Class 6 Mathematics

Class 6 Maths Chapter 3 Playing With Numbers

Directions: In questions 1 to 14, out of the four options, only one is correct. Write the correct answer.

1 Sum of the number of primes between 16 to 80 and 90 to 100 is

(1) 20
(2) 18
(4) 16
(3) 17

Solution: (3): Prime numbers between 16 to 80 are 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.

So, there are 16 prime numbers between 16 to 80.

Also, 97 is the only prime number between 90 to 100.

So, there is only a prime number between 90 to 100.

Required sum = 16 +1 = 17

Read and Learn More Class 6 Maths Exemplar Solutions

 

2. Which of the following statements is not true?

(1) The HCF of two distinct prime numbers is 1
(2) The HCF of two coprime numbers is 1
(3) The HCF oftwo consecutive even numbers is 2
(4) The HCF of an even and an odd number is even

Solution: (4): The HCF of an even and an odd number is always an odd number.

3. The number of distinct prime factors of the largest 4-digit number is

(1) 2
(3) 5
(2) 3
(4) 11

Solution: (2): The largest 4-digit number is 9999.

Now,

9999 = 3 x 3×11 x101

So, distinct prime factors of 9999 are 3, 11 and 101, i.e., 3 distinct prime factors.

4. The number of distinct prime factors of the smallest 5-digit number is

(1) 2
(2) 4
(4) 8
(3) 6

Solution: (1): The smallest 5-digit number is 10000.

Now,

10000 = 2x2x2x2x5x5x5x5

So, distinct prime factors of10000 are 2 and 5, i.e., 2 distinct prime factors.

5. If the number 7254*98 is divisible by 22, the digit at* is

(1) 1
(2) 2
(3) 6
(4) 0

Solution: (3): 7254*98 is divisible by 22, if it is divisible by both 2 and 11.

Given number is even, therefore it is divisible by 2.

7254*98 is divisible by 11, if (7 + 5 + * + 8) – (2 + 4 + 9) or (20 + *)- 15 or 5 + * is divisible by 11.

The digit at * should be filled by 6.

6. The largest number which always divides the sum of any pair of consecutive odd numbers is

(1) 2
(2) 4
(3) 6
(4) 8

Solution:(2): The sum of any pair of consecutive odd numbers comesin the form ofmultiple of 4.

The required largest number is 4.

7. A number is divisible by 5 and 6.It may not be divisible by

(1) 10
(3) 30
(2) 15
(4) 60

Solution: (4): The LCM of 5 and 6 is 30.

And 30 is divisible by 10, 15 and 30 but not by 60.

8. The sum of the prime factors of 1 729 is

(1) 13
(2) 19
(4) 39
(3) 32

Solution: (4): We have,

The prime factors of 1729 = 7 x 13 x 19.

The sum of the prime factors of 1729 = 7+ 13 +19 = 39

9. The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is

(1) 6
(2) 4
(3) 16
(4) 8

Solution: (2): Since the odd natural numbers other than1 are 3, 5, 7, 9 and so on.

Now, the predecessor and successor of 3 are 2 and 4 respectively, and their product is 2×4 = 8

Similarly, the predecessor and successor of 5 are 4 and 6 respectively and their product is 4 x 6 = 24 and so on.

Thus, the above shows that the greatest number which always divides the product of the predecessor and successor of an odd natural other than 1 is 4.

10. The number of common prime factors of 75, 60,105 is

(1) 2
(2) 3
(3) 4
(4) 5

Solution: (1): We have,

75 = 3 x 5 x 5

60 = 2 x 2 x 3 x 5

105 = 3 x 5 x 7

The common prime factors of 75, 60 and 105 are 3 and 5 i.e., 2 in number.

11. Which of the following pairs is not coprime?

(1) 8,10
(3) 1,3
(2) 11, 12
(4) 31,33

Solution: (1): 8 and 10 are not co-prime numbers.

Their common factor other than 1 is 2

12. Which of the following numbers is divisible by 11?

(1) 1011011
(2) 1111111
(3) 22222222
(4) 3333333

Solution: (3): The difference of the sum of digits of 22222222 at even and odd places is 0. It must be divisible by 11.

13. LCM of10, 15 and 20 is

(1) 30
(2) 60
(4) 180
(3) 90

Solution: (2): We have,

The LCM of 10, 15 and 20 is 2x2x3x5=60

14. LCM of two numbers is 1 80. Then which of the following is not the HCF of the numbers?

(1) 45
(2) 60
(3) 75
(4) 90

Solution:(3): We have,

The factors of 180 = 2 x 2 x 3 x 3 x 5

75 does not divide 180

75 can not be the HCF of the numbers whose LCM is 180.

Directions: In questions 15 to 32 state whether the given statements are true (T) orfalse (6).

15. Sum of two consecutive odd numbers is always divisible by 4.

Solution: True

16. If a number divides three numbers exactly, it must divide their sum exactly.

Solution: True

17. If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately.

Solution: False

18. If a number is divisible both by 2 and 3, then it is divisible by 12.

Solution: False

Since, a number is divisible by both 2 and 3 implies that it is also divisible by 6.

19. A number with three or more digits is divisible by 6, if the number formed by its last two digits (i.e., ones and tens) is divisible by 6.

Solution: False

A number is divisible by 6 if the number is divisible by both 2 and 3.

20. A number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

Solution: True

21. If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 9.

Solution: False

If the sum of the digits of a number is divisible by 3, then the number it self is divisible by 3.

22. All numbers which are divisible by 4 may not be divisible by 8.

Solution: True

Since, 12 is divisible by 4 but not by 8.

23. The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.

Solution: False

HCF of two or more numbers is smaller than their LCM.

24. LCM of two or more numbers is divisible by their HCF.

Solution: True

25. LCM of two numbers is 28 and their HCF is 8.

Solution: False

The LCM of two numbers i.e., 28 is not divisible by their HCF, i.e., 8.

Their HCF cannot be 8.

26. LCM of two or more numbers may be one of the numbers.

Solution: True

27. HCF of two or more numbers may be one of the numbers.

Solution: True

28. Any two consecutive numbers are coprime.

Solution: True

The common factor of any two consecutive numbers is 1.

29. If the HCF of two numbers is one of the numbers, then their LCM is the other number.

Solution: True

The product of two numbers =HCF * LCM.

30. The HCF of two numbers is smaller than the smaller ofthe numbers.

Solution: False

Since, HCF of two numbers is less than or equal to the smaller of the numbers.

31. The LCM of two numbers is greater than the larger ofthe numbers.

Solution: False

Since, LCM of two numbers is greater than or equal to the larger of the numbers.

32. The LCM of two coprime numbers is equal to the product ofthe numbers.

Solution: True

33. A number is a _________ of each ofits factor.

Solution: Multiple

34.__________ is a factor of every number.

Solution:1

35. The number of factors of a prime number is_______

Solution: 2

36. A number for which the sum of all its factors is equal to twice the number is called a___________ number.

Solution: Perfect

37. The numbers having more than two factors are called___________

Solution: Composite

38. 2 is the only__________

Solution: Prime

39. Two numbers having only 1 as a common factor are called___________numbers.

Solution: Co-prime

40. Number of primes between 1 to 100 is_______.

Solution: 25 : Prime numbers between I to 100,are 2. 3, 5. 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So, there are 25 primes between 1 to 100.

41. If a number has ________in ones place, then it is divisible by 10.

Solution: 0

42. A number is divisible by 5, if it has____________ or in its ones place.

Solution: 0, 5

43. A number is divisible by________if it has any of the digits 0, 2, 4, 6, or 8 in its ones place. 1,1,1

Solution: 2

44. If the sum of the digits in a number is a __________ of 3, then the number is divisible by

Solution: Multiple

45. If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by_____________, then the number is divisible by 11.

Solution: 11

46. The LCM of two or more given numbers is the lowest of their common________.

Solution: Multiple

47. The HCF of two or more given numbers is the highest of their common____________.

Solution: Factors

48. Find the LCM of 80, 96, 125, 160.

Solution: We have,

The LCM of 80, 96, 125 and 160 is = 2x2x2x2x2x3x5x5x5 = 12000

49. Find the LCM of 160, 170 and 90.

Solution: We have,

The LCM of 160, 170 and 190

=2x2x2x2x2x3x3x5x17

= 24480

50. Determine the sum of the four numbers as given below:

(1) successor of 32
(2) predecessor of 49
(3) predecessor of the predecessor of 56
(4) successor of the successor of 67

Solution:

Since, successor of32is 33, predecessor of 49 is 48, predecessor of the predecessor of 56 is 54 and successor of the successor of 67 is 69.

The required sum = 33 + 48 + 54 + 69 = 204

51. Determine the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

Solution: We have,

Since, the LCM of 3, 4 and 5 is 2 x 2 x 3 x 5 = 60.

The required number is 60 + 2 = 62

So, 62 is the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

52. A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Solution:

Given:

merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity.

A merchant has 3 kinds of oil with different quantities like 120 litres, 180 litres and 240 litres

Since, he wants lo soli the oil by filling the three kinds of oil in tins of equal capacity, so the greatest capacity of such a tin is the HCF of 120, 180 and 240.

Now, 120 = 2x2x2x3x5

180 = 2x2x3x3x5

240 = 2x2x2x2x3x5

The required greatest capacity of a tin

= (2 x 2 x 3 x 5) litres

= 60 litres

The greatest capacity of a tin = 60 litres

53. Find a 4-digit odd number using each of the digits 1, 2, 4 and 5 only once such that when the first and the last digits are interchanged, it is divisible by 4.

Solution: By using the digits 1, 2, 4 and 5 only once, we get a 4 digit odd number 4521.

When we interchanged its first and last digits we get a new number, i.e., 1524 which is divisible by 4.

Thus, the required number is 4521.

54. Using each of the digits 1, 2, 3 and 4 only once, determine the smallest 4-digit number divisible by 4.

Solution: By using the digits 1, 2, 3 and 4 only once, the smallest 4-digit number which is divisible by 4 is 1324.

55. Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively. If she wants to buy stamps only of one denomination, what is the greatest denomination of stamps she must buy to mail the three parcels?

Solution:

Given

Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively.

The postal charges to mail three parcels are 20, 28 and 36 respectively.

Also, Fatima wants to buy stamps only of one denomination.

So, to find the greatest denomination of stamps, we find the HCF of 20, 28 and 36.

 

Now, 20 = 2 x 2 x 5

28 = 2 x 2 x 7

36 = 2 x 2 x 3 x 3

The HCF of 20, 28 and 36 is 2×2 = 4

So, ? 4 is the greatest denomination of stamps, she must buy to mail the three parcels.

56. Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand, what is the minimum number of packets of each brand, he should buy?

Solution:

Given

Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand

A shopkeeper has three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively.
Also, a shopkeeper wants to buy equal number of biscuits of each brand for that we need to find the LCM of 12, 15 and 21.

 

The LCM of 12, 15 and 21 = 2x2x3x5x7 = 420

Thus, the required number of packets of

brand A =420/12= 35,

brand B =420/15 =28 and

brand C =420/21 = 20

57. The floor of a room is 8 m 96 cm long and 6 m 72 cm broad. Find the minimum number of square tiles of the same size needed to cover the entire floor.

Solution:

Given

The floor of a room is 8 m 96 cm long and 6 m 72 cm broad.

Length of floor of a room = 8 m 96 cm = 896 cm

Breadth of floor of the room = 6 m 72 cm = 672 cm

To find the minimum number of square tiles of same size needed to cover the entire floor, we find the LCM of 896 cm and 672 cm.

LCM of 672 and 896 is 2 X 2 X 2 x 2 X 2 X 2 X 2 X 3 X 7 = 2688

Number of square tiles =( 2688/672×2688/896)

= 4×3 = 12

58. In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject. What should be the minimum number of books in each shelf?

Solution:

Given

In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject.

Number of books of English = 780

Number of books of Science = 364

 

780 = 2x2x3x5x13

364 = 2 x 2 x 7 x 13

The HCF of 780 and 364 = 2 x 2 x 13 = 52

Thus, the minimum number of books in each shelf = 52

59. In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block. On which stops will both of them stop if they start from the entrance of the colony?

Solution:

Given

In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block.

There are 100 blocks in a colony numbering 1 to 100.

A school van stops at every sixth block and a school bus stops at every tenth block.

We have to find the common stops at which they both stop if they start from a same entrance.

We need to find the LCM of 6 and 10.

The LCM of 6 and 10 is 2 * 3 * 5 = 30. i.e.

Firstly both will stop at 30th block, then at 60th block and lastly at 90th block.

60. Test the divisibility of following numbers by 11

(1) 5335
(2) 9020814

Solution: (1) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number 5335 is (5 + 3) = (3 + 5) = 8- 8 = 0, which is divisible by 11.

5335 is divisible by 11.

(2) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from
the right) of the number 9020814 is (4 + 8 + 2 + 9) -(1 + 0 + 0) = 23-1 = 22, which is divisible by 11.

9020814 is divisible by 11.

61. Using divisibility tests, determine which of the following numbers are divisible by 4?

(1) 4096
(2) 21084
(3) 31795012

Solution: (1) We have, 4096

Since, the last two digits 96 is divisible by 4.

4096 must be divisible by 4.

(2) We have, 21084

Since, the last two digits 84 is divisible by 4.

21084 must be divisible by 4.

(3) We have, 31795012

Since, the last two digits 12 is divisible by 4.

31795012 must be divisible by 4.

62. Using divisibility test determine which of the following numbers are divisible by 9?

(1) 672
(2) 5652

Solution: (1) We have, 672

Since, the sum of all the digits of 672 is 15, which is not divisible by 9.

672 is not divisible by 9.

(2) We have, 5652

Since, the sum of all the digits of 5652 is 18, which is divisible by 9.

5652 must be divisible by 9.

 

 

Class 6 Maths Chapter 8 Decimals

Question 1. 0.7499 lies between

1. 0.7 and 0.74
2. 0.75 and 0.79
3. 0.749 and 0.75
4. 0.74992 and 0.75

Solution: (3) On observing the given options, we notice that 0.7499 lies between 0.749 and 0.75.

[As 0.7499 is the successor of 0.749 and the predecessor of 0.75]

Question 2. 0.023 lies between

1. 0.2 and 0.3
2. 0.02 and 0.03
3. 0.03 and 0.029
4. 0.026 and 0.024

Solution: (2): On observing the given options, we notice that 0.023 lies between 0.02 and 0.03.

Question 3. 0.07 + 0.008 is equal to

1. 0.15
2. 0.015
3. 0.078
4. 0.78

Solution: (3): We have to find the value of 0.07 + 0.008

Read and Learn More Class 6 Maths Exemplar Solutions

First converting the given decimals into like decimals, we get 0.070 and 0.008

By writing these decimals in column form and adding them, we get

⇒ \(\begin{array}{r}
0.070 \\
+0.008 \\
\hline 0.078 \\
\hline
\end{array}\)

Hence, 0.07 + 0.008 is 0.078

Question 4. Which of the following decimals is the greatest?

1. 0.182
2. 0.0925
3. 0.29
4. 0.038

Solution: (3): First converting the given decimals into like decimals, we get 0.1820, 0.0925, 0.2900 and 0.0380.

Thus, on comparing the above, 0.2900 is the greatest among the given decimals

Question 5. Which of the following decimals is the smallest?

1. 0.27
2. 1.5
3. 0.082
4. 0.103

Solution: (3): First converting the given decimals into like decimals, we get 0.270, 1.500, 0.082, and 0.103.

Thus, on comparing the above, 0.082 is the smallest among the given decimals.

Question 6. 1 3.572 correct to the tenth place is

1. 10
2. 13.57
3. 14.5
4. 13.6

Solution: (4): When 13.572 is estimated correctly to the tenth place, we get 13.6

Question 7. 15.8 -6.73 is equal to

1. 8.07
2. 9.07
3. 9.13
4. 9.25

Solution: (B): We have given, 15.8- 6.73. Firstly converting the given decimals into like fractions, then writing the decimals in column form and subtracting, we get

∴ \(\begin{array}{r}
15.80 \\
-6.73 \\
\hline 9.07 \\
\hline
\end{array}\)

Question 8. The decimal 0.238 is equal to the fraction

1. \(\frac{119}{500}\)
2. \(\frac{238}{25}\)
3. \(\frac{119}{25}\)
4. \(\frac{119}{50}\)

Solution:

(A): We have given 0.238, which can be written in the form of

⇒ \(\frac{238}{1000}=\frac{238 \div 2}{1000 \div 2}\left[\begin{array}{l}
\text { On dividing the numerator } \\
\text { and denominator by } 2
\end{array}\right]\)

∴ \(=\frac{119}{500}\)

Question 9. \(9+\frac{2}{10}+\frac{6}{100}\) is equal to the decimal number______.
Solution:

9.26: The L.C.M. of 10 and 100 is 100.

⇒ \(9+\frac{2}{10}+\frac{6}{100}\)

⇒ \(=9 \times \frac{100}{100}+\frac{2}{10} \times \frac{10}{10}+\frac{6}{100}\)

⇒ \(\frac{900+20+6}{100}=\frac{926}{100}=9.26\)

Question 10. Decimal 1 6.25 is equal to the fraction______.
Solution:

⇒ \(\frac{65}{4}:\) We have

⇒ \(16.25=\frac{1625}{100}=\frac{1625 \div 25}{100 \div 25}\)

⇒ \(\frac{65}{4} \text { or } 16 \frac{1}{4}\)

Question 11. Fraction \(\frac{7}{25}\) is equal to the decimal number _______.
Solution:

0.28: We have, \(\frac{7}{25}=\frac{7}{25} \times \frac{4}{4}=\frac{28}{100}=0.28\)

Question 12. 4.55 + 9.73 = _______.
Solution: 14.28 : 4.55 + 9.73 = 14.28

Question 13. 8.76-2.68 = _______.
Solution: 6.08: 8.76- 2.68 = 6.08

Question 14. The value of 50 coins of 50 paisa = ₹ _______.
Solution:

25 : The value of 50 coins of 50 paisa = 50 x 50 paisa

= 2500 paisa \(=₹ \frac{2500}{100}\)

= ₹ 25

Question 15. 3 Hundredths + 3 tenths = _______.
Solution: 0.33 : 3 hundredths + 3 tenths = 0.33

Question 16. The place value of a digit at the tenth place is 10 times the same digit at the one’s place.
Solution: False

Since the place value of a digit at the tenth place is \(\frac{1}{10}\) times the same digit at the one’s place

∴ \(\frac{1}{10}=10x\)

Question 17. The place value of a digit at the hundredth place is \(\frac{1}{10}\) times the same digit at the tenth place.
Solution: True

Question 18. The decimal 3.725 is equal to 3.72 correct to two decimal places.
Solution: False

Since the digit at the thousandth place of 3.725 is 25

3.725 can be written as 3.73 (after rounding off at hundredths place)

Question 19. In the decimal form, fraction \(\frac{25}{8}=3.125\)
Solution: True

⇒ \(\frac{25}{8}=\frac{25 \times 125}{8 \times 125}\)

On multiplying the numerator and denominator by 125

∴ \(\frac{3125}{1000}=3.125\)

Question 20.The decimal 23.2 = \(23 \frac{2}{5}\)
Solution: False

We have \(23.2=\frac{232}{10}=\frac{116}{5}\)

⇒ \(23 \frac{1}{5}\)

Question 21. 3.03 + 0.016 = 3.019
Solution: False

Firstly convert 3.03 and 0.016 into like fractions, writing the decimals in column form and finally by adding we get,

⇒ \(\begin{array}{r}
3.030 \\
+0.016 \\
\hline 3.046
\end{array}\)

Question 22. 42.28-3.19 = 39.09
Solution: True

⇒ \(\begin{array}{r}
42.28 \\
-3.19 \\
\hline 39.09 \\
\hline
\end{array}\)

Question 23. 19.25 <19.053
Solution: False

Since, the digit at a tenth place of 19.25 is 2 and the digit at a tenth place of 19.053 is 0, where 2>0.

19.25 > 19.053

Question 24. 13.730 = 13.73
Solution: True

Since, after converting the given decimals into like decimals we get, 13.730 = 13.73.

Question 25. 3.25…3.4
Solution:

<: Converting the given decimals into like decimals, they become 3.25 and 3.40. The whole number part of these is the same. On comparing, we get their tenth digits 2 < 4

3.25 < 3.4

Question 26. 6.25…\( \frac{25}{4}\)
Solution:

=: \(6.25=\frac{625}{100}=\frac{625 \div 25}{100 \div 25}=\frac{25}{4}\)

∴ \(6.25\frac{25}{4}\)

Question 27. Arrange 12.142, 12.124, 12.104, 12.401 and 12.214 in ascending order.
Solution:

Given

12.142, 12.124, 12.104, 12.401 and 12.214

The digits are already given in the form of like decimals

Clearly,

12.104 < 12.124 < 12.142 < 12.214 < 12.401

Question 28. Write the largest four-digit decimal number less than 1 using the digits 1, 5, 3 and 8 once.
Solution:

The required number is 0.8531, which is the largest four-digit decimal number less than 1.

Question 29. Using the digits 2, 4, 5 and 3 once, write the smallest four-digit decimal number.
Solution:

The required number is 0.2345, which is the smallest four-digit decimal number

Question 30. Express \(\frac{11}{20}\) as a decimal.
Solution:

Given

latex]\frac{11}{20}[/latex]

We have, \(\frac{11}{20}=\frac{11 \times 5}{20 \times 5}=\frac{55}{100}=0.55\)

Question 31. Express \(3 \frac{2}{5}\) as a decimal.
Solution:

Given

\(3 \frac{2}{5}\)

We have \(3 \frac{2}{5}=\frac{(3 \times 5)+2}{5}=\frac{15+2}{5}=\frac{17}{5}\)

Now, \(\frac{17}{5}=\frac{17 \times 2}{5 \times 2}=\frac{34}{10}=3.4\)

Question 32. Express 0.041 as a fraction.
Solution:

Given 0.041

We have, \(0.041=\frac{41}{1000}\)

Question 33. Express 6.03 as a mixed fraction.
Solution:

Given 6.03

We have, \(6.03=\frac{603}{100}\)

⇒ \(6 \frac{\mathrm{3}}{100}\)

Question 34. Convert 5201 g to kg.
Solution:

Given

5201

We have ,\(5201 \mathrm{~g}=\frac{5201}{1000} \mathrm{~kg}\)

[1kg = 1000 g]

= 5.201 kg

Question 35. Convert the 2009 paise to rupees and express the result as a mixed fraction.
Solution:

We have 2009 paise \(=₹ \frac{2009}{100}=₹ 20.09\)

∴ \(\text { or } ₹ 20 \frac{9}{100}\)

36. Convert 1 537 cm to m and express the result as an improper fraction.
Solution:

We have, 1537 cm \(=\frac{1537}{100} \mathrm{~m}\)

[1 m = 100 cm]

= 15.37 m

Question 37. Convert 2435 m to km and express the result as a mixed fraction
Solution:

Given 2435

we have, 2453 m \(=\frac{2435}{1000} \mathrm{~km}\)

[1 km = 1000 m]

= 2.435 km

Firstly, convert the fraction \(\frac{2435}{1000}\) into the simplest form, for dividing the numerator and denominator by 5, we get

⇒ \(\frac{2435 \div 5}{1000 \div 5} \mathrm{~km}=\frac{487}{200} \mathrm{~km}\)

∴ \(2 \frac{87}{200} \mathrm{~km}\)

Question 38. Round off 20.83 to the nearest tenths.
Solution: The estimated value of 20.83 to the nearest tenths is 20.8

Question 39. Round off 75.1 95 to the nearest hundredths.
Solution: The estimated value of 75.195 to the nearest hundredths is 75.20

Question 40. Round off 27.981 to the nearest tenths.
Solution: The estimated value of 27.981 to the nearest tenths is 28.0

Question 41. What should be added to 25.5 to get 50?
Solution: According to question, 25.5 +? = 50

⇒ \(\begin{array}{r}
50.0 \\
-25.5 \\
\hline 24.5 \\
\hline
\end{array}\)

? = 50 – 25.5

= 24.5

Thus, the required number is 24.5

Question  42. Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.
Solution:

Given

Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes.

Alok purchased,

Potatoes =1 kg 200 g = 1.200 kg

Dhania = 250 g = 0.250 kg

Onion = 5 kg 300 g = 5.300 kg

Palak = 500 g = 0.500 kg

Tomatoes = 2 kg 600 g = 2.600 kg

The total weight of the above purchases

⇒ \(\begin{array}{r}
1.200 \mathrm{~kg} \\
0.250 \mathrm{~kg} \\
5.300 \mathrm{~kg} \\
0.500 \mathrm{~kg} \\
+2.600 \mathrm{~kg} \\
\hline \underline{9.850 \mathrm{~kg}} \\
\hline
\end{array}\)

Question 43. Arrange in ascending order: 0.011, 1.001, 0.101, 0.110
Solution:

Given

0.011, 1.001, 0.101, 0.110

Since, all the decimals are already given in like fractions, i.e., 0.011, 1.001, 0.101, 0.110.

Arranging them in ascending order, we get 0.011, 0.101, 0.110, 1.001

Question 44. Add the following: 20.02 and 2.002
Solution:

We have, 20.02 and 2.002 To add the above decimals, we must convert them into like decimals first.

Writing 20.020 and 2.002 in a column

So,

⇒ \(\begin{array}{r}
20.020 \\
+2.002 \\
\hline 22.022 \\
\hline
\end{array}\)

which is the required sum.

Question 45. The energy content of different foods is as follows:

Which food provides the least energy and which provides the maximum? Express the least energy as a fraction of the maximum energy.
Solution:

Milk provides the least energy and rice provides the maximum energy

The required fraction = \(\frac{3.0}{5.3}=\frac{3.0 \times 10}{5.3 \times 10}\)

∴ \(=\frac{30}{53}\)

Question 46. Which one is greater? 1 metre 40 centimetres + 60 centimetres or 2.6 metres.
Solution:

1 metre 40 centimetres + 60 centimetres = 1.40 metres + 0.60 metres

[ 100 centimetres =1 metre]

= 2.00 metres

Since, 2.6 >2.00

2.6 metres is greater.

Class 6 Maths Chapter 2 Whole Numbers

Directions: In questions 1 to 11, out of the four options, only one is correct. Write the correct answer.

1. The number of whole numbers between 38 and 68 is

(1) 31
(2) 30
(3) 29
(4) 28

Solution (3): There are 29 whole numbers between 38 and 68.

2. The product of the successor and predecessor of 999 is

(1) 999000
(2) 998000
(3) 989000
(4) 1998

Solution (2): Successor of 999 = 999 +1 = 1000

Predecessor of 999 = 999-1 = 998

Now, their product = 998 x 1000 = 998000

Read and Learn More Class 6 Maths Exemplar Solutions

3. The product of a non-zero whole number and its successor is always

(1) an even number
(2) an odd number
(3) a prime number
(4) divisible by 3

Solution (1): The product of a non-zero whole number (even/odd) and its successor (odd/ even) is always an even number.

4. A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is

(1) 0
(2) 25
(3) 50
(4) 75

Solution(3): Let the whole number be x.

According to question,

Required sum = (x + 25) + (25- x)

= x + 25 + 25- x = 50

5. Which of the following is not true?

(1) (7 + 8) + 9 = 7 + (8 + 9)
(2) (7 x 8) x 9 = 7 x (8 x 9)
(3) 7 + 8 x 9 = (7 + 8) x (7 + 9)
(4) 7 x (8 + 9) = (7 x 8) + (7 x 9)

Solution (3): 7 + 8 x 9 = 7 + 72 = 79,

(7 + 8) x (7 + 9)- 15 × 16 = 240 and 79 x 240

6. By using dot (•) patterns, which ofthe following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?

(1) 9
(2) 10
(3) 11
(4) 12

Solution (2): As we know that every number can be arranged as a line.

The number 10 can be shown as

Also, the number 10 can be shown as a triangle as given below :

And, the number 10 can also be shown as a rectangle, as given below :

7. Which of the following statements is not true?

(1) Both addition and multiplication are associative for whole numbers.
(2) Zero is the identity for multiplication of whole numbers.
(3) Addition and multiplication both are commutative for whole numbers.
(4) Multiplication is distributive over addition for whole numbers.

Solution (2): Zero is the identity for addition of whole numbers.

8. Which of the following statements is not true?

(1) 0 + 0 = 0
(3) 0x0 = 0
(2) 0-0 = 0
(4) 0-0 = 0

Solution (4): 0 + 0 is not defined.

9. The predecessor of 1 lakh is

(1) 99000
(2) 99999
(4) 100001
(3) 999999

Solution (2): 1 lakh = 100000

Predecessor of 100000 = 100000-1 = 99999

10. The successor of 1 million is

(1) 2 millions
(2) 1000001
(3) 100001
(4) 10001

Solution(2): 1 million = 1000000

Successor of 1000000 = 1000000 +1 = 1000001

11. Number of even numbers between 58 and 80 is

(1) 10
(2) 11
(3) 12
(4) 13

Solution (1): Even numbers between 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

So, here are 10 even numbers between 58 and 80.

Directions: In questions 12 to .37 stole whether the given statements are true (T) or false (F)

12. Successor of a one digit number is always a one digit number.

Solution: False

The successor of a one digit number 9 is a two digit number, i.c., 10.

13. Successor of a 3-digit number is always a 3-digit number.

Solution: False

The successor of a 3-digit number 999 is a 4-digit number, i.e., 1000.

14. Predecessor of a two digit number is always a two digit number.

Solution: False

The predecessor of a two digit number 10 is a one digit number, i.e., 9.

15. Every whole number has its successor.

Solution: True

16. Every whole number has its predecessor.

Solution: False

0 is a whole number and it does not have any predecessor.

17. Between any two natural numbers, there is one natural number.

Solution: False

Since, 1 and 2 are two natural numbers and there is no natural number between1 and 2.

18. The smallest 4-digit number is the successor of the largest 3-digit number.

Solution: True

The smallest 4-digit number is 1000, i.e., 999 +1 and the largest 3-digit number is 999.

19. Of the given two natural numbers, the one having more digits is greater.

Solution: True

20. Natural numbers are closed under addition.

Solution: True

The sum of two natural numbers is also a natural number.

21. Natural numbers are not closed under multiplication.

Solution: False

The multiplication of two natural numbers is also a natural number.

22. Natural numbers are closed under subtraction.

Solution: False

2 and 5 are natural numbers, but their subtraction, 2-5 =-3 isnot anatural number.

23. Addition is commutative for natural numbers.

Solution: True

Since, a + b = b + a, where ‘a’ and ‘V are natural numbers.

24. 1 is the identity for addition of whole numbers.

Solution: False

0 is the identity for addition of whole numbers.

As 0 + a = a + 0 = a, where a is any whole number.

25. 1 is the identity for multiplication of whole numbers.

Solution: True

a×1 = 1 x a = a, where ‘a’ is any whole number.

26. There is a whole number which when added to a whole number, gives the number itself.

Solution: True

27. There is a natural number which when added to a natural number, gives the number itself.

Solution: False

28. If a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.

Solution: True

29. Any non-zero whole number divided by itself gives the quotient 1.

Solution: True

30. The product of two whole numbers need not be a whole number.

Solution: False

The product of any two whole numbers will always be a whole number.

31. A whole number divided by another whole number greater than 1 never gives the quotient equal to the former.

Solution : True

32. Every multiple of a number is greater than or equal to the number.

Solution: True

33. The number of multiples of a given number is finite.

Solution: False

Since, the number of multiples of a given number is infinite.

34. Every number is a multiple of itself.

Solution: True

35. Every whole number is the successor of another whole number.

Solution: False

0 is a whole number but it is not a successor of any whole number.

36. Sum of two whole numbers is always less than their product.

Solution: False

Product of two whole numbers may or may not be greater than their sum.

37. If the sum of two distinct whole numbers is odd, then their difference also must be odd.

Solution: True

Directions:In questions 38 to 59,fill in the blanks to make the statements true.

38. The smallest whole number is_________.

Solution 0 : 0 is the smallest whole number.

39. Successor of 1061 59 is_________.

Solution 106160: Successor of 106159 is 106159 +1, i.e., 106160

40. Predecessor of 1 00000 is_________.

Solution 99999: Predecessor of 100000 is 100000-l, i.e., 99999

41. 400 is the predecessor of__________.

Solution 401: 400 is the predecessor of 400 + 1, i.e., 401

42.________ is the successor of the largest 3 digit number.

Solution 1000: Largest 3 digit number = 999

Successor of 999 is 999 + 1, i.e., 1000

43. If 0 is subtracted from a whole number, then the result is the_________ itself.

Solution: Number

44. Whole numbers are closed under and under________.

Solution: Addition, multiplication

45. Natural numbers are closed under and under_______.

Solution: Addition, multiplication

46. Division of a whole number by ___________ is not defined.

Solution: 0

47. Multiplication is distributive over__________ for whole numbers.

Solution: Addition

48. 2395 x_________ = 6195×2395

Solution 6195: Since, multiplication is commutative for whole numbers

49. 1001 x 2002 = 1001 x (1001 +_________)

Solution: 1001

1001 x 2002 = 1001 x (1001 +1001)

50. 10001 x 0 = __________

Solution: 0.

10001 x 0 = 0

51. 2916 x_________ = 0

Solution: 0.

2916 x 0 = 0

52. 9128 x_______= 9128

Solution 1 : Since,1 is the multiplicative identity for whole numbers.

53. 125 + (68 + 17) = (125 +_______)+ 17

Solution 68: Since, addition is associative for whole numbers.

54. 8925x 1 =_____

Solution: 8925

8925x 1= 8925

55. 19 x 12 + 19 = 19 x (12 +_______)

Solution 1: Since multiplication is distributive over addition for whole numbers.

56. 24 x 35 = 24 x 1 8 + 24 X_______

Solution: 17

24 x 35 = 24 x 1 8 + 24 X 17

57. 32x(27×19) = (32x__________)x19

Solution 27: Since, multiplication is associative for whole numbers.

58. 786×3 + 786×7 =_______

Solution: 7860 : 786 x 3 + 786 x 7 = 786 * (3 + 7)

= 786 x 10 = 7860

786×3 + 786×7 =7860

59. 24 x 25 = 24 X □÷4= 600

Solution: 100

60. Given below are two columns – Column I and Column II. Match each item of Column I with the corresponding item of Column II.

 

 

Class 6 Maths Chapter 1 Knowing Our Numbers

Directions: In questions 1 to 13, out of the four options, only one is correct. Write the correct answer.

1. The product of the place values of two 2’s in 428721 is

(1) 4
(2) 40000
(3) 400000
(4) 40000000

Solution : (3) Place values of 2’s in 428721 are 20000 and 20

The required product = 20000 x 20 = 400000

2. 3x 1 0000 + 7x 1 000 + 9×1 00 + 0x10 + 4 is the same as

(1) 3794
(2) 37940
(3) 37904
(4) 379409

Read and Learn More Class 6 Maths Exemplar Solutions

Solution (3) : 3 x 10000 + 7 x 1000 + 9 x 100 + 0 x 10 + 4

= 30000 + 7000 + 900 + 4 = 37904

3. If 1 is added to the greatest 7-digit number, it will be equal to

(1) 10 thousand
(2) 1 lakh
(3) 10 lakh
(4)1,00,00,000

Solution (4): The greatest 7-digit number = 99,99,999

Now, 99,99,999 +1 = 1,00,00,000

4. The expanded form of the number 9578 is

(1) 9×10000 + 5×1000 + 7×10 + 8×1
(2) 9x 1000 + 5x 100 + 7x 10 + 8×1
(3) 9x 1000 + 57x 10 + 8×1
(4) 9x 100 + 5x 100 + 7x 10 + 8×1

Solution (2) : Expanded form of 9578 = 9 x 1000 + 5 x 100 + 7 x 10 + 8×1

5. When rounded off to the nearest thousands, the number 85642 is

(1) 85600
(2) 85700
(3) 85000
(4) 86000

Solution (4): Round off 85642 to the nearest thousands = 86000

6. The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is

(1) 9652
(2) 9562
(3) 9659
(4) 9965

Solution: (4): The largest 4-digitnumber formed by 5, 9, 2 and 6 using digit ‘9’ twice = 9965

7. In Indian System of Numeration, the number 58695376 is written as

(1) 58,69,53,76
(2) 58,695,376
(3) 5,86,95,376
(4) 586,95,376

Solution: (3): In Indian System of Numeration, 58695376 can be written as 5,86,95,376

8. One million is equal to

(1) 1 lakh
(3) 1 crore
(2) 10 lakh
(4) 10 crore

Solution: (2): One million = 1,000,000

Also, 10,00,000 = 10 lakh

9. The greatest number which on rounding off to nearest thousands gives 5000, is

(1) 5001
(2) 5559
(3) 5999
(4) 5499

Solution: (4) : (1) Rounding off 5001 to nearest thousands = 5000

(2) Rounding off 5559 to nearest thousands = 6000

(3) Rounding off 5999 to nearest thousands = 6000

(4) Rounding off 5499 to nearest thousands = 5000

And 5499 > 5001

10. Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is

(1) 6975430
(2) 6043579
(3) 6034579
(4) 6034759

Solution (3): The new number formed = 6034579

11. Which of the following numbers in Roman numerals is incorrect?

(1) LXXX
(2) LXX
(3) LX
(4) LLX

Solution: (4) : LLX is incorrect because L can be used exactly once in Roman numerals.

12. The largest 5-digit number having three different digits is

(1) 98978
(2) 99897
(3) 99987
(4) 98799

Solution: (3) : The largest 5-digit number with three different digits is 99987

13. The smallest 4-digit number having three different digits is

(1) 1102
(2) 1012
(3) 1020
(4) 1002

Solution: (4) : The smallest 4-digit number with three different digits is 1002.

Directions: In questions 14 to 29 state whether the given statements are true (T) or false (F)

14. In Roman numeration, a symbol is not repeated more than three times.

Solution: True

15. In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs.

Solution: False

If a symbol is repeated, then its value is added as many times as it occurs in Roman numeration

16. 5555 = 5×1000 + 5×100 + 5×10 + 5×1

Solution: True

17. 39746 = 3 × 10000 + 9 × 1000 + 7 × 100 + 4×10 + 6

Solution: True

18. 82546 = 8 x 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6

Solution: False

Since, 82546 = 8 × 10000 + 2× 1000 + 5×100 + 4×10 + 6

19. 532235 = 5 X 100000 + 3 x 10000 + 2 x 1000 + 2X 100 + 3X 10 + 5

Solution: True

20. XXIX = 31

Solution: False

Since, XXIX = 29

21. LXXIV = 74

Solution: True

22. The number LIV is greater than LVI.

Solution: False

Since, LIV = 54 and LVI = 56

LVI is greater than LTV

23. The numbers 4578, 4587, 5478, 5487 are in descending order.

Solution: False

Since, the numbers 4578, 4587, 5478, 5487 are in ascending order.

24. The number 85764 rounded off to nearest hundreds is written as 85700.

Solution: False

The number 85764 rounded off to nearest hundreds is 85800.

25. Estimated sum of 7826 and 1 2469 rounded off to hundreds is 20,000.

Solution: False

Round off 7826 to hundreds is 7800.

Round off 12469 to hundreds is 12500.

Required sum = 7800 + 12500 = 20300

26. The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403.

Solution: False

The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.

27. The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen.

Solution: False

The number 81652318 will be read as eight crore sixteen lakh fifty two thousand three hundred eighteen.

28. The largest 4-diglt number formed by the digits 6, 7, 0, 9 using each digit only once is 9760.

Solution: True

29. Among kilo, milli and centi, the smallest is centi.

Solution: False

Among kilo, milli and centi, the smallest is milli.

Directions: In questions 30 to 44, fill in the blanks to make the statements true

30. (1) 10 million =__________ crore

Solution:1

(2) 10 lakh =_______ million.

Solution: 1

31. (1) 1 metre = ____________ millimetres

Solution: 1000

(2) 1 centimetre =___________ millimetres

Solution: 10

(3) 1 kilometre =__________millimetres

Solution: 10,00,000

33. 100 thousands =________ lakh.

Solution: 1

34. Height of a person is 1 m 65 cm. His height in millimetres is_________.

Solution: 1650 : 1 m 65 cm = (1000 + 650) mm = 1650mm

35. Length of river ‘Narmada’ is about 1290 km. Its length in metres is___________.

Solution: 1290000 : 1290 km = (1290 x 1000) m = 1290000m

36. The distance between Srinagar and Leh is 422 km. The same distance in metres is___________

Solution: 422000: 422 km= (422 x 1000) m = 422000 m

37. Writing of numbers from the greatest to the smallest is called an arrangement in__________ order.

Solution: Descending

38. By reversing the order of digits of the greatest number made by five different non zero digits, the new number Is the________ number of five digits.

Solution: Smallest*

By reversing the order of digits of the greatest number made by five different non zero digits, the new number is the smallest number of these digits.

39. By adding 1 to greatest_______ digt number, we get ten lakh.

Solution: 6: Greatest 6-digit number = 999999 By adding1 to 999999, we get 1000000.

40. The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian
System of Numeration as_______.

Solution: 5,23,78,401

41. In Roman Numeration, the symbol X can be subtracted from ________, M and C only.

Solution: L

42. The number 66 in Roman numerals is__________.

Solution: LXVI: 66 =LXVI

43. The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was____________.

Solution: 2,538,000

44. The smallest 6 digit natural number ending in 5 is__________

Solution: 100005

45. Arrange the following numbersin descending order: 8435, 4835, 13584, 5348, 25843

Solution: Descending order of given numbers is, 25843, 13584, 8435, 5348, 4835

46. Of the following numbers which is the greatest? Which is the smallest? 38051425, 30040700, 67205602

Solution: The greatest number is 67205602 and the smallest number is 30040700.

47. Write In expanded form :

(1) 74836
(2) 574021
(3) 8907010

Solution:(1) 74836- 7 x10000) x 4 x 1000 + 8 x 100 +3x 10+6×1

(2) 574021 =5x 1 00000 + 7x 1 0000 + 4x 1000 + 0x 100 + 2x 10+ 1 x1

(3) 8907010 = 8 x 1000000 + 9x 1 00000 + 0 x 10000 + 7 x 1000 + 0 x 100 + 1 x 10 + 0 x 1

48. As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.

(1) Maharashtra 96878627
(2) Andhra Pradesh 76210007
(3) Bihar 82998509
(4) Uttar Pradesh 166197921

Solution: Ascending order -» (2), (3), (1), (4)

Descending order —» (4), (1), (3), (2)

49. The diameter of Jupiter is 142800000 metres. Insert commas suitably and write the diameter according to International System of Numeration.

Solution:

Given

The diameter of Jupiter is 142800000 metres.

The diameter of Jupiter is 142,800,000 metres.

50. India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.

Solution:

Given

India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001.

Total increaseinpopulation from 1961 to 2001= 1028 millions- 439 millions = 589 millions

According toIndian System of Numeration, the increase in population = 58,90,00,000

51. Radius of the Earth is 6400 km and that of Mars is 4300000 m. Whose radius is bigger and by how much?

Solution:

Given

Radius of the Earth is 6400 km and that of Mars is 4300000 m.

Radius of the Earth = 6400 km = 6400 x 1000 m = 6400000 m

And radius of the Mars = 4300000 m

Radius of the Earth is greater than the radius of the Mars by (6400(X)0- 4300000) m = 2100000 m.

52. In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively. Write the populations of these two states in words.

Solution:

Given

In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively.

The population of Tripura = 3,199,203 i.e., Three million one hundred ninety-nine thousand two hundred three.

And the population of Meghalaya = 2,318,822, i.e., Two million three hundred eighteen thousand eight hundred twenty two.

53. In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.

Solution:

Given

In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month.

Number of children getting polio drops in March 2008 = 2,12,583

Number of children getting polio drops in April 2008 = 2,16,813

The required difference of the number of children getting polio drops in the two months = 2,16,813- 2,12,583 = 4,230

54. A person had ? 1000000 with him. He purchased a colour T.V. for ? 16580, a motor cycle for ? 45890 and a flat for ? 870000. How much money was left with him?

Solution:

Given :

A person had 1000000 with him. He purchased a colour T.V. for 16580, a motor cycle for  45890 and a flat for  870000.

Total amount a personhad = 1000000

The amounthe spent on a colour T.V. = 16580

The amount he spent on a motorcycle = 45890

The amount he spent on a flat =870000

Total amount he spent = (16580 + 45890 + 870000) = 932470

Thus, the amount left with him = 1000000- 932470 = 67530

55. Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district. Find the number of the remaining vitamin tablets.

Solution:

Given:

Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district.

Total tablets of Vitamin A= 180000

Number of tablets distributed among the students in a district = 18734

The number of remaining vitamin tablets -180000-18734-161266

56. Chinmay had ? 610000. He gave 87500 to Jyoti, 126380 to Javed and 350000 to John. How much money was left with him?

Solution:

Chinmay had total amount = 610000

The amount he gave to Jyoti = 87500

The amount he gave to Javed = 126380

The amount he gave to John = 350000

Total amount given by Chinmay = (87500 + 126380+ 350000) = 563880

Thus, the amount left with him = 610000- 563880 = 46120

57. Find the difference between the largest number of seven digits and the smallest number of eight digits.

Solution: The smallest number of eight digits = 10000000

The largest number of seven digits = 9999999

The required difference = 10000000-9999999 =1

58. A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number. What are these digits?

Solution:

Given

A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number.

A mobile number consists of 10 digits.

If the first four digits of the number are 9, 9, 8 and 7 and the last three digits of the number are 3, 5 and 5.

Thus, for the greatest possible number, the remaining distinct digits are 6, 4 and 2.

59. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.

Solution:

Given

A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9.

A mobile number consists of 10 digits.

If the first four digits are 9, 9, 7 and 9.

Thus, the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0 is 9979003568.

60. In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place. Write the number.

Solution:

Given

In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place.

A number consists of 5 digits.

Now, the digit at ten’s place = 4,

the digit at unit’s place = 1/4×4=1,

tine digit at hundred’s place = 0,

the digit at thousand’s place = 5 x1 = 5

the digit at ten thousand’s place = 2×4 =8

Therefore, the number is 85041.

61. Find the sum of the greatest and the least six digit numbers formed by the digits 2, 0, 4, 7,6, 5 using each digit only once.

Solution: By using the digits 2, 0, 4, 7, 6, 5

The greatest number formed = 765420,

and the least number formed = 204567

The required sum= 765420 + 204567= 969987

62. A factory has a container filled with 35874 litres of cold drink. In how many bottles of 200 ml capacity each can it be filled?

Solution:

Given:

A factory has a container filled with 35874 litres of cold drink.

Quantity of cold drink in a container = 35874 litres = 35874 x 1000 ml = 35874000 ml

The capacity of one bottle = 200 ml

The required number of bottles = 35874000 + 200 = 179370

Therefore, 179370 bottles can be filledby cold drink.

63. The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate. In all how many illiterate persons are there in the town?

Solution:

Given:

The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate.

Total population of a town = 450772

Since, one out of every 14 personsis illiterate.

The number of illiterate persons in the town = 450772 + 14 = 32198

Therefore, 32198 persons are illiterate in the town.

64. Make the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten’s place.

Solution: By using the digit 5 at ten’s place, the greatest 5-digit number is 98756, and the smallest 5-digitnumber is 10253

65. How many grams should be added to 2 kg 300 g to make it 5 kg 68 g?

Solution: 5 kg 68 g = (5 x 1000 + 68) g = 5068 g and 2 kg 300 g = (2 x 1000 + 300) g = 2300 g

The required number of grams should be added = 5068 g- 2300 g = 2768 g or 2kg 768 g

2768 grams should be added to 2 kg 300 g to make it 5 kg 68 g

66. A box contains 50 packets of biscuits each weighing 1 20 g. How many such boxes can be loaded in a van which cannot carry beyond 900 kg?

Solution:

Given

A box contains 50 packets of biscuits each weighing 1 20 g.

The total weight of a box containing 50 packets ofbiscuits each weighing 120 g = 50 x 120 g = 6000 g

The capacity of a van = 900 kg = 900 x 1000 g = 900000 g

The required number of boxes = 900000/6000 = 150

Therefore, 150 boxes can be loaded in the van.

67. How many lakhs make five billions?

Solution: 50000 lakhs make 5 billions.

68. How many millions make 3 crores?

Solution: 30 millions make 3 crores.

69.Estimate each of the following by rounding off each number to nearest hundreds:

(1) 874 + 478
(2) 793 + 397
(3) 11244 + 3507
(4) 17677 + 13589

Solution:(1) 874 rounded off to The nearest hundreds-900

478 rounded off to the nearest hundreds -500

Estimated sum = 900 + 500 – 1400

(2) 793 rounded off to the nearest hundreds = 800

397 rounded off to the nearest hundreds = 400

Estimated sum = 800 + 400 = 1200

(3) 11244 rounded off to the nearest hundreds= 11200

3507 rounded off to the nearest hundreds = 3500

Estimated sum = 11200 + 3500 = 14700

(4) 17677roundedofftothenearesthundreds = 17700

13589 rounded off to the nearest hundreds = 13600

Estimated sum = 17700 + 13600 = 31300

70. Estimate each of the following by rounding off each number to nearest tens:

(1) 11963-9369
(2) 76877-7783
(3) 10732-4354
(4) 78203-16407

Solution: (1) 11963 rounded off to the nearest tens- 11960

9369 rounded off to the nearest tens = 9370

Estimated difference=11960- 9370= 2590

(2) 76877 rounded off to the nearest tens = 76880

7783 rounded off to the nearest tens = 7780

Estimated difference= 76880- 7780= 69100

(3) 10732 rounded off to the nearest tens = 10730

4354 rounded off to the nearest tens = 4350

Estimated difference = 10730- 4350 = 6380

(4) 78203 rounded off to the nearest tens = 78200

16407 rounded off to the nearest tens = 16410

Estimated difference = 78200 – 16410 = 61790

71. Estimate each of the following products by rounding off each number to nearest tens:

(1) 87×32
(2) 311 × 113
(3) 3239 × 28
(4) 1385×789

Solution: (1) 87 rounded off to the nearest tens = 90

32 rounded off to the nearest tens = 30

Estimated product = 90 * 30 = 2700

(2) 311 rounded off to the nearest tens = 310

113 rounded off to the nearest tens = 110

Estimated product = 310 * 110 = 34100

(3) 3239 rounded off to the nearest tens= 3240

28 rounded off to the nearest tens = 30

Estimated product = 3240 * 30 = 97200

(4) 1385 rounded off to the nearest tens=1390

789 rounded off to the nearest tens = 790

Estimated product= 1390 * 790= 1098100

72. The population of a town was 78787 in the year 1991 and 95833 in the year 2001. Estimate the increase in population by
rounding off each population to nearest hundreds.

Solution:

Given

The population of a town was 78787 in the year 1991 and 95833 in the year 2001.

78787 rounded off to the nearest hundreds = 78800

95833 rounded off to the nearest hundreds = 95800

The estimated increase in population = 95800-78800 = 17000

73. Estimate the product 758 x 6784 using the general rule.

Solution: 758 can be rounded off to 800 and 6784 can be rounded off to 7000 Estimated product= 800 × 7000 = 5600000

The product 758 x 6784 using the general rule = 5600000

74. A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year. What is the total production of all the three items in that year?

Solution:

Given

A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year.

Number of shirts produced by the factory = 216315

Number of trousers produced by the factory = 182736

Number of jackets produced by the factory = 58704

Total production of the factory = 216315 + 182736 + 58704 = 457755

Total production of the factory = 457755

75. A vessel has 13 litres 200 mL of fruit juice. In how many glasses each of capacity 60 mL can it be filled?

Solution:

Given:

A vessel has 13 litres 200 mL of fruit juice.

Quantity of fruit juice in a vessel = 13 L 200 mL

= (13 x 1000 + 200) mL = 13200 mL

Capacity of one glass = 60 mL

The required number of glasses = 13200 ÷ 60 = 220

Therefore, 220 glasses can be filled by fruit juice.

76. A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight. Find the total weight that can be carried by both the vehicles.

Solution:

Given

A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight.

Total weight can be carried by a tempo= (482×15) kg = 7230kg

and the total weight can be carried by a van = (518×15) kg = 7770 kg

Thus, the total weight that can be carried by both the vehicles = (7230 + 7770) kg = 15000 kg

The total weight that can be carried by both the vehicles = 15000 kg

77. In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items. Find the total amount spent by her on the above items.

Solution:

Given :

In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items.

Amount spent by Leela on food and decoration = 216766

Amount spent by her on jewellery = 122322

Amount spent by her on furniture = 88234

Amount spent by her on kitchen items = 26780

Total amount spent by her = (216766 + 122322 + 88234 + 26780) = 454102

Total amount spent by her = 454102

78. A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.

Solution:

Given :

A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule.

Quantity of medicine in one capsule = 500 mg

Quantity ofmedicinein 12 capsules or 1 strip = (500 x 12) mg = 6000 mg- 6 g

Quantity of medicine in 5 strips or 1 box = (6 x 5) g = 30 g

Quantity of medicinein 32 boxes = (30 x 32)g = 960g

The total weight in grams of medicine in 32 such boxes = 960g

 

• Chapter 1 Knowing Our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing With Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shapes
• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio and Proportion