to exist forthwith as a corporate body.
  • Amalgamated into a new Producer Company under the provisions of sub-section (2) and the Producer Company so formed is duly registered by the Registrar, the registration of each of the amalgamating companies shall stand canceled forthwith on such registration and each of the Companies shall thereupon cease to exist as a corporate body.

Where a Producer Company divides itself into two or more Producer Companies under the provisions of clause (b) of sub-section (1) and the new Producer Companies are registered under the provisions of this Chapter, the registration of the erstwhile Producer Company shall stand canceled forthwith and that Company shall be deemed to have been dissolved and cease to exist as a corporate body.

  • The amalgamation, merger, or division of companies under the foregoing sub-sections shall not in any manner whatsoever affect the pre-existing rights or obligations and any legal proceedings that might have been continued or commenced by or against any erstwhile company before the amalgamation, merger, or division, may be continued or commenced by, or against, the concerned resulting company, or merged company, as the case may be.
  • The Registrar shall strike off the names of every Producer Company deemed to have been dissolved under sub-sections (11) to (14).
  • Any member creditor or employee aggrieved by the transfer of assets, division, amalgamation, or merger may, within thirty days of the passing of the resolution, prefer an appeal to the Tribunal.
  • The Tribunal shall, after giving a reasonable opportunity to the person concerned, pass such orders thereon as it may deem fit.
  • Where an appeal has been filed under sub-section (16), the transfer of assets, division, amalgamation, or merger of the Producer Company shall be subject to the decision of the Tribunal.

Disputes PartX

Resolution of Disputes

378Z-0:

Where any dispute relating to the formation, management, or business of a Producer Company arises—

  • Amongst Members, former Members or persons claiming to be Members or nominees of deceased Members; or
  • Between a Member, former Member, or a person claiming to be a Member, or nominee of deceased Member and the Producer 26 of 1996 Company,
  • Its Board of Directors, office-bearers, or liquidator, past or present; or (c) between the Producer Company or its Board, and any director,
  • Office-bearer or any former director, or the nominee, heir, or legal representative of any deceased director
  • The Producer Company, such dispute shall be settled by conciliation or by arbitration as provided under the Arbitration and Conciliation Act, 1996 as if the parties to the dispute have consented in writing for determination of such disputes by conciliation or by arbitration and the provisions of the said Act shall apply accordingly.
  • Dispute shall include
  • A claim for any debt or other amount due;
  • A claim by surety against the principal debtor, where the Producer Company has recovered from the surety amount in respect of any debtor or other amount due to it from the principal debtor as a result of the default of the principal debtor whether such debt or amount due be admitted or not;
  • A claim by Producer Company against Member for failure to supply produce as required of him;
  • A claim by a Member against the Producer Company for not taking goods supplied by him.
  • If any question arises whether the dispute relates to the formation, management, or business of the Producer Company, the question shall be referred to the arbitrator, whose decision thereon shall be final.

Strike off the name Of the Producer Company

Part XI

Miscellaneous Provisions

378ZP:

Where a Producer Company fails to commence business within one. year of its registration or ceases to transact business with the Members or if the Registrar is satisfied, after making such inquiry as he thinks fit, that the Producer Company is no longer carrying on any of its objects specified in section 378B, he shall make an order striking o the name of the Producer Company, which shall thereupon cease to exist forthwith:

  • Provided that no such order canceling the registration as aforesaid shall be passed until a notice to show cause has been given by the Registrar to the Producer Company with a copy to all its directors on the proposed action and reasonable opportunity to represent its case has been given.
  • Where the Registrar has reasonable cause to believe that a Producer Company is not maintaining any of the mutual assistance principles specified, he shall strike its name off the register by the provisions contained in section 248.
  • Any Member of a Producer Company, who is aggrieved by an order made under subsection (1), may appeal to the Tribunal within sixty days of the order.
    Where an appeal is filed under sub-section (3), the order of striking off the name shall not take effect until the appeal is disposed of.

Provisions of this Chapter to override other laws 378ZQ:

The provisions of this Chapter shall have effect notwithstanding anything inconsistent therewith contained in this Act or any other law. for the time being in force or any instrument having effect under any such law, but the provisions of any such Act or law or instrument in so far as the same are not varied by, or are inconsistent with, the provisions of this Chapter shall apply to the Producer Company.

Application Of Provisions Relating To Private Companies:

378ZR: All the limitations, restrictions, and provisions of this Act, other than those specified in this Chapter, applicable to a private company, shall, as far as may be, apply to a Producer Company, as if it is a private limited company under this Act in so far as they are not in conflict with the provisions of this Chapter

Re-conversion of Producer Company to inter-State co-operative society:

378ZS:

Any Producer Company, being an erstwhile inter-State co-operative society, formed and registered under this Chapter, may make an
application—

  • After passing a resolution in the general I meeting by not less than two-thirds of its] Members present and voting; or
  • Request by its creditors representing] three-fourths value of its total creditors, to the Tribunal for its re-conversion to the inter-State co-operative society.
  • The Tribunal shall, on the application made under] sub-section (1), direct holding meeting of its Members or such creditors, as the case may be, to] be conducted in such manner as it may direct.
  • If a majority in number representing three-fourths in value of the creditors, or Members, as the case may be, present and voting in person at the meeting conducted in pursuance of the directions of the Tribunal under sub-section (2), agree for re-conversion, if sanctioned by the Tribunal, be binding on all the Members and all the creditors,’ as the case may be, and also on the company which is being converted:
  • Provided that no order sanctioning re-conversion shall be made by the Tribunal unless the Tribunal! is satisfied that the company or any other person! by whom an application has been made under sub-section (1) has disclosed to the Tribunal, by affidavit or otherwise, all material facts relating to the company, such as the latest financial position of the company, the latest report of the auditor on the accounts of the company, the pendency of any investigation proceedings about the company under and the like.
  • An order made by the Tribunal under sub-section
  • shall have no effect until a certified copy of the order has been filed with the Registrar.

A copy of every such order shall be annexed to every copy of the memorandum of the company issued after the certified copy of the order has been filed as aforesaid, or in the case of a company not having a memorandum, to every copy so issued of the instrument constituting or defining the constitution of the company.

  • If the default is made in compliance with sub-section (4), the company, and every officer of the company who is in default, shall be punishable with a fine which may extend to one hundred rupees, for each copy in respect of which default is made.
  • The Tribunal may, at any time after an application has been made to it under this section, stay the commencement or continuation of any suit or proceeding against the company on such terms as the Tribunal thinks fit until the application is finally disposed of.
  • Every Producer Company, which has been sanctioned re-conversion by the Tribunal, shall make an application under the Multi-State Co-operative Societies Act, 2002, or any other law for the time being in force for its registration as a multi-State co-operative society or co-operative society, as the case may be, within six months of sanction by the Tribunal and file a report thereof to the Tribunal and the Registrar of Companies and to the Registrar of the Co-operative Societies under which it has been registered as a multi-State co-operative society or co-operative society, as the case may be.

Power to modify Act in its application to Producer Companies

378ZT: The Central Government may, by notification, direct that any of the provisions of this Act (other than those contained in this Chapter) specified in the said notification

Shall not apply to the Producer Companies or any class or category thereof; or

  • Shall apply to the Producer Companies or any class or category thereof with such exception or adaptation as may be specified in the notification.
  • A copy of every notification proposed to be issued under sub-section (1), shall be laid in draft before each House of Parliament, while it is in session, for a total period of thirty days which may be comprised of one session or two or more successive sessions, and if, before the expiry of the session immediately following the session or the successive sessions aforesaid both
  • Houses agree in disapproving the issue of the notification or both Houses agree in making any modification in the notification, the notification shall not be issued or, as the case may be, shall be issued only in such modified form as may be agreed upon by both the Houses.

Power to make rules 378ZU: The Central Government may make rules for carrying out the purposes of this Chapter.

Amendment of section 379:  In section 379 of the principal Act, in sub-section (1), the proviso shall be omitted.

Amendment of section 392: In section 392 of the Principal Act,—

  • The words “with imprisonment for a term which may extend to six months or” shall be omitted;
  • For the words “five lakh rupees, or with both”, the words “five lakh rupees” shall be substituted.
  • Insertion of new section 393A Exemptions under this. Chapter 55. After section 393 of the principal Act, the following section shall be inserted, namely:—
  • “393A. The Central Government may, by notification, exempt any class of— Foreign companies”;
  • companies incorporated or to be incorporated outside India, whether the company has or has not been established, or when formed may or may not be established.

A place of business in India, as may be specified in the notification, from any of the provisions and a copy of every such notification shall, as soon as may be after it is made, be laid before both Houses of Parliament.

  • Amendment of section section 403 of the principal Act, in sub-section (1), for the third proviso, the following proviso shall be substituted, namely:—
  • “Provided also that where there is default on two or more occasions in submitting, filing, registering or recording of such document, fact or information, as may be prescribed, it may, without prejudice to any other legal action or liability under this Act, be. submitted, filed, registered or recorded, as the case may be, on payment of such higher additional fee, as may be prescribed.”

Amendment of section 405:  In section 405 of the principal Act, for sub-section (4), the following sub-section shall be substituted, namely:—

  • “If any company fails to comply with an order made under subsection (1) or sub-section (3), or furnishes any information or statistics which is incorrect or incomplete in any material respect,
  • The company and every officer of the company who is in default shall be liable to a penalty of twenty thousand rupees and in case of continuing failure, with a further penalty of one thousand rupees for each day after the first during which such failure continues, subject to a maximum of three lakh rupees.”

Amendment of section 410: In section 410 of the principal Act,— v (i) in the opening portion, the words “not exceeding eleven” shall be omitted;

  • In clause (b), for the word, figures, and letter “section 53N”, the word, figures, and letter “section 53A” shall be substituted.
  • insertion of new section 418A Benches of Appellate Tribunal 59. After section 418 of the Principal Act, the following section shall be inserted, namely:—
  • “418A. (1) The powers of the Appellate Tribunal may be exercised by the Benches thereof to be constituted by the Chairperson:”
  • Provided that a Bench of the Appellate Tribunal shall have at least one Judicial Member and one Technical Member 12 of 2003. 1 of 1956. 31 of 2016.
  • The Benches of the Appellate Tribunal shall ordinarily sit at New Delhi or such other places as the Central Government may, in consultation with the Chairperson, notify:

Provided that the Central Government may, by notification, after consultation with the Chairperson, establish such number of Benches of the Appellate Tribunal, as it may consider necessary, to hear appeals against any direction, decision, or order referred to in section 53A of the Competition Act, 2002 and under section 61 of the Insolvency and Bankruptcy Code, 2016.”.

Amendment of section 435: In section 435 of the principal Act, in sub-section (1), for the words “offenses under this Act, by notification”, the words and figures “offenses under this Act, except under section 452, by notification” shall be substituted.

Amendment of section 441: In section 441 of the principal Act, for sub-section (5), the following sub-section shall be substituted, namely:—

  • “If any officer or other employee of the company who fails to comply with any order made by the Tribunal or the Regional Director or any officer authorized by the Central Government under sub-section (4),
  • The maximum amount of fine for the offense proposed to be compounded under this section shall be twice the amount provided in the corresponding section in which punishment for such offense is provided.”
  • Substitution of new section for section 446B Lesser penalties for certain companies:
  • For section 446B of the principal Act, the following section shall be substituted, namely:
  • ‘446B. Notwithstanding anything contained in this Act, if the penalty is payable for non-compliance with any of the provisions of this Act

One Person Company, small company, start-up company or Producer Company, or by any of its officers in default, or any other person in respect of such company, then such company, its officer in default or any other person, as the case may be, shall be liable to a penalty which shall not be more than one-half of the penalty specified in such provisions subject to a maximum of two lakh rupees in case of a company and one lakh rupees in case of an officer who is in default or any other person.

Explanation: For this section,—

1 of 1956:

“Producer Company” means a company as defined in clause (I) of section 378A;

“start-up company” means a private company incorporated under this Act or the Companies Act, 1956 and recognized as a start-up under the notification issued by, the Central Government in the Department for Promotion of Industry and Internal Trade.”.

Amendment of section 450:

In section 450 of the principal Act, for the words “punishable with fine which may extend to ten thousand rupees, and where the contravention is continuing one, with a further fine which may extend to one thousand rupees for every day after the first during which the contravention continues”, the words “liable to a penalty of ten thousand rupees, and in case of continuing contravention, with a further penalty of one thousand rupees for each day after the first during which the contravention continues, subject to a maximum of two lakh rupees in case of a company and fifty thousand rupees in case of an officer who is in default or any other person” shall be substituted.

Amendment of section 452: In section 452 of the principle. Act, in sub-section (2), the following proviso shall be inserted, namely:—

  • “Provided that the imprisonment of such officer or employee, as the case may be, shall not be ordered for wrongful possession or withholding of  a dwelling unit, if the court is satisfied that the company has not paid to that officer or employee, as the case may be, any amount relating to—
  • Provident fund, pension fund, gratuity fund, or any other fund for the welfare of its officers or 19 of 1923 employees, maintained by the company;
  • Compensation or liability for compensation under the Workmen’s Compensation Act, .1923 in respect of death or disablement.”.

Amendment of section 454:

In section 454 of the principal Act, in sub-section (3), the following proviso shall be inserted, namely:-

“Provided that in case the default relates to non-compliance of sub-section (4) of section 92 or sub-section (1) or sub-section (2) of section 137 and such default has been rectified either before, or within thirty days of, the issue of the notice by the adjudicating officer, no penalty shall be imposed in this regard and all proceedings under this section in respect of such default shall be _ deemed to be concluded.”.

Amendment of section 465: In section 465 of the principal Act, in sub-section

  • The first proviso shall be omitted;
  • In the second proviso, for the words “Provided further that”, the words
  • “Provided that” shall be substituted;
  • In the third proviso, for the words “Provided also that”, the words “Provided further that” shall be substituted.