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NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Introduction

• The counting numbers 1, 2, 3, 4,………… are called natural numbers.
• The smallest natural number is 1. There is no greatest natural number.
• If we add 0 to the collection of natural numbers, what we get is called the collection of whole numbers. are called natural numbers. are whole numbers.
• Every natural number is a whole number but every whole number is not a natural number, e.g., 0 is a whole number but not a natural number.
• The positive integers are the same as natural numbers.
• We get negative integers if we put a negative sign before each positive integer. Thus, – 1, – 2, – 3, – 4,
• The smallest positive integer is 1. The greatest negative integer is -1.
• The number ‘O’ is neither positive nor negative. It is greater than every negative integer and smaller than every positive integer. It is simply an integer.
• Thus, 0, 1, 2, 3, and 4, are negative integers.÷
• Several p/q where p and q are integers and q ≠ 0 is called a rational number.
• All the above types of numbers are needed to solve various kinds of simple algebraic equations  

Read and Learn More NCERT Solutions For Class 8 Maths

Properties Of Rational Numbers

The list of properties of rational Numbers can be given as follows :

1. Closure
2. Commutativity
3. Associativity
4. The role of zero
5. The role of 1
6. Negative (or additive inverse) of a number
7. Reciprocal (or multiplicative inverse)
8. Distributivity of multiplication over addition for rational numbers

Closure

Whole numbers

Question 1. 4+7 =________________ Is it a whole number
Solution:

4 + 7=11

Yes, It is a whole number.

Question 2. 3×7 = _______________Is it a whole number?
Solution:

3 × 7 = 21

Yes, It is a whole number

Question 3. Check for closure property under all four operations for natural numbers
Solution:

Integers

Question 1. Is – 7 + (- 5) an integer?
Solution:

– 7 + (- 5)

– 7 + (- 5) = – 12

Yes,  Which is an integer.

Question 2. Is 8 + 5 an integer?
Solution:

8 + 5

8 + 5 = 13

Yes,  Which is an integer

Question 3. Is 5 – 7 an integer?
Solution:

5 – 7

5 – 7 = -2

Yes, Which is an integer.

Question 4. Is 8 – (- 6) an integer?
Solution:

8 – (- 6)

8 – (- 6) = 8 + 6

= 14,

Yes, Which is an integer.

Question 5. Is – 5 × 8= do integer?
Solution:

– 5 × 8

– 5 × 8= – 40

Yes,  Which is an integer

Rational numbers

Question 1.\(\frac{-3}{8}+\frac{(-4)}{5}\)= ________ Is it a rational number ?
Solution:

⇒ \(\frac{-3}{8}+\frac{(-4)}{5}\)

=\(\frac{-15+(-32)}{40}\)

= \(\frac{-47}{40}\)

Yes, It is a rational number

Question 2. \(\frac{5}{8}-\frac{4}{5}=\frac{25-32}{40}\) = _________________ it is a ralional number?
Solution:

⇒ \(\frac{5}{8}-\frac{4}{5}\)

= \(\frac{25-32}{40}\)

= \(\frac{-7}{40}\)

Yes, It is a rational number.

Question 3. \(\frac{-5}{9}+\frac{(-7)}{6}\) _______________ Is it a rational number?
Solution:

⇒ \(\frac{3}{7}-\left(\frac{-8}{5}\right)=\frac{3}{7}+\frac{8}{5}\)

= \(\frac{15+56}{35}\)

=\(\frac{71}{35}\) .

Yes, It is a rational number.

Question 4. \(\frac{4}{7}+\frac{6}{11}\) Is it a rational number ?
Solution:

⇒  \(\frac{4}{7}+\frac{6}{11}\)

= \(\frac{44+42}{77}=\frac{86}{77}\)

Yes, It is a rational number

Question 5. \({4}{5} \times \frac{-6}{11}\)___________Is it a rational number?
Solution:

⇒ \(\frac{4}{5} \times \frac{-6}{11}\)

= \(\frac{24}{55}\)

Yes, It is a rational number.

Question 6. \(\frac{2}{7}+\frac{5}{3}\)= _______________Is it a rational number?
Solution:

⇒ \(\frac{2}{7} \div \frac{5}{3}\)

=\(\frac{2}{7} \times \frac{3}{5}\)

= \(\frac{6}{35}\)

Yes, It is a rational number

Question 7. \(\frac{-3}{8}+\frac{-2}{9}\)= ______________ It is a rational numbers
Solution:

⇒ \(\frac{-3}{8} \div \frac{-2}{9}\)

= \(\frac{-3}{8} \times \frac{-9}{2}\)

= \(\frac{27}{16}\)

Yes, it is a rational number.

Question 8. Check it for a few more pairs of rational numbers

1. \(\frac{3}{5}+\frac{7}{13}\)
Solution: 

⇒ \(\frac{3}{5}+\frac{7}{13}\)

= \(\frac{39+35}{65}\)

= \(\frac{74}{65}\)

Rational number

2. \(\frac{3}{7}+\frac{(-4)}{9}\)
Solution:

⇒ \(\frac{3}{7}+\frac{(-4)}{9}\)

= \(\frac{27+(-28)}{63}\)

Rational number

3. \(\frac{-5}{9}+\frac{(-7)}{6}\)
Solution:

⇒ \(\frac{-5}{9}+\frac{(-7)}{6}\)

= \(\frac{-10+(-21)}{18}\)

Rational number

Question 9. Try this for some more pairs of rational numbers

1. \(\frac{-3}{8}-\frac{4}{5}\)
Solution:

⇒ \(\frac{-3}{8}-\frac{4}{5}\)

= \(\frac{-15-32}{40}\)

=\(\frac{-47}{40}\)

Rational number

2. \(\frac{2}{9}-\frac{3}{7}\)
Solution:

⇒ \(\frac{2}{9}-\frac{3}{7}\)

= \(\frac{14-27}{63}\)

= –\(\frac{13}{63}\)

Rational number

3. \(\frac{-3}{4} \times \frac{7}{8}\)
Solution:

⇒ \(\frac{-3}{4} \times \frac{7}{8}\)

= –\(\frac{21}{32}\)

Rational number

4. \(\frac{5}{8} \times \frac{3}{7}\)
Solution:

⇒ \(\frac{5}{8} \times \frac{3}{7}\)

= \(\frac{15}{56}\)

Rational number

Question 10. Fill in the blanks in the following table

Solution:

Commutativity

Whole numbers

Question 1. Recall the commutativity of different operations for whole numbers by filling the following table.

Check whether the commutativity of the. operations hold for natural numbers also.
Solution:

Commutativity of the operations for natural numbers also.

Integers

Question 1.  Fill in the following table and check the commutativity of different operations for integers:

Solution:

Rational numbers

Question 1. \(\frac{-6}{5}+\left(\frac{-8}{3}\right)\) = \(\ldots \ldots \text { and } \frac{-8}{3}+\left(\frac{-6}{5}\right)=\ldots\) Is \(\frac{-6}{5}+\left(\frac{-8}{3}\right) \)= \(\left(\frac{-8}{3}\right)+\left(\frac{-6}{5}\right) \)
Solution:

⇒ \(\frac{-6}{5}+\left(\frac{-8}{3}\right)\)

= \( \frac{-18+(-40)}{15}=\frac{-58}{15}\)

= \(\frac{-8}{3}+\left(\frac{-6}{5}\right)\)

= \(\frac{-40+(-18)}{15}=\frac{-58}{15}\)

So, \(\frac{-6}{5}+\left(\frac{-8}{3}\right)=\left(\frac{-8}{3}\right)+\left(\frac{-6}{5}\right)\)

Question 2. Is \(\frac{3}{8}+\frac{1}{7}=\frac{1}{7}+\left(\frac{-3}{8}\right)\)?
Solution.

⇒ \(\frac{-3}{8}+\frac{1}{7}=\frac{-21+8}{56} \)

= \(\frac{-13}{56}\)

= \(\frac{1}{7}+\left(\frac{-3}{8}\right)=\frac{8+(-21)}{56}\)

= \(\frac{-13}{56}\)

So, \(\frac{-3}{8}+\frac{1}{7}=\frac{1}{7}+\left(\frac{-3}{8}\right)\).

Question 3. Is \(\frac{2}{3}-\frac{5}{4}=\frac{5}{4}-\frac{2}{3}\)
Solution:

⇒ \(\frac{2}{3}-\frac{5}{4}=\frac{8-15}{12}=\frac{-7}{12} \)

⇒ \(\frac{5}{4}-\frac{2}{3}=\frac{15-8}{12}=\frac{7}{12}\)

⇒  \(\frac{-7}{12} \neq \frac{7}{12} \)

∴ \(\frac{2}{3}-\frac{5}{4} \neq \frac{5}{4}-\frac{2}{3} \)

⇒  \(\frac{1}{2}-\frac{3}{5}=\frac{3}{5}-\frac{1}{2} \)

⇒  \(\frac{1}{2}-\frac{3}{5}=\frac{5-6}{10}=\frac{-1}{10} \)

⇒  \(\frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10} \)

⇒  \(\frac{-1}{10} \neq \frac{1}{10} \)

⇒  \(\frac{1}{2}-\frac{3}{5} \neq \frac{3}{5}-\frac{1}{2}\)

Question 4. \(\frac{-8}{9} \times\left(\frac{-4}{7}\right)=\frac{32}{63} \)
Solution:

⇒ \(\frac{-4}{7} \times\left(\frac{-8}{9}\right)=\frac{32}{63}\)

So, \(\frac{-8}{9} \times\left(\frac{-4}{7}\right)=\frac{-4}{7} \times\left(\frac{-8}{9}\right)\)

Question 5. \(\frac{-5}{9} \times \frac{7}{8}\)
Solution:

= \(\frac{7}{8} \times\left(\frac{-5}{9}\right)\)

= \(\frac{-35}{72}\)

Question 6. \(\frac{-6}{11} \times\left(\frac{-3}{5}\right)\)
Solution:

⇒ \(\frac{-6}{11} \times\left(\frac{-3}{5}\right)\)

=\(\frac{-3}{5} \times\left(\frac{-6}{11}\right)\)

= \(\frac{18}{55}\)

⇒ \(\frac{2}{3} \times \frac{4}{7}=\frac{4}{7} \times \frac{2}{3}=\frac{8}{21}\)

Question 7. Is \(\frac{-5}{4} \div \frac{3}{7}=\frac{3}{7}\div\left(\frac{-5}{4}\right)\)
Solution:

⇒ \(\frac{-5}{4} \div \frac{3}{7}=\frac{3}{7} \div\left(\frac{-5}{4}\right)\)

= \(\frac{-35}{12}\)

And \(\frac{3}{7} \div \frac{-5}{4}=\frac{3}{7} \times\left(\frac{-4}{5}\right) \)

=  \(\frac{-12}{35}\)

Since \(\frac{-35}{12} \neq \frac{-12}{35} \)

∴ \(\frac{-5}{4} \div \frac{3}{7} \neq \frac{3}{7} \div\left(\frac{-5}{4}\right)\)

Question 8. Complete the following table:

Solution: 

Associativity

Whole numbers

Question 1. Recall the associativity of the four operations for whole numbers through this table

Fill in this table and. verify the remarks given in the last column. Check for yourself the associativity of different operations for natural numbers
Solution:

Integers

Question 1. Is (-2) + [3 + (-4)] = [(-2) + 3] + (-4)?
Solution:

Yes; (-2) + [3 + (-4)]

= [(-2) + 3] + (-4)

= -3.

Addition is associative

Question 2. -Is (-6) +[(-4) +(-5)] = (-6) + (-4)] + (-5)?
Solution:

Yes; (-6) + [(-4) + (-5)]

= [(-6) + (-4)] + (-5)

= -15

Addition is associative

Question 3. Is 5 – (7 – 3) = (5 – 7) – 3?
Solution:

5 – (7 – 3) = 5 – 4

= 1

(5 – 7) – 3 = -2 – 3

= -5

1 ≠ – 5

∴  No; 5 – (7 – 3) ≠ (5 – 7) – 3.

Subtraction is not associative

Question 4. Is 5× [(- 7) × (-8)] = [5× (- 7)] × (- 8) ?
Solution:

Yes; 5 × [(-7) × (-8)]

= [5× (-7)] ×(-8)

= 280.

Multiplication is associative

Question 5. Is (- 4) × [(- 8)×(- 5)]= (-4) x (-8)]× (-5)
Solution:

Yes; (-4) × [(-8) × (-5)]

= [(-4) × (-8)] × (-5)

= -160

Multiplication is associative

Question 6.  Is [(-10) ÷ 2] ÷ (- 5)= (-10) ÷[2 ÷ (-5)]?
Solution:

[(-10) ÷ 2] ÷ (-5)= (-5) ÷ (-5) = 1

(-10) ÷ [2 ÷ (-5)]= (-10) ÷ \(\frac{-2}{5}\)

= (-10) ×\(\frac{-5}{2}\)

= 25

∴ 1 ≠ 25

∴ No; [(-10) ÷ 2] ÷ (-5)≠ (-10) ÷ [2 ÷ (-5)]

The division is not associative

Rational numbers

Question 1. Find \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right]\) and \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right) \text {. }\) Are the two sums equal?
Solution:

⇒ \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right]\)

= \(\frac{-1}{2}+\frac{9+(-28)}{21}=\frac{-1}{2}+\left(\frac{-19}{21}\right)\)

= \(\frac{-21+(-38)}{42}=\frac{-59}{42}\)

And \( {\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)}\)

= \(\frac{-7+6}{14}+\left(\frac{-4}{3}\right)=\frac{-1}{14}+\left(\frac{-4}{3}\right)\)

= \(\frac{-3+(-56)}{42}=\frac{-59}{42}\)

= \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right] \)

= \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)\)

= \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)\)

Question 2. \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)
Solution:

⇒ \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)

= \(\frac{-3}{4}+\frac{14+(-18)}{21}=\frac{-3}{4}+\left(\frac{-4}{21}\right)\)

= \(\frac{-63+(-16)}{84}=\frac{-79}{84} \)

= \(\left[\frac{-3}{4}+\frac{2}{3}\right]+\left(\frac{-6}{7}\right) \)

= \(\frac{-9+(8)}{12}+\left(\frac{-6}{7}\right)=\frac{-1}{12}+\left(\frac{-6}{7}\right) \)

= \(\frac{-7+(-72)}{84}=\frac{-79}{84}\)

So,yes \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)

= \(\left[\frac{-3}{4}+\frac{2}{3}\right]+\left(\frac{-6}{7}\right) \)

Question 3. \( \frac{-1}{4}+\left[\frac{2}{9}\right.\left.+\left(\frac{-5}{11}\right)\right] \)
Solution:

⇒  \( \frac{-1}{4}+\left[\frac{2}{9}\right.\left.+\left(\frac{-5}{11}\right)\right] \)

= \(\frac{-1}{4}+\left(\frac{22-(45)}{99}\right) \)

= \(\frac{-1}{4}+\left(\frac{-23}{99}\right) \)

⇒ \(\left[\frac{-1}{4}\right.\left.+\frac{2}{9}\right]+\left(\frac{-5}{11}\right)\)

= \(\frac{-99+(-92)}{396}=\frac{-191}{396}\)

= \(\frac{-9+8}{36}+\left(\frac{-5}{11}\right)\)

= \(\frac{-1}{36}+\left(\frac{-5}{11}\right)=\frac{-11+(-180)}{396}\)

= \(\frac{-191}{396} \)

⇒ \(\frac{-1}{4}+\left[\frac{2}{9}+\left(\frac{-5}{11}\right)\right] \)

= \(\left[\frac{-1}{4}+\frac{2}{9}\right]+\left(\frac{-5}{11}\right)\)

Question 4. \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\) Check for yourself
Solution:

= \(\left[\frac{-2}{3}-\left(\frac{-4}{5}\right)\right]-\frac{1}{2}\)

= \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\)

= \(\frac{-2}{3}-\left(\frac{-8-5}{10}\right)=\frac{-2}{3}-\left(\frac{-13}{10}\right)\)

= \(\frac{-2}{3}+\frac{13}{10}=\frac{-20+39}{30}=\frac{19}{30}\)

And \({\left[\frac{-2}{3}-\left(\frac{-4}{5}\right)\right]-\frac{1}{2}=\left[\frac{-2}{3}+\frac{4}{5}\right]-\frac{1}{2}} \)

= \(\left[\frac{-10+12}{15}\right]-\frac{1}{2}=\frac{2}{15}-\frac{1}{2} \)

= \(\frac{4-15}{30}=\frac{-11}{30}\)

So, No ; \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\)≠ \(\frac{-2}{3}\)– (\(\frac{-4}{5}\)) – \(\frac{1}{2}\)

Question 5. \(\left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}\)___________
Solution:

⇒\(\left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}=\frac{-35}{12} \times \frac{2}{9}\)

=\(\frac{-35}{54}\)

So, Yes ;

Question 6. Is \(\frac{2}{3} \times\left(\frac{-6}{7} \times \frac{4}{5}\right)=\left(\frac{2}{3} \times \frac{-6}{7}\right) \times \frac{4}{5} \)
Solution:

⇒ \(\frac{2}{3} \times\left(\frac{-6}{7} \times \frac{4}{5}\right)\)

= \(\frac{2}{3} \times \frac{-24}{35}=\frac{-48}{105}\)________ (1)

=  \(\frac{2}{3}\)× \(\frac{-6}{7}\)× \(\frac{4}{5}\)

= \(\frac{-12}{21} \times \frac{4}{5}=\frac{-48}{105}\)________ (2)

So, Yes; A = B

Question 7. \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)\)
Solution:

⇒ \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)=\frac{-6}{5} \times \frac{1}{12}\)

=\(\frac{-1}{10}\)

And  \(\left(\frac{-6}{5} \times \frac{2}{3}\right) \times \frac{1}{8}=\frac{-4}{5} \times \frac{1}{8}=\frac{-1}{10}\)

So, Yes

⇒ \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)=\left(\frac{-6}{5} \times \frac{2}{3}\right) \times \frac{1}{8}\)

Question 8. \(\frac{2}{7} \times\left(\frac{-5}{9} \times \frac{2}{3}\right) \)
Solution:

⇒ \(\frac{2}{7} \times\left(\frac{-5}{9} \times \frac{2}{3}\right) \)

=\(\frac{2}{7} \times\left(\frac{-10}{27}\right)=\frac{-20}{189} \)

= \(\left(\frac{2}{7} \times \frac{-5}{9}\right) \times \frac{2}{3}=\frac{-10}{63} \times \frac{2}{3}\)

=\(\frac{-20}{189}\)

So, Yes; A=B

Question 9. \(\frac{1}{2}+\left[\frac{-1}{3}+\frac{2}{5}\right]=\left[\frac{1}{2}+\left(\frac{-1}{3}\right)\right]+\frac{2}{5}\),Is L.H.S. =R.H.S. ? Check for yourself
Solution:

L.H.S.\(\frac{1}{2}+\left(\frac{-1}{3}+\frac{2}{5}\right)\)

= \(\frac{1}{2} \div\left(\frac{-1}{3} \times \frac{5}{2}\right)\)

= \(\frac{2}{5} is \frac{5}{2}\)

= \(\frac{1}{2}+\left(-\frac{5}{6}\right) \)

= \(\frac{1}{2} \times \frac{-6}{5}=\frac{-3}{5} \)

R.H.S= \(\left[\frac{1}{2}+\left(\frac{-1}{3}\right)\right]+\frac{2}{5}\)

= \(\left(\frac{1}{2} \times \frac{-3}{1}\right) \div \frac{2}{5}=\frac{-3}{2}+\frac{2}{5} \)

= \(\frac{-3}{2} \times \frac{5}{2}=\frac{-15}{4}\)

So, No; L.H.S ≠R. H.S.

Question 10. Complete the following table:

Solution:

Question 11. Do you think the properties commutativity and associativity made the calculations
Sol.

Yes! The properties of commutativity and associativity made the calculations easier

The Rolie Of Zero (0)

Question 1. \(-5+0=\ldots \ldots. \ldots \ldots .=-5\)
Solution:

-5+0=0+(-5)=-5

Question 2.  \(\frac{-2}{7}+\ldots \ldots .=0+\left(\frac{-2}{7}\right)=\frac{-2}{7}\)

⇒ \(\frac{-2}{7}+0=0+\left(\frac{-2}{7}\right)=\frac{-2}{7}\)

Question 3. Do a few more such additions. What do you observe?
Solution:

1. 3 + 0 = 0 + 3 = 3

2. -9 +0 = 0 + (-9) = -9

3.\(\frac{-3}{4}+0=0+\left(\frac{-3}{4}\right)=\frac{-3}{4}\)

We observe that when we add 0 to a whole number, the sum is again that whole number. This happens for integers and rational numbers also.

The Role Of 1

Question 1. \(\frac{-2}{7} \times 1=\ldots \ldots . . \times \ldots . .=\frac{-2}{7}\)
Solution:

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

Question 2. \(\frac{3}{8} \times \ldots \ldots=1 \times \frac{3}{8}=\frac{3}{8}\)
Solution:

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

= \(\frac{3}{8}\)

Question 3.  What do you find?
Solution:

We find that when we multiply any rational number by 1, we get back the same rational number as the product

Question 4. Check for a few more rational numbers

1. 7 × 1
Solution:

7 × 1 = 7 = 1 × 7

2.\(\frac{-3}{5} \times 1\)
Solution:

⇒ \(\frac{-3}{5} \times 1\)

=\(1 \times \frac{-3}{5}=\frac{-3}{5}\)

3. \(\frac{7}{9} \times 1\)
Solution:

⇒ \(\frac{7}{9} \times 1\)

= \(1 \times \frac{7}{9}=\frac{7}{9}\)

Question 5. Is 1 the multiplicative identity for integers? For whole numbers?
Solution:

Yes, 1 is the multiplicative identity for integers as well as for whole numbers.

Question 6. If a property holds for rational numbers, will it also hold for integers?

1. For whole numbers?
2. Which will?
3. Which will not?

Solution:

1. Except for the following property, the properties of the rational numbers will also hold good for integers :

If a and b are rational numbers, then a÷b is also a rational number if b ≠ 0 but a÷ b is not necessarily an integer where a and b are integers

Eample: \(\frac{4}{5}\)is a rational number. Here 4 and 5 are integers, while \(\frac{4}{5}\) is not an integer

2. Except for the following properties, the properties of the rational numbers will also hold good for whole numbers:

If a and b are rational numbers, then, a- b is also a rational number. But if a and b are whole numbers, then A b is not necessarily a whole number

Let 4 and 5 be whole numbers, but 4 – 5 = -1 is not a whole number.

2. If a and b are rational numbers, then, a + b is a rational number if b  but if a and b are whole numbers, then a + b is not necessarily a whole number.

Example: 4 and 5 are whole numbers but \(\frac{4}{5}\) is not a whole number

Disberstivity Of Multiplication Over Addition For Rational

For all rational numbers a, b, and c,

a(b + c) = ab + ac.

a(b-c) = ab – ac

Question 1. \(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
Solution:

⇒ \(\left\{\frac{7}{\mathbf{5}} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{\mathbf{5}}{\mathbf{1 2}}\right\}\)

= \(\frac{7}{5} \times\left\{\left(\frac{-3}{12}\right)+\frac{5}{12}\right\} \)

= \(\frac{7}{5} \times\left\{\frac{(-3)+5}{12}\right\}=\frac{7}{5} \times \frac{2}{12} \)

= \(\frac{7}{5} \times \frac{1}{6}=\frac{7}{30} \)

Question 2. \(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
Solution:

⇒ \(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\} \)

= \(\frac{9}{16} \times\left\{\frac{4}{12}+\left(\frac{-3}{9}\right)\right\}\)

= \(\frac{9}{16} \times\left\{\frac{1}{3}+\left(\frac{-1}{3}\right)\right\} \)

= \(\frac{9}{16} \times\left\{\frac{1-1}{3}\right\}=\frac{9}{16} \times 0=0 .\)

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1

Question 1.  Name the property under multiplication used in each of the following

1. \( \frac{-4}{5} \times 1=1 \times \frac{-4}{5}=-\frac{4}{5} \)

2. \(\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)

3. \(\frac{-19}{29} \times \frac{29}{-19}=1\)

Solution:

1.1 is the multiplicative identity

2. Commutativity of multiplication

3. Multiplicative inverse.

Question 2. Tell what property allows you tocompute :\(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3} \text {. }\)
Solution:

Associativity of multiplication

Question 3. The product of two rational numbers is always a
Solution:

The product of two rational numbers is always a rational number.

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Multiple Choice Questions

Question 1. Which of the following statements is false?

1. Natural numbers are closed under addition
2. Whole numbers are closed under the addition
3. Integers are closed under the addition
4. Rational numbers are not closed under addition.

Solution: 4. Rational numbers are not closed under addition.

Question 2. Which of the following statements is false?

1. Natural numbers are closed under subtraction
2. Whole numbers are not closed under subtraction
3. Integers are closed under subtraction
4. Rational numbers are closed under subtraction.

Solution: 1. Natural numbers are closed under subtraction

Question 3. Which of the following statements is true?

1. Natural numbers are closed under multiplication
2. Whole numbers are not closed under multiplication
3. Integers are not closed under multiplication
4. Rational numbers are not closed under multiplication.

Solution: 1. Natural numbers are closed under multiplication

Question 4. Which of the following statements is true?

1. Natural numbers are closed under the division
2. Whole numbers are not closed under the division
3. Integers are closed under the division
4. Rational numbers are closed under division.

Solution: 2.  Whole numbers are not closed under the division

Question 5. Which of the following statements is false?

1. Natural numbers are commutative for addition
2. Whole numbers are commutative for addition
3. Integers are not commutative for addition
4. Rational numbers are commutative for addition.

Solution: 3. Integers are not commutative for addition

Question 6. Which of the following statements is true?

1. Natural numbers are commutative for subtraction
2. Whole numbers are commutative for subtraction
3. Integers are commutative for subtraction
4. Rational numbers are not commutative for subtraction.

Solution: 4. Rational numbers are not commutative for subtraction.

Question 7. Which of the following statements is false?

1. Natural numbers are commutative for multiplication
2. Whole numbers are commutative for multiplication
3. Integers are not commutative for multiplication
4. Rational numbers are commutative for multiplication.

Solution: 3. Integers are not commutative for multiplication

Question 8. Which of the following statements is true?

1. Natural numbers are commutative for division
2. Whole numbers are not commutative for division
3. Integers are commutative for division
4. Rational numbers are commutative for division.

Solution: 2. Whole numbers are not commutative for division

Question 9. Which of the following statements is true?

1. Natural numbers are associative for addition
2. Whole numbers are not associative for addition
3. Integers are not associative for addition
4. Rational numbers are not associative for addition.

Solution: 1. Natural numbers are associative for addition

Question 10. Which of the following statements is true?

1. Natural numbers are associative for subtraction
2. Whole numbers are not associative for subtraction
3. Integers are associative for subtraction
4. Rational numbers are associative for subtraction

Solution: 2. Whole numbers are not associative for subtraction

Question 11. Which of the following statements is true?

1. Natural numbers are not associative for multiplication
2. Whole numbers are not associative for multiplication
3. Integers are associative for multiplication
4. Rational numbers are not associative for multiplication.

Solution: 3. Integers are associative for multiplication

Question 12. Which of the following statements is true?

1. Natural numbers are associative for division
2. Whole numbers are associative for division
3. Integers are associative for division
4. Rational numbers are not associative for division

Solution: 4. Rational numbers are not associative for division

Question 13. 0 is not

1. A natural number
2. A whole number
3. An integer
4. A rational number.

Solution: 1. A natural number

Question 14. ½ is

1. A natural number
2. A whole number
3. A n integer
4. A rational number.

Solution: 4.  A rational number.

Question 15. a + 6 = 6 + a is called

1. Commutative law of addition
2. Associative law of addition
3. Distributive law of addition
4. None of these.

Solution: 1. Commutative law of addition

Question 16. a × b = b × a is called

1. Commutative law for addition
2. Commutative law for multiplication
3. Associative law for addition
4. Associative law for multiplication.

Solution: 2. Commutative law for multiplication

Question 17. (a + b) + c = a + (b + c) is called

1. Commutative law for multiplication
2. Commutative law for
3. Additionassociative law for addition
4. Associative law for multiplication.

Solution: 3. Additionassociative law for addition

Question 18. a × (b × c) = (a × b) × c is called

1. Associative law for addition
2. Associative law for multiplication
3. Commutative law for addition
4. Commutative law for multiplication.

Solution: 2. Associative law for multiplication

Question 19. a (6 + c) = ab + ac is called

1. Commutative law
2. Associative law
3. Distributive law
4. None of these

Solution: 3. Distributive law

Question 20. The additive identity for rational numbers is

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

Solution: 3. 0

Question 21. The multiplicative identity for rational numbers is

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

Solution: 2. 1

Question 22. How many rational numbers are there between any two given rational numbers?

1. Only one
2. Only two
3. Countless
4. Nothing can be said.

Solution: 3. Countless

Question 23. If a and 6 are two continuous rational numbers, then

1. \(\frac{a+b}{2}<a\)
2. \(\frac{a+b}{2}<b\)
3. \(\frac{a+b}{2}=a\)
4. \(\frac{a+b}{2}>b\)

Solution: 2.  \(\frac{a+b}{2}<b\)

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers True Or False

1. a – 0 and 0 – a represent the same rational number where a ≠ 0. – False

2. 0 is a rational number – True

3. Between any two given rational numbers, there are countless rational numbers – True

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Fill In The Blanks

1. A number of the form p/q, where p and q are integers and q ≠0 is called a → Rational number

2. Write the number 0 reduced by 1 → (-1)

3. Are the two rational numbers p/q and q≠0 equivalent    →Yes

4. In the rational number p/q, why q≠0  → Since division by zero is not defined

5. Which rational number on the number line is equidistant from 0 and -1  → (\(\frac{-1}{2}\)).

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Introduction

Hardy – Ramanujan Number

Look at the following relations :

1729 = 1728 + 1 = 12³ + 1³

1729 = 1000 + 729 = 10³ + 9³

1729 is the smallest Hardy – Ramanujan Number (a number expressed as a sum of two cubes in two different ways is known as a Hardy- Ramanujan Number). There are infinitely many such numbers.

Few are 4104 (2, 16; 9, 15), 13832 (18, 20; 2, 24

Q. Check it with the numbers given in the brackets. 4104 (2, 16; 9, 15) and 13832 (18, 20; 2, 24)

2³ + 16³  = 8 + 4096 = 4104

9³ + 15³ = 729 + 3375 = 4104

18³ + 20³ = 5832 + 8 000 = 13832

2³ + 24³ = 8 + 13824 = 13832

Read and Learn More NCERT Solutions For Class 8 Maths

Cubes

Numbers obtained when a number is multiplied by itself three times are known as cube numbers or perfect cubes.

For example: 1, 8, 27,

The cube of a natural number m is denoted by m3 and is expressed as

m³ = m x m x m.

Thus, 1³ = 1 x 1 x 1 = 1

2³ = 2 × 2× 2 = 8

3³ = 3 ×  3 × 3 = 27, and so on

Question 1. How many cubes of side? 1 cm will \ make a cube of side 2 cm?
Solution:

2 × 2 × 2 = 8 cubes of side 1 cm will make a cube of side 2 cm.

Question 2. How many cubes of side 1 cm will make a cube of side 3 cm?
Solution:

3 × 3 × 3 = 27 cubes of side 1 cm will make a cube of side 3 cm

Question 3. Following are the cubes of the numbers from 11 to 20
Solution:

Question 4. Complete the following cubes of numbers  From 1 to 10

Solution:

Question 5. Are there only ten perfect cubes from 1 to 1000?
Solution:
Yes :

There are only ten perfect cubes from 1 to 1000.

These are: 1, 8, 27, 64, 126, 216, 343, 512, 729 and 1000

Question 6. How many perfect cubes are there from 1 to 100?
Solution:

There are only four perfect cubes from 1 to 100.

 These are: 1, 8, 27, and 64.

Question 7. Observe the cubes of even numbers. Are they all even? What can you say about the cubes of odd numbers?
Solution:

1. The cubes of even numbers are all even.
2. The cubes of odd numbers are all odd.

Question 8. Consider a few numbers having 1 as the one’s digit (or unit’s). Find the cube of each of them. What can you say about the one digit of the cube of a number having 1 as the one’s digit? Similarly, explore the one’s digit of cubes of numbers ending in 2, 3, 4,
Solution:

We observe that the one digit of the cube of a number having 1 as the one’s digit is 1. Similarly, the one’s digit of cubes of numbers ending in 0, 2, 3, 4, 5, 6, 7, 8, 9 are 0, 8, 7, 4, 5, 6, 3, 2, 9 respectively

Question 9. Find the one digit ofthe cube ofeach of the following numbers.

1. 3331
2. 8888
3. 149
4. 1005
5. 1024
6. 77
7. 5022
8. 53

Solution:

1. 3331

Unit digit of the number = 1

Unit digit of the cube of the number = 1

∴  1 × 1 × 1 = \(\underline{1}\)

2.  8888

Unit digit of the number = 8

Unit digit of the cube of the number = 2.

∴  8 × 8 × 8 = 51\(\underline{2}\)

3.  149

Unit digit of the number = 9

Unit digit of the cube of the number = 9

∴  9 × 9 × 9 = 72\(\underline{9}\)

4.  1005

Unit digit of the number = 5

Unit digit of the cube of the number = 5.

∴  5 × 5 × 5 = 12\(\underline{5}\)

5. 1024

Unit digit of the number = 4

Unit digit of the cube of the number = 4

∴  4 × 4 × 4 × 4 = 6\(\underline{4}\)

4. 77

Unit digit of the number = 7

Unit digit of the cube of the number = 3.

∴  7 × 7 × 7 = 34\(\underline{3}\)

8. 5022

Unit digit of the number = 2

Unit digit of the cube of the number = 8

∴  2 × 2 × 2 = \(\underline{8}\)

8. 53

Unit digit of the number = 3

Unit digit of the cube of the number = 7

∴  3 × 3 × 3 = 2\(\underline{7}\)

Some Interesting Patterns

If in the prime factorization of any number, each prime factor appears three times, then the number is a perfect cube.

For example, 216 =\(\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)

= 2³ ×  3³ = (2× 3)³ = 6³ which is a perfect cube.

Question 1. How many consecutive odd numbers will be needed to obtain the sum as 10³?
Solution:  Ten.

Question 2. Express the following numbers as the sum of odd numbers using the above pattern.

1. 6³
2. 8³
3. 7³

Solution: 

1. 6³

Here, n = 6

∴ (n – 1) = 5

We start with n × (n – 1) + 1

= 6× 5 +1

= 31

6³ = 31 + 33 + 35 + 37+ 39 + 41

2. 8³

Here, n = 8

∴ (n – 1) = 7

We start with n × (n – 1) + 1

= 8 × 7  + 1

=  56+1

=  57

8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

3. 7³

Here, n = 7

(n -1) = 6

We start with n × (n – 1) + 1,

7 × 6 + 1,

= 42+1

= 43

7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55

Question 3. Consider the following pattern.

1. 2³ – 1³ = 1 + 2 × 1 × 3
2. 3³ – 2³ = 1 + 3 × 2 × 3
3. 4³ – 3³= 1 + 4 × 3 × 3

Using the above pattern, find the value of the following:

1. 7³ – 6³
2. 12³ – 11³
3. 20³ – 19³
4. 51³  – 50³ 

Sol.

1. 7³ – 6³ 

= 1 + 7 × 6 ×3

=1+126

= 127

2. 12³ – 11³

= 1 + 12 × 11 × 3

=1 + 396

= 397

3.  20³- 19³

T = 1 + 20 × 19 × 3

= 1 + 1140

= 1141

4.  51³ – 50³

1 + 51 × 50 × 3

= 1 + 7650

= 7651

Question 4. Which of the. following are perfect cubes

1. 400
2. 3375
3. 8000
4. 15625
5. 90000
6. 6859
7. 2025
8. 10648

1.  400

By prime factorisation,

Grouping the factors in triplets

400 = \(\underline{2 \times 2 \times 2} \times\)  2× 5 × 5.

= 2³ × 2× 5× 5

Bylaws of exponents

In the above factorization, 2 and 5 × 5 remain after grouping 2’s in triplets.

∴ 400 is not a perfect cube

2.33375

Therefore, 400 is not a perfect cube

By prime factorisation,

3375 = \(\underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 3³× 5³

Bylaws of exponents

= (3 × 5)³

= 15³, which is a perfect cube.

∴  3375 is a perfect cube.

3. 80000

By prime factorisation,

8000 =  \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³× 2³ × 5³

Bylaws of exponents

= (2 × 2 ×5)³

= 20³, which is a perfect cube.

∴   8000 is a perfect cube

4.15625

By prime factorisation,

15625 = \(\underline{5 \times 5 \times 5} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 5³× 5³

Bylaws of exponents

= (5 × 5)³

Bylaws of exponents

25³, which is a perfect cube.

∴  15625 is a perfect cube

5. 9000

By prime factorisation,

9000 = \(\underline{2 \times 2 \times 2} \times\) 3× 3 × \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

In the above factorization, 3×3 remains after grouping 2’s and 5’s in triplets.

∴  9000 is not a perfect cube.

6. 6859

By prime factorisation

6859 = \(\underline{19 \times 19 \times 19}\)

Grouping the factors in triplets

= 19³

Bylaws of exponents which is a perfect cube

∴  6859 is a perfect cube

7. 2025

By prime factorisation,

2025  = \(\underline{3 \times 3 \times 3} \times\) 3 × 5 × 5

Grouping the factors in triplets

= 3³ × 3 × 5 × 5

Bylaws of exponents

In the above factorization, 3 and 5 ×  5 remain after grouping 3’s in triplets.

∴ 2025 is not a perfect cube

8. 10648

By prime factorisation

10648 = \(\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11} \times \)

Grouping the factors in triplets

= 2³ × 11³

Bylaws of exponents

= (2 × 11)³

Bylaws of exponents

= 22³, which is a perfect cube.

∴  10648 is a perfect cube

Smallest Multiple That Is A Perfect Cube

Sometimes we have to find the smallest natural number by which a number is multiplied or divided to make it a perfect cube

Question 1. Check which of the following are perfect cubes

1. 2700
2. 6000
3. 64000
4. 500
5. 125000
6. 50000
7. 21600
8. 10000
9. 27000000
10. 1000

What pattern do you observe in these perfect cubes
Solution:

1. 2700

By prime factorisation,

2700 = 2× 2× \(\underline{3 \times 3 \times 3}\)×  5 ×  5

Grouping the factors in triplets

= 2  × 2 × 3³× 5× 5

Bylaws of exponents

In the above factorization, 2 × 2 and

5× 5 remain after grouping 3’s in triplets.

∴ 2700 is not a perfect cube

2. 16000

By prime factorisation,

16000 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)  × 2 × \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2³ × 2 ×  5³

Bylaws of exponents

In the above factorization, 2 remains after grouping 2’s and 5’s in triplets.

∴ 16000 is not a perfect cube

3. 64000

By prime factorisation,

64000 = \(\underline{2 \times 2 \times 2 } \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ ×2³ × 2³× 5³

By laws of exponents

= (2 × 2 × 2 ×5)³

= 40³, which is a perfect cube.

∴  64000 is a perfect cube

4.  900

By prime factorisation,

900 = 2 ×2 × 3 × 3 × 5 × 5

In the above factorization, 2 × 2, 3 × 3, and 5 × 5 remain when we try to group the factors in triplets.

∴ 900 is not a perfect cube.

5. 125000

By prime factorisation,

125000 = \(\underline{2 \times 2 \times 2 } \times \underline{5 \times 5 \times 5} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 5³ × 5³

Bylaws of exponents

= (2 × 5 × 5)³

= 50³, which is a perfect cube.

∴  125000 is a perfect cube.

6. 36000

By prime factorisation,

3600 = \(\underline{2 \times 2 \times 2 } \times \underline{2 \times 2 } \times \underline{3 \times 3} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2 × 2 × 3 × 3 × 5³

Bylaws of exponents

In the above factorization, 2 × 2 and 3 ×3 remain after grouping 2’s and 5’s in  triplets

∴  36000 is not a perfect cube.

7. 21600

By prime factorisation

21600 = \(\underline{2 \times 2 \times 2 }\) × 2 × 2 ×  \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2 × 2×3³×5 × 5

Bylaws of exponents

In the above factorization, 2×2 and 5×5 remain after grouping 2’s and 3’s in triplets

∴  21600 is not a perfect cube.

8. 10000

By prime factorisation,

1000 =\(\underline{2 \times 2 \times 2}\) ×  2 × \(\underline{5 \times 5 \times 5}\) × 5

Grouping the factors in triplets

= 2³ × 2 × 5³ × 5

Bylaws of exponents

In the above factorization, 2 and 5 remain after grouping 2’s and 5’s in triplets.

∴ 10000 is not a perfect cube

9. 27000000

By prime factorisation,

27000000

=\(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}\times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplet

= 2³×2³ × 3³× 5³ × 5³

Bylaws of exponents

= 300³, which is a perfect cube.

∴  27000000 is a perfect cube.

10. 1000

By prime factorization,

1000 = \( \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 5³

Bylaws of exponents

= (2 × 5)³ I By laws of exponents

= 10³, which is a perfect cube.

Therefore, 1000 is a perfect cube.

We observe the following pattern in these perfect cubes:

1. If in the end of a number, the number of zeros is not 3 or a multiple of 3, then that number cannot be a perfect cube
2. If in the end of a number, the number £ of zeros is 3 or a multiple of 3, then JJJ that number may be a perfect cube.
3. Thus, the number of zero at the end of a perfect cube (if it is so) must essentially be 3 or a multiple of3, failing which the number cannot be a perfect cube.

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Exercise 6.1

Question 1. Which of the following numbers are not perfect cubes?

1. 216
2. 129
3. 1000
4. 100
5. 46656

Solution:

1. 216

By prime factorisation

216 =  \(\underline{2 \times 2 \times 2}\times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³ × 3³

Bylaws of exponents

= (2 × 3)³

= 6³ which is a perfect cube.

∴  216 is a perfect cube

2. 128

By prime factorisation

128 = \(\underline{2 \times 2 \times 2}\times \underline{2 \times 2 \times 2}\) × 2

Grouping the factors in triplets

= 2³ × 2³ × 2

Bylaws of exponents

In the above factorization,2remainsafter grouping the 2 ‘s in triplets

∴ 128 is not a perfect cube

3. 1000

By prime factorisation

= \(\underline{2 \times 2 \times 2}\times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³× 5³

Bylaws of exponents

= (2 × 5)³ = 10³

Bylaws of exponents

Which is a perfect cube.

∴ 1000 is a perfect cube

4. 100

By prime factorisation

100 =  \(\underline{2 \times 2 \times }\times \underline{5 \times 5}\)

In (he above factorization, 2 × 2,5  × 5 remains when we try to group the factors in triplets

Therefore, 100 is not a perfect, cube.

5.46656

By prime factorisation,

46656 = \(\underline{2 \times 2 \times 2 \times 2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\).

Grouping the factors in triplets

= 2³ × 2³×3³ × 3³

Bylaws of exponents

= (2 ×2 × 3 × 3)³

Bylaws of exponents

= 36³, which is a perfect cube.

∴  46656 is a perfect cube

Question 2. Find the smallest number by which each of the following numbers must be multi¬ plied to obtain, a perfect cube

1. 243
2. 256
3. 72
4. 675
5. 100

Solution:

1. 243

By prime factorisation

243 =\(\underline{3 \times 3 \times 3}\) × 3 × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three.

Therefore, 243 is not a perfect cube.   To make it a cube, we need one more 3. In that case

243 = \(\underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\)

= \(3^3 \times 3^3\)

= (3×3)³

729, which is a perfect cube.

Hence, the smallest number by which 243 should be multiplied to obtain a perfect cube is 3.

The resulting perfect cube is 3.

2. 256

By prime factorisation

256 =\(\underline{2 \times 2 \times 2}\underline{2 \times 2 \times 2}\) 2 x 2

Grouping the factors in triplets

In the above factorisation 2× 2  after grouping 2‘s in Iriplols. Therefore, 128 is not a perfect cube. To make it a perfect, cube, we need one 2 more. In that, case

256× 2 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times 2 \times 2\)

= 2³×2³× 2³

Bylaws of exponents

= (2×2×2)³

Bylaws of exponents

8³ = 512, which is a perfect cube.

Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2

The resulting perfect cube is 512 (= 8³)

3. 72

By prime factorisation,

72 = 2 × 2 × 2 × 3 × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three.

Therefore, 72 is not a perfect cube. To make it a perfect cube, we need one more 3. In that case,

72 × 3 = \(\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\)

= 2³ ×3³

Bylaws of exponent

= (2 × 3)³

Bylaws of exponents

6³ = 216 which is a perfect cube.

Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

The resulting perfect cube is 216 (= 6³)

4. 675

By prime factorisation,

675 = 5 × 3 × 3 × 5 × 5

Grouping the factors in triplets

The prime factor 5 does not appear in a group of three. Therefore, 675 is not a perfect cube.

To make it a perfect cube, we need one more 5. In that case

675 ×  5 = \(\underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5} \)

= 3³× 5³

Bylaws of exponents

= (3×5)³

Bylaws of exponents

= 15³

∴ 3375 which is a perfect cube.

Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5.

The resulting perfect cube is 3375 (= 15³)

5. 100

By prime factorisation,

100 = 2 × 2 × 5 × 5

Grouping the factors in triplets

The prime factors 2 and 5 do not appear in a group of three. Therefore, 100 is not a perfect cube.

To make it a perfect cube, we need one 2 and one 5 more. In that case,

100 × 2 × 5 = \(\underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5} \)

= 2³ × 5³

Bylaws of exponents 2

= (2 × 5)³

Bylaws of exponents

= 10³

∴ 1000 which is a perfect cube.

Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is 2× 5 = 10

The resulting perfect is 1000 (= 10³ )

Question 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

1. 81
2. 128
3. 135
4. 192
5. 704

Solution:

1. 81

By prime factorisation

81 =\(\underline{3 \times 3 \times 3}\times\) 3

Grouping the factors in triplets

In the above factorization, 3 remains after grouping the 3’s in triplets. Therefore,

81 is not a perfect cube. If we divide the number by 3, then in the prime factorization of the quotient, this 3 will not remain. In that case,

81 ÷ 3  = \(\underline{3 \times 3 \times 3}\)

= 3³

Bylaws of exponents which is a perfect cube

Hence, the smallest whole number by which 81 must be divided to obtain a perfect cube is 3.

The resulting ideal cube is 27 (= 3³)

2. 128

By prime factorisation,

128 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

Grouping the factors in triplets

In the above factorization, 2 remains after grouping’ the 2’s in triplets.

Therefore, 128 is not a perfect cube. If we divide the number by 2, then in the prime factorization ofthe quotient, this 2 will not remain. In that case,

128 ÷ 2 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

= 2³ × 2³

= (2 × 2)³

Bylaws of exponents

= 4³, which is a perfect cube.

Hence, the smallest whole number by  which 128 must be divided to obtain a perfect cube is 2

The resulting ideal cube is 64 (= 4³)

3. 135

By prime factorisation,

135 = \(\underline{3 \times 3 \times 3} \times 5\)

Grouping the factors in triplets.

The prime factor 5 does not appear in a group of three. So 135 is not a perfect cube

In the factorisation 5 appears only once. If we (u) 704 divide 135 by 5, then the prime factorization of the quotient, will not contain 5.

In that case

135 ÷ 5 = \(\underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 27

Bylaws of exponents, which is a perfect cube.

Hence, the smallest whole number by  which 135 must be divided to obtain a perfect cube is 5

The resulting perfect cube is 27 (= 33)

4. 192

By prime factorisation,

192  = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three. So 192 is not a perfect cube. In the factorization of 192, 3 appears only once. So if we divide the number by 3, then the prime factorization of the quotient will not contain 3.

.In that case,

192 ÷ 3 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 3

= 2³ × 2³

= (2 × 2)³

Bylaws of exponents

= 4³ = 64

Which is a perfect cube.

Hence, the smallest whole number by which 192 must be divided to obtain a perfect cube is 3.

The resulting perfect cube is 64 (= 4³)

5. 704

By prime factorisation,

704  =  \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 11

Grouping the factors in triplets

The prime factor 11 does not appear in the group of three.

So, 704 is not a perfect cube.

In the factorization 11 appears only one time.

So if we divide 704 by 11, then the prime factorization of the quotient will not contain 11. In that case

704 ÷ 11 = 2 × 2 × 2 × 2 × 2 × 2

Grouping the factors in triplets

2³ × 2³

Bylaws of exponents

= (2 × 2)³

= 4³

= 64

Which is a perfect cube.

Hence, the smallest whole number by which 704 must be divided to obtain a perfect cube is 11.

The resulting perfect cube is 64 (= 4³).

Question 4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2  How many such cuboids will he need to form a cube? cm, 5 cm.
Solution:

Volume of a cuboid

= 5 × 2 × 5 cm³

Since there is only one 2 and only two 5’s in the prime factorization, so, we need 2 × 2 × 5, i.e.t 20 to make a perfect cube.

Therefore, we need 20 such cuboids to make a cube.

Cube Roots

The cube root is the inverse operation of finding a cube.

2³ = 8 ⇒  2 is the cube root of 8.

The symbol denotes the cube root. Thus,\(\sqrt[3]{8}=2\)

Cube Root Through Prime Factorisation Method

We express the given number into product of its prime factors and make triplets (groups of three) of similar factors. Then, we take one factor from each triplet and multiply. The product so obtained gives the cube root of the given number.

Question 1. State true or false: for any integer m, m²< m³. Why?
Solution:

False, if m is a negative integer.

True, if m is a positive integer (natural number).

Verification. Let us take a negative integer m =-1. Then,

m²= (- 1)2 = (- 1)² ×  (-1) = 1

m³ = (_ 1)3 = (- 1)³ × (-1) × (-1)

= – 1

Clearly, m² > m³

Hence, the given statement is false if m is a negative integer.

We get the same inference for m = – 2, -3,-4, etc.

Again, let us take a positive integer(natural number) m = 2. Then,

m² = (2)² = 2 × 2 = 4

m³ = (2)³ = 2 × 2 × 2 = 8

Clearly, m² < m³

Hence, the given statement is true if m is a positive integer (natural number).

We get the same inference for m = 3, 4,5… etc

Cube Root of a Cube Number Steps

1. Obtain the given number. Start making groups of three digits starting from the rightmost digit of the number.
2. The first group will give one’s (unit’s) digit of the required cube root.
3.  Then, take another group. Find the two closest cube numbers between which this group lies. Take the one’s place of the smaller number as the ten’s place of the required cube root

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Exercise 6.2

Question 1. Find the cubit root of each of the foUouiu# numbers by prime factorization method :

1. 64
2. 512
3. 10648
4. 27000
5. 15625
6. 13824
7. 110592
8. 46656
9. 175616
10. 91125

Solution:

1. 64

The prime factorization of 64 is

= \(\underline{2 \times 2 \times 2} \times \underline {2\times 2 \times 2}\)

Grouping the factors in triplets

= 2³× 2³

= (2 × 2)³

= 4³

Bylaws of exponents

∴ \(\sqrt[3]{64}\) = 2 × 2

2.512

The prime factorisation of 512 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

Grouping the factors in triplets

= 2³× 2³ × 2³

= (2 × 2 ×2)³

= 8³

Bylaws of exponents

∴ \(\sqrt[3]{512}\) = 2 × 2 ×2

= 8³

3. 10648

Prime factorization of 10648 is

⇒ [late]\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11} [/latex]

Grouping the factors in triplets

= 2³ x 11³

= (2 × 11)³

∴ \(\sqrt[3]{10648}\) = 2 × 11

= 22³

4. 27000

Prime factorisation of 27000 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{3 \times 3  \times 3 } \times \underline{5 \times 5 \times 5 } \times \underline{5 \times 5}\)

Grouping the factors in triplets

= 2³ × 3³ × 5³

= (2 × 3 × 5)³

Bylaws of exponents

= 30³

∴ \(\sqrt[3]{27000}\) = 2×3 × 5

= 30

5. 15625

The prime factorization of 15625 is

⇒ \(\underline{5 \times 5 \times 5} \times \underline{5 \times 5  \times 5 }\)

Grouping the factors in triplets

= 5³ × 5³

= (5 × 5)³

Bylaws of  exponents

= 25³

∴ \(\sqrt[3]{15625}\) = 5 × 5

= 25

6. 13824

Prime factorization of 13824 is

⇒  \(\underline{2 \times 2 \times 2}\times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³ × 2³× 2³ × 3³

=(2× 2 × 2 × 3)³

Bylaws of exponents

= 24³

∴ \(\sqrt[3]{13824}\)  = 2× 2 × 2 × 3

= 24

7. 1110592

Prime factorisation of 110592 is

⇒  \(\underline{2 \times 2 \times 2}\)× \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

Grouping the factors into triplets

= 2³ × 2³ × 2³ × 2³ × 3

= (2 × 2 × 2 × 2 ×3)³

Bylaws of exponents

= 48³

∴ \(\sqrt[3]{110592}\) =2 × 2 × 2 × 2 ×3

8. 46656

Prime factorization of 46656 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³× 2³ × 3³ × 3³

= (2 × 2 × 3 × 3)³

Bylaws of exponents

= 36³

∴ \(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3

=36

9. 175616

Prime factorisation of 175616 is

⇒  \(\underline{2 \times 2 \times 2 \times} \times \underline{2 \times 2 \times 2 \times} \times \underline{2 \times 2 \times 2 \times} \times \underline{7 \times 7 \times 7}\)

Grouping the factors in triplets

= 2³ × 2³ × 2³ × 7³

= (2 × 2 × 2 × 7)³

Bylaws of exponents

= 56³

∴ \(\sqrt[3]{175616}\) = 2 × 2 × 2 × 7

= 56

10. 91125

Prime factorization of 91125 is

⇒ \(\underline{3 \times 3 \times 3 \times} \times \underline{3 \times 3 \times 3 \times} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 3³ ×3³ × 5³

= (3 × 3× 5)³

Bylaws of exponents

= 45³

∴ \(\sqrt[3]{91125}\) =3 × 3× 5

= 45

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Multiple Choice Questions

Question 1. Which of the following numbers is a perfect cube?

1. 125
2. 36
3. 75
4. 100

Solution: 1. 125

125 = 5 × 5 × 5 = 5³

Question 2. Which of the following numbers is a cube number?

1. 1000
2. 400
3. 100
4. 600.

Solution: 1.1000

1000 = 10 × 10 × 10 = 10³

Question 3. Which of the following numbers is not a perfect cube?

1. 1331
2. 512
3. 343
4. 100

Solution: 4. 100

100 = 2 ×2 × 5 ×5

= 2² × 5².

Question 4. Which of the following numbers is not a cube number?

1. 10000
2. 3375
3. 64
4. 729.

Solution: 1. 10000

10000 = 2 × 2 ×2 × 2 ×5 ×5× 5 ×5

= 2⁴ × 5⁴

= 2³× 2 × 5³ × 5

Question 5. The cube of an odd natural number is

1. Even
2. Odd
3. Maybe even, maybe odd
4. Prime number.

Solution: 2. Odd

3 × 3 × 3 = 27 (odd)

Question 6. The cube of an even natural number is

1. Even
2. Odd
3. Maybe even, maybe odd
4. Prime number.
5. Solution: 1. Even

6 × 6 × 6 = 216 (even)

Question Question 7. The one digit of the cube of the number 111 is

1. 1
2. 2
3. 3
4. 9.

Solution: 1. 1

1 × 1 × 1 =  1

Question 8. The one digit of the cube of the number 242 is

1. 2
2. 4
3. 6
4. 8.

Solution: 4. 8

2 × 2 × 2= 8.

Question 9. The one digit of the cube of the number 123 is

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

Solution: 4. 7

3 × 3 × 3 =2\(\underline{7}\)

Question 10. The one’s digit of the cube of the number 144 is

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

Solution: 4. 4

4 × 4 ×4 = 6\(\underline{4}\)

Question 11. The one’s digit of the cube of the number 50 is

1. 1
2. 0
3. 5
4. 4

Solution: 2. 0

0 × 0 × 0 = 0

Question 12. The one digit of the cube of the number 326 is

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

Solution: 3. 3

6 × 6 × 6 = 21\(\underline{6}\)

Question 13. The one digit of the cube of the number 325 is

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

Solution: 2. 5

5× 5× 5= 12\(\underline{5}\)

Question 14. The one digit of the cube of the number 347 is

1. 3
2. 4
3. 7
4. 1

Solution: 1. 3

7×7×7 = 34\(\underline{3}\)

Question 15. The one digit of the cube of the number 68 is

1. 1
2. 2
3. 6
4. 8

Solution: 2. 2

8× 8 ×8 = 51\(\underline{2}\)

Question 16. The one digit of the cube of the number 249 is

1. 2
2. 4
3. 9
4. 1

Solution: 3. 9

9 × 9 × 9 = 72\(\underline{9}\)

Question 17. What is the one’s digit in the cube root of the cube number 1331?

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

Solution: 1. 1

1 × 1 × 1 = 1

Question 18. What is the one digit in the cube root of the cube number 1000000?

1. 0
2. 1
3. 2
4. 9

Solution: 1. 0

0 ×0 × 0 = 0

Question 19. What is the one’s digit in the cube root of the cube number 1728?

1. 1
2. 2
3. 3
4. 9

Solution: 2. 2

2 × 2 × 2 = \(\underline{8}\)

Question 20. What is the one’s digit in the cube root, of the cube number 2197′?

1. 1
2. 2
3. 3
4. 7

Solution: 3. 3

3× 3 × 3 = 2\(\underline{7}\)

Question 21. What is t lie one’s digit, in the cube root, of the cube number 2744?

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

Solution: 4. 4

4 × 4 × 4 = 6\(\underline{4}\)

Question 22. What is the one’s digit in the cube root of the cube number 3375?

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

Solution: 3. 5

5 × 5 × 5 = 12\(\underline{5}\)

Question 23. What is the one’s digit in the cube root of the cube number 4096? 

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

Solution: 2. 6

6 × 6 × 6 = 21\(\underline{6}\).

Question 24. What is the one’s digit in the cube root of the cube number 4913?

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

Solution: 1. 7

7 × 7 ×7 = 34\(\underline{3}\)

Question 25. What is the one’s digit in the cube root of the cube number 5832?

1. 2
2. 4
3. 6
4. 8

Solution: 4. 8

8 × 8 × 8 = 51\(\underline{2}\)

Question 26. What is the one’s digit in the cube root of the cube number 6859?

1. 7
2. 8
3. 9
4. 6

Solution: 3. 9

8 × 8 × 8 = 51\(\underline{2}\)

Question 27. What, is the one’s digit in the cube root of the cube number 8000 ?

1. 0
2. 2
3. 4
4. 8

Solution: 1. 0

0 × 0 × 0 = 0.

Question 28. The number of zeroes at the end of the cube of the number 20 is

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

Solution: 3. 3

Number of zeroes at the end of the number 20 = 1

Number of zeroes at the end of its cube = 3 × 1 = 3

Question 29. The number of zeroes at the end of the cube root of the cube number 1000 is

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

Solution: 1. 1

The number of zeroes at the end of the cube = 3

Number of zeroes at the end of the cube root = \(\frac{3}{3}\)

= 1

Question 30. The number of zeroes at the end of the cube of the number 100 is

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

Solution: 4. 6

Number of zeroes at the end of the number 100 = 2

Number of zeroes at the end of its cube = 3 × 2

= 6

Question 31. The number of zeroes at the end of the cube root Of the cube number 8000000 in

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

Solution: 2. 2

The number of zeroes at the end of the cube = 6

Number of zeroes at the end of the cube root =\(\frac{6}{3}\)

= 2

Question 32. Find the smallest number by which the number 108 must be multiplied to obtain a perfect, cube. ,

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

Solution: 1. 2

108 = 2 × 2 × 3 × 3 × 3

= 2 × 2 × 3³

Question 33. Find the smallest number by which the number 250 must be divided to obtain a perfect cube.

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

Solution: 1. 3

250 = 5 ×5 × 5 × 2 = 5³ × 2

Question 34. Find the smallest number by which the number 72 must be multiplied to obtain a perfect cube.

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

Solution: 2. 3

72 = 2 × 2 × 2 × 3 × 3

= 2³ × 3 × 3

Question 35. Find the smallest number by which the number 375 must be divided to obtain a perfect cube.

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

Solution: 2. 3

375 = 3 × 5 × 5 × 5

= 3× 5³.

Question 36. Find the smallest number by which the number 100 must be multiplied to obtain a perfect cube.

1. 5
2. 2
3. 4
4. 10

Solution: 4. 10

100 = 2 × 2 ×5 × 5.

Question 37. Find the smallest number by which the number 10000 must be divided to obtain a perfect cube.

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

Solution: 3. 10

10000 = 2 × 2 × 2 × 2 ×5×5 × 5 × 5

= 2³ × 2 × 5³× 5

Question 38. Find the smallest number by which the number 200 must be multiplied to obtain a perfect cube.

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

Solution: 3. 5

200 =2 × 2 × 2 × 5 × 5

= 2³ ×5 × 5

Question 39. Find the smallest number by which the number 625 must be divided to obtain a perfect cube.

1. 3
2. 5
3. 25
4. 125

Solution: 2. 5

625 = 5 × 5 × 5 ×5

= 5³ × 5.

Question 40. Kind the smallest number by which (bo number 128 must the multiplied to obtain a perfect cube.

1. 2
2. 4
3. 3
4. 8

Solution: 2. 4

128 =2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ ×2³ × 2

Question 41. Find the smallest number bv which the number 250 must be divided to obtain a perfect cube.

1. 2
2. 4
3. 8
4. 16

Solution: 2. 4

256 = 2 × 2× 2× 2 × 2× 2 ×  2×2

= 2³ × 2³ × 2 × 2.

Question 42. Find the smallest number by which the number 30 must be multiplied to obtain a perfect cube.

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

Solution: 1. 6

36 = 2 × 2 × 3 × 3

Question 43. Find the smallest number by which the number 1290 must be divided to obtain a perfect cube.

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

Solution: 1. 6

1296 = 2 × 2× 2 × 2 × 3 × 3

=  2³× 2 ×3³× 3

Question 44. Find the smallest number by which the number 392 must be multiplied to obtain a perfect cube.

1. 3
2. 5
3.  7
4. 6

Solution: 3.  7

392 = 2 × 2 × 2 × 7× 7

Question 45. Find the smallest number by which the number 2401 must be divided to obtain a perfect cube.

1. 1
2. 0
3. 5
4. 9.

Solution: 1. 1

2401 = 7 × 7 × 7 × 7

= 7³ × 7

Question 46. Find the smallest number by which the number 121 must be multiplied to obtain a perfect cube.

1. 1
2. 9
3. 11
4. 5

Solution: 3.11

121 = 11× 11

Question 47. Find the smallest number by which the number 88 must be divided to obtain a perfect cube.

1. 11
2. 5
3. 7
4. 9

Solution: 1.11

88 = 2 × 2 ×2 × 11

= 2³ ×11

Question 48. The volume of a cube is 64 cm3. The edge of the cube is

1. 4 cm
2. 8 cm
3. 16 cm
4. 6 cm

Solution: 1. 4 cm

Edge = \(\sqrt[3]{64}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2}\)

= \(\sqrt[3]{2^3 \times 2^3}\)

= 2 × 2

= 4.

Question 49. Apala makes a cuboid of plasticine with sides 5 cm, 4 cm, 2 cm. How many such cuboids will be needed to form a cube?

1. 20
2. 25
3. 10
4. 16

Solution: 2. 25

Volume = 5 × 4 × 2

= 5 × 2× 2 × 2

= 5 ×2³

Question 50. Which of the following is false?

1. A Cube of any odd number is odd
2. A perfect cube does not end with two zeroes
3. The cube of a single-digit number may be a single-digit number
4. There is no perfect cube that ends with 8.

Solution: 4.   There is no perfect cube that ends with 8.

1728 = 12 × 12  × 12

= 12³

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube RootsTrue Or False

There are four perfect cubes between 1 and 100 – False

2. Ifa2 ends in 9, then a3 ends in 9  – False

3. Ifa2 ends in 5, then a3 ends in 25  – False

4. 999 is not a perfect cube  – True

5. Cube roots of 1 are + 1 and -1   – False

6. Cube of any odd number is even – False

7. A perfect cube does not end with two zeros – True

8. If the square of a number ends with 5, then its cube ends with 25. – False

⇒  15² = 225, 15³= 3375

9. There is no perfect cube that ends – False

⇒   12³ = 1728

10. The cube of a two-digit number may be a three-digit number – False 

⇒   10³ = 1000, 99³= 970299

11. The cube of a two-digit number may have seven or more digits -False 

⇒  10³ = 1000, 99³ = 970299

12. The cube of a single-digit number may be a single-digit number – True

⇒  1³ = 1; 2³ = 8

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Fill In The Blanks

1. \(\sqrt[3]{x}\)represents the Cube  root of the number x

2. 1 cm³  10³ mm³

3. The cube of a number ending in 9 will end in the digit → 9

4. The cube of 100 will have zeros 6 at the end

5. There are  8 perfect cubes between 1 and 1000.

6. Find the value of \(\sqrt[3]{125} \times \sqrt[3]{-64}\) → 20

7. Find the least number by which 72 should be multiplied to make it a perfect cube → 3

8. Find the least number by which 72 should be divided to make it a perfect cube → 9

9. Find(1.1)³→ 1.331

10. Find the number whose cube is 64000 →  40

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Recalling Ratios And Percentages Important Points

1. We usually compare two quantities by division, i.e., by using fractions. Comparison by division is called ratio.

Note that two quantities can be compared only when they have the same units. A ratio has no unit.

However, if the two quantities are not in the same units, we convert them into the same units before comparing.

2. Two quantities can also be compared using percentages. By percentage, we mean a fraction where the denominator is 100. The numerator of the fraction is called the rate percent.

For example:  \(\frac{1}{5}\) means 5%. The symbol % is often used to express ‘percent’ (p. c.).

3. To convert the ratio into a percentage, we convert it into a fraction whose denominator is 100. [or we multiply by 100 and employ the % sign.]

4. To convert percentages in \(\frac{5}{100}}\) to fractions, we divide the numerator by 100 and express it in the lowest form.

For example : 5% =\(\frac{1}{20}\)

Read and Learn More NCERT Solutions For Class 8 Maths

.
5. In the unitary method, we find the value of one unit from the given value of some units and then we find the value of the required number of units.

Question 1. In primary school, the parents were asked about the number of hours they spend per day helping their children to do homework. 90 parents helped for hour to \(\frac{1}{2}\) hours to 1 \(\frac{1}{2}\) . The distribution of parents according to the time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) ho per day; 30% helped for \(\frac{1}{2}\)  an hour to 1 \(\frac{1}{2}\) hours,  50% did not help at all.

Using this, answer the following:

1. How many parents were surveyed?
2. How many said that they did not help
3. How many said that they helped for more than 1\(\frac{1}{2}\) hour?

Solution:

1. According to the question

⇒ \(\frac{30}{100}\) of number of parents surveyed = 90

Number of parents surveyed = 90

Number of parents surveyed

= \(\frac{90 \times 100}{30}\)

= 300

Hence, 300 parents were surveyed.

2. A number of those parents said that they did not help

= 50% of 300

= \(300 \times \frac{50}{100}\)

= 150

Hence, 150 parents said that they did 30% not help.

3. A number of those parents said that they helped for more than 1\(\frac{1}{2}\) hour

= 20% of 300 = \(300 \times \frac{20}{100}\)

= 60

Hence, 60 parents said that they helped for more than 1/[latxe]\frac{1}{2}[/latex] hours

Note: ‘Of’ means multiplication

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.1

Question 1. Find the ratio of the following:

1. The speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.
2. 5 m to 10 km
3. 50 paise to 5

Solution:

1. The speed ofa a cycle 15 km per hour to the speed of a scooter 30 km per hour

= 15 km per hour: 30 km per hour

= 15: 30

= \(\frac{15}{30}=\frac{1}{2}\) Or 1:2

2. 5 m to 10 km

10 km = 10 × 1000 m = 10000 m

∴Ratio of 5 m to 10 km

= 5 m: 10000 m

= 5: 10000

⇒ \(\frac{5}{10000}=\frac{1}{2000}\)

= 1:20000

3. 50 paise to ₹ 5

₹ 5 = 5× 100 = 500 paise

A Ratio of 50 paise to ₹ 5

= 50 paise : 500 paise

= 50: 500

= \(\frac{50}{500}=\frac{1}{10}\) Or 1:10

Question 2. Convert the following ratios to percentages

1. 3:4
2. 2: 3.

Solution:

1. 3: 4

= \(\frac{3}{4}=\frac{3}{4} \times \frac{25}{25}\)

Making denominator 100

= \(\frac{75}{100}=75 \%\)

Aliter:

⇒  \(3: 4=\frac{3}{4} \times 100 \%\)

= 75 %

2. 2:3

= \(\frac{2}{3}=\frac{2}{3} \times \frac{100}{100}\)

= \(66 \frac{2}{3} \%\)

Aliter:

2: 3 = \(\frac{2}{3} \times 100 \%\)

= \(\frac{200}{3} \%=66 \frac{2}{3} \%\)

Question 3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?
Solution:

Total number of students = 25

Percentage of students interested in Mathematics s = 72%

Percentage of students who are not interested in Mathematics

= (100 – 72)% = 28%

Number of those students who are not interested in Mathematics

= 28% of 25

= \(\frac{28}{100} \times 25\)

= 7

Hence, 7 students are not interested in or 1: 10 Mathematics.

Question 4. A football team won 10 matches out ofthe total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all?
Solution:

If 40 matches were won, then the total number of matches played = 100

∴ If 1 match was won, then the total number of matches played

=\(\frac{100}{40}\)

If 10 matches were won, then the total number of matches played

= \(\frac{100}{40} \times 10\)

= 25

Hence, they played 25 matches in all.

Aliter:

According to the question,

40% of (total number of matches) = 10

⇒  \(\frac{40}{100} \times \text { (total number of matches) }\)=10

Total number of matches = \(\frac{10 \times 100}{40}\) = 25

Hence, they played 25 matches in all.

Question 5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning ?’
Solution:

Percentage of money spent = 75%

Percentage of money left = (100- 75)% = 25%

∴ If Chameli had ₹ 25 left, then the money she had in the beginning

= 100

∴  If Chameli had ₹1 left, then the money she had in the beginning

= \(₹ \frac{100}{25}\)

∴ If Chameli has v600 left, then the money she had in the beginning

= \(₹ \frac{100}{25} \times 600\)

=  ₹ 24000

Hence, the money she had in the beginning was ₹ 2400.

Aliter :

According to this question,

25% of total money = ₹ 600

⇒ \(\frac{25}{100}\) of total money = ₹ 600

⇒  \(₹ \frac{600 \times 100}{25}\)

= ₹ 2400.

Hence, the money she had in the beginning was? 2400

Question 6. If 60% of people in a city like cricket, 30% like football and, the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
Solution:

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

∴Percentage of people who like other games

= [100 -(60 +30)]%

= (100-90)% = 10%.

Total number of people = 50 lakh

= 50,00,000

∴ Number of people who like cricket

= 60% of 50,00,000

= \(50,00,000 \times \frac{60}{100}\)

= 30,00,000 = 30 lakh

Number of people who like football

= 30% of 50,00,000

= \(50,00,000 \times \frac{30}{100}\)

Number of people who like the other games

= 10% of 50,00,000

= \(50,00,000 \times \frac{10}{100}\)

= 5,00,000 = 5 lakh

Finding Discounts

Discount = Marked Price – Sale Price

Discount per cent = \(\frac{\text { Discount }}{\text { Marked Price }} \times 100 \%\)

Question 1. 4 shop gives 20% discount. What Would the sale price ofeach of these be?

1. A dress marked at.₹ 120
2. A pair of shoes marked at ₹ 750
3. A bag marked at₹ 250.

Solution:

1. A dress marked at ₹120

Marked price ofthe dress = ₹120

Discount rate= 20%
,
∴ Discount = 20% of ₹ 120

20 MarkedDiscountprice × 100

=\(₹ \frac{20}{100} \times 120\)

= ₹ 24

∴ Sale price ofthe dress x 100 % = 4%.

= Marked price – Discount

= ₹ 120 – ₹ 24 = ₹ 96

2. A pair of shoes marked at ₹ 750

Marked price ofthe pair ofshoes= ₹ 750

Discount rate = 20%

Discounts 20% of ₹ 750

= \(₹ \frac{20}{100} \times 750\)

= ₹150

∴ Sale price ofthe pair of shoes

= Marked price – Discount

= ₹ 750 -₹ 150

= ₹600

3. A bag marked at ₹ 250

The marked price of the bag =₹ 250

Discount rate = 20%

Discount = 20% of  ₹ 250

= \(₹ \frac{20}{100} \times 250\)

= ₹50

∴ The sale price of the bag

= Marked price – Discount

=₹250 – ₹ 50 =₹200.

Question 2. If the table marked at f 15,000 is * available for f14, 400. Find the discount given £ and the discount percent.
Solution:

Marked price ofthe table =₹ 15,000

Sale price ofthe table =  ₹ 14,400

∴  Discount given

= Marked price – Sale price

= ₹ 15,000 – ₹14,400

= ₹ 600

∴  Discount per cent

= \(\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \% \)

= \(\left(\frac{600}{15000} \times 100\right) \%=4 \%\)

Question 3. An almirah is sold at ₹5,225 after allowing a discount of 5%. Find its marked price.
Solution:

Let the marked price bet x.

Discount rate = 5%

Discount = 5% of ₹ x

=\(₹ \frac{5}{100} \times x=₹ \frac{x}{20}\)

∴ Sale price

= Marked price -Discount

= \(₹ x-₹ \frac{x}{20}=₹ \frac{19 x}{20}\)

According to the question,

∴ \(\frac{19 x}{20}=5225\)

x = \(\frac{5225 \times 20}{19}\)

x = ₹ 5500

Hence, the marked price of the almirah is ₹5500

Question 4. Try estimating 20% of the same bill amount.

1. Round of the bill to the nearest tens.
2. Find the amount of discount.
3. Reduce the bill amount by the discount amount

Solution:

Your bill in a shop is ₹  577.80.

1. Round off the bill to the nearest tens of ₹ 577.80, i.e., to ₹ 580.

2. Find 20% of this,

⇒  \(₹ \frac{20}{100} \times 580\)

= ₹  116

3. Therefore estimated 20% of the same bill amount is ₹116 or ₹120 (rounded off to the nearest tens).

Question 5. Try finding 15% of  ₹ 375.
Solution:

Find 10% of ₹ 375, i.e.

⇒ ₹\(₹ \frac{10}{100} \times 375\)

= ₹ 37.5

= ₹  40

(rounded off to the nearest tens)

2. Take half of this, i.e \(\frac{1}{2} \times ₹ 40\)

= ₹ 20

(3) Add (2) and (1) to get ₹ 60.

Therefore, an estimated 15% of 7 375 is

= 60

Sales Tax/Value Added Tax/Goods And Services Tax

Sales tax was charged at a specified rate on the sale price of an item by the state government and was added to the bill amount. It is different for different items and also for different states.

• Amount of Sales Tax = Tax% of bill amount
• These days, the prices include the tax known as Value Added Tax (VAT).
• From July 1, 2017, the Government of India introduced GST which stands for Goods and Services Tax which is levied on the supply of goods or services or both

Question 1. Find the buying price of each of the following when 5%STis added to the purchase

1. A towel at ₹ 50
2. Two bars of soap at ₹ 35 each
3. 5 kg of flour at ₹15 per kg.

Solution:

1. Cost of the towel

Rate of ST = 5%

∴ ST =5% of 7 50

=₹ \(\frac{5}{100} \times 50\)

= ₹ 2.50

∴ Buying price of the towel

= Cost of the towel + ST

= ₹ 50 + ₹ 2.50 =₹ 52.50.

2. The cost of two bars of soap

= ₹35 × 2 = ₹ 70

Rate of  ST= 5%

ST = 5% of ₹ 70

= \(₹ \frac{5}{100} \times 70\)

= ₹ 3.50

∴ Buying price of two bars of soap

= Cost of two bars of soap + ST

= ₹ 70 +₹ 3.50 = ₹ 73.50

3. Cost of 5 kg of flour

= ₹ 15 × 5 = ₹ 75

Rate of ST = 5%

ST = 5% of ₹ 75

= \(₹ \frac{5}{100} \times 75\)

= ₹ 3.75

∴ The buying price of 5 kg of flour + ST

= Cost of 5 kg of flour + ST

= ₹ 75 +₹ 3.75

=₹ 78.75.

Question 2. If 8% is included in the prices, find the original price of

1. A TV bought for ₹13,500
2. A shampoo bottle was bought for ₹ 180.
   

Solution:

1. When the price of a TV including

VAT is ₹108,

original price = ₹ 100

When the price of the TV including VAT is

₹ 1, original price = ₹ \( \frac{100}{108}\)

The price of a TV including VAT is ₹ 13500,

Original price = \(₹ \frac{100}{108} .\)

= ₹ 12500

Hence, the original price of the TV Is ₹ 12500.

Aliter:

Price of TV including VAT

= ₹ 13500

Rate of VAT = 8%

Let, the original price of the TV be ₹x

Then,

x+8 %  of x =₹ 13500

⇒ \(x+\frac{8}{100} x =₹ 13500 \)

⇒ 1+ \(\frac{8}{100}\)x =₹ 13500

⇒  \(\frac{108}{100} x =₹ 13500\)

= ₹ \(\frac{13500 \times 100}{108}\)

Hence, the original price of the T.V. is ₹12500.

2. When the price of a shampoo bottle including VAT is ₹ 108,

Original price = ₹ 100

When the price of a shampoo bottle including VAT is ₹1, the original price

= \(₹ \frac{100}{108} .\)

∴ The price of a shampoo bottle including VAT is? 180,

Original price = \(=₹ \frac{100}{108} \times 180 \)

⇒ \(₹ \frac{100 \times 5}{3}=₹ \frac{500}{3}\)

=\(₹ 166 \frac{2}{3}\)

Hence, the original price ofthe shampoo Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)

Alitor: Price of shampoo bot t lo including

VAT = ₹ 180

Rate of VAT = 8%

Let the original price of shampoo bottle be ₹x

Then,

x+ 8%of x = ₹180

⇒ \(x+\frac{8}{100} x=₹ 180\)

⇒  \(\left(1+\frac{8}{100}\right) x=₹ 180\)

⇒  \(\frac{108}{100} x=₹ 180\)

x =\(₹ \frac{180 \times 100}{108}\)

x = \(₹ \frac{500}{3}=₹ 166 \frac{2}{3}\)

Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.2

Question 1. During a sale, a shop offered a discount of 10% on the marked prices of all the items. W7iat would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at f850 each
Solution:

For a pair of jeans

Marked price = ₹ 1450

Discount rate = 10%

Discount = 10% of marked price

= 10% of  ₹ 1450

= \(₹ \frac{10}{100} \times 1450\)

= ₹ 145

Sale price – Marked price – Discount

= ₹1450-₹ 145

= ₹ 1305

For two shirts

Marked price =₹ 850 × 2 = ₹ 1700

Discount rate = 10%

Discount = 10% of marked price

= 10% of  ₹ 1700

=₹ \(\frac{10}{100} \times 1700=₹ 170\)

Sale price

= Marked price – Discount

= ₹1700 – ₹70 =₹1530

Total payment made by the customer

= ₹ 1305 + ₹ 1530 =₹ 2835

Hence, the customer will have to pay. 2835 for a pair of jeans and two shirts.

Question 2. The price ofa W is f13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it
Solution: 

Price of TV = ₹ 13,000

Sales tax charged on it

= 12% of ₹ 13,000

= \(₹ \frac{12}{100} \times 13,000\)

= ₹ 1560

Amount to be paid

= Price + Sales Tax

= ₹ 13,000 + ₹ 1,560

= ₹14,560.

Hence, the amount that Vinod will haveto pay for the TV if he buys it is ₹14,560

Question 3. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is  ₹ 1600, find the marked price
Solution:

Let the marked price of the pair of 6  shoes be ₹ 100.

Kate of discount = 20

Discount = 20% of ₹ 100

= ₹ \(\frac{12}{100} \times 100\)

=₹ 20

Amount paid = Marked price -Discount

= ₹ 100 – ₹20 – = ₹ 80

If the amount paid is ₹80,

Then the marked price – \(₹ \frac{100}{80}\)

If the amount paid is ₹ 1, then the marked price = \(₹ \frac{100}{80} \times 1600\)

= ₹ 2000

Hence, the marked price of the pair of shoes is ₹ 2000.
.
Question 4. I purchased a hair dryer for f 5,400 including 8% VAT. Find the price before it was added,
Solution:

Price of hair-dryer including VAT

=  ₹ 5400

VAT rate = 8%

Let the original price be ₹ 100

∴ VAT = ₹ 8

Price including VAT = Original price + VAT

= ₹100 + ₹ 8

= ₹ 108

If the price including VAT is ₹108, then

Original price =  ₹ 100

If the price including VAT is ₹ 1, then

Orginalprice = \(₹ \frac{100}{108}\)

If the price including VAT is  ₹5,400, then

Original price = ₹ \(\frac{100}{108} \times 5,400\)

= ₹ 5000

Hence, ‘the price before VAT was added is ₹5,000

Question 5. An article was purchased for ₹1239 including a GST of 18%. Find the price of the article before GST was added,
Solution:

Let the original price of the article be 100. GST Rate = 18% ,

Price after GST is included

= ₹(100 + 18) = ₹118

If the selling price is? 118 then original price = ₹100

If the selling price is ₹ 1, then

Original Price = \(₹ \frac{100}{108} \times 5,400\)

If the selling price is ₹ 1239, then

original price = \(₹ \frac{100}{118} \times 1239\)

= 100 × 10.5

= ₹ 1050

Hence, the price ofthe article before GST was added is? 1050

Compound Interest

1. Interest:

Interest is the extra money paid by institutions like banks or post offices on money that is deposited (kept) with them. Interest is also paid by people when they borrow money. The money deposited or borrowed is called the principal. Interest is generally given in percent for one year.

2. Simple interest (SI):

The interest is called simple when the principal does not change.

3. The formula for simple interest:

Simple interest on a principal of 1 P at R% rate of interest per year for T years is given by

Simple Interest = \(\frac{\text { Principal } \times \text { Rate } \times \text { Time }}{100}\)

SI =  \(\frac{\text { P R T }}{100}\)

4. Amount :

Amount (A)= Principal (P) + Simple Interest(SI)

Question 1.   Find interest and amount to be paid on 115,000 at 5% per annum after 2 years
Solution:

Interest on₹ 100 for 1 year is = 5

Interest on ₹ for 1 year is

= \(₹ \frac{5}{100}\)

Interest On ₹ for 1 year is

= \(₹ \frac{5}{100} \times 15,000\)

Interest on ₹ 15000 for 2 years is

= \(₹ \frac{5 \times 15,000 \times 2}{100}\)

= ₹  1,500

∴ Amount = Principal + Interest

=  ₹ 15,000 + ₹ 1,500

= ₹ 16,500

Aliter:

Here, P = ₹ 15,000

R = 5% per annum

T = 2 years

∴ \(\text { S.I. }=\frac{\text { PRT }}{1000}\)

= \(₹ \frac{15000 \times 5 \times 2}{100}\)

₹ 1500

Amount = Principal (P) + SimpleInterest (SI)

= ₹ 15000 + ₹ 1500

= ₹ 1,500 = ₹ 16500

Deducing A Formula For Compound Interest

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

where P = Principal

R = Rate of interest per annum compounded annually

n = Number of years

A = Amount

CI = A-P

Question 1. Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.
Solution:

Here,

P =₹ 8000

R = 5%p.a.

n = 2 years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

= \(8000\left(1+\frac{5}{100}\right)^2\)

= \(8000\left(1+\frac{1}{20}\right)^2\)

= \(8000\left(\frac{21}{20}\right)^2 \)

= \(8000 \times \frac{21}{20} \times \frac{21}{20}\)

= 20 × 21 × 21

=₹ 8820

Cl = A – P

= ₹ 8820 – ₹ 8000

= ₹ 820.

∴ Hence, the compound interest is ₹ 820

Applications Of Compound Interest Formula

We use the compound interest formula to find

1. Increase (or decrease) in population.
2. The growth of a bacteria if the rate of growth is known.
3. The value of an item, if its price increases or decreases in the intermediate years.

Note: For increase, R is positive and for decrease, R is negative.

Question 1. A machinery worth f 10,500 depreciated by 5%. Find its value after one year.
Solution:

P = ₹ 10,500

R = -5% per annum

There is depreciation

n = year

A = \(P\left(1-\frac{R}{100}\right)^n \)

= \(10,500\left(1-\frac{5}{100}\right)\)

= \(10,500\left(1-\frac{1}{20}\right) \)

= \(10,500 \times \frac{19}{20}\)

= 9,975

Hence value after 1 year is ₹ 9,975.

Question 2. Find the population ofa city after 2 years, which is at present 12 lacks, if the rate of increase is 4%.
Solution:

P = 12,00,000

R = 4% per annum

n = 2years

A = \(P\left(1+\frac{R}{100}\right)^n\)

= \(12,00,000\left(1+\frac{4}{100}\right)^2 \)

= \(12,00,000\left(1+\frac{1}{25}\right)^2\)

= \(12,00,000\left(\frac{26}{25}\right)^2 \)

= \(12,00,000 \times \frac{26}{25} \times \frac{26}{25} \)

= 12,97,920

Hence, the population of the city after 2 years is 12,97,920

Question 3. Calculate the amount and compound interest on

1. 10,800 for 3 years at \(2 \frac{1}{2} \)% perannum compounded annually.
2. ₹ 18,000 for  \(2 \frac{1}{2} \)% years at  perannum compounded annually.
3. ₹  62,500 for \(1 \frac{1}{2} \)% years at  perannum compounded half-yearly.
4. ₹  8,000 for 1 year at   9% per annum compounded half yearly.You could use the year-by-year calculation to verify).
5. ₹  10,000 for a year at 8% per annum compounded halfyearly

Solution:

1. By using year-by-year calculation

Here, P =₹ 10800

R = ₹ 12% per annum

T = ₹ year

SI on ₹ 10,800 at \(12 \frac{1}{2}\) per annum for

= \(10,800 \times \frac{25}{2} \times \frac{1}{100}\)

= ₹ 1,350

∴ Amount at the end of 1st year

= ₹ 10,800 + ₹ 1,350 (A = P + SI)

= ₹ 12,150

= Principal for 2nd year.

SI on ₹ 12,150 at \(12 \frac{1}{2} \% \)per annum

For 1 year = \(\frac{\text { PRT }}{100}\)

=’\(12,150 \times \frac{25}{2} \times \frac{1}{100}\)

= ₹  1,518.75

∴Amount at the end of the 2nd year

= ₹ 12,150+ ₹ 1,518.75

= ₹ 13,608.75

= Principal for 3rd year

SI on  ₹ 13,668.75 at 12\(\frac{1}{2}\)% per annum

For 1year =\(\frac{P R T}{100}\)

=\(13,668.75 \times \frac{25}{2} \times \frac{1}{100} \)

=\(13,668.75 \times \frac{1}{8}\)

= ₹1,708.59

Amount at the end of the year

= ₹13,668.75 + ₹ 1,708.59

=₹ 15,377.34

This is the required amount. Now, CI = ₹15,377.34 – ₹ 10,800

= ₹ 4,577.34

By using the compound interest formula

P =  ₹ 10,800

R = 12\(\frac{1}{2}\) per annum

= \(\frac{25}{2}\)% per annum

n = 3 years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

= \(10,800\left(1+\frac{25}{2 \times 100}\right)^3 \)

= \(10,800\left(1+\frac{1}{8}\right)^3\)

= \(10,800\left(\frac{9}{8}\right)^3\)

= \(10,800 \times \frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\)

= ₹  15,377.34

2. By using year-by-year calculation

Here

P = ₹ 18000

R = ₹ 10% per annum

T = ₹ year

SI on ₹ 18,000 at 10% p.a. for 1 year

= \(\frac{\mathrm{PRT}}{100}\)

= \(\frac{18,000 \times 10 \times 1}{100}\)

= ₹ 1,800

∴ Amount at the end of the year

= ₹ 18,000 + ₹ 1,800

= ₹ 19,800

= Principal for 2nd year

SI on ₹ 19,800 at 10% p.a. for 1 year

= \(\frac{19,800 \times 10 \times 1}{100}\)

= ₹ 1,980

Amount at the end of 2nd year

= ₹ 19,800 + ₹1,980

= ₹ 21,780

= Principal for 3rd year

SI on ₹ 21,780 at 10% p.a. for \(\frac{1}{2}\)year

⇒ \(\frac{\mathrm{PRT}}{100}=\frac{21,780 \times 10 \times 1}{2 \times 100}\)

= ₹1089

∴ Amount at the end of \(2 \frac{1}{2}\) year

= ₹ 21,780 +₹1,089

= ₹ 22,869

This is the required amount.

Now, CI = ₹ 22,869 – ₹ 18,000

= ₹ 4,869

By using the compound interest formula

P = ₹ 18,000

R = 10% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n \)

= \(18,000\left(1+\frac{10}{100}\right)^2 \)

= \(18,000\left(1+\frac{1}{10}\right)^2 \)

= \(18,000\left(\frac{11}{10}\right)^2\)

=18,000  × \(\frac{11}{10} \times \frac{11}{10}\)

=₹ 21,780

SI on? 21,780 at 10% p.a. for \(\frac{1}{2}\) year

=\(\frac{21,780 \times 10 \times 1}{2 \times 100}\)

= ₹ 1089

Amount at the end of \(\frac{1}{2}\) years

= ₹21,780 + ₹ 1,089

= ₹ 22,869

Cl = A- P

= ₹22,869- ₹ 18,000

= ₹ 4,869

3. By using half-year by half-year calculation

Here, P = ₹ 62500

R = 8% per annum

= \(\frac{8}{2}\) per half-year

T = 1 year

∴ SI on ₹ 62,500 at 8% p.a. for 1st half year

= \(\frac{P R T}{100} \)

= \(\frac{62,500 \times 8 \times 1}{2 \times 100}=\frac{62500}{25}\)

= ₹ 2,500

∴  Amount at the end of 1st half-year

= ₹ 62,500 +₹ 2,500

= ₹65,000

= Principal for 2nd half year ₹ 65,000 at 8% p.a. for 2nd half-year

= \(\frac{P R T}{100}\)

= \(\frac{65,000 \times 8 \times 1}{2 \times 100}\)

= ₹2,600

Amount at the end of 2nd halfyear

= ₹ 65,000 + ₹ 2,600

= ₹ 67,600

= Principal for 3rd halfyear

SI on ₹ 67,600 at 8% p.a. for 3rd half-year

= \(\frac{P R T}{100} \)

⇒ \(\frac{67,600 \times 8 \times 1}{2 \times 100}\)

= ₹ 24704

∴ Amount at the end of 3rd half-year

= ₹ 67,600 + ₹ 2,704

= ₹ 70,304

This is the required amount.

Now, Cl = Amount – Principal

= ₹ 70,304 – ₹ 62,500

= ₹7,804

Cl = ₹ 2,500 + ₹ 2,600 +₹ 2,704

= ₹ 7,804

By using the compound interest 2 formula

P =₹ 62,500

= \(\frac{1}{2} \times 8 \% \text { per half year } \)

= 4% per half-year

n = \(1 \frac{1}{2} \text { year } \)

= \(1 \frac{1}{2} \times 2 \text { half years }\)

3 half years

A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

= \(62,500\left(1+\frac{4}{100}\right)^3 \)

= \(62,500\left(1+\frac{1}{25}\right)^3 \)

= \(62,500\left(\frac{26}{25}\right)^3 \)

= \(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)

= \(4\times 26 \times 676\)

= 4 × 26 × 676

= ₹ 70,304

Cl = A- P = ₹ 70,304- ₹ 62,500

=₹ 7804.

By using the compound interest formula

P =₹  62,500

R = 8%p.a

= \(\frac{1}{2} \times 8 \% \text { per half year }\)

1% per half-year

n =\(\frac{1}{2}\)

= 1 \(\frac{1}{2} \times 2\)

= 3 half years

A = \(P\left(1+\frac{R}{100}\right)^n\)

= \(62,500\left(1+\frac{4}{100}\right)^3 \)

= \(62,500\left(1+\frac{1}{25}\right)^3\)

= \(62,500\left(\frac{26}{25}\right)^3\)

=\(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)

4. By using the compound interest formula

P = ₹ 8,000

R = 9% p.a.

= \(\frac{9}{2}\) % per half-year

= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)

= \(8,000\left(1+\frac{9}{200}\right)^2\)

=\(8,000\left(\frac{209}{200}\right)^2 \)

= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)

\(=\frac{209 \times 209}{5}\)

= 4 ×  26× 676

= ₹  70,304

Cl = A- P = ₹ 70,304 – ₹ 62,500

= ₹ 7804

By using compound interest formula

P=8000

R=9%p.a

=\(\frac{9}{2} \%\)per halfyear

n=1 year = 1 × 2 half-year

= 2 half years

∴ A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)

= \(8,000\left(1+\frac{9}{200}\right)^2 \)

= \(8,000\left(\frac{209}{200}\right)^2\)

= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)

=\(\frac{209 \times 209}{5}\)

= ₹8,736.20

Cl = A-P

= ₹ 8,736.20 – ₹8,000

=₹ 736.20

5. By using the compound interest formula

P = ₹ 10,000

= 8% per annum

= \(\frac{8}{2}\) per half-year

= 4 % per half-year

n = year = 1 × 2 half years

= 2 half years

= \(P\left(1+\frac{R}{100}\right)^n \)

= \(10,000\left(1+\frac{4}{100}\right)^2 \)

=\(10,000\left(1+\frac{1}{25}\right)^2 \)

= \(10,000 \times \frac{26}{25} \times \frac{26}{25}\)

16 × 26 × 26

= ₹ 10,816

∴ Cl = A-P

= ₹ 10,816 – ₹ 10,000

= ₹ 816

Question 4.  Kamala borrowed ₹26,400from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of the year and 4 months to clear the loan 

(Hint: Find A for 2 years, interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\)
Solution:

P = ₹ 26,400

R = 15% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n \)

= \(26,400\left(1+\frac{15}{100}\right)^2 \)

= \(26,400\left(1+\frac{3}{20}\right)^2 \)

= \(26,400\left(\frac{23}{20}\right)^2\)

= \(26,400 \times \frac{23}{20} \times \frac{23}{20}\)

= \(66 \times 23 \times 23 \)

= ₹ 34,914

S.I. on X 34,914 at 15% p.a. for 4 months

=\(\text { (i.e., } \left.\frac{4}{12} \text { year, i.e., } \frac{1}{3} \text { year }\right) \)

= \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \)

= \(34,914 \times \frac{15}{100} \times \frac{1}{3} \)

= \(\frac{34,914 \times 15 \times 1}{3 \times 100}\)

= 1,745.70

Required amount

= ₹ 34,914 +₹ 1,745.70

= ₹ 36,659.70

Hence, the amount that Kavita will pay is ₹ 36,659.70.

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3

Question 1. The population of the place increased to 54,000 in 2003 at a rate of 5% per annum. Find the population in 2001. What, would be its population in 2005?
Solution:

1. Let the population in 2001 be P.

R = 5% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n=P\left(1+\frac{5}{100}\right)^2 \)

= \(P\left(1+\frac{1}{20}\right)^2=P\left(\frac{21}{20}\right)^2\)

According to the question

⇒ \(\mathrm{P}\left(\frac{21}{20}\right)^2=54,000\)

P = \(54,000\left(\frac{20}{21}\right)^2\)

= \(54,000 \times \frac{20}{21} \times \frac{20}{21} \)

= 48,980(approx.)

Hence, the population in 2001 was 48,980 (approx.)

Initial population in 2003

(P) = 54,000

R = 5% p.a.

n = 2 years (2005 – 2003 = 2)

A=\(\left(1+\frac{R}{100}\right)^n\)

= \(54,000 \times \frac{20}{21} \times \frac{20}{21}\)

= \(54,000\left(1+\frac{1}{20}\right)^2\)

= \(54,000\left(\frac{21}{20}\right)^2\)

= \(54,000 \times \frac{21}{20} \times \frac{21}{20}\)

=\(135 \times 21 \times 21=59,535\)

Hence, the population in 2005 would be 59,535. .1

Question 2. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of2 hours, if the count was initially 5,06,000
Solution:

The initial count of bacteria

(P) = 5,06,000

R = 2.5% per hour

n = 2 hours

=\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

=\(5,06,000\left(1+\frac{2.5}{100}\right)^2 \)

= \(5,06,000\left(1+\frac{1}{40}\right)^2 \)

= \(5,06,000\left(\frac{41}{40}\right)^2 \)

= \(5,06,000 \times \frac{41}{40} \times \frac{41}{40} \)

= \(\frac{1265}{4} \times 41 \times 41\)

= 531616.25

= 5,31,616 (approx.)

Hence, the bacteria count at the end UJ of 2 hours is 5,31,616 (approx.).

Question 3. A scooter was bought at ₹42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:

Initial value of the scooter

P =₹ 42,000

R = -8% per annum

I v There is depreciation

A = \(P\left(1-\frac{R}{100}\right)^n \)

=\(42,000\left(1-\frac{8}{100}\right)^1\)

= \(42,000\left(1-\frac{2}{25}\right)\)

= \(42,000 \times \frac{23}{25}\)

= \(1680 \times 23=₹ 38,640\)

= 1680 × 23 = ₹ 38,640

Hence, its value after 1 year is ₹ 38,640.

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Multiple Choice Questions

Question 1. The ratio of 50 paise to  ₹ 1 is

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

Solution: 1. 1:2

1 = 100 paise

50paise : 100 paise or 50 : 100 or 1:2

Question 2. The ratio of 10 m to 1 km is

1. 1:10
2. 10:1
3. 1:100
4. 100 :1.

Solution: 3.1:100

1 km = 1000 m

10 m: 1000 m or 10: 1000 or 1: 100

Question 3. The ratio of 10 km per hour to 30 km per hour is

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

Solution: 3. 1: 3

10 : 30 = 1:3

Question 4. The ratio 1: 4 converted to percentage is

1. 50%
2. 25%
3. 75%
4. 4%.

Solution: 2. 25%

1: 4 = \(\frac{1}{4} \times 100 \%=25 \% \text {. }\)

Question 5. The ratio 4:25 converted to percentage is

1. 8%
2. 4%
3. 16%
4. 25%.

Solution: 3. 16%

4: 25 = \(\frac{4}{25} \times 100 \%=16 \%\)

Question 6. The fraction ^ converted to a percentage is

1. 20%
2. 30%
3. 40%
4. 50%.

Solution: 3.40%

⇒ \(\frac{2}{5}\) = \(\frac{2}{5} \times 100 \%=40 \% \text {. }\)

Question 7. The fraction\(\frac{1}{8}\) converted to percentage

1. 8
2. 12%
3. 25%
4. 8%
5. 16%.

Solution: 1. 8

⇒ \(\frac{1}{8}=\frac{1}{8} \times 100 \%=12 \frac{1}{2} \%\)

Question 8. Out of 40 students in a class, 25% passed. How many students passed?

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

Solution: 2.10

Out of 100, passed = 25

Out of40, passed =\(\frac{25}{100} \times 40\)

⇒ \(25 \% \text { of } 40=\frac{25}{100} \times 40\)

= 10

Question 9. Out of 100 students in a class, 30% like to watch T.V. How many students like to watch T.V.?

1. 70
2. 50
3. 60
4. 30.

Solution: 4.30

Number of students who like to watch

TV =30% of\(100=\frac{30}{100} \times 100\)

= 30

Question 10. There are 50 students in a class of which 40 are boys and the rest are girls. The ratio of the number of boys and the number of girls is

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

Solution: 3. 4:1

Number of Girls = 50 – 40 – 10

Required ratio = 40 : 10 = 4 : 1

Question 11. 40% of 50 students in a class are good at Science. How many students are not good at Science?

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

Solution: 2. 30

Out of 100, good at Science = 40

Out of 50, good at Science

Number of students good at science

= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)

Or

Number of students good at science

= 40% of 50

= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)

Number of students not good at science

= 50-20 = 30

Question 12. Apala has ₹ 200 with her. She spent 80% amount she had. How much money is left with her?

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

Solution: 4.40

Out of  100, money spent =? 80

Out of? 200, money spent

= \(\frac{80}{100} \times 200\)

Money spent = 80% of₹ 200

= \(200 \times \frac{80}{100}\)

= ₹ 160

Money left = ₹ 200 – ₹  160

= ₹ 40.

Question 13. A toy marked at t 40 is available for 132. What percent discount is given on the marked price?

1. 10%
2. 20%
3. 25%
4. 40%.

Solution: 2.20%

Discount = ₹  40 – ₹ 32 = ?8

Discount on ₹  40 = ? 8

Discount on ₹  100

⇒\(\frac{80}{40}\) x100

= ₹ 20

Question 14. Find the simple interest on? 1000 for 2 years at 8% per annum.

1. ₹ 80
2. ₹ 40
3. ₹ 120
4. ₹ 60.

Solution: 4. ₹ 60

S.I = \(\frac{1000 \times 2 \times 8}{100}\)

= ₹ 160

Question 15. A sofa-set was bought for l 10000. Its value depreciated at the rate of 10% per annum. Find its value after one year.

1. ₹11000
2. ₹19000
3. ₹10000
4. ₹1000.

Solution: 2. ₹19000

Depreciation in 1 year =\(10000 \times \frac{10}{100}\)

= 1000

Value after 1 year

= 10000 – 1000 =₹ 9000.

Question 16. There are 1275 trees in Chaudhary Farm. Out of these 36% trees are of fruits. How many trees of fruits are there in Chaudhary Farm?

1. 459
2. 549
3. 945
4. 954.

Solution: 1.459

Required number = \(\frac{1275 \times 36}{100}\)

= 459

Question 17. The quantity of protein in a particular variety of pulse is 25%. Find the amount of protein in 4 kg of pulse.

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

Solution: 1. 1 kg

Amount of protein =\(4 \times \frac{25}{100}\)

= 1kg

Question 18. In a mixture, the amount of zinc is 45%. Find the amount of zinc in the 400 g mixture.

1. 60 g
2. 120 g
3. 180 g
4. 200 g.

Solution: 3. 180 g

Amount of zinc =\(400\times \frac{45}{100}\)

= 80kg

Question 19. In a school out of 340 students, 55% of students are of Science. The remaining students are of Commerce. Find the number of students of Commerce.

1. 135
2. 153
3. 315
4. 140.

Solution: 2. 153

Number of Science students

= \(340 \times \frac{55}{100}\) = 187

Number of students of Commerce

= 340 – 187 = 153

Question 20. The salary of Manish is ₹ 10000. His salary gets increased by 10%. Find his increased salary.

1. ₹ 9000
2. ₹ 11000
3. ₹ 8000
4. ₹ 12000.

Solution: 2.₹ 11000

Increase in salary

= ₹  \(10000 \times \frac{10}{100}=₹ 1000\)

Question 21. A shopkeeper purchased 2 refrigerators for ₹ 9800 and  ₹ 8200 respectively. He sold them for ₹ 16920. Find loss%.

1. 2%
2. 4%
3. 5%
4. 6%.

Solution: 4.6%

Total C.P. =₹ 9800 + ₹ 8200 = ₹18000

S.P. = ₹ 16920

Loss = ₹18000 – ₹16920 = ₹1080

=\(\frac{1080}{18000} \times 100 \%=6 \%\)

Question 22. In selling a plot of land for ₹ 61200, a profit of 20% is made. The cost price of the plot is

1. ₹ 51000
2. ₹ 50000
3. ₹  49000
4. ₹  52000.

Solution: 1. ₹ 51000

If S.P. is ₹120(100 + 20),

C.P. =₹ 100

If S.P. is ₹ 61200,

C.P =\(\frac{100}{120} \times 61200\)

=  ₹ 51000

Question 23. The simple interest on X 2000 for 4 years is ? 400. The rate percent of interest is

1. \(\frac{2000 \times 100}{400 \times 4}\)

2. \(\frac{400 \times 4}{2000 \times 10} \)

3. \(\frac{400 \times 100}{2000 \times 4}\)

4. \( \frac{2000 \times 4}{400 \times 100} \)

Solution: 3.\(\frac{400 \times 100}{2000 \times 4}\)

⇒ \(\frac{2000 \times 4 \times \mathrm{R}}{100}=400\)

R= \(\frac{400 \times 100}{2000 \times 4}\)

Question 24. The simple interest of ₹ 500 at the rate of 5% is ₹ 100. This interest is of the time

1. 1 year
2. 4 years
3. 10 years
4. 20 years.

Solution: 2. 4 years

⇒ \(\frac{500 \times 5 \times \mathrm{T}}{100}=100\)

⇒ \(\mathrm{T}=\frac{100 \times 100}{500 \times 5}=4\)

Question 25. The S.I. of ₹ 100 for 1 year at the rate of 3 paise per rupee per month is

1. ₹ 30
2. ₹ 36
3. ₹ 24
4. ₹ 8.

Solution: 2.₹ 36

S.I. = 100 × 1 × (3 ×  12)

= 3600 paise

= ₹ 36

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities True-False

1. VAT is always calculated on the selling price – True

2. A machinery worth .₹ P is depreciated by R% per annum. Its value after 1 year will be \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)\)– False

3. Comparison of two quantities by division is called ratio-True

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Fill In the Blanks

1. The overhead changes are added to → CP

2. If the conversion period is not specified, then it is taken as → Oneyear

3. If a discount of ₹ 4y is available on the market price of ₹ 8y, then the discount percent – 50%

4. If — % of a number is 154, then the number is → 4900

5. Find 10% of [100- 20% of 300] → 4

6. Explain 0.1234 as a percentage → 12.34

7 A jacket work for? 4000 is offered for sale at? 3600. What per cent discount is offered during the sale →  10%

8. Apala got 100 marks out of 200 and Meenu got 120 works out of 300. Whose performance is better → Apalas.

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Important Points

1. The numbers expressed as the product of the number with itself are called square numbers or perfect squares.

For example, 1 = 1×1 = 1²

4 = 2 ×2 = 2²

9 = 3 × 3 = 3²

16 = 4 × 4 = 4²

25 = 5 × 5 = 5²………………..

2. If a natural number m can be expressed as n², where n is also a natural number, then, m is called a square number.

For example, 100 is the square of 10, as

Read and Learn More NCERT Solutions For Class 8 Maths

100 – 10 × 10 – 10².

So, 100 is a square number.

121 is the square of 11, as

121 – 11 × 11 = n²

So, 121 is a square number.

Question 1. Consider the following numbers and their squares.

From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers up to 100 left out ? From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers up to 100 left out?
Solution:

From, the above table, we can enlist the square numbers between 1 and 100.

Such numbers are listed below:

4, 9, 16, 25, 36, 49, 64, and 81 Also, we see that no natural square numbers up to 100 are left out.

Question 2. Find the perfect square numbers between

1. 30 and 40
2. 50 and 60.

Solution:

1. We know that  5²= 5 × 5 = 25,  6²= 6 ×  6 = 36 and 7² = 7 ×  7 = 49.

So, the perfect square number between 30 and 40 is 36.

2. We know that 7²2 = 7 × 7 = 49, and 8² = 8 × 8 = 64.

So there is no perfect square number between 50 and 60.

Properties Of Square Numbers

1. All square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place, i.e., all square numbers have the unit’s digit as 0, 1, 4, 5, 6, or 9 only.
2. No square number ends with 2, 3, 7 or 8 at the unit’s place.
3. Square numbers can have only an even number of zeros at the end.
4. No square number ends with an odd number of zeros at the end.
5. If a square number ends in 6 then the number whose square it is will end in 4 or 6.

Question 1. Can we say that if a number ends in 0, 1, 4, 5, 6, or 9, then it must be, a square number? Think about it,
Solution:

If a number ends with the digit 0, 1, 4, 5, 6, or 9 at its unit’s place, then it is not necessarily a square number.

For example:

The numbers 10, 21, 34, 325, 146, and 209 are not square numbers although they end with 0, 1, 4, 5, 6, and 9 respectively at their unit’s places.

Note 1: The non-perfect, square numbers end with the digit 2, 3, 7, or 8. i.e., if a number ends with the digit 2, 3, 7, or 8 at its unit place, then it is necessarily a non-perfect square number.

Examples:  32, 42, 43, 53, 63

Note 2: If a number ends with an odd number of zeros at the end, then it is necessarily a non-perfect square number.

Examples: 70, 690, 1430

Question 2.  Can we say whether the following numbers are perfect squares? How do we know?

1. 1057
2. 23453
3. 7928
4. 222222
5. 1069
6. 2061.

Write five numbers which you can decide by looking at their unit’s digit that they are not square numbers.
Solution:

1. 1057 

The number 1057 is not a perfect square because it ends with 7 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

2. 23453

The number 23453 is not a perfect square because it ends with 3 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

3. 7928

The number 7928 is not a perfect square because it ends with 8 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

4. 222222

The number 222222 is not a perfect square because it ends with 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

5. 1069

The number 1069 ends with 9 at the unit’s place. But we cannot say that 1069 is a perfect square number. It may or may not be a square number.

Also, 30 ×  30 = 900

31 × 31 = 961

32 × 32 = 1024

33  × 33 = 1089

This shows that there is no natural number lying between 1024 and 1089 which is a square number.

As 1024 < 1069 < 1089, i.e., the number 1069 lies between 1024 and 1089, so 1069 is not a square number.

4. The number 2061 ends with 1 at the unit’s place. But we cannot say that 2061 is a perfect square number. It may or may not be a perfect square number.

Also, 45 × 45 = 2025

46 × 46 = 2116

This shows that there is no natural number lying between 2025 and 2116 which is a square number. Since 2025 < 2061 < 2116, i.e., the number 2061 lies between 2025 and 2116, therefore 2061 is not a square number.

The number 2061 ends with 1 at & the unit’s place. But we cannot say that 2061 is a j perfect square number. It may or may not be an O-perfect square number.

Also, 45 × 45 = 2025

46 ×46 = 2116

This shows that there is no natural number lying between 2025 and 2116 which is a square number. Since 2025 < 2061 < 2116, i.e., the number 2061 lies between 2025 and 2116, therefore 2061 is not a square number.

Question 3. Write five numbers that you can decide by looking at their unit’s digit that they are not square numbers.
Solution:

The five numbers that we can decide by looking at their one digit that they are not square numbers are 1032, 453, 5567, 13258, and 13293.

[Note that the non-perfect square numbers end with the digits 2, 3, 7, or 8 at the unit’s place]

Question 4.  Write five numbers that you can not decide just by looking at their unit digit (or unit place) whether they are square numbers or not
Solution:

100000, 3246, 56556, 12331 and 799

Note that the numbers ending with the digits 0, 1, 4, 5, 6, or 9 at the unit’s place may or may not be perfect square numbers.

Question 5. Which of 123², 77², 82², and 161², 109²would end with the digit 1?
Solution:

We know that if a number has 1 or 1 is 9 in the unit’s place, then its square ends with the digit 1.

So, 161² and 109² would end with the digit 1

Question 6. Write the next two square numbers respectively. after 441 which ends in 1 and their corresponding numbers
Solution:

From Table 1 on page 91 ). We find that 441 = 21 x 21= 21s. So, the next two square numbers that end in are

29² (= 841) and 31² ( = 961)and their corresponding numbers are 29 and 31respectively.

[Note that if a number has 1 or 9 in the unit’s place, then its square ends with the digit 1.]

Question 7. Which of the following numbers has the digit 6 at the unit place ?

1. 19²
2. 26²
3. 36²
4. 34².

Solution:

Note that when a square number ends with digit 6, then the number whose square it is, will have either 4 or 6 at the unit’s place. So,

1. The digit at the unit’s place = 9

∴ 19² would not have 6 in place.

2. The digit at the unit’s place = 4

∴ 24² would have 6 at unit’splace.

3. The digit at the unit’s place = 6

∴ 26² would have 6 at the unit place

Question 8. Can you find more such rules by observing the numbers and their squares
Solution:

Yes; We can find more such rules by observing the numbers and their squares

These are as follows:

• A square number will end with digit 4 only and only when the number, of which it is the square, ends either with digit 2 or with digit 8.
• A square number will end with the digit 9 only and only when the number of which, it is square, ends either with digit 3 or with the digit 7.
• Asquare number will end with the h digit fi only and only when Ilie number, of which it is square, ends with digit f>.
• A square number will end with digit 1 only mid only when ( ho number, of which it is the Hqunro, ends will) digit 4 or 0,
• A square number will end with a digit Only and only when the number, of which it is the square, ends with the digit 0.

Question 9. What will be the one digit in the square of the following number?

1. 1234
2. 26387
3. 52698
4. 99880
5. 21222
6. 9106

Solution:

1. 1234

Ending digit = 4 and 4² = 4× 4 = 16

∴ The one digit in the square of the number 1234 will be 6.

2. 26387

Ending digit = 7 and 7² = 7 × 7 = 49

∴ The one digit in the square ofthe number 26387 will be 9.

3. 52698

Ending digit = 8 and 8² = 8×8 = 64

∴ The one digit in the square of the number 52698 will be 4.

4. 99880

Ending digit = 0 and 0² = 0×0 = 0

∴ The one digit in the square ofthe number 99880 will be 0.

5.  21222

Ending digit = 2 and 2² = 2×2 = 4

∴ The one’s digit in the square of the number 21222 will be 4

6. 9106

Ending digit = 6 and 6² = 6 × 6 = 36

∴ The one’s digit in the square of the number 9106 will be 6

Question 10.  If a number contains 3 zeros at the end, how many zeros will its square have? What do you notice about the number of zeros at the end ofthe number and the number of zeros at the end of its square? Can we say that square numbers can only have an even number of zeros at the end?
Solution:

If a number contains 3 zeros at the end, then its square will have 3 × 2 = 6 zeros at the end

The number of zeros at the end of a number

=\(\frac{1}{2}\) ×The number of zeros at the end of its square.

Yes; we can say that square numbers can only have an even number of zeros at the end.

Question 11.  With numbers and their squares. What can you say about the squares of even numbers and squares of odd numbers? 
Solution:

1. The squares of even numbers are always even numbers.
2. The squares of odd numbers are always odd numbers.

Question 12. Square of which of the following numbers would be an odd number/an even number? Why?

1. 727
2. 158
3. 269
4. 1980

Solution:

1. 727

The square of 727 will be an odd number because the number 727 is odd

2. 158

The square of 158 will be an oven number because be number

∴  158 is an even number

3. 269

The square of 269 will be an odd number because the number ‘

∴ 269 is an odd number

4. 1980

The squares of 1980 will be an even number because the number

∴ 1980 is an even number

Question 13. What will be the number of zeros in the square of the following numbers?

1. 60
2. 400

Solution:

1. 60

Number of zeros at the end of (the number 60 = 1)

∴ The number of zeros in the square of (be number 60 will be 2 ×1 = 2.

2. 400

Number of zeros at (the end of the number 400 = 2

∴  The number of zeros in the square of the number 400 will be 2 × 2 = 4

Note that the number of zeros at the end of the square of a number = 2 × number of zeros at the end of that number

Some More Interesting Patterns

1. A triangular number is one whose dot patterns can be arranged as triangles. If we combine two consecutive triangular numbers, we get a square number.

2. There are 2n non-perfect square numbers between the squares of the numbers n and (n + 1) which is 1 less than the difference of two squares.

3. The sum of the first n odd natural numbers is n².

4. We can express the square of any odd number as the sum of two consecutive positive integers

Note: \(n^2=\frac{n^2-1}{2}+\frac{n^2+1}{2}\)

Example: \((11)^2=\frac{121-1}{2}+\frac{121+1}{2}\)

= 60 + 61

=121

5. If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) x (n – 1) = n² – 1

Question 1. How many numbers are there between 6² and 7²
Solution:

Let n = 6

Then, n + 1 = 6 + 1 = 7

∴ There are 2n = 2 × 6 = 12

Nonperfect square numbers between 62 and 72

Question 2. Check for n = 5, n = 6, n = 7 etc., and verify
Solution:

1. n = 5

.’. n + 1 = 6

There are 2n = 2 × 5 = 10 non perfect square numbers between 5² and 6²

These are 26, 27, 28, 29, 30, 31, 32, 33, 34 and 35.

2. n = 6

n + 1 = 7

There are 2n = 2× 6 = 12 nonperfect square numbers between 62 and 72.

These are 37, 38, 39, 40, 41, 42, 43,44, 45, 46, 47 and 48

3. n = 7

n + 1 = 8

There are 2n = 2 × 7 = 14

Nonperfect square numbers between 7² and 8².

These are 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62 and 63

Question 3. How many natural numbers lie between 9² and 10² Between 11² and 12²
Solution:

1. Here, n = 9

2n = 2 × 9 = 18.

So, 18 natural numbers lie between 9² and 10².

2. Here, n = 11

2n = 2 × 11 = 22.

So, 22 natural numbers lie between 11² and 12²

Question 4. How many nonsquare numbers lie between the following pairs of numbers?

1. 100² and 101²
2. 90² and 91²
3. 1000² and 1001²

Solution:

1. Here, n = 100

∴ 2n = 2 × 100 = 200.

So, 200 nonsquare numbers lie between the pair of numbers 100² and 101²

2. Here, n = 90

∴ 2n = 2×90 = 180.

So, 180 nonsquare numbers lie between the pair of numbers 90² and 91².

3. Here, n = 1000

∴ 2n = 2 × 1000 = 2000.

So, 2000 nonsquare numbers lie between the pair of numbers 1000² and 1001²

Question 5. 1 + 3 + 5 + 7 […] = 16 = 42 1+3 + 5 + 7+ 9[..]= 25 = 52 1+3 + 5+7 + 9 + 11 […] =36 = 62
Solution:

⇒ 1 + 3 + 5 + 7

[Sum of first four odd numbers]

= 16 = 4²

⇒  1 +3 + 5 + 7 + 9

[Sum of first five odd numbers]

= 25

= 5²

1 + 3 + 5 + 7 + 9+11

[Sum of first six odd numbers]

= 36

= 36

Question 6. Find whether each of the following numbers is a perfect square or not.

1. 121
2. 55
3. 81
4. 49
5. 69

Solution:

1. Successively subtract consecutive odd natural numbers starting with l. i.e .. 1. 3, 5, 7. 5), … from 121.

121-1 = 120

120-3 = 117

117-5 = 112

112-7 = 105

105-9 = 96

96-11 = 85

85- 13 = 72

72-15 = 57

57 – 17 = 40

40 – 19 = 21

21 – 21 = 0

This means that

121=1 + 3 + 5 + 7 + 9+11 + 13+15 + 17+19 + 21

i.e., we can express 121 as the sum of consecutive odd numbers starting with 1.

∴ 121 is a perfect square number. number.

2. Successively subtract consecutive odd natural numbers starting with 1, i.e., 1, 3, 5, 7, 9, … from 55.

55 – 1 = 54

54 – 3 = 51

51 – 5 = 46

46 – 7 = 39

39 – 9 = 30

30 – 11 = 19

19 – 13 = 6

6 – 15 = -9

This shows that we are not able to express 55 as the sum of consecutive odd numbers starting with 1

∴ 55 is not a perfect square number

3. Successively subtract consecutive odd natural numbers starting with 1, i.e.,  1, 3, 5, 7, 9, … from 81

81 – 1 = 80

80 – 3 = 77

77 – 5 = 72

72 – 7 = 65

65 – 9 = 56

56 – 11 = 45

45 – 13 = 32

32 – 15 = 17

17 – 17 = 0

This means that

81 =1 + 3 + 5 + 7 + 9+11 + 13 + 15 + 17.

i.e., we can express 81 as the sum of consecutive odd numbers starting with 1.

∴  81 is a perfect square number

4. Successively subtract consecutive odd natural numbers starting with 1, i.e., 1, 3, 5, 7, 9, … from 49

49 – 1 = 48

48 – 3 = 45

45 – 5 = 40

40 – 7 = 33

33 – 9 = 24

24 – 11 = 13

13 – 13 = 0

This means that

9 =1 + 3 + 5 + 7 + 9+11 + 13

We can express 49 as the sum of consecutive odd numbers starting with 1.

∴  49 is a perfect square number.

5. Successively subtract consecutive odd natural numbers starting with 1, i.e. 3. 5. 7. 9. … from 69.

69 – 1 = 68

68 – 3 = 65

65 – 5 = 60

60 – 7 = 53

53 – 9 = 44

44-11= 33

33 – 13 = 20

20 – 15 = 5

5 – 17 = -12

This shows that we are not able to express 69 as the sum of consecutive odd numbers starting with 1
.
∴ 69 is not a perfect square number

Question 7. Express the following as the sum of two consecutive integers.

1. 21²
2. 13²
3. 11²
4. 19²

Solution:

1. Here,n=21

∴  \(\frac{n^2-1}{2}=\frac{21^2-1}{2}=\frac{441-1}{2} \)

= \(\frac{440}{2}=220 \\\)

And \(\frac{n^2+1}{2}=\frac{21^2+1}{2}=\frac{441+1}{2}\)

= \(\frac{442}{2}=221 \)

∴ 21² = 220+221

= 441

2. Here n= 13

∴ \(\frac{n^2-1}{2}=\frac{13^2-1}{2}=\frac{169-1}{2} \)

=\(\frac{168}{2}=84 \)

And \(\frac{n^2+1}{2}=\frac{13^2+1}{2}=\frac{169+1}{2} \)

= \(\frac{170}{2}=85 \)

13² = 84 + 85

= 169

3. Here n= 11

∴  \(\frac{n^2-1}{2}=\frac{11^2-1}{2}=\frac{121-1}{2}\)

= \(\frac{120}{2}=60 \)

And \(\frac{n^2+1}{2}=\frac{11^2+1}{2}=\frac{121+1}{2} \)

= \(\frac{122}{2}=61 \)

11² = 60+61

= 121 .

4. Here n= 19

⇒ \(\frac{n^2-1}{2}=\frac{19^2-1}{2}=\frac{361-1}{2} \)

=\(\frac{360}{2}=180 \)

And\(\frac{n^2+1}{2}=\frac{19^2+1}{2}=\frac{361+1}{2} \)

= \(\frac{362}{2}=181\)

19²  =180+181

=361 .

Question 8. Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is perfect square ofa number? Give an example to support your answer
Solution:

No; The reverse is not always true.

For example :

1. Consider two consecutive positive integers 3 and 4. Their sum is 3 + 4 = 7 which is not a perfect square ofa number.

2. Consider two consecutive positive integers 8 and 9. Their sum is 8 + 9 = 17 which is not a perfect square ofa number

Question 9. Write the square, making use of the above pattern.

1. 111111²
2. 1111111²

Solution:

1. 11111²

11111²= 1234565432

2. 1111111²

1111111² = 1234567654321

Question 10. Can you find the square, of the following numbers using the above pattern?

1. 6066667²
2. 66666667²

Solution:

1. 6666667²

66666672 = 44444448888889

2. 66666667²

66666667² = 4444444488888889

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.1

Question 1. What will be the unit digit of the squares of the following numbers?

1. 81
2. 799
3. 1234
4. 52698
5. 12796
6. 272
7. 3853
8. 26387
9. 99880
10. 55555

Solution:

1. Unit’s digit of 81 = 1

⇒  1×1= 1

∴ The unit digit of the square of the number 81 will be 1.

2. Unit’s digit of 272 = 2

⇒ 2×2= 4

∴  The unit digit of the square of the number 272 will be 4.

3. Unit’s digit of 799 = 9

⇒  9×9= 81

∴  The unit digit of the square of the number 799 will be 1.

4.  Unit’s digit of 3853 = 3

⇒   3×3= 9

∴  The unit digit of the square of the number 3853 will be 9.

5. Unit’s digit of 1234 = 4

⇒ 4×4 = 16

∴  The unit digit of the square of the number 1234 will be 6.

6. Unit’s digit of 26387 = 7

⇒   7×7 = 49

∴  The unit digit of the square of the number 26387 will be 9.

7. Unit’s digit of 52698 = 8

⇒  8× 8 = 64

∴  The unit digit of the square of the number 52698 will be 4.

8. Unit’s digit of 99880 = 0

⇒  0 × 0 = 0

∴  The unit digit of the square of the number 99880 will be 0.

9. Unit’s digit of 12796 = 6

⇒  6 × 6 = 36

∴  The unit digit of the square of the number 12796 will be 6.

10. Unit’s digit of 55555 = 5

⇒ 5× 5 = 25

∴  The unit digit of the square of the number 55555 will be 5.

Question 2. The following numbers are not perfect squares. Give reasons.

1. 1057
2. 23453
3. 798
4. 22222
5. 64000
6. 89722
7. 222000
8. 505050

Solution:

1. 1057 

The number 1057 is not a perfect square because it ends with 7 at the unit’s place whereas the square numbers end with 0. I, 4. 5, 6 or 9 only at unit’s place.

2.  23453

The number 23453 is not a perfect square because it ends with 3 at the unit’s place whereas the square numbers end with 0, 1. 4, 5, 6, or 9 only at the unit’s place.

3.7928 

The number 7928 is not a perfect square because it ends with 8 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or only 9 at the unit’s place.

4. 222222 

The number 222222 is not a perfect square because it ends with 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 only at the unit’s place.

5. 64000

The number 64000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even

6. 89722

The number 89722 is not a square number because it ends in 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

7. 222000

The number 222000 is not a square number because it has 3 (an odd number) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

8. 505050

The number 505050 is not a square number because it has 1 (an odd number) zeros i at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

Question 3. The squares of which of the following would be odd numbers?

1. 431
2. 2826
3. 7779
4. 82004

Solution:

1. 431

431 is an odd number

∴ Its square will also be an odd number

2.  2826

2826  is an oven number

∴  Its square will not be an odd number.

3. 779

779 is an odd number

∴  Its square will be an odd number.

4. 82004

82004  is an even number

∴  Its square will not be an odd number.

[Note that the squares of odd numbers are odd numbers and the squares of even numbers are even numbers

Question 4. Observe the following pattern and find the missing digits :

1. 112²= 121
2. 101² = 10201
3. 1001² = 1002001
4. 100001² = 1…………2 ……………..1
5. 10000001² =………………………………

Solution:

4. 100001² = 10000200001

5. 10000001² = 100000020000001

Question 5. Observe the following pattern and supply the missing numbers

1. 11² =  121
2. 101² = 10201
3. 10101² = 102030201
4. 1010101² = ………………………
5. …………………………..² = 10203040504030201

Solution:

4. 1010101² = 1020304030201

5. 101010101² = 10203040504030201

Question 6. Using the given pattern, find the jJJ missing numbers

1. 1² + 2²+ 2² = 3²
2. 2² + 3² + 6² = 7²
3. 3² + 4² + 12² = 13²
4. 4² + 5² + ………………²= 21²
5. 5²+ ………………..² +  30²= 31²
6. 6² + 7² + …………….² = …………..²

Solution:

4. 4² + 5² + 20² = 21²

5. 5² + 6²+ 30² = 31²

6. 6² + 7² + 4²= 43²

Question 7. Without adding, find the sum :

1. 1+3 + 5 + 7 + 9
2. 1+3 + 5 + 7 + 9 + 11 + 13 + 15+ 17 + 1
3. 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21 + 23

Solution:

1. 1 + 3 + 5 + 7 + 9

= Sum of first five odd natural numbers

= 5² = 25

2. 1 + 3 + 5+ 7 + 9 + 11 + 13 + 15+17+19

= Sum of first ten odd natural numbers

= 10² = 100

1 + 3 + 5 + 7 + 9+11+13 + 15+17+19 + 21 + 23

= Sum of first twelve odd natural

= 12² = 144.

[Note that the sum of the first n odd natural numbers is n².]

Question 8.

1. Express 49 as the sum of 7 odd numbers.
2.  Express 121 as the sum of 11 odd numbers

Solution:

1. 49 =  7 ×  7 = 7²

= 1 + 3 + 5 + 7 + 9+11 + 13

2. 121 = 11 × 11 = 11²

= 1 + 3 + 5 + 7 + 9+11 + 13 + 15 + 17 + 19 + 21

Question 9. How many numbers lie between squares of the following numbers?

1. 12 and13
2. 25 and 26
3. 99 and 100.

Solution:

1. Here, n = 12 and n + 1 = 13

∴ 2n, = 2 × 12 = 24

So, 24 numbers lie between squares of the numbers 12 and 13.

2. Here, n = 25 and n + 1 = 26

∴ 2n = 2 × 25 = 50

So, 50 numbers lie between squares of the numbers 25 and 26.

3. Here, n = 99 and 11+ 1 = 100

∴ 2n = 2 x 99 = 198

So, 198 numbers he between squares of the numbers 99 and 100
.
[Note that 2n natural numbers lie between the squares of the numbers n and (n + 1

Finding The Square Of A Number

23s = (20 + 3)² = (20 + 3) (20 + 3)

= 20 (20 + 3) + 3 (20 + 3)

= 202 + 20 × 3 + 3 × 20 + 32

= 400 + 60 + 60 + 9 = 529

Other Patterns In Squares

Let o5 be a number with the unit digit 5.

Then, (a.5)² = a (a + 1) hundred + 25

Question 1. Now can you find the square of 95?
Solution:

(95)² = 9 (9 + 1) ×100 + 25

= 9 × 10 × 100 + 25

= 9000+ 25

= 9025

Question 2. Find the the. squares of Ike following numbers containing 5 in unit’s place:

1. 15
2. 95
3. 105
4. 205

Solution:

1. 15² = (1×2) hundred + 25

= 200 + 25 = 225

2. 952 = (9 ×10) hundred + 25

= 9000 + 25 = 9025

3.  1052 = (10 × 11) hundred + 25

= 11000 + 25 = 11025

4. 2052 = (20 × 21) hundred + 25

= 42000 + 25 = 42025

Pythagorean Triplets Points

1. If a, b, and c are three numbers such that any one of the following three relations holds :

1. a² + b² = c²
2. b² + c² = a²
3. c² + a² = b²

Then the numbers a, b, and c are said to form a Pythagorean triplet.

For example : 3, 4, 5 is a Pythagorean triplet because

3² + 42 = 9 + 16 = 25 = 5² .

Can you find more such triplets?

Yes; we can find more such triplets.

For example: 8, 15, 17; 9, 12, 15; 12, 35, 37, etc.

2. For any natural number m > 1, (2m, m² – 1, m² + 1) forms a Pythagorean triplet.

Try to find some more Pythagorean triplets using this form.

Some more Pythagorean triplets using this form are as follows :

6, 8, 10; taking m = 3

8, 15, 17; taking m. = 4

10, 24, 26; taking m = 5

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.2

Question 1. Find the square of the following numbers

1. 32
2. 35
3. 86
4. 93
5. 71
6. 46

Solution:

1. 32 = 30 + 2

⇒ 32 = 30 + 2

Therefore, 32² = (30 + 2)²

= (30 + 2) (30 + 2)

= 30(30 + 2) +2(30 + 2)

= 900 + 60 + 60 + 4

= 1024

2. 35 =30+5

⇒ 35 =30+5

Therefore, 35² = (30 + 5)²

= (30 + 5) (30 + 5)

= 30 (30 + 5) + 5 (30 + 5)

= 900+ 150+ 150 + 25

= 1225

3. 86 = 80 + 6

⇒  86 = 80 + 6

Therefore, 86² = (80 + 6)²

= (80 + 6) (80 + 6)

= 80 (80 + 6) + 6 (80 + 6)

= 6400 + 480 + 480+36

= 7396

4. 93 = 90 + 3

⇒  93 = 90 + 3

Therefore, 93² = (90 + 3)²

= (90 + 3) (90 + 3)

= 90 (90 + 3) + 3 (90 + 3)

= 8100 + 270+270+9

= 8649

5. 71 = 70+1

⇒   71 = 70+1

Therefore, 71² = (70 + 1)²

= (70+ 1) (70+ 1)

= 70(70+ 1)+ 1(70+ 1)

= 4900+70 + 70+1

= 5041

6. 46 =40 + 6

⇒ 46 =40 + 6

Therefore, 46² = (40 + 6)²

= (40 + 6) (40 + 6)

= 40 (40 + 6) + 6 (40 + 6)

= 1600 + 240 + 240 + 36

= 2116

Question 2. Write a Pythagorean triplet whose one number is

1. 6 
2. 14
3. 16
4. 18.

Solution:

1. Let 2m. = 6

⇒  m = – = 3

m² – 1 = 3² – 1 = 9 – 1 = 8

And m² + 1 = 3² + 1 = 9 + 1 = 10

So, a Pythagorean triplet, whose one member is 6, is 6, 8, 10.

2. Let 2m = 14

⇒  m = \(\frac{14}{2}\) = 7

∴ m²- 1=7² -1 = 49 – 1= 48

and m² + 1 = 7² + 1 = 49 + 1 = 50

So, a Pythagorean triplet, whose one member is 14, is 14, 48, 50.

3. Let 2m = 16

⇒  m =  \(\frac{16}{2}\)

∴  m²- 1= 8² – 1 = 64 – 1 = 63

m² + 1 = 8² + 1 = 64 + 1 = 65

So, a Pythagorean triplet, whose one number is 16, is 16, 63, 65.

4.  Let 2m =18

⇒  m =  \(\frac{18}{2}\)

∴  m²- 1= 9² – 1 = 81- 1 = 80

m² + 1 = 9² + 1 = 81 + 1 = 82

So, a Pythagorean triplet, whose one the number is 18, is 18, 80, 82.

Square Roots

The square root of a number ‘a is that number which when multiplied by itself gives that number ‘a as a product.

Thus, if b is the square root of a;

Then b × b = a or b² = a

Symbolically, we write

b = \(\sqrt{a}\)

Note: b = \(\sqrt{a}\) ⇔  b² = a i.e., is the square root iofaaisthesquare ofb

Finding Square Roots

To /meet a number whose square is known is known as finding the squarewot.

Finding the square root is the inverse (opposite) operation of squaring.

There are two integral square roots of a perfect square number.

For example : 4 = (2)² = (- 2)²

∴ \(\sqrt{4}\) = 2 and – 2 both. Here, we shall take up only the positive square rootof a natural number.

Thus, \(\sqrt{4}\) = 2 (not- 2)

The positive square root of a number is denoted by the symbol V.

For example,

3² = 9

⇒ \(\sqrt{9}\)

= 3

Question 1. 11² = 121. What is the square root
Solution:

The square root of 121 is 11.

Question 2. 14² = 196. What is the square root of
Solution:

The square root of 196 is 14

Question 3.

1. (- 1)² =1. Is-  1 a square root of 1?
2. (- 2)² = 4.Is – 2 a square root of 4?
3. (- 9)²= 81. Is – 9a square root of 81?

Solution:

1. Yes; – 1 is a square root of 1.
2. Yes; – 2 is a square root of 4.
3. Yes; -9 is a square root of 81.

Finding Square Root Through Repeated Subtraction

We subtract successive odd numbers starting from 1 from the given square number till we get zero. The number of times, we have to make subtractions, gives the square root of the given square number.

Question 1. By repeated subtraction of odd (iii) 0) numbers starting from 1, find whether the following numbers are. perfect squares or not? If the number is a perfect square then find its square root.

1. 121
2. 55
3. 36
4. 49
5. 90.

Solution:

1.  121

1. 121 – 1 = 120
2. 120 – 3 = 117
3. 117 – 5 = 112
4. 112- 7 = 105
5. 105 – 9 = 96
6. 96 – 11 = 85
7. 85 – 13 = 72
8. 72 – 15 = 57
9. 57 – 17 = 40
10. 40 – 19 = 21
11. 21 – 21 = 0

Since from 121 we subtracted successive odd numbers starting from 1 and obtained 0 at the 11th step,

∴ \(\sqrt{121}\) = 11.

2. 55

1. 55 – 1 = 54
2. 54 – 3 = 51
3. 51 – 5 = 46
4. 46 – 7 = 39
5. 39 – 9 = 30
6. 30 – 11 = 19
7. 19 – 13 = 6
8. 6 – 15 = -9

This shows that we are not able to get zero at any step while subtracting successive odd numbers starting from 1 to 55.

∴  55 is not a perfect square

3.36

1. 36 – 1 = 36
2. 35 – 3 = 32
3. 32 – 5 = 27
4. 27 – 7 = 20
5. 20 – 9 = 11
6. 11 – 11 = 0

Since from 36 we subtracted successive odd numbers starting from 1 and obtained 0 at the 6th step, therefore, = 6.

4. 49

1. 49 – 1 = 48
2. 48 – 3 = 45
3. 45 – 5 = 40
4. 40 – 7 = 33
5. 33 – 9 = 24
6. 24 – 11 = 13
7. 13 – 13 = 0

Since from 49 we subtracted successive odd numbers starting from 1 and obtained 0 at the 7th step,

∴ \(\sqrt{49}\) =7

1. 90 – 1 = 89
2. 89 – 3 = 86
3. 86 – 5 = 81
4. 81 – 7 = 74
5. 74 – 9 = 65
6. 65 – 11 = 54
7. 54 – 13 = 41
8. 41 – 15 = 26
9. 26 – 17 = 9
10. 9 – 19 = – 10

This shows that we are not able to get zero at any step while subtracting successive odd numbers starting from 1 to 90.

∴ 90 is not a perfect square.

Finding Square Root Through Prime Factorisation

We find the prime factors of the given perfect square and arrange in pairs. Then, we choose one factor from each pair and multiply it together. The product thus obtained gives the required square root.

Note: A square number has complete pairs of prime factors.

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.3

Question 1. What could be the possible ‘one’s’digits ofthe square root of each ofthe following numbers?

1. 9801
2. 99856
3. 99856
4. 657666025.

Solution:

1. 9801

∴1 × 1 = 1 and 9 × 9 = 81

The possible one’s digit of the square root ofthe number 9801 could be1 or 9.

2. 99856

∴ 4 × 4= 16 and 6 × 6 = 36

The possible one digit of the square root ofthe number 99856 could be 4 or 6.

3. 99856

∴ 1 × 1 = 1 and 9 × 9 = 81

The possible one digit of the square root of the number 998001 could be 1 or 9.

4. 657666025

∴  5 × 5 = 25

The possible one digit of the square root of the number 657666025 could be 5

Question 2. Without doing any calculation, find the numbers which are surely not perfect squares.

1. 153
2. 408
3. 257
4. 441.

Solution:Solution:

1. 153

The number 153 is surely not a perfect square because it ends in 3 whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only.

2. 257

The number 257 is surely not a perfect square because it ends in 7 whereas the square numbers end with 0, 1, 4, 5, 6 or 9

3. 408

The number 408 is surely not a perfect square because it ends in 8 whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only.

4. 441

The number 441 may be a perfect square as the square numbers end with 0, 1, 4, 5, 6, or 9 only.

Question 3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:

1.100

1. 100 – 1 = 99
2. 99 – 3 = 96
3. 96 – 5 = 91
4. 91 – 7 = 84
5. 84 – 9 = 75
6. 75 – 11 = 64
7. 64 – 13 = 51
8. 51 – 15 = 36
9. 36 – 17 = 19
10. 19 – 19 = 0

Since from 100, we subtracted successive odd numbers starting from 1 and obtained 0 at the 10th step, therefore, \(\sqrt{100}\)= 10.

2. 169

1. 169 – 1 = 168
2. 168-3=165
3. 165-5=160
4. 160 – 7 = 153
5. 153-9 = 144
6. 144-11=133
7. 133 – 13 = 120
8. 120 – 15 = 105
9. 105 – 17 = 88
10. 88 – 19= 69
11. 69 – 21 = 48
12. 48 – 23 = 25
13. 25 – 25 = 0

Since from 169, we subtracted successive odd numbers starting from 1 and obtained 0 at I the 13th step, therefore,\(\sqrt{169}\)= 13.

Question 4. Find the square roots of the following numbers by the Prime Factorisation Method

1. 729
2. 400
3. 1764
4. 4096
5. 7744
6. 9604
7. 5929
8. 9216
9. 529
10. 8100

Solution:

1. The prime factorization of 729 is

729 = 3 × 3 × 3 × 3 × 3 × 3.

By pairing the prime factors, we get

729 = \(3 \times 3 \times 3 \times 3 \times 3 \times 3\)

So, \(\sqrt{729}=3 \times 3 \times 3\) = 27

2. The prime factorization of 400 is

400 = 2× 2× 2× 2× 5× 5.

By pairing the prime factors we get

400 =  \(2 \times 2 \times 2 \times 2 \times 5 \times 5\)

∴ \(\sqrt{400}=2 \times 2 \times 5\)

= 20

3.   The prime factorisation 2 1764of 1764 is 2

1764 = 2 × 2 × 3×3 × 7 × 7

By pairing the prime factors, we get

1764 = \(2 \times 2 \times 3 \times 3 \times 7 \times 7\)

So,  \(\sqrt{1764}=2 \times 3 \times 7\)

= 42

4. The prime factorization 2 512 of 4096 is

4096 = 2 × 2 ×  2 ×  2 ×  2 ×  2× 2 × 2× 2×2× 2× 2

By pairing the prime factors, we get

4096 =  \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\) × \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\)

So, \(\sqrt{4096}=2 \times 2 \times 2 \times 2 \times 2 \times 2\)

= 64

5. The prime factorisation of 7744 is

7744 = 2 × 2× 2 × 2 × 2 × 2 ×11× 11.

By pairing the prime factors, we get

7744 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\)

So,\(\sqrt{7744}=2 \times 2 \times 2 \times 11\)

6. The prime factorisation of 9604 is

9604 = 2 × 2 × 7 × 7 × 7 ×7.

By pairing the prime factors, we get

9604 = \(\underline{2 \times 2} \times \underline{7 \times 7} \times \underline{7 \times 7} \)

So, \(\sqrt{9604}=2 \times 7 \times 7=98\)

7. The prime factorization of 5929 is

5929 =  7 × 7 × 11 × 11

By pairing the prime factors, we get

5929 = \(\underline{7 \times 7} \times \underline{11 \times 11}\)

So, \(\sqrt{5929} =7 \times 11=77\)

8. The prime factorisation of 9216 is

9216 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2× 2 × 2 × 3 × 3

By pairing the primary factors get

9216 =  \( \underline{2 \times 2} \times \underline{2 \times 2} \)

So, \(\times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3}\)

⇒ \(\sqrt{9216}=2 \times 2 \times 2 \times 2 \times 2 \times 3\)

= 96

9. The prime factorization of 529 is

529 = 23 × 23.

By pairing the prime factors, we get

529 = \(\underline{23 \times 23} \)

So,\(\sqrt{529}\) = 23

10. The prime factorisation of 8100 is

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5.

By pairing the prime factors, we get

8100 =\(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)

So, \(\sqrt{8100}\) = 2×3 × 3 × 5

= 90.

Question 5. For each ofthe following numbers, find the smallest whole number by which it should be multiplied to get a perfect square number. Also, find the square root ofthe square number so obtained

1. 252
2. 180
3. 10008
4. 2028
5. 1458
6. 768

Solution:

1. The prime factorization of 252 is

252 = 2 × 2 × 3× 3× 7

By pairing the prime  factors, we get

252= \(\underline{2 \times 2} \times \underline{3 \times 3} \times 7\)

As the prime factor 7 has no pair, 252 is not a perfect square.

If 7 gets a pair, then the number will be a perfect square. So, we multiply 252 by 7 to get

252 × 7 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{7 \times 7}\)

Now each prime factor has a pair. Therefore, 252× 7 = 1764 is a perfect square. Thus the required smallest number is 7.

Also,\(\sqrt{1764}\) =2 × 3× 7

= 42

2. The prime factorisation of 180 is

180 = 2 × 2 × 3 × 3×5

By pairing the prime factors, we get

180 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \) x 5

As the prime factor 5 has no pair, 180 is not a perfect square.

If 5 gets a pair, then the number will be a perfect square. So, we multiply 180 by 5 to get

180 × 5= \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)

Now each prime factor has a pair. Therefore, 180 × 5 = 900 is a perfect square. Thus the required smallest number is 5

Also, \(\sqrt{900}\)

= 2 ×  3× 5 = 30

3. The prime factorization of 1008 is

1008 = 2 × 2 × 2×2 × 3 × 3 × 7

By pairing the prime factors, we get

1008 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3} \times 7\)

As the prime factor 7 has no pair, 1008 is not a perfect square.

If 7 gets a pair, then the number wil be a perfect square. So, we multiply 1008 by 7 to get

1008 × 7 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{7 \times 7}\)

Now each prime factor has a pair, Therefore, 1008 x 7 = 7056 is a perfect square,  Thus the required smallest number is 7

Also, \(\sqrt{7056}\)

= 2 × 2 × 3× 7 = 84

4. The prime factorization of 2028

2028 = 2 × 2 × 3 × 13 × 13.

By pairing the prime factors, we get

2028 = \(\underline{2 \times 2}\) × 3 ×  \(\underline{13 \times 13}\)

As the prime factor 3 has no pair, 2028 is not a perfect square

If 3 gels a pair, then (.ho number will be a perfect square. So, we multiply 2028 by 3 to got

2028 × 3 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{13 \times 13}\)

Now each prime factor has a pair. Therefore, 2028 x 3 = 6084 is a perfect square. Tims the required smallest number is 3.

Also,\(\sqrt{6084}\)

= 2 × 3 × 13 = 78.

5. The prime factorisation of 1458 is

1458 =2 × 3 × 3×  3  × 3 × 3 × 3

By pairing the prime 3 9 factors, we get

1458 = 2 ×\(\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3}\)

As the prime factor 2 has no pair,1458 is not a perfect square.

If 2 gets a pair, then the number will be a perfect square. So, we multiply 1458 by 2 to get

1458 × 2 = \(\underline{2 \times 2} \times\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3}\)

Now each prime factor has a pair. Therefore, 1458 x 2 = 2916 is a perfect square. Thus the required smallest number is 2

Also,\(\sqrt{2916}\)= 2 × 3 × 3 × 3

= 54

6. The prime 768 factorisation of  768 is

768 = 2 ×  2 ×  2 × 2 × 2 × 2 × 2 × 2 × 3.

By pairing the prime factors, we get

786= \(\underline{2 \times 2} \times\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\) x 3

As the prime factor 3 has no pair, 768 is not a perfect square.

If 3 gets a pair, then the number will be a perfect square. So, we multiply 768 by 3 to get

768 ×  3 = \(\underline{2 \times 2} \times\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3}\)

Now (the inch prime factor has a pair. Therefore, 768 x 3 = 2304 to a perfect square. Thus the required smallest number is 3.

Also,\(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3

= 48

Question 6. For each ofthe following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root ofthe square number so obtained

1. 252
2. 2925
3. 396
4. 2645
5. 2800
6. 1620

Solution:

1. The prime factorisation of 252 is

252 = 2 ×  2 × 3 × 3 × 7

By pairing the prime factors, we get

252 = \(\underline{2 \times 2} \times\underline{3 \times 3} \times\)7

We see that the prime factor 7 has no pair. So, if we divide 252 by 7, then we get

252 ÷ 7 = \(\underline{2 \times 2} \times\underline{3 \times 3}\)

Now each prime factor has a pair.

∴  252 ÷ 7 = 36 is a perfect square.

Thus, the required smallest number is 7

Also,\(\sqrt{36}\) = 2 ×  3

= 6

2. The prime factorisation of 2925 is

2925 = 3 × 3  ×5 × 5 × 13.

By pairing the prime factors, we get

2925 = \(\underline{3 \times 3} \times\underline{5 \times 5} \times\) 13

We see that the prime factor 13 has no pair.

So, if we divide 2925 by 13, then we get

2925 ÷ 13 =\(\underline{2 \times 2} \times\underline{5 \times 5}\)

Now each prime factor has a pair.

∴ 2925 ÷ 13 = 225 is a perfect square.

Thus, the required smallest number

Also,\(\sqrt{224}\)= 3 × 5

= 15

3. The prime factor ion of 306 is

306 = 2 × 2 × 3 × 3 × 11

By pairing the prime 2 factors, we get

396 = \(\underline{3 \times 3} \times\underline{5 \times 5}\) ×11

We see that the prime factor 11 has no pair. So, if we divide 396 by 11, then we get

= \(\underline{2 \times 2} \times\underline{3 \times 3}\)

Now each prime factor has a pair.

∴ 396 ÷ 11 = 36 is a perfect square.

Thus, the required smallest number is 11.

Also , \(\sqrt{2}{3}\)

= 6

4. The prime factorisation of 2645 is

2645 = 5 × 23 × 23.

By pairing the prime factors, we get

2645 =  \(\underline{23 \times 23}\)

Now the only prime factor 23 has a pair.

Therefore, 2645 ÷ 5 = 529 is a perfect square.

Thus, the required smallest number is 5

Also,\(\sqrt{529}\) =23

5. The prime factorisation of 2800 is

2800 = 2  × 2 × 2 × 2 × 5 × 5 × 7.

By pairing the prime factors, we get

2800 = \(\underline{2 \times 2} \times\underline{2 \times 2} \times \times\underline{5\times 5} \times\)7

We see that the prime factor 7 has no pair. So, if we divide 2800 by 7, then we get

2800 ÷ 7  \(\underline{2 \times 2} \times\underline{2 \times 2} \times \times\underline{5\times 5}\)

Now each factor has a pair. Therefore, 2800 + 7 = 400 is a perfect square. Thus, the required smallest number is 7.

Also.\(\sqrt{400}\)

= 2 × 2 ×5 = 20

6. The prime factorisation of 1620 is

1620 = 2 × 2 × 3×3 × 3 × 3 × 5

By pairing the prime factors, we get

1620 ÷ 5 =  \(\underline{2 \times 2} \times\underline{3 \times 3} \times \times\underline{3\times 3}\)

Now each factor has a pair. Therefore, 1620 ÷ 5 = 324 is a perfect square. Thus, the required smallest number is 5.

Hence, \(\sqrt{324}\)= 2 × 3 × 3

= 18.

Question 7. The students of Class 8 of a school donated  ₹  2401 in all, for Prime Minister’s National ReliefFund, Each student donated as many rupees as the number of students in the class. Find the number of students in the class
Solution:

Let the number of students in the

Then rupees donated by each student = ₹ x

Rupees denoted by x students

= ₹ x  × x

= ₹ x²

∴ The students of class 8th of a school donated ₹ 2401 for the Prime Minister’s National ReliefFund

⇒ \(x^2= 2401 \)

= \(\sqrt{2401}\)

The prime factorisation of 2401 is

x = \(\sqrt{2401} \)

= \(\sqrt{7 \times 7} \times 7 \times 7\)

x = 7 × 7 = 49

Hence, the number of students in the class is 49

Question 8. 2025 plants arc to be planted in a Now prime factorization of 2 180 pardons in such a nay that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row
Solution:

Let the number of rows 3 2025 be x. 3 675

Then, the number of plants in each row = x.

Number of plants to be planted

= x × x

=x²

But 2025 plants are to be planted in the garden.

⇒ \(x^2=2025 \Rightarrow x=\sqrt{2025}\)

The prime factorization of 2025 is

2025= \(\underline{3 \times 3} \times\underline{3 \times 3} \times \times\underline{5\times 5}\)

x = \(\sqrt{2025}\)

x = \(\sqrt{3 \times 3} \times 3 \times 3 \times 5 \times 5\)

r = 3 × 3 × 5

x = 45

Hence, the number of rows is 45 and the number of plants in each row is 45

Question 9. Find the smallest square number that is divisible by each of the numbers 4, 9, and l0
Solution:

The least number divisible by each one of 4, 9, and  10 is their L.C.M

The L.C.M. of 4, 9 and 10 is

2 × 2 × 3 × 3 × 5 = 180

Now prime factorisation of 180 is

180 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)

The prime factor 5 is not in pairs. Therefore 180 is not a perfect square.

To get a perfect square, each factor of 180 must be paired. So, we need to make a pair of 5

Therefore, 180 should be multiplied by 5.

Hence, the required smallest square

number is 180 ×5 = 900.

Question 10. Find the smallest square number that is divisible by each of the numbers 8, 15, and 20.
Solution:

The least number divisible by each one of 8, 15, and 20 is their L.C.M.

The L.C.M. of 8, 15 and

2 × 2 × 2 × 3 × 5 = 120

Now prime factorisation of 120 is

120 = \(\underline{2 \times 2}\) × 2 × 3 × 5

The prime factors 2, 3, and  5 are not in pairs.

∴ 120  is not a perfect square.

To get a perfect square, each factor of 120 must be paired. So, we need to make pairs of 2, 3 and 5. Therefore 120 should be multiplied by 2 × 3 × 5; ie. 30.

Hence, the required smallest square number is 120 × 30 = 3600

Finding Square Root By Division Method

Steps

• Obtain the number whose square root is to be found. Place a bar for every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar. Each pair and the left-most single digit (if any) is called a period.
• Think of the largest, number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide and get the remainder.
• Bring down the number under the next bar to the right of the remainder. This becomes the new dividend.
• Double the divisor and enter it with a blank on its right. Guess the largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied by the new quotient, the product is less than or equal to the new dividend obtained in step 3.
• Continue this process till the remainder is 0 and no digits are left in the given number.
• The quotient thus obtained is the required square root of the given number.

Question 1.  Can we say that if the perfect square is of n-digits, then its square root will have n/2 digits if n is even or \(\frac{(n+1)}{2}\)
Solution:

Yes. The following examples will establish the truth ofthe above fact.

1. Consider the perfect square 729. Here, n = 3 (which is odd)

∴ Number of digits in the square root

=\(\frac{n+1}{2}=\frac{3+1}{2}=2\)

= 2

Also =\(\sqrt{729}\)

= 27 (which has 2 digits)

2. Consider the perfect square 2025.

Here, n = 4 (which is even)

∴ Number of digits in the square root

= \(\frac{n}{2}=\frac{4}{2}\)

= 2

Also\(\frac{n}{2}=\frac{4}{2}\) =45(which has 2 digits)

Question 2. Without calculating square roots, find. the number of digits in the square root of the following numbers.

1. 25600
2. 100000000
3. 36864

Solution:

1. By placing bars, we get  \(\overline{2}\) \(\overline{56}\) \(\overline{00}\) Since there are 3 bars, the square root will be of 3 digits

Aliter:

Here, n = 5 (which is odd)

Number of digits in the square root

= \(\frac{n+1}{2}\)

= \(\frac{5+1}{2}\)

= \(\frac{6}{2}\)

= 3

2. By placing bars, we get \(\overline{3}\) \(\overline{68}\)\(\overline{64}\)

⇒ \(\overline{1} \overline{00} \overline{00} \overline{00} \overline{00}\)

Since there are 5 bars, the square root will be 5 digits.

Aliter:

Here, n = 9 (which is odd)

Number of digits in the square root

=\(\frac{n+1}{2}=\frac{9+1}{2}\)

= \(\frac{10}{2}=5\)

3. By placing bars, we get \(\)

Since there are 3 bars, the square root will be of 3 digits.

Aliter:

Here, n = 5 (which is odd)

Number of digits in the square root

=\(\frac{n+1}{2}=\frac{5+1}{2}\)

= \(\frac{6}{2}\)

= 3

Square Roots Of Decimals

Put bars on the integral part of the number in the usual manner. Place bars in the decimal part on every pair of digits beginning with the first decimal place. Proceed as usual to find the square root.

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise  5.4

Question 1. Find the square root of each of the following numbers by the Division method:

1. 2304
2. 4489
3. 3481
4. 529
5. 3249
6. 1369
7. 5776
8. 7921
9. 576
10. 1024
11. 3136
12. 900

Solution:

1. 2304

∴ \(\sqrt{2304}\)

= 48

2. 4489

∴ \(\sqrt{4489}\)

= 67

3. 3481

∴\(\sqrt{3481}\)

= 59

4. 529

∴ \(\sqrt{529}\)

= 23

5.3249

∴ \(\sqrt{3249}\)

= 57

6. 1369

∴ \(\sqrt{1369}\)

= 37

7. 5776

∴ \(\sqrt{5776}\)

= 76

8. 7921

∴ \(\sqrt{7921}\)

= 89

9. 576

∴ \(\sqrt{7921}\)

= 89

10. 1024

∴ \(\sqrt{1024}\)

= 32

11.  3136

∴ \(\sqrt{3136}\)

= 56

12. 900

∴ \(\sqrt{900}\)

= 30

Question 2. Find the number of digits in the square root of each of the following numbers (without any calculation)

1. 64
2. 144
3. 4489
4. 27225
5. 390625

Solution:

1. Number (n) of digits in 64

= 2 which is even.

∴ Number of digits in the square root of 64 = \(\frac{n}{2}=\frac{2}{2}\)

= 1

2. Number (n) of digits in 144

= 3 which is odd.

∴ Number of digits in the square root of

144 = \(\frac{n+1}{2}=\frac{3+1}{2}\)

= \(\frac{4}{2}\)

= 2

3. Number (n) of digits in 4489

= 4 which is even.

∴Number of digits in the square root of

4489 = \(\frac{n}{2}=\frac{4}{2}\)

= 2

4. Number (n) of digits in 27225

= 5 which is odd.

∴ Number of digits in the square root of

27225 = \(\frac{n}{2}=\frac{6}{2}\)

= 3

= \(\frac{6}{2}\)

5. Number (n) of digits in 390625

= 6 which is even.

∴ Number of digits in the square root of

=  \(\frac{n+1}{2}=\frac{5+1}{2}\) = \(\frac{6}{2}\)

= 3

Question 3. Find the square root of the following decimal number

1. 2.56
2. 7.29
3. 51.84
4. 42.25
5. 31.36

Solution:

1. 2.56

Hence, \(\sqrt{2.56}=1.6\)

2. 7.29

Hence, \(\sqrt{7.29}=2.7\)

3. 51.84

Hence, \(\sqrt{51.84}=7.2\)

4.  42.25

Hence, \(\sqrt{42.25}=6.5\)

5. 31.36

Hence, \(\sqrt{3136}=5.6\)

Question 4. Find the least number that must be subtracted from each of the following numbers to get a perfect square. Also, the square root of the perfect square  so obtained

1. 402
2. 1989
3. 3250
4. 825
5. 4000

Solution:

1.  We have 402

This shows that 20² is less than 40² by 2. This means, that if we subtract the remainder of 2 from the number, we get a perfect square. So, the required least number is 2.

∴The required perfect square is 402-2 = 40

Hence,\(\sqrt{400}\)=20.

2. We have 1989

This shows that 44² is less than 1989 by 53. This means that if we subtract the remainder of 53 from the number, we get a perfect square. So, the required least number is 53

∴The required perfect square is 1989 – 53 = 1936.

Hence, \(\sqrt{1936}\) = 44

3. We have,3250

This shows that 57² is less than 3250 by

This means if we subtract the remainder (1) from the number, we get a perfect square. So, the required least number is 1.

∴ The required perfect square is

3250-1 = 3249

Hence,  \(\sqrt{43249}\)= 57

4. We have, 825

This shows that 28²  is less than 825 by 41. This means if we subtract the remainder (41) from the number, we get a perfect square. So, the required least number is 41.

∴ The required perfect square is 825-41 = 784

Hence,\(\sqrt{784}\) = 28

5.  We have, 4000

This shows that 63² is less than 4000 by 31

This means if we subtract the remainder (31) from the number, we get a perfect square. So, the required least number is 31.

∴ The required perfect square is 4000 – 31 = 3969

Hence, \(\sqrt{3969}\) = 63

Question 5. Find the least number that must be added to each of the following numbers to get a perfect square. Also, find the square root of the perfect square to obtain

1. 525
2. 1750
3. 252
4. 1825
5. 6412

Solution:

1. We have 525

This shows that 22² < 525.

The next perfect square is 23² = 529.

Hence, the least number to be added is

23² – 525 = 529 – 525 = 4

∴ The perfect square so obtained is 525 + 4 = 529

Hence,\(\sqrt{529}\) = 23

2. We have 1750

This shows that 41² < 1750.

The next perfect square is 42² = 1764.

Hence, the least number to be added is

42² – 1750 = 1764 – 1750

= 14

∴ The perfect square so obtained is 1750 + 14 = 1764

Hence \(\sqrt{1764}\)

= 42

3. We have, 252

This shows that 15² < 252.

The next perfect square is 16² = 256.

Hence, the least number to be added is

16² – 252 = 256 – 252 = 4.

∴ The perfect square so obtained is 252 + 4 = 256

Hence \(\sqrt{256}\)

= 16

4. We have, 6412

This shows that 42² < 1825

The next perfect square is 43² = 1849.

Hence, the least number to be added is

43² – 1825 = 1849 – 1825

= 24.

Therefore, the perfect square so obtained is 1825 + 24 = 1849

Hence, \(\sqrt{1849}\)

= 43

5. We have 1825

This shows that 80² < 6412.

The next perfect square is 81² = 6561.

Hence, the least number to be added is

81² – 6412 = 6561 – 6412 = 149.

Therefore, the perfect square so obtained is 6412+ 149 = 6561

Hence,\(\sqrt{6561}\)

= 81

Question 6. Find the length of the side of a square whose area is 441 m.2.
Solution:

Area of the square = 441 m

Length ofthe side of the square

⇒  \(\sqrt{441}\)m

∴  \(\sqrt{441}\) = 21

Hence, the length ofthe side ofthe square is 21 m

Question 7. In a right triangle ABC, ∠B – 30°.

1. If AB = 6 cm, BC = 8 cm, find AC
2.  If AC =13 cm, BC = 5 cm, find AB.

Solution:

1. In the right triangle ABC,

∠B = 90°

By Pythagoras Theorem

AC² = AB² + BC²

AC² = 6² + 8²

AC² = 36 + 64

AC² = 100

AC = \(\sqrt{100}\)

⇒ \(\sqrt{100}\) = 10

Hence, AC is equal to 10 cm.

In the right triangle ABC,

∠B = 90°

By Pythagoras Theorem,

AC² = AB² + BC²

13² = AB² + 5²

169 = AB² + 25

AB² = 169-25

AB² = 144

AB = \(\sqrt{144}\)

Therefore,= \(\sqrt{144}\) 12.

Hence, AB is equal to 12 cm

Question 8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs for this
Solution:

Let the number of rows be x. Then the number of columns is x. Let us find out the square root of1000 by the division method

This shows that 31² < 1000

32² = 1024.

Hence, the minimum number of plants needs more for this  = 1024 – 1000 = 24

So, number of rows

= number of columns = 32

Question 9. There are 500 children in a school. For a. P. T. drill they have to stand so that the number of rows is equal to several columns. How many children would be left out in this arrangement
Solution:

Let us find out the square root of the 500 division method.

We get the remainder of 16. It shows that 222 is less than 500 by 16.

This means that 16 children would be left out of this arrangement

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Multiple-Choice Questions

Question 1. The perfect, square number out of 2, 3, 4, and 5 is

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

Solution: 3. 4

4 = 2 × 2 = 2²

Question 2. A perfect square number between 30 and 40 is

1. 36
2. 32
3. 33
4. 39

Solution: 1. 36

36 = 6 × 6 = 6²

Question 3. Between 50 and 60, the perfect square number is

1. 56
2. 55
3. 54
4. None

Solution: 4. None

None of 51,52,……59 is a perfect square

Question 4. Which ofthe following is a perfect square number?

1. 1067
2. 7828
3. 4333
4. 625

Solution: 4. 625

Perfect square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

Question 5. Which ofthe following is a perfect square number?

1. 2222
2. 32543
3. 888
4. 10000

Solution: 4. 10000

Perfect square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.

6. Which of 132², 87², 72² and 209² would end with digit 1?

1. 132²
2. 87²
3. 72²
4. 209²

Solution: 4. 209²

If a number has 1 or 9 in the unit’s place, then its square ends in 1

Question 7. Which of 105², 216², 333² and 111² would end with the digit 1

1. 105²
2. 216²
3. 333²
4. 111²

Solution: 4. 111²

If a number has 1 or 9 in the unit’s place, then its square ends in 1

Question 8. Which of 17², 34²,  25², and 49² would have 6 at unit place?

1. 17²
2. 34²
3. 25²
4. 49²

Solution: 2. 34²

If a number has 4 or 6 in the unit’s place, then its square ends in 6.

Question 9. Which of 21², 33², 47², and 36² would have 6 at unit place?

1. 21²
2. 33²
3. 47²
4. 36²

Solution: 4. 36²

If a number has 4 or 6 in the unit’s place, then its square ends in 6.

Question 10. What will be the number of zeros in the square of the number 100

1. 2
2. 4
3. 6
4. 8

Solution: 2. 4

Number of zeros at the end of the number 100 = 2

Number of zeros at the end of the square ofthe number 100 = 2 ×2 = 4

Question 11. What will be the number of zeros in the square of the number 50?

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

Solution: 2. 2

Number of zeros at the end ofthe number 50 = 1

Number of zeros at the end of the square of the number 50 = 2×1 = 2

Question 12. What will be the number of zeros in the square ofthe number 9000?

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

Solution: 4. 6

Number of zeros at the end ofthe number 9000 = 3

Number of zeros at the end of the square ofthe number 9000 = 2 × 3 = 6

Question 13. The square of which of the following numbers will be even?

1. 11
2. 111
3. 1111
4. 112

Solution: 4. 112

∴ 112 is even 3.

∴ Its square will be even

Question 14. The square of which of the following numbers will be odd?

1. 10
2. 100
3. 1000
4. 99

Solution: 4. 99

∴ 99 is odd

∴  Its square will be odd.

Question 15. The square of which of the following numbers will be even?

1. 21
2. 27
3. 35
4. 50

Solution: 4. 50

∴ 50 is even.

∴  Its square will be even.

Question 16. The square of which of the following numbers will be odd?

1. 42
2. 54
3. 66
4. 81

Solution: 4. 81

∴ 81 is odd

∴  Its square will be odd

Question 17. How many natural numbers is he between 8² and 9²

1. 16
2. 17
3. 18
4. 19

Solution: 1. 16

2 × 8 = 16

Question 18. How many natural authors lie between 12² and 13²?

1. 20
2. 22
3. 24
4. 6

Solution: 3. 24

2× 12 = 24

Question 19. How many nonsquare numbers lie between the pair of numbers 80² and

1. 162
2. 160
3. 161
4. 154

Solution: 2. 160

2 × 80 = 160

Question 20. How many nonsquare numbers lie between the pair of numbers 36² and 37²?

1. 36
2. 37
3. 74
4. 72

Solution: 4. 72

2 × 36 = 72

Question 21. How many nonsquare numbers lie between the pair of numbers 500² and 501²

1. 1000
2. 999
3. 1001
4. 1002

Solution: 1. 1000

2 × 500=1000

Question 22. Express the square number 5² as the sum of two consecutive integers.

1. 12+13
2. 10+ 15
3. 9+16
4. 20 + 5

Solution: 1. 12+13

⇒  \(\frac{5^2-1}{2}=12, \quad \frac{5^2+1}{2}=13\)

Question 23. Express 9² as the sum of two consecutive integers.

1. 40 + 41
2. 50+31
3. 36 + 45
4. 72 + 9

Solution: 1. 40 + 41

⇒ \(\frac{9^2-1}{2}=40, \quad \frac{9^2+1}{2}=41\)

Question 24. Express 7² as the sum of two consecutive integers.

1. 40 + 9
2. 24+25
3. 36+13
4. 32+17

Solution: 2. 24+25

⇒ \(\frac{7^2-1}{2}=24, \quad \frac{7^2+1}{2}=25 .\)

Question 25. The unit digit in the square ofthe number 132 is

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

Solution: 4. 4

2 × 2 = 4

Question 26. The unit digit in the square ofthe number 1000 is

1. 1
2. 0
3. 2
4. None of these

Solution: 2. 0

0 × 0 = 0

Question 27. The unit digit in the* square of the number 1111 is

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

Solution: 1. 1

1 × 1 = 1

Question 28. The unit digit in the square of the number  1333 is

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

Solution: 2. 6

3 × 3 = 9

Question 29. The unit digit in the square of the number 2644 is

1. 4
2. 6
3. 8
4. 2

Solution: 2. 6

4 × 4 = 1\(\underline{6}\)

Question 30. The unit digit in the square of the number 125 is

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

Solution: 3. 5

5 × 5 = 2 \(\underline{5}\)

Question 31. The unit digit in the square of the number 166 is

1. 2
2. 4
3. 6
4. 8.

Solution: 3. 6

6 × 6 = 3\(\underline{6}\)

Question 32. The unit digit in the square of the number 27 is

1. 7
2. 6
3. 5
4. 9.

Solution: 4. 9

7 × 7 =4 \(\underline{9}\)

Question 33. The unit digit in the square of number 78 is

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

Solution: 3. 3

8 × 8= 6\(\underline{4}\)

Question 34. The unit digit in the square of number 209 is

1. 1
2. 2
3. 0
4. 3

Solution: 1. 1

9× 9 = 8\(\underline{1}\)

Question 35. If 10² = 100, then the square root of 100 is

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

Solution: 2. 10

⇒ \(\sqrt{100}\) = 10

Question 36. If 25² = 625, (hen the square root of 625 is

1. 5
2. 26
3. 125
4. 625.

Solution: 2. 26

⇒  \(\sqrt{625}\) = 25

Question 37. What could be the possible one digit of the square root of 625?

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

Solution: 4. 5

5 × 5 = 25

Question 38. What could be the possible one digit of the square root of 121?

1. 1, 9
2. 3, 4
3. 6, 7
4. 7, 8.

Solution: 1. 1, 9

1 × 1 = 1

9 × 9 = \(1 \underline{81}\)

Question 39. What could be the possible one digit of the square root of 361?

1. 1, 9
2. 3, 4
3. 6, 7
4. 7, 8.

Solution: 1. 1, 9

1 × 1 = 1

9 × 9 = \(8 \underline{1}\)

Question 40. What could be the possible one digit of the square root of 576?

1. 1, 9
2. 5, 7
3. 1, 8
4. 2, 9.

Solution: 1. 1, 9

4 × 4 = \(1 \underline{6}\)

6 × 6 = \(1 \underline{36}\)

Question 41. What could be the possible one digit of the square root of 676?

1. 4, 6
2. 5, 7
3. 1, 8
4. 2, 9.

Solution: 1. 4, 6

4 × 4 = \(1 \underline{6}\)

6 × 6 = \(1 \underline{36}\)

Question 42. The smallest number by which 32 should be multiplied to get a perfect square is

1. 2
2. 3
3. 4
4. 8

Solution: 1. 2

32 × 2 = 64 = 8²

Question 43. The smallest number by which 48 should be multiplied to get a perfect square is

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

Solution: 2. 3

48 ×  3 = 144 = 12²

Question 44. The smallest number by which 45 should be multiplied to get a perfect square is

1. 2
2. 3
3. 5
4. 7

Solution: 3. 5

45 ×  5 = 225 = 15²

Question 45. The smallest number by which 54 should be multiplied so that we get a perfect square is

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

Solution: 1. 2

54 × 6 = 324 = 18²

Question 46. The smallest number by which 28 should be multiplied to get a perfect square is

1. 2
2. 4
3. 3
4. 7

Solution: 4. 7

28 × 7 = 196 = 14²

Question 47. The smallest number by which 1000 should be multiplied to get a perfect square is

1. 5
2. 10
3. 4
4. 8.

Solution: 2. 10

1000 × 10 = 10000 = 100²

Question 48. The smallest number by which 128 should be divided to get a perfect square is

1. 2
2. 3
3. 4
4. 8.

Solution: 1. 2

128 ÷ 2 = 64 = 8²

Question 49. The smallest number by which 48 should be divided to get a perfect square is

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

Solution: 2. 3

48 ÷ 3 = 16 = 4²

Question 50. The smallest number by which 125 should be divided so as to get a perfect square is

1. 3
2. 5
3. 25
4. 125

Solution: 2. 5

125 ÷ 5 = 25 = 5²

Question 51. The smallest number by which 150 should be divided so as to get a perfect square is

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

Solution: 4. 6

150 ÷ 6 = 25 = 5²

Question 52. The smallest number by which 112 should be divided so as to get a perfect square is

1. 6
2. 4
3. 3
4. 7

Solution: 4. 7

112 ÷ 7 = 16 = 4²

Question 53. The smallest number by which 1000 should be divided so as to get a perfect square is

1. 5
2. 10
3. 100
4. 1000.

Solution: 2. 10

1000 ÷ 10 = 100 = 10²

Question 54. The smallest 3-digit, perfect square is

1. 999
2. 100
3. 961
4. 125

Solution: 2. 100

100 = 10²

Question 55. The number of digits in the square root of 62500 is

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

Solution: 3. 3

n = 5, \(\frac{n+1}{2}=3\)

Question 56. The number of digits in the square root of 441 is

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

Solution: 2. 2

n = 3, \(\frac{n+1}{2}=2\)

Question 57. The number of digits in the square root of 100 is

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

Solution: 2. 2

n = 3,  \(\frac{n+1}{2}=\) = 2

Question 58. Find the length of the side of a square whose area is 100 cm2.

1. 5 cm
2. 10 cm
3. 100 cm
4. 4 cm

Solution: 2. 10 cm

⇒ \(\sqrt{100}=10\)

Question 59. The students of class VIII of a school donated? 10000 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. The number of students in the class is

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

Solution:  2. 100

⇒ \(\sqrt{10000}=100\)

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots True or False

1. That sum of two perfect squares is a perfect square – False

2. The product of two perfect squares is a perfect square – True

3. 1000 is a perfect square – False

4. All numbers of a Pythagorean triplet are even – False

5. There is only one square number between 20 and 30 – True

NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Fill In The Blanks

1. 11² = 60 +→ 61

2. 112 + 113 =→ (15)²

3. There are non-square numbers between 42 and 52. → 8

4. The square of 1.1 is → 1.21

5. For every natural number m (> 1), 2m, m² – 1 and m² + 1 form a -Pythagorea

1. Write the greatest two-digit square number → 81

2. The smallest number of a Pythagorean triplet is 3. Find its other two numbers → 4,5

3. Which letter best represents the location of 716 on a number line →  E

4. The area of a square board is 144 square units. Find the length ofthe side ofthe square board  →12 units.

5. Find the value of \(\sqrt{17+\sqrt{64}}\) →  5

NCERT Solutions For Class 8 Maths Chapter 4 Data Handling Introduction

The information collected in the form of numbers in the context of a situation under study is called data.

For example :

Marks obtained by the students of your class in the monthly test in English, weights of students of your class, etc.

We systematically organise the data and then interpret it.

Sometimes, we represent the data graphically to get a clear idea of what it represents.

Question 1. A pictograph

It is a pictorial representation of data using symbols

Read and Learn More NCERT Solutions For Class 8 Maths

1. How many cars were produced in July?
Solution:

250 cars were produced in July

2. In which month was the maximum number of cars produced?
Solution:

The maximum number (= 400) was produced in September’

Question 2. A bar graph

It is a display of information using bars of uniform width, with equal gaps in between. Their heights are proportional to the respective values.

1. What is the information given by the bar graph?
2. In which year is the increase in the number of students maximum?
3. In which year is the number of students maximum?
4. Slate whether true or false :

‘The number of students during 2005-2006 is twice that of 2003-04’.

Solution:

1. The bar graph gives information about the number of students in class VIII in various academic years in the school.

2. Increase in the number of students in

2004 – 05 = 200 – 100 = 100

Increase in the number of students in

2005- 06 = 250 – 200 = 50

Increase in the number of students in

2006-07 = 300 – 250 = 50

Increase in the number of students in

2007-08 = 350 – 300 = 50

Hence, the increase in the number of X students is maximum in the year 2004-2005.

3. The number of students is maximum in the year 2007-08.

4. The number of students during

2005-06 = 250

The number of students during

2003-04 = 100

Twice the number of students during

2003-04 = 2 xl00 = 200* 250

( = The number of students during 2005- 06)

Since the number of students during 2005-06 is not twice that of 2003-04, therefore the given statement is false.

Question 3. Double bar graph

Bar graph allowing Iwo sola of data simultaneously. It is useful for the comparison of the data.

1. What is the information given by the double bar graph?
2. In which subject has the performance improved the most?
3. In which subject has the performance deteriorated?
4. In which subject is the performance at par?

Solution:

1. The double bar graph gives information about the marks obtained by a student in different subjects in the academic years 2005-06 and 2006-07.

2. The performance has improved the most in the subject of Mathematics.

3. The performance has deteriorated in the subject of English.

4. The performance is at par in the subject of Hindi

Question 4. If we change the position of any of the bars of a bar graph, would it change the information being conveyed? Why
Solution:

No, because bar heights give the quantity for each category. Since by changing the position of any of the bars of a bar graph, the height of the bar remains unchanged, the information conveyed regarding the quantity remains unchanged

Question 5. Draw an appropriate graph to represent the given information.

Solution:

Question 6.

Solution:

Question 7. Percentage wins in ODI by 8 top cricket teams

Solution:

Circle Graph Or Pie Chart

A circle graph shows the relationship between a whole and its parts. We divide the whole circle into various sectors. The size of each sector is proportional to the information or activity it represents.

Question 8. Each of the following pie charts gives you a different piece of information about your class. Find the fraction ofthe circle representing each of these information

Solution:

1. Fraction ofthe circle representing the ‘girls’

= \(\frac{\text { Percentage of girls }}{\text { Percentage of all students }}\)

= \(\frac{50}{100}=\frac{1}{2}\)

Dividing the numerator and denominator by 50 [= HCF (50, 100)]

Fraction of the circle representing the boys

= \(\frac{\text { Percentage of boys }}{\text { Percentage of all students }}\)

= \(\frac{50}{100}=\frac{1}{2}\)

2. Fraction of the circle representing walk

= \(\frac{\text { Percentage of walk }}{\text { Total percentage of transport }} \)

= \(\frac{40}{100}=\frac{2}{5}\)

Dividing the numerator and denominator by 20 [= HCF (40, 100)]

Fraction ofthe circle representing‘cycle’

= \(\frac{\text { Percentage of cycle }}{\text { Total percentage of transport }} \)

= \(\frac{20}{100}=\frac{1}{5}\)

Fraction of the circle representing (Bus or car)

= \(\frac{\text { Percentage of Bus or } \mathrm{Car}}{\text { Total percentage of transport }}\)

= \(\frac{40}{100}=\frac{2}{5}\)

Dividing the numerator and denominator by 20 [= HCF (40, 100)]

3. The fraction of the circle represents those who hate Mathematic

= \(\frac{\text { Percentage of students who hate Mathematics }}{\text { Percentage of all students }}\)

= \(\frac{15}{100}=\frac{3}{20}\)

Dividing the numerator and denominator by 20

[= HCF (40, 100)]

Dividing the numerator and denominator by 5

[= HCF (15, 100)

The fraction ofthe circle representing those who whose Mathematic

= \(\frac{\begin{array}{c}
\text { Percentage of those who love Mathematics } \\
\end{array}}{\text { Percentage of all students }}\)

= \(\frac{100-15}{100}\)

= \(\frac{85}{100}=\frac{17}{20}\)

Question 9. Answer the following questions based on the pie chart given

1. Which type of programmes are viewed the most?
2. Which two types of programmes have several viewers equal to those watching sports channels

Solution:

From the given pie chart, we prepare the following table

1. Since the percentage of entertainment viewers is the highest, therefore, entertainment programmes are viewed the most.

2. Percentage of viewers watching the news

= 15% r

Percentage of viewers watching X informative = 10%

∴ Some of the percentages of viewers watching news and informative

= (15 + 10)% = 25%

= Percentage of viewers watching spo ts

Hence, news and information programmes have many viewers equal m those watching sports channels.

Drawing Pie Charts

The total angle at the centre of a circle is 360°. The central angles of sectors are fractions of 360°.

The central angle of a sector

= \(\left(\frac{\text { Value of the component represented by the sector }}{\text { Sum of the values of all the components }} \times 360^{\circ}\right)\)

Question 1. Draw a pie chart of the data given below. The time spent, by a child during a day.

1. Sleep – 8 hours
2. School – 6 hours
3. Homework – 4 hours
4. Play – 4 hours
5. Others – 2 hours

Solution:

Now, we make the pie chart pie chart

NCERT Solutions For Class 8 Maths Chapter 4 Data Handling  Exercise 4.1

Question 1. A survey was made to find the type of music Il\al a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey

1. From this pie chart answer the following
2. If 20 people liked classical music, how many young people were surveyed?
3. Which type of music is liked by the maximum number of people?
4. If a cassette company were to make 1000 CD’s, how many of each type would they make?

Solution:

1. Suppose that x young people were surveyed. Then, the number of young people who liked classical music = 10% of x

= \(\frac{10}{100} \times x=\frac{x}{10}\)

According to the question

⇒ \(\frac{x}{10}=20\)

x = 20 × 10

Multiplying both sides by 10

X = 200

Hence 200 young people were surveyed

2. Since the percentage of young people who liked light music is the highest, therefore, light music is liked by the maximum number of people

3. Total number of CD’s = 1000

Number of CD’s of semi-classical music

= 20% of 1000

= \(\frac{20}{100} \times 1000\)

= 200

Number of CDs of classical music 10% of 1000

= \(\frac{10}{100} \times 1000\)

=100

Number of CD’s of folk music

= 30% of 1000

= \(\frac{30}{100} \times 1000\)

= 300

Number of CDs of light music

= 40% of 1000

= \(\frac{40}{100} \times 1000\)

= 400

Question 2. A group of 360 people were asked to vote for their favourite season from the h “ee Semi seasons rainy, winter and summer.

1. Which season got the most votes?
2. Find the central angle ofeach sector.
3. Draw a pie chart to show this information.

Solution:

1. Since the no. of votes corresponding to the winter season is the maximum, therefore, winter season got the most votes.

2. Total votes = 90+ 120+ 150 = 360.

The central angle of the sector corresponds to the summer season

= \(\frac{\begin{array}{c}\text { Number of people who vote for summer season } \\\end{array}}{\text { Total number of people }} \times 360^{\circ} \)

= \(\frac{90}{360} \times 360^{\circ}=90^{\circ}\)

The central angle of the sector corresponds to the rainy season

= \(\frac{\begin{array}{c}\text { Number of people who vote for rainy season } \\\end{array}}{\text { Total number of people }} \times 360^{\circ}\)

= \(\frac{120}{360} \times 360^{\circ}=120^{\circ}\)

The central angle of the sector corresponding to the winter season

= \(\frac{\begin{array}{c}\text { Number of people who vote for winter season } \\\end{array}}{\text { Total number of people }} \times 360^{\circ}\)

= \(\frac{150}{360} \times 360^{\circ}=150^{\circ}\)

Pie Chart

Question 3. Draw a pie chart showing the following information. The table shows the colours Preferred  by a group of people

Find the proportion of each sector. For example,

Blue is – \(\frac{18}{36}=\frac{1}{2}\)

Green – \(\frac{9}{36}=\frac{1}{4}\)  and so on

Use this to find the corresponding and

Solution:

Pie chart:

Question 4. The adjoining pie chart gives the marks scored in an examination by students in Hindi, English, Mathematics, SocialScienceand Science. If the total marks obtained by the students were 540, answer the following questions:

1. In which subject did the student score 105 marks?

Hint: For 540 marks, the central angle

= 360°. So, for 105 marks, what is the central angle?

2. How many more malts were obtained by the student in Mathematics than in Hindi?

3. Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi

Solution:

1. Total marks = 540

∴ Central angle corresponding to 540 marks

= 360°

∴  Central angle corresponding to 105 marks

= \(\frac{360^{\circ}}{540} \times 105=70^{\circ}\)

Since the sector having a central angle of 70° corresponds to Hindi, therefore, the student scored 105 marks in Hindi.

2. Central angle corresponding to the sector of mathematics = 90°

∴ Marks obtained by the student in Mathematics

= \(\frac{90^{\circ}}{360^{\circ}} \times 540\)

= 135

Marks obtained by the student in Hind =105

Hence, the student obtained 135 – 105 = 30 marks more in Mathematics than in Hindi.

3. Sum ofthe Central Angles for Social

Science and Mathematics

= 65° + 90° = 155°

Some of the central angles for Science and Hindi

= 80° + 70° = 150°

Since the sum of the central angles for Social Science and Mathematics is greater than the sum of the central angles for Science and Hindi, therefore the sum of the marks obtained in social science and mathematics is more than that in Science and Hindi.

Question 5.The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart

Solution:

Now, we make the pie chart.

Chance And Probability

We face many situations where we take a chance and it does not go as per our wish. For example, on tossing a coin, we may get a head or we may get a tail; but simply by chance. There exist many situations when the chances of a certain thing happening or not happening are not equal.

Getting A Result

A random experiment is one whose outcome cannot be predicted exactly in advance.

Question 1. If you try to start a scooter, what arc the possible outcomes?
Solution:

If we try to start, a scooter, then the possible outcomes are

1. The scooter may start ;
2. The scooter may not start.

Question 2. When a die is thrown, what are the six possible outcomes?
Solution:

When a die is thrown, the six possible outcomes are 1, 2, 3, 4, 5 and 6.

Question 3. When you spin the wheel shown, what are the possible outcomes?  List them. (Outcome here means the sector at which the pointer stops).

Solution:

Since in the given wheel, there are three sectors A, B, and C, so, when we spin the wheel, the possible outcomes are A, B and C.

Question 4. You have a bag with five identical balls of different colours and you are to pull out (draw) a ball without looking at it; list the outcomes you would
Solution:

Since the given bag contains five identical balls of different colours, W, R, B, G and Y, therefore, when we pull out (draw) a ball without looking at it, the possible outcomes are W, R, B, G and Y.

Question 5. In throwing a die

1. Does the first player have a greater chance of getting a six?
Solution: No

2. Would the player who played after him have a lesser chance of getting a six?
Solution: No

3. Suppose the second player got a six. Does it mean that the third player would not have a chance of getting a six?
Solution: No

Equally Likely Outcomes

Outcomes of an experiment are equally likely if each has the same chance of occurring.

Linking Chances To Probability

Chances of the happening of an event lead us to find the probability of that event. Of course, we have to take into account the total number of chances related to a particular event under consideration

Coin And Die

Question 1. What is the probability of getting a tail?
Solution: The probability of getting a tail is \(\frac{1}{2}\)

Question 2. What is the probability of getting Ike number 5?
Solution: The probability of getting the number 5 is \(\frac{1}{6}\)

Question 3. What is the probability of getting the number 7?
Solution: The probability of getting the number 7 is 0.

Question 4. What is the probability of getting a number 1 through 6?
Solution: The probability of getting a number 1 through 6 is \(\frac{1}{6}\)

Probability of an event =\(\frac{\text { Number of outcomes that make an event }}{\text { Total number of outcomes of the experiment }} \text {, }\) when the outcomes are equally likely.

Outcomes As Events

One or more outcomes of an experiment make an event

Question 1. List the number of outcomes of getting a green sector and not getting a green sector on this wheel

1. Find the probability of getting a green sector.
2. Find the probability of not getting a green sector.

Solution:

1. Number of outcomes getting a green sector on this wheel = 5

The number of outcomes of not getting a green sector on this wheel = 3

2. Total number of equally likely outcomes = 8

Number of outcomes getting a green sector = 5

∴ Probability of getting a green sector

= \(=\frac{\begin{array}{l}
\text { Number of outcomes of getting a green sector } \\
\end{array}}{\text { Total number of outcomes }}\)

= \(\frac{5}{8}\)

Total number of equally likely outcomes = 8

Number of outcomes of not getting a green sector = 3

The probability of not getting a green sector

= \(\frac{\begin{array}{l}\text { Number of outcomes of getting a green sector }\end{array}}{\text { Total number of outcomes }}\)

= \(\frac{3}{8}\)

Chance And Probability Related To Real Life

Chances and probability are related to real life. The UHO of probability i.s ma(te in various cases in real life. For example, during elections ‘an exit poll’ is taken which gives a rough idea of the chance of winning of each candidate and predictions regarding the poll are made based on it. Here, we have used a small part of the group to find the characteristics of a large group.

NCERT Solutions For Class 8 Maths Chapter 4 Data Handling Exercise 4.2

Question 1. List the outcomes you can see in these experiments.

1. Spinning wheel

2. Tossing two coins together

Solution:

1. The outcomes we can see in spinning the given wheel are A, B, C and D.
2. The outcomes we can see in tossing two coins together are HT, HH, TH, and TT (HereIT means Head on first coin and Tail on the end coin and so on).

Question 2. When a die is thrown, list the outcomes of an event of getting

1.

1. A prime number
2. Not a prime number.

2.

1. A number greater than 5
2. A number not greater than 5

Solution:

Possible outcomes are :

1, 2, 3, 4, 5, and 6.

Out of these, prime numbers are 2, 3 and 5.

The outcomes of an event of getting a prime number are :

2, 3 and 5

The outcomes of an event of not getting a prime number are 1, 4 and 6.

The outcome of an event of getting a number greater than 5 is 6.

Outcomes of an event of getting a number not greater than 5 are1, 2, 3, 4 and 5

Question 3. Find the

1. The probability of the pointer stopping on D in
2. Probability of getting an AE from a well-shuffled deck of 52 playing cards?
3.  Probability of getting a red apple.

Solution:

1. There are 5 sectors on the spinning wheel represented by A, A, B, C and D. The pointer stopping on D has m only 1 outcome, i.e., D

∴ Probability of the pointer

Stopping on D =\(\frac{1}{5}\)

2. Total number of playing cards = 52

Number of possible outcomes = 52

Number of aces in a deck of playing cards = 4

∴Probability y of getting an ace from a well-shuffled deck of 52 playing cards =\(\frac{4}{52}\)

= \(\frac{1}{3}\)

3. Total number of apples = 7

Number of red apples = 4

The probability of getting a red apple

= \(\frac{4}{7}\)

Question 4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well One slip is chosen from the box without looking into it. What is the probability of this?

1. Getting a number 6?
2. Getting a number less than 6?
3. Getting a number greater than 6?
4. Getting a 1-digit number?

Solution:

Total number of outcomes of the event (1, 2. 3, 4, 5, 6, 7, 8, 9 and 10) = 10

1. Number of outcomes of getting a number 6 = 1

∴ The probability of getting the number 6

= \(\frac{1}{10}\)

2. There are 5 numbers (1, 2, 3, 4 and 5) less than 6.

∴  Number of outcomes getting a number less than 6 = 5

∴  Probability of getting a number Less than

6 = \(\frac{5}{10}\)

= \(\frac{1}{2}\)

3. There are ‘I number (7, 8, 9nnd 10) greater than

Number of outcomes of ting a number greater than G = 4

The probability of getting a number greater than 6 = 4 is greater than

6 = \(\frac{4}{10}\)

= \(\frac{2}{5}\)

There are 9 1-digit numbers(1, 2, 3, 4, 5, 6, 7, 8 and 9)

∴  Number of outcomes of getting a1-digit number = 9

∴  Probability of getting a 1-digit number = \(\frac{9}{10}\)

Question 5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:

Number of green sectors = 3

Number of blue sectors = 1

Number of red sectors = 1

∴  Total number of sectors

= 3+1+1 = 5

∴ Total number of outcomes of the event = 5

Number of outcomes of getting green sector = 3

Probability of getting a green sector

= \(\frac{3}{5}\)

Number of outcomes of getting non-blue sector

= Number of green sectors + Number of red sectors

= 3+1 = 4

∴ The probability of getting a non-blue

= \(\frac{4}{5}\)

Question 6 Find the probabilities ofthe (rents I have given in
Solution:

Total number of outcomes of the event (1. 2. 3, 4, 5 and 6) = 6

1.

1. Number of prime numbers

(2. 3 and 5) = 3

Number of outcomes of getting a prime number = 3

∴ Probability of getting a prime number

= \(\frac{3}{6}\)

= \(\frac{1}{2}\)

2. Number of non-prime numbers (1, 4 and 6) = 3

Number of outcomes of getting a prime number = 3

∴ Probability of gelling a non-prime

= \(\frac{3}{6}\)

= \(\frac{1}{2}\)

2.

1. Number greater than 5 = 6, i.e., only one.

∴ The probability of getting a number greater than 5 =\(\frac{1}{6}\)

2. Number of numbers not greater than 5 (1, 2, 3, 4 and 5) = 5

The number of outcomes of getting a number not greater than 5 = 5

Probability of getting a number not greater than 5 = \(\frac{5}{6}\)

NCERT Solutions For Class 8 Maths Chapter 4 Data Handling Multiple Choice Questions

Observe the following bar graph and answer the following questions :

Question 1. On which item has the maximum expenditure been done?

1. Conveyance
2. Rent
3. Fee
4. Servant’s salary.

Solution:  1. Conveyance

The height of the bar corresponding to conveyance is the maximum.

Question 2. On which item has the minimum expenditure been done?

1. Servant’s salary
2. Food
3. Rent
4. Conveyance.

Solution: 1. Servant’s salary

The height of the bar corresponding to the servant’s salary is the minimum.

Question 3. What is the expenditure on food?

1. ₹ 1000
2. ₹ 2000
3.  ₹3000
4.  ₹5000

Solution: 4. ₹5000

Expenditure done on food = 5 × 1000 = 5000.

Question 4. What is the difference between expenditures done on conveyance and rent?

1.  ₹ 1000
2.  ₹ 2000
3.  ₹ 3000
4.  ₹ 4000

Solution: 2. ₹ 2000

Expenditure done on conveyance

= 6 × 1000 = ₹ 6000

Expenditure done on rent = 4 × 1000

= ₹ 4000

.-. Difference = ₹ 6000 – ₹ 4000

= ₹ 2000

Question 5. ₹ 5000 is the expenditure done on

1. Rent
2. Food
3. Fee
4. Recreation

Solution: 2. Food

⇒ \(\frac{5000}{1000}\) = 5 cm is the height of the bar corresponding to food

Question 6. ₹60000 is the expenditure done on

1. Fee
2. Rent
3. Conveyance
4. Food

Solution:  3. Conveyance

⇒ \(\frac{6000}{1000}\) = 6 cm is the height of the bar corresponding to conveyance

Question 7. How much expenditure h;m been done in all?

1. 21000
2. 18000
3. 15000
4. 20000

Solution: 1. 21000

Total expenditure

= (1 + 6 + 4+ 5 + 2 + 3) × 1000

= ₹ 21000

Observe the following paragraph and answer the following questions:

Question 8. Of which subject are there the maximum books?

1. Hindi
2. English
3. Maths
4. Science

Solution: 1. Hindi

The height of the bar corresponding to Hindi is maximum.

Question 9. How many books are there on ofthe subject whose books are maximum?

1. 100
2. 200
3. 300
4. 400

Solution: 4.  400

Hindi  → 400

Question 10. Of which subject are there the minimum books?

1. Social
2. Science
3. Hindi
4. English

Solution: 1. Social

The height of the bar corresponding to Social Science is the minimum

Question 11. How many books are there on the subject whose books are minimum?

1. 100
2. 200
3. 30
4. 400

Solution: 1. 100

Social Science →  100

Question 12. Which two subjects have the same number of books?

1. Maths and Hindi
2. Hindi and English
3. English and Science
4. Science and Social Science

Solution: 3. English and Science

English → 200

Science → 200

Question 13. 300 books on the subject

1. Maths
2. English
3. Hindi
4. Science

Solution:  1. Maths

300 → Maths

Question 14. The difference of the number of books in English and Science is

1. 200
2. 100
3. 400
4. 0

Solution: 4. 0

200-200 = 0

Question 15. The difference of the number of books of Hindi and Social Science is

1. 200
2. 300
3. 400
4. 100

Solution: 2. 300

400-100 = 300

Question 16. The total number of books is

1. 1200
2. 1400
3. 1600
4. 1800

Solution: 1. 1200

300+400 + 200 + 200+ 100 = 1200

Question 17. The total number of books of English and Science is

1. 200
2. 100
3. 400
4. 0

Solution: 3. 400

200 + 200 = 400

Observe the pie chart given below and answer the following questions:

Question 18. The central angle for sector A is

1. 108°
2. 144°
3. 72°
4. 150°

Solution: 1. 108°

Central angle for sector A

= \(\frac{30}{100} \times 360^{\circ}\)

= 108°

Question 19. The central angle for sector B is

1. 108°
2. 144°
3. 72°
4. 150°

Solution: 2. 144°

Central angle for sector A

= \(\frac{40}{100} \times 360^{\circ}\)

= 144°

Question 20. Which sector has the greatest angle?

1. A
2. B
3. C
4. None of these

Solution: 2. B

Greatest percentage = 40% → B

Question 21. What is the difference between the central angles for Sector B and Sector C?

1. 36°
2. 72
3. 9°
4. 81°

Solution: 1. 36°

144° – 108° = 36°

Observe the pie chart and answer the English following questions

Question 22. Which two colours have the same central angles?

1. Red, yellow
2. Red, green
3. Yellow, green
4. Blue, red.

Solution:  1. Red, yellow

Red →  45% ; Yellow → 45%.

Question 23. Which colour has the greatest central angle?

1. Red
2. Yellow
3. Green
4. Blue

Solution: 4. Blue

Blue →  180°

Question 24. The proportion of the sector in red is

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{8}\)
4. \(\frac{1}{3}\)

Solution: 3. \(\frac{1}{8}\)

⇒ \(\frac{45^{\circ}}{360^{\circ}}=\frac{1}{8}\)

Question 25. The difference between the central angles for green and blue is

1. 45°
2. 90°
3. 180°
4. 22 \(\frac{1}{2}\)

Solution: 2. 90°

Central angle for blue = 180°

Central angle for green = 90°

Difference = 180° – 90° = 90°.

Question 26. A child has a block in the shape of a cube with one letter written on each face as shown below

The cube is thrown once. What is the probability of getting A

1. \(\frac{1}{3}\)
2. \(\frac{1}{6}\)
3. \(\frac{1}{2}\)
4. \(\frac{1}{4}\)

Solution:  1. \(\frac{1}{3}\)

Probability =\(\frac{2}{6}\)

= \(\frac{1}{3}\)

Question 27. A die is thrown. What is the probability of getting an even prime number

1. \(\frac{1}{6}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{2}\)

Solution: 1. \(\frac{1}{6}\)

Even prime number = 2

Probability =\(\frac{1}{6}\)

NCERT Solutions For Class 8 Maths Chapter 4 Data HandlingTrue Or False

1. In the n chart, a whole circle is divided into various sectors – True

2. The number of times a particular observation that occurs in a given data is called its frequency – True

3. The probability of a sure event is 0– False

4. The probability of an impossible event is 1 – False

5. In a throw of a die, the outcomes are equally likely – True

NCERT Solutions For Class 8 Maths Chapter 4 Data Handling Fill In The Blanks

1. The difference between the highest and the lowest values ofthe observations in a data is called the Range of the data.

2. A geometric representation showing the relationship between a whole and its parts is called a → Pie chart

3. The probability that it will rain tomorrow is 0. 75. What is the probability that it will not rain tomorrow  → 0.25

4. Find the range ofthe marks obtained by 10 students in class as follows   →  11, 9, 13, 18, 20, 18, 42, 41, 13,

5. In the throw of die, what is the probability of getting a number greater than 6  →  0

6. What is the total number of outcomes, when a coin is tossed →  2

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Introduction

A paper is a very common example of a plane surface. The curve obtained by joining several points consecutively without lifting the pencil from the paper is called a plane curve or simple curve or simply curve.

Some examples of simple curves are as follows

A circle is a very common example of a plane curve.

Some types of plane curves are as follows :

1. Open curve: A curve that does not cut itself is called an open curve.

Read and Learn More NCERT Solutions For Class 8 Maths

For example:

   is an open curve.

2. Closed curve:  A curve that cuts itself is called a closed curve.

For example:

 is a closed curve.

3. Simple closed curve: A closed curve that does not pass through a point more than once is called a simple closed
curve.

For example: 

is a simple closed curve.

Note: A curve can be straight also, i.e., A straight line is also a curve. It is an open

Question 1. Match the following: (Caution A figure may match more than one type)
Solution:

Compare your matchings with those of your friends. Do they agree?
Solution:

1-C

2-B

3-A

4-D

Please compare yourself your matchings with those of friends and find out whether they agree.

Polygons

A polygon is a simple closed-curve formed of only line segments. A triangle is a very common example of a polygon

Question 1. Try to give a few more examples and non-examples for a polygon.
Solution:

The following curves are examples of a polygon:

The following curves are non-examples of a polygon:

Question 2. Draw a rough figure of a polygon and identify its sides and vertices.
Solution:

Sides : AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL and LA.

Vertices: A, B, C, D, E, F, G, H, I, J, KandL.

Convex And Concave Polygons

A polygon is said to be convex if it has no portion of its diagonals in its exterior otherwise it is said to be a concave polygon.

Question 1. Can you find how these types of polygons differ from one another

Solution:

Yes, some of these are convex polygons while the others are concave polygons.

These polygons differ in the sense that in a convex polygon, the line segment joining any two points inside it lies completely inside the polygon whereas this property does not hold well in a concave polygon.

Question 2. Polygons that are convex have no portions of their diagonals in their exteriors. Is it true with concave polygons?
Solution:  No, it is not true with concave poly polygons

Question 3. Study the figures given below

Then try to describe in your own words what we mean by a convex polygon and what we mean by a concave polygon. Give two rough sketches of each kind
Solution:

By a convex polygon, we mean a polygon that has no portion of its diagonals in its exterior. On the contrary, by a concave polygon, we mean a polygon that has some portion of its diagonals in its exterior

1. Two rough sketches of a convex polygon

2. Two rough sketches of a concave polygon

Regular And Irregular Polygons

A polygon that is both ‘equiangular’ (has all angles of equal measure) and ‘equilateral’ (has all sides of equal measure) is called a regular polygon,

For example: A square, an equilateral triangle.

A polygon which is equiangular but not equilateral is called an irregular polygon. For example: a rectangle

Question 1. Is a rectangle a regular polygon? my?
Solution: No, A rectangle is not a regular polygon because it is equiangular but not equilateral

Question 2. Is an equilateral triangle a regular polygon? Why?
Solution: Yes, An equilateral triangle is a regular polygon because it is both equiangular and equilateral.

Question 3. Is there a triangle that is equilateral but not equiangular?
Solution:

No, there is no triangle which is equilateral but not equiangular

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

Question 1.  Given Here are some figures

Classify each of them based on the following;

1. Simple curve
2. Simple closed curve
3. Polygon
4. Convex polygon
5. Concave polygon

Solution:

1. Simple curve-1,2,5,6,7
2. Simple closed curve-1,2,5,6,7
3. Polygon – 1,2
4. Convex polygon – 2
5. Concave polygon – 1

Question 2. What is a regular polygon? State the name of a regular polygon of

1. 3 sides
2. 4 sides
3. 6 sides.

Solution:

A polygon, both ‘equilateral’ and ‘equiangular’, is called a regular polygon.

1. 3 sides: The name of the regular polygon of 3 sides is an equilateral triangle.
2. 4 sides: The name of the regular polygon of 4 sides is square.
3. 6 sides: The name of the regular polygon of 6 sides is a regular hexagon.

Sum Of The Measures Of The Exterior Angles Of A Polygon

The sum of the measures of the exterior angles of a polygon is 360°

Draw a polygon on the floor, using a piece of chalk. (In the figure, a pentagon ABCDE is shown)

We want to know the total measure of angles, i.e.

m∠1 + m∠2 + m∠3 + m∠4 + m∠5

Start at A. Walk along \(\overline{\mathrm{AB}} \text {. }\). On reaching B, we need to turn through an angle of m∠1, to walk along \(\overline{\mathrm{BC}} \text {. }\)

When we reach C, we need to turn E D through an angle of m∠2 to walk along CD.

We continue to move in this manner until we return to side AB. We would have made one complete turn.

∴  m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°

This is true whatever the number of E sides ofthe polygon.

Therefore, the sum of the measures of y the external angles of any polygon is 360°

Take a regular hexagon

Question 1. What is the sum of the measures of its exterior angles r, y, z, p, q, r?
Solution:

The sum of the measures of its exterior angles x, y, z, p , q, r is 360°.

Question 2. Is x = y = z = p = q = r? Why?
Solution:

Yes, x = y = z = p = q = r because

x + a = y + a = z + a = p + a = q + a = r + a

= 180°

Linear pair property

Question 3. What is the measure of each?

1. Exterior angle
2. Interior angle

Solution:

1. The measure of each exterior angle is \(\frac{360^{\circ}}{6}=60^{\circ}.\) = 60°

The sum of the measures of the exterior angles of a polygon is 360°. Also, all interior angles are of equal Consequently, all exterior angles are of equal measure (by Unear pair property).

2. The measure of each interior angle is 180°-60° = 120°

Note:

1. Each exterior angle of an n–sided regular polygon

⇒ \(\frac{360^{\circ}}{6}=60^{\circ} .\)

2. Number of sides of a regular polygon

= \(\frac{360^{\circ}}{n} .\)

3. Each interior angle of an n-sided regular polygon

= \(\frac{(n-2) 180^{\circ}}{n} \text { Or } \frac{(2 n-4) 90^{\circ}}{n}\)

=  \(\frac{(2 n-4)}{n}\)right angles.

Question 4. Repeal, this activity in the cases of

1. A regular octagon
2. A regular 20-gon.

Solution:

1. A regular octagon

1. The measure of each exterior angle = \(\frac{360^{\circ}}{8}=45^{\circ} .\)
2. The measure of each interior angle is 180° -45° =135°

2. A regular 20-gon

1. The measure of each exterior angle \(\frac{360^{\circ}}{20}=18^{\circ} .\)
2. The measure of each interior angle is 180° -18° =162°

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2

Question 1. Find x in the following figures

Solution:

1.  x°+ 125° + 125° = 360°.

∵ The sum of the measures of the exterior angles of a polygon is 360°

x + 250° = 360°

x = 360° -250°

x  = 110°

2. x + 70° + 90° + 60° + (180°- 90°)

= 360°

x + 160° + 60° + 90° = 360°

x + 220° + 90° = 360°.

The sum of the measures of the exterior angles of a polygon is 360° and the linear pair property.

x+ 310° = 360°

x = 360° – 310°

x = 50°.

Question 2. Find the measure ofeach exterior angle of a regular polygon of

1. 9 sides
2. 15sides

Solution:

1. 9 sides:

Here, n = 9

Measure of each exterior angle

= \(\frac{360^{\circ}}{n}=\frac{360^{\circ}}{9}\)

= \(40^{\circ}\)

2. 15 sides:

No, since 22 is not a factor of

Hero, n = 15

Measure of each exterior angle

= \(\frac{360^{\circ}}{n}=\frac{360^{\circ}}{15}\)

= \(24^{\circ}\)

Question 3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:

Let the number of sides be n. Then

n = \(\frac{360^{\circ}}{\text { Each exterior angle }} \text {. }\)

n = \(\frac{360^{\circ}}{24^{\circ}}\)

= 15

Hence, the number of sides is 15.

Question 4. How many sides does a regular polygon have if each of its interior angles is 165°?
Solution: 

Each interior angle = 165°

∴  Each exterior angle

= 180° – 165° = 15°

∴ Linear pair property

Let the number of sides be n. Then

n = \(\frac{360^{\circ}}{\text { Each exterior angle }}\)

n = \(\frac{360^{\circ}}{15^{\circ}}\)

= 24

Hence, the number of sides is 24.

Question 5.

1. Is it possible to have a regular polygon with a measure of each exterior angle as 22°?
2. Can it be an interior angle of a regular polygon? Why?

Solution:

1. No, since 22 is not a factor of m 360°

2. No, because by linear pair property, each exterior angle is 180° -22°= 158°

Which 360°n – 360°15 = 24°. is not. a factor of 360

Question 6.

1. What is the minimum interior angle possible for a regular polygon? Why ?
2. What is the maximum exterior angle possible for a regular polygon

Solution:

1. Minimum number of sides of a regular polygon = 3.

• A regular polygon of 3 sides is an equiangular triangle.
• Each interior angle of an equilateral triangle

n = \(\left(\frac{n-2}{n}\right) 180^{\circ}\)

= \(\left(\frac{3-2}{3}\right) 180^{\circ}\)

= 60°

Hence, the minimum interior angle possible for a regular polygon is 60°.

2. We know that the measure of an interior angle + the measure of a corresponding exterior angle

= 180°

By linear pair property

∵ Minimum interior angle possible for a regular polygon = 60°

∴ Maximum exterior angle possible for a regular polygon = 180°- 60° = 120°

Kinds Of Quadrilaterals

The important types of quadrilaterals are as follows :

• Trapezium
• Kite
• Parallelogram
• Rhombus
• Rectangle
• Square

Trapezium

A quadrilateral, which has only one pair of parallel sides, is called a trapezium.

Question 1. Study the above, [inures and discuss with your friends why some of them are trapeziums while some are not,(Note: The arrow marks indicate parallel lines).
Solution:

Some of them are trapeziums because they have only one pair of parallel sides. Some of them are not trapeziums because they have no pair of parallel sides.

Note: One of these figures is not a trapezium although it has only one pair of parallel  sides because a trapezium is essentially a quadrilateral whereas this figure is a pentagon

Question 2. Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, and 5 cm. Arrange them as shown

You get a trapezium. (Check it!) Which are the parallel sides here? Should the non-parallel sides be equal? You can get two more trapeziums using the same set of triangles. Find them out and discuss their shapes
Solution:

DE = CB (= 3 cm)

DC = EB (= 5 cm)

∴ Quadrilateral DCBE parallelogram

∴ DC || EB => DC|| AB

∴ ABCD is a trapezium. Its parallel sides are AB and DC.

Its non-parallel sides DA and CB are D 5 cm c not equal.

Thus, the non-parallel sides need not to be equal.

We get two more trapeziums using the same set of triangles. These are shown below

Question 3. Take four sol squares from your and your friend’s instrument boxes. Use different numbers of them to place side-by-side and obtain different trapeziums.

If the non-parallel sides of a trapezium arc are of equal length, we call it an isosceles trapezium. Did you get an isosceles trapezium in any of your investigations?

Solution:

Taking the different numbers of set-squares and pinning (hem side-by-side, we obtain the following trapeziums

No, we did not get an isosceles trapezium in any of our investigations.

Kite

A quadrilateral, which has exactly two pairs of equal consecutive sides, is called a kite.

Question 4. Take a thick white sheet. Fold the paper once. Draw two line segments of different lengths as shown in the figure

Cut along the line segments and open up. You have the shape of a kite. Has the kite any line symmetry?

Fold both the diagonals of the kite. Use the set square to check if they cut at right angles. Are the diagonals equal in length?

Verify by (paper-folding or measurement) if the diagonals bisect each other, i j By folding an angle of the kite on its opposite, check for angles of equal measure.

Observe the diagonal folds; do they indicate any diagonal being an angle bisector? Share your findings with others and list them.

Solution:

The kite has only one line of symmetry shown by a dotted line segment (a diagonal) AC and ∠B = ∠D

The diagonal AC is an angle bisector as it bisects ∠A and ∠C both.

Yes, the diagonals are cut at right angles.

No, the diagonals are not equal in length.

Yes, one of the diagonals bisects the other.

Question 5. Show that ΔABC mid ΔADC congruent, What do we infer from this?
Solution:

In ΔABC and  ADC

AB = AD

One pair of consecutive sides

BC = DC

Another pair of consecutive sides common

AC = AC

Δ ABC = Δ ADC

SSS Congruence Axiom

∴ ∠AC = ∠DAC

∠BCA = ∠DCA

i-e., diagonal AC bisects ∠BAD and ∠BCD both.

Parallelogram

A quadrilateral whose opposite sides are parallel is called a parallelogram

Question 1.  Study these figures and try to describe in your own words what we mean by a parallelogram. Share your observations with your friends
Solution:

By a parallelogram, we mean a quadrilateral whose opposite sides are parallel.

We observe that in a parallelogram,

Opposite sides are parallel and equal

Opposite angles are equal

Adjacent angles are supplementary

Diagonals bisect each other but the not equal

Elements Of A Parallelogram

The elements of a parallelogram are as follows :

1. Two pairs of opposite sides
2. Four pairs of adjacent sides
3. Two pairs of opposite equal angles
4. Four pairs of adjacent angles.

Question 2.  Are \(\overline{B C} \text { and } \overline{C D}\) adjacent sides too? Try to find two more pairs of adjacent sides.

Solution:

Yes, \(\overline{B C} \text { and } \overline{C D}\) are also adjacent sides. Two more pairs of adjacent sides are CD and DA; DA and AB.

Question 3. Identify other pairs of adjacent angles of the parallelogram
Solution:

The other pairs of adjacent angles of the parallelogram are∠C, ∠D, and ∠D, ∠A.

Question 4.

1.  Place \(\overline{A^{\prime} B^{\prime}}\) over \( \overline{D C}\). Do they coincide?
Solution:

Yes, they coincide.

2. What can you say about the lengths \( \overline{A B}\) and \( \overline{D C}\)
Solution:  These lengths are equal

3. Examine the lengths \( \overline{A D}\) and \( \overline{B C}\) What do you find?
Solution: These lengths are equal

Question 5.  Take two identical set squares with angles 30°- 60°- 90°andplace them adjacently to form a parallelogram as shown in the figure. Does this help you to verify the above property?

Solution: 

Yes, this helps us to verify the above property. We can further strengthen this idea through a logical argument.

Consider a parallelogram ABCD. Draw any one diagonal, say \(\overline{\mathrm{AC}}\). Looking at the angles,

∠1 =∠2 and ∠3 = ∠4

Alternate interior angles

Since in triangles ABC and ADC,

∠1 = ∠2, ∠3 =∠4 (alternate interior angles), and \(\overline{\mathrm{AC}}\) is common;

So, by ASA congruency condition,

Δ ABC≅ Δ CDA

This gives AB = DC and BC = AD

Question 6. Does this tell you anything about the measures of angles A and C ? Examine the same for angles B and D. State your findings.
Solution:

Findings: Angles A and C are equal.

Similarly, angles B and D are equal.

Also, pairs of consecutive angles (adjacent angles) ∠A and ∠B, ∠B and ∠C, ∠C and ∠D, ∠D and ∠A are supplementary.

Angles Of A Parallelogram

Note point

Property :

The opposite angles of a parallelogram are of equal measure

Question 1.  Take two identical 30°- 60°- 90° set squares and form a parallelogram as before. Does the figure obtained help us to confirm the above property

Solution:

Yes, the figure obtained helps us to confirm the above property. We can further justify this idea through logical arguments.

If \( \overline{A C}\) and \( \overline{B D}\) are the diagonals of the parallelogram, we find that

∠ 1 = ∠2 and ∠3 = ∠4

Alternate interior angles

Studying Δ ABC and Δ ADC separately, helps us to see that, by ASA-congruency condition,

Δ ABC ≅  Δ CDA

1 ∠1 = ∠2, ∠3 = ∠4, \( \overline{A C}\) and \( \overline{B D}\). This shows that ∠B and ∠D have the same measure. In the same way, we can get m ∠A = m∠C.

Question 2. Identify two more pairs of supplementary angles from the figure

Solution:

Two more pairs of supplementary angles from the figure are ∠B, ∠C, and ∠C, ∠D

Note Point

Property: The adjacent angles in a parallelogram are supplementary.

Question 3.  After showing m ∠R = m ∠N = 70°, can you find m ∠I and m∠G by any other method?
Solution:

RG || IN and RI is a transversal

∴ ∠R + ∠ I = 180°

Consecutive interior angles

70° + ∠ I = 180°

∠l  = 180° – 70° = 110°

Again,  RI || GN and RG is a transversal

∴∠R + ∠ G = 180°

Consecutive interior angles

70° +  ∠G = 180°

∠G = 180° -70° = 110°

∠I = ∠G = 110°.

Question 4. Is the mid-point, the same as O?
Solution: Yes

Question 5. Does this show that diagonal \(\overline{D B}\) bisects the diagonal \(\overline{A C}\)at point O ? Discuss it with your friends
Solution: Yes

Question 6. Repeat this activity to find where the I mid-point of \(\overline{D B}\) could lie.
Solution:

The midpoint of \(\overline{D B}\) will also lie at

Diagonals Of A Parallelogram

Note Point

Property:

The diagonals of a parallelogram bisect each other at the point. of their intersection.

Proof: Consider the parallelogram given below:

In Δ AOB and Δ COD

∠1 = ∠2

∠3 = ∠ 4

Alternate interior ∠ s

AB = DC

Opp. sides of parallelogram

∴  AOB ≅ COD

ASA congruence criterion

This gives AO = CO and BO = DO

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Question 1. Given a parallelogram ABCD. x + 100° = 180° Complete each statement along with the definition or property used

AD  = _______________

∠DCB =_______________

OC =_______________

m∠DAB + m∠CDA = _______________

Solution:

1. AD = BC

∵ Opposite sides of a supplementary parallelogram are equal

2. ∠DCB =∠DAB

∵ Opposite angles of a parallelogram are equal

3. OC = OA

∵ Diagonals of a parallelogram bisect each other

m∠DAB + m∠CDA = 180°

∵ Adjacent angles of a parallelogram are supplementary.

Question 2. Consider the following parallelo¬ grams. Find the values ofthe unknowns x, y, z.

Solution:

1. y = 100° 

∵ Opposite angles of a parallelogram are equal

x + 100° = 180°

Adjacent angles in a parallelogram are supplementary

x = 180° – 100°

x = 80°

z = x = 80°

∵ Opposite angles of a parallelogram are of equal measure

Thus, x = 80°, y = 100° and z = 80°

2. x + 50° = 180°

Adjacent angles in a parallelogram are supplementary

x= 180°- 50° = 130°

y = x = 130°

∵ Opposite angles of a parallelogram are of equal measure

180° – z = 50°

∵ Linear pair property and opposite angles of a parallelogram are of equal measure

z = 180° – 50° = 130°

Thus, x = 130°, y = 130° and z = 130°

3. x = 90°

x + y 4- 30° = 180°

Vertically opposite angles are equal

X + y + 30° = 180°

By angle sum property of a triangle

90° + y + 30° = 180°

120° + y = 180°

y = 180°- 120° = 60°

∵ AD|| BC and AC is a transversal

∴ y = z

Alternate interior angles

But y = 60°

z = 60°

Thus, .x = 90°, y = 60° and z = 60°

4. y = 80°

∵ Opposite angles of a parallelogram are of equal measure

x + 80° = 180°

Adjacent angles in a parallelogram are supplementary

x = 180° – 80°

x = 100°

180° – z + 80° = 180°

∵ Linear pair property and adjacent angles in a parallelogram are supplementary. Solution:

z = 80°

Thus, x = 100°, y = 80° and z = 80°

5. y = 112°

∵ Opposite angles of a parallelogram are equal

x+ y + 40° = 180°

∵ By angle sum property of a triangle

⇒  x + 112° + 40° = 180°

⇒  x + 152° = 180°

x = 180° – 152°

x = 28°

Z = X

Alternate interior angles

But x = 28°

∴ z = 28°

Thus, x= 28°, y = 112° and z = 28°

Question 3. Can a quadrilateral ABCD be a parallelogram if

1. ∠D + ∠B = 180°?
2. AB = DC = 8 cm, AD = 4 cm, and BC = 4.4 cm?
3. ∠A = 70° and ZC = 65° ?

Solution:

1. Can be, but needs not be.

2. No, in a parallelogram, opposite sides are equal; but here, AD ≠ BC.

3. No, in a parallelogram, opposite angles are of equal measure; but here ∠A ≠∠ C.

Question 4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure
Solution:

 

ABCD is a quadrilateral that is not a x parallelogram. Here,  ∠A = ∠ C. It is a kite. °

Question 5. The measures of two adjacent angles of a parallelogram are in the ratio 3: 2. Find the measure of each of the angles of the parallelogram.
Solution:

Let the two adjacent angles be 3x° and 2x

Then,

3x° + 2x° = 180°

∵ The sum of the two adjacent angles of a parallelogram is 180°

5x° = 180°

x° = \(\frac{180^{\circ}}{5}\)

x° = 36°

3x° = 3 × 36° = 108°

2x° = 2 × 36° = 72°

Since the opposite angles of a parallelogram are of equal measure, therefore the measures of the angles of the parallelogram are 72°, 108°; 72°, and 108°.

Question 6. Two adjacent angles of a parallelogram have equal measure. Find the immure of each of the angles of the parallelogram.
Solution:

Let the two adjacent, angles of a parallelogram be .v° each.

Then, x° + x° = 180°

The sum of the two adjacent angles of a parallelogram is 180°

2x° = 180°

x° =  \(\frac{180^{\circ}}{2}\)

x° = 90°

Since the opposite angles of a parallelogram are of equal measure, therefore, the measure of each of the angles of the parallelogram is 90°, i.e., each angle of the parallelogram is a right angle

Question 7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y, and z. State the properties you use to find them
Solution:

x = ∠HOP = 180° – 70° = 110°

The opposite angles of a parallelogram are of equal measure

∵ HOPE is a ||gm

∴ HE || OP

And HP is a transversal

∴ y = 40°

An alternate interior angles

40° + z + x = 180°

The adjacent angles in a parallelo¬ gram are supplementary

40° + z + 110° = 180°

z + 150° = 180°

z = 180° – 150°

z = 30°.

Thus, x =110°, y = 40°  and z = 30°

Question 8. The following figures GUNS mid RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:

1. For GUNS:

Since the opposite sides of a parallelogram are of equal length, therefore, 3x = 18

x = \(\frac{18}{3}\) = 6

And, – 1 = 26

3y = 26 + 1

3y = 27

x = \(\frac{27}{9}\)

Hence, x = 6; y = 9.

2. For RUNS:

Since the diagonals of a parallelogram bisect each other,

∴ x + y = 16______________ (1)

y + 7 = 20 ______________ (2)

From (2), y = 20 – 7 = 13

Putting y = 13 in (1), we get

x + 13 = 16 ⇒  x = 16 – 13

= 3.

Hence, x = 3; y = 13

Question 9. In the below figure, both RISK and CLUE are parallelograms. Find the value of x

Solution:  ∠RISK is a parallelogram

∴ ∠RIS = ∠RKS = 120°

The opposite angles of a parallelo¬ gram are of equal measure

Also, ∠RIS + ∠ISK = 180°.

The adjacent angles in a parallelo¬ gram are supplementary

120° + ∠ISK = 180°

∠ISK = 180° – 120°

∠ISK = 60°_____________ (1)

CLUE is a parallelogram

∠CES = ∠CLU = 70°_____________ (2)

The opposite angles of a parallelo¬ gram are of equal measure

In triangle EST,

x° + ∠TSE + ∠TES = 180°

By angle sum property of a triangle

⇒ x° + ∠ISK + ∠CES = 180°

⇒ x° + 60° + 70° = 180°

From (1) and (2)

x° + 130° = 180°

x° = 180° – 130°

x° = 50°.

Question 10. Explain how this figure is a trapezium. Which of its two sides is parallel

Solution:

∠KLM + ∠NML = 80°+ 100°

= 180°

∴ KL || NM

∵ The sum of consecutive interior angles is 180°

∴ KLMN is a trapezium.

Its two sides \(\overline{\mathrm{KL}} \text { and } \overline{\mathrm{NM}}\) are parallel.

Question 11. Find m∠C in the figure, if \(\overline{A B} \| \overline{D C}\)

Solution:

∴ \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\)

m∠C + m∠B = 180°

The sum of consecutive interior angles is 180

m∠C + 120° = 180°

m∠C = 180° – 120° = 60

Question 12. Find the measure of ∠P and ∠S, if \(\overline{S P} \| \overline{R Q}\) in the figure. (If you find m∠R, is there more than one method to find m∠P
Solution: 

⇒ \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) and \(\overline{\mathrm{SR}}\)isatransversal

∴m ∠P + m ∠Q = 180°

∴ The sum of consecutive interior angles is 180°

130° = 180°

m ∠P = 180° – 130°

m∠P = 50°

Again,

∵ \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) and \(\overline{\mathrm{SR}}\) and SR is a transversal.

m∠R + m ∠S = 180°

The sum of consecutive interior angles is 180°

90° + m∠S = 180°

m∠S = = 180° – 90°

= 90°

Yes; there is one more method of finding m ∠P if m ∠R is given and that is by using the angle sum property of a quadrilateral.

We have,

m∠P+m∠Q+m∠R+m∠S = 360°

m∠P+ 130° + 90° + 90° = 360°

∠S = 90° as above

m∠P + 310° = 360°

m∠P = 360°-310°

= 50°

Some Special Parallelograms

Rhombus

A quadrilateral whose all four sides are of equal length is called a rhombus.

Note Point

Property: The diagonals of a rhombus are perpendicular bisectors of one other.

A Rectangle

A rectangle is a parallelogram with equal angles.

Question 1. What Is The Full Meaning Of This Definition? Discuss With Your Friends.
Solution:

A rectangle is a parallelogram in which every angle is a right angle.

Note point:

Property: The diagonals of a rectangle are of equal length.

A Square

Property: A square is a rectangle whose all four sides have equal length.

The diagonals of a square are perpendicular bisectors of each other

Question 2. Take a square sheet, say PQRS. Fold along both the diagonals. Are their mid-points the same?

Check if the angle at 0 is 90° by using a set square. see that This verifies the property stated above.

Solution:

Yes, their mid-points are the same.

Yes, the angle at 0 is 90°.

We can justify this also by arguing logically

ABCD is a square whose diagonals meet at 0.

(Since the square is a parallelogram)

By the SSS congruency condition, we now see that

Δ AOD =Δ  COD

OA = OC, OD = OD, AD = CD

m ∠AOD = m∠COD

These angles being a linear pair, each is a right angle

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

Question 1. State whether True or False :

1. All rectangles are squares
2. All rhombuses are parallelograms
3. All squares are rhombuses and also rectangles
4. All squares are not parallelograms
5. All kites are rhombuses
6. All rhombuses are kites
7. Allparallelograms are trapeziums
8. All squares are trapeziums.

Solution:

(2), (3), (6), (7), (8) are true

(1), (4), (5) are false

Question 2. Identify all the quadrilaterals that have.

1. Four sides of equal length
2. Four right angles

Solution:

The quadrilaterals that have four sides of equal length are square and rhombus.

The quadrilaterals that have right angles are square and rectangle

Question 3. Explain how a square is

1. A quadrilateral
2. A parallelogram
3. A rhombus
4. A rectangle.

Solution:

1.  A quadrilateral: A square is 4 sides, so it is a quadrilateral.
2. A parallelogram: A square has its opposite sides parallel; so it is a parallelogram.
3. A rhombus: A square is a parallelogram with all 4 sides equal, so it is a rhombus.
4. A rectangle: A square is a parallelogram with each angle a right angle, so it is a rectangle

Question 4. Name the quadrilaterals whose diagonals:

1. Bisect each other
2. Are perpendicular bisectors of each other
3. Are equal.

Solution:

1. Bisect each other: The names of the quadrilaterals whose diagonals bisect each other are parallelogram; rhombus, square, and rectangle.
2. Are perpendicular bisectors of each other: The names of the quadrilaterals whose diagonals are perpendicular bisectors of each other are rhombus and square.
3. Are equal: The names of the quadrilaterals whose diagonals are equal are square and rectangle

Question 5. Explain why a rectangle is a convex quadrilateral.
Solution:

A rectangle is a convex quadrilateral because both of its diagonals he wholly in its interior.

Question 6. ABC is a right-angled triangle and AC = BD 0 is the mid-point of the side opposite to the right angle. Explain why is equidistant from A, B, and C. (The dotted lines are drawn addi¬ tionally to help you).

Solution:

Construction: Produce 130 to D such that BO = OD. Join AD and CD.

Proof: AO = OC

∵ O is the mid-point of AC by construction

BO=OD

∴ Diagonals of quadrilateral ABCD bisect each other.

∴  Quadrilateral ABCD isparallelogram.

Now, ∠ ABC = 90°

ABCD is a rectangle.

Since the diagonals of a rectangle bisect each other, therefore,

O is the mid-point of AC and BD both.

But AC = BD

The diagonals of a rectangle are equal

∴  \(\mathrm{OA}=\mathrm{OC}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \mathrm{BD}=\mathrm{OB}\)

OA = OB = OC

Question 7. A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular
Solution:

Different ways:

1. If each angle at the corner is 90° in
2. If the diagonals are equal in length.

Question 8. A square was defined as a rectangle with all sides equal. Can we define it as a rhombus with equal angles? Explore this idea
Solution:

Yes, a square can be defined as a rhombus with equal angles. If a rhombus has equal angles

Then the measure o each angle of the rhombus will be \(\frac{360^{\circ}}{4^{\circ}}\) = 90°

Thus, then it becomes a quadrilateral with all sides equal and each angle of measure 90° and hence a square.

Question 9. Can a trapezium have all angles equal? Can it have all sides equal? Explain.
Solution:

If a trapezium has all angles equal, then either it becomes a rectangle or a square.

If a trapezium has all sides equal, then either it becomes a rhombus or a square

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Multiple Choice Questions

Question 1. What, is the number of sides of a triangle

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

Solution: 3. 3

Question 2. What is the number of vertices of a triangle?

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

Solution: 3. 3

Question 3. What is the number of sides of a quadrilateral?

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

Solution: 4. 4

Question 4. What is the number of vertices of a quadrilateral?

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

Solution: 4. 4

Question 5. How many diagonals does a quadrilateral have?

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

Solution: 2. 2

Question 6. How many diagonals does a triangle have?

1. 0
2. 1
3. 2
4. 4

Solution: 1. 0

Question 7. How many diagonals does a regular hexagon have?

1. 2
2. 0
3. 4
4. 9

Solution: 4. 9

Question 8. What is the name of a regular polygon of 3 sides?

1. Equilateral triangle
2. Square
3. Regular hexagon
4. Regular octagon.

Solution: 1. Equilateral triangle

Question 9. What is the name of a regular polygon of 6 sides?

1. Square
2. Equilateral triangle
3. Regular hexagon
4. Regular octagon.

Solution: 3. Regular hexagon

Question10. What is the name of a regular polygon of 4 sides?

1. Regular hexagon
2. Regular octagon
3. Square
4. Equilateral triangle.

Solution: 3. Square

Question 11. The sum of the measures of the exterior angles of any polygon is

1. 90°
2. 180°
3. 360°
4. 720°

Solution: 3. 360°

Question 12. The number of sides of a regular polygon, whose exterior angle has a measure of 45°. is

1. 4°
2. 6°
3. 8°
4. 10°

Solution: 3. 8°

Number of sides =\(\frac{360^{\circ}}{45^{\circ}}\)

= 8°

Question 13. The measure of each exterior angle of a regular polygon of 9 sides is

1. 30°
2. 40°
3. 60°
4. 45°

Solution: 2. 40°

Required measure =\(\frac{360^{\circ}}{9^{\circ}}\)

= 40°

Question 14. The measure of each exterior angle of a regular polygon of 15 sides is

1. 30°
2. 45°
3. 60°
4. 24°

Solution: 4. 24°

Required measure =\(\frac{360^{\circ}}{24^{\circ}}\)

= 24°

Question 15. How many sides does a regular polygon have if the measure of an exterior angle is 24°?

1. 6°
2. 9°
3. 15°
4. 12°

Solution: 3. 15°

Number of sides =\(\frac{360^{\circ}}{15^{\circ}}\)

Question 16. How many sides does a regular polygon have if each of its interior angles is 165°

1. 12°
2. 24°
3. 9°
4. 6°

Solution: 2. 24°

Each exterior angle = 180° -165°= 15°

Number of sides = \(\frac{360^{\circ}}{15^{\circ}}\)

= 24°

Question 17. Which of (ho following statements is false?

1. All the angles of a rectangle are equal
2. No angle of a rectangle can be obtuse
3. The diagonals of a rectangle bisect each other
4. The opposite sides of a rectangle are not equal.

Solution: 4. The opposite sides of a rectangle are not equal.

Question 18. Which of the following statements is false?

1. Asquare is a rectangle whose adjacent sides are equal
2. A square is a rhombus whose one angle is a right angle
3. The diagonals of a square bisect each other at right angles
4. The diagonals of a square do not divide the whole square into four equal parts.

Solution: 4. The diagonals of a square do not divide the whole square into four equal parts.

Question 19. Which of the following statements is false?

1. All the rectangles are parallelograms
2. All the squares are rectangles
3. All the parallelograms are rectangles
4. All the rhombuses are parallelo¬ grams.

Solution: 3. All the parallelograms are rectangles

Question 20. Which ofthe following statements is true?

1. All the rectangles are squares
2. All the parallelograms are rhombuses
3. All the squares are rhombuses
4. Each parallelogram is a trapezium.

Solution: 3. All the squares are rhombuses

Question 21. Which ofthe following statements is true?

1. All the rhombuses are squares
2. Each square is a parallelogram
3. Each parallelogram is a square
4. Each trapezium is a parallelogram.

Solution: 2. Each square is a parallelogram

Question 22. One angle of a parallelogram is a right angle. The name ofthe quadrilateral is

1. Square
2. Rectangle
3. Rhombus
4. Kite

Solution: 2. Rectangle

Question 23. Two adjacent sides of a rod angle are equal. The name of the quadrilateral is kite

1. Square
2. Kite
3. Rhombus
4. None of these.

Solution: 1. Square

Question 24. Which of the following statements is false?

1. All the four sides of a parallelogram are equal.
2. The opposite angles of a parallelogram are equal
3. The diagonals of a parallelogram bisect each other
4. All the four sides of a rhombus are equal

Solution: 1. All the four sides of a parallelogram are equal.

Question 25. Which of the following statements is false?

1. All the four angles of a rhombus are equal
2. The diagonals of a rhombus bisect each other at right angles
3. A rectangle is a parallelogram
4. All squares are rectangles.

Solution: 1. All the four angles of a rhombus are equal

Question 26. If one angle of a parallelogram is of 65°, then the measure of the adjacent angle is

1. 65°
2. 115°
3. 25°
4. 90°.

Solution: 2. 115°

The measure of the adjacent angle

= 180° – 65° = 115°

Question 27. If ∠A of a parallelogram ABCD is of60°, then the measure of the opposite angle ∠C is

1. 60°
2. 120°
3. 30°
4. None of these.

Solution: 1. 60°

∠C =∠A = 60°.

Question 28. If all four sides of a parallelogram are equal and the adjacent angles are 120° and 60°, then the name of the quadrilateral is

1. Rectangle
2. Square
3. Rhombus
4. Kite

Solution: 3. Rhombus

Question 29. If the length of a side of a rhombus is 6 cm, (when the perimeter of the rhombus 

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

Solution: 3. 24 cm

Perimeter = 4 side = 4× 6 = 24 cm

Question 30. In a kite, what is false?

1. The diagonals are perpendicular to each other
2. One ofthe diagonals bisects the other
3. Both the pairs of opposing angles are equal
4. All the four sides are equal.

Solution: 4. All the four sides are equal.

Question 31. ABCD is a rectangle. Its diagonals meet at O.

OA = 2x-1, OD = 3x-2. Find x 

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

Solution: 1. 1

3x – 2 = 3x – 1

3x -2x = -1+ 2

x = 1

Question 32. Find the perimeter of the rectangle ABCD

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

Solution: 2. 12 cm

Perimeter = 2 (4 + 2) cm = 12 cm

Question 33. The four angles of a quadrilateral are in the ratio 1: 2 : 3: 4. The measure of its smallest angle is

1. 120°
2. 36°
3. 18°
4. 10°.

Solution: 2. 36°

Sum of the ratios =1 + 2+3 + 4=10

∴ Smallest angle =\(\frac{1}{10} \times 360^{\circ}\)

Question 34. In a parallelogram ABCD, ∠A : ∠B = 1: 2. Then, ∠A =

1. 30°
2. 60°
3. 45°
4. 90°.

Solution: 2. 60°

∠A + ∠B = 180°

∠A : ∠B = 1:2

Sum of the ratios =1 + 2 = 3

∴ \(\angle \mathrm{A}=\frac{1}{3} \times 180^{\circ}=60^{\circ}\)

Question 35. Two adjacent angles of a parallelogram are of equal measure. The measure of each angle of the parallelogram is

1. 45°
2. 30°
3. 60°
4. 90°.

Solution: 4. 90°.

x° + x° = 180°

2x° = 180°.

x° = \(\frac{180^{\circ}}{2^{\circ}}\)

x° = 90°

Question 36. ABCD is a parallelogram as shown. Find x and y

1. 1,7
2. 2,6
3. 3,5
4. 4,4

Solution: 3. 3,5

x + y = 8

y + 5 = 10

y=5

x + 5 = 8

x = 3.

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals True Or False

1. The number of diagonals in a decagon is 35 – True

2. The measure of each interior angle of a regular pentagon is 108° – True

3. The minimum number of sides a polygon can have is 4 – False

4. A regular polygon is both ‘equiangular’ and ‘equilateral’ – True

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Fill In The Blanks

1. The name of a three-sided regular polygon is → An equilateral triangle

2. The sum of the adjacent angles of a parallelogram is  →  180°

3. The angle between the diagonals of a rhombus is → 90°

4. The maximum number of obtuse angles that a quadrilateral can have is → 3

5. Apala has a beautiful garden in her flat. The shape ofthe garden is triangular. The sum of all the exterior angles taken in order of the triangular garden is → 360°

6. The sides of a rectangle are 24 cm and 10 cm. Find the length of one of the diagonals → 26 cm

7. ABCD is a parallelogram. Find ∠A – ∠C →0

8. Two adjacent angles of a parallelogram are (3x + 3)° and (lx + 7)°. Find x → 17

9. In the following figure, find the value of x → 70°

10. In the following figure, find the measure of ∠ABC – 130°