## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.

Fill in the blanks to make each of the following a true statement:

(i) 378 + 1024 = 1024 + …….

(ii) 337 + (528 + 1164) = (337 + ……..) + 1164

(iii) (21 + 18) + ……….. = (21 + 13) + 18

(iv) 3056 + 0 = ……….. = 0 + 3056

Solution:

(i) 378 + 1024= 1024 + 378 (Commutative property of addition)

(ii) 337 + (528 + 1164) = (337 + 528) + 1164 (Associative law of addition)

(iii) (21 + 18) + 13 = (21 + 13) + 18 (Associative law of addition)

(iv) 3056 + 0 = 3056 = 0 + 3056

Question 2.

Add the following numbers and check by reversing the order of addends :

(i) 3189 + 53885

(ii) 33789 + 50311.

Solution:

(i) 3189 + 53885 = 57074

Check 53885 + 3189 = 57074

∴57074

(ii) 33789 + 50311 = 84100

Check 50311 + 33789 = 84100

∴ 84100

Question 3.

By suitable arrangements, find the sum of:

(i) 311,528,289

(ii) 723, 834, 66, 277

(iii) 78, 203, 435, 7197, 422.

Solution:

(i) 311, 528, 289

Sum (311 +289)+ 528

= 600+ 528= 1128

(ii) 723 + 834 + 66 + 277

= (723 + 277) + (834 + 66)

= 1000 + 900 = 1900

(iii) 78, 203, 435, 7197, 422

Sum = (78 + 422) + (203 + 7197) + 435

= 500 + 7400 + 435

= 7900 + 435 = 8335

Question 4.

Fill in the blanks to make each of the following a true statement:

(i) 375 × 57 = 57 × ……….

(ii) (33 × 16) × 25 = 33 × (…….. × 25)

(iii) 37 × 24 = 37 × 18 + 37 × …………

(iv) 7205 × 1 = …………. = 1 × 7205

(v) 366 × 0 =

(vi) …………… × 579 = 0

(vii) 473 × 108 = 473 × 100 + 473 × ………….

(viii) 684 × 97 = 684 × 100 – …………… × 3

(ix) 0 ÷= 5 =

(x) (14 – 14) ÷ 7 = ………….

Solution:

(i) 375 × 57 = 57 × 375 (Commutative property of multiplication)

(ii) 33 × 16) × 25 = 33 × (16 × 25) (Associative law of multiplication)

(iii) 37 × 24 = 37 × 18 + 37 × 6 (Distributive law of multiplication)

(iv) 7205 × 1 = 7205 = 1 × 7205

(v) 366 × 0 = 0

(vi) 0 × 579 = 0

(vii) 473 × 108 = 473 × 100 + 473 × 8

(viii) 684 × 97 = 684 × 100 – 684 × 3

(ix) 0 ÷ 5 = 0

(x) (14 – 14) ÷ 7 = 0

Question 5.

Determine the following products by suitable arrangement:

(i) 4 × 528 × 25

(ii) 625 × 239 × 16

(iii) 125 × 40 × 8 × 25

Solution:

(i) 4 × 528 × 25 = 4 × 25 × 528

= 100 × 528 = 52800

(ii) 625 × 239 × 16 = 625 × 16 × 239

= 10000 × 239 = 2390000

(iii) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25

= 1000 × 1000 = 1000000

Question 6.

Find the value of the following:

(i) 54279 × 92 + 54279 × 8

(ii) 60678 × 262 – 60678 × 162

Solution:

(i) 54279 × 92 + 54279 × 8

= 54279 (92 + 8)

= 54279 × 100 = 5427900

(ii) 60678 × 262 – 60678 × 162

= 60678 (262 – 162)

= 60678 × 100 = 6067800

Question 7.

Find the following products by using suitable properties:

(i) 739 × 102

(ii) 1938 × 99

(iii) 1005 × 188

Solution:

(i) 739 × 102

= 739 × (100 + 2)

= 739 × 100 + 739 × 2

= 73900 + 1478 = 75378

(ii) 1938 × 99

= 1938 × (100- 1)

= 1938 × 100 – 1938 × 1

= 193800 – 1938 = 191862

(iii) 1005 × 188

= (1000 + 5) × (100 + 88)

= 1000 × 100 + 1000 × 88 + 5 × 100 + 88 × 5

= 100000 + 88000 + 500 + 440 = 188940

Question 8.

Divide 7750 by 17 and check the result by division algorithm.

Solution:

7750 ÷ 17

On dividing 7750 by 17, we get

Quotient = 455 and Remainder = 15

Check by division algorithm:

Divident = Divisior × Quotient + Remainder

= 17 × 455 + 15 = 7750

Question 9.

Find the number which when divided by 38 gives the quotient 23 and remainder 17.

Solution:

Divisor = 38,Quotient = 23

Remainder = 17

Dividend = divisor × quotient + remainder

= 38 × 23 + 17 = 874 + 17 = 891

Question 10.

Which least number should be subtracted from 1000 so that the difference is exactly divisible by 35.

Solution:

On dividing 1000 by 35

we get quotient = 28 and remainder 20

So, 20 should be subtracted from 1000.

Question 11.

Which least number should be added to 1000 so that 53 divides the sum exactly.

Solution:

On dividing 1000 by 53, we get quotient = 18 and remainder = 46. To get the remainder 0, we should add 53 – 46 = 7 to 1000.

∴ 7

Question 12.

Find the largest three-digit number which is exactly divisible by 47.

Solution:

Largest three digit no. = 999

On dividing 999 by 47, we get

Quotient = 21 and Remainder =12

So on subtracting 12 from 999, we get

999 – 12 = 987

Question 13.

Find the smallest five-digit number which is exactly divisible by 254.

Solution:

Smallest 5 digit number = 10000

On dividing 10000 by 254, we get

Remainder = 94

So 254 – 94 = 160 should be added to 10000 to get the smallest 5 digit number divisible by 254.

∴ 10000 + 160 = 10160

Question 14.

A vendor supplies 72 litres of milk to a student’s hostel in the morning and 28 litres of milk in the evening every day. If the milk costs?39 per litre, how much money is due to the vendor per day?

Solution:

Supply of milk in morning = 72 litres

Supply of milk in evening = 28 litres

Cost of per litre milk = ₹ 39

Money of per day = ₹ 39 (72 l + 28 l)

= ₹ 39 × 100 = ₹ 3900

Question 15.

State whether the following statements are true (T) or false (F):

(i) If the product of two whole numbers is zero, then atleast one of them will be zero.

(ii) If the product of two whole numbers is 1, then each of them must be equal to 1.

(iii) If a and b are whole numbers such that a ≠ 0 and b ≠ 0, then ab may be zero.

Solution:

(i) True

(ii) True

(iii)False

Question 16.

Replace each * by the correct digit in each of the following:

Solution: