# ML Aggarwal Whole Number Exe-2.2 Class 6 ICSE Maths Solutions

ML Aggarwal Whole Number Exe-2.2 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-2.2 Questions for Whole Number as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Whole Number Exe-2.2 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-2 Whole Number Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-2.2 Questions Edition 2023-2024

### Whole Number Exe-2.2

ML Aggarwal Class 6 ICSE Maths Solutions

Page-36

#### 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

(i)  378

(ii) 528

(iii) 13

(iv) 3056

### Whole Number Exe-2.2

ML Aggarwal Class 6 ICSE Maths Solutions

Page-37

#### Question 2. Add the following numbers and check by reversing the order of addends :

(i) 3189 + 53885
(ii) 33789 + 50311.

(i) 3189 + 53885 = 57074

(ii) 33789 + 50311 = 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.

(i) 311 + 528 + 289 = 1128

(ii) 723 + 834 + 66 + 277 = 1900

(iii) 78 + 203 + 435 + 7197 + 422 = 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 = ………….

(i) 375 x 57 = 57 x 375

(ii) (33 x 16) x 25 = 33 x (16 x 25)

(iii) 37 x 24 = 37 x 18 + 37 x 6

(iv) 7205 x 1 = 7205 = 1 x 7205

(v) 366 x 0 = 0

(vi) 0 x 579 = 0

(vii) 473 x 108 = 473 x 100 + 473 x 8

(viii) 684 x 97 = 684 x 100 – 684 x 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

(i) 4 x 528 x 25

= 52800

(ii) 625 x 239 x 16

= 2390000

(iii) 125 x 40 x 8 x 25

= 1000000

#### Question 6. Find the value of the following:

(i) 54279 × 92 + 54279 × 8
(ii) 60678 × 262 – 60678 × 162

(i) 54279 x 92 + 54279 x 8

= 4993668 + 434232

= 5427900

(ii) 60678 x 262 – 60678 x 162

= 15897636 – 9829836

= 6067800

(i) 739 × 102
(ii) 1938 × 99
(iii) 1005 × 188

(i) 739 x 102

= 75378

(ii) 1938 x 99

= 191862

(iii) 1005 x 188

= 188940

7750 / 17

= 455.8823529412

38 x 23 + 17

= 874 + 17

= 891

#### Question 10. Which least number should be subtracted from 1000 so that the difference is exactly divisible by 35.

1000 / 35

= Quotient is 28 Remainder is 20

So, If we subtract 20 from 1000 then the number divided by  1000

Ans. 20

#### Question 11. Which least number should be added to 1000 so that 53 divides the sum exactly.

1000 / 53

=> Quotient is = 17 Remainder = 99

So we have to add 99 with 1000.

Ans. 99

#### Question 12. Find the largest three-digit number which is exactly divisible by 47.

47 x 21 = 987

The largest 3 digit number is 987, which is exactly divisible by 987

#### Question 13. Find the smallest five-digit number which is exactly divisible by 254.

10000 / 254

=> Quotient =39, Remainder = 94

Now, 10000 + (254 – 94)= 10160

Ans. 10,160

#### 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?

In morning, 72 x 39

= 2808 Rs.

In evening, 28 x 39

= 1092 Rs.

Money is due to the vendor per day = 2808 + 1092 = 6708 Rs.

#### 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.

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

Ans. True

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

Ans. True

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

Ans. False