NCERT Solutions Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

NCERT Solutions Class 6 Maths
Chapter – 2 (Whole Numbers)

The NCERT Solutions in English Language for Class 6 Mathematics Chapter – 2 Whole Numbers Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.

Chapter 2: Whole Numbers

Exercise – 2.2

1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647

Solutions:
(a) Given 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408

(b) Given 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600

2. Find the product by suitable rearrangement:
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25

Solutions:

(a) Given 2 × 1768 × 50
= 2 × 50 × 1768
= 100 × 1768
= 176800

(b) Given 4 × 166 × 25
= 4 × 25 × 166
= 100 × 166
= 16600

(c) Given 8 × 291 × 125
= 8 × 125 × 291
= 1000 × 291
= 291000

(d) Given 625 × 279 × 16
= 625 × 16 × 279
= 10000 × 279
= 2790000

(e) Given 285 × 5 × 60
= 285 × 300
= 85500

(f) Given 125 × 40 × 8 × 25
= 125 × 8 × 40 × 25
= 1000 × 1000
= 1000000

3. Find the value of the following:
(a) 297 × 17 + 297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218

Solutions:

(a) Given 297 × 17 + 297 × 3
297 × (17 + 3)
= 297 × 20
= 5940

(b) Given 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8
= 54279 × (92 + 8)
= 54279 × 100
= 5427900

(c) Given 81265 × 169 – 81265 × 69
81265 × (169 – 69)
= 81265 × 100
= 8126500

(d) Given 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218
= 3845 × 5 × 782 + 3845 × 5 × 218
= 3845 × 5 × (782 + 218)
= 19225 × 1000
= 19225000

4. Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168

Solutions:
(a) Given 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3 (using distributive property)
= 73800 + 2214
= 76014

(b) Given 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 (using distributive property)
= 85400 + 1708
= 87108

(c) Given 258 × 1008
= 258 × (1000 + 8)
= 258 × 1000 + 258 × 8 (using distributive property)
= 258000 + 2064
= 260064

(d) Given 1005 × 168
= (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (using distributive property)
= 168000 + 840
= 168840

5. A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?

Solutions:
Petrol quantity filled on Monday = 40 litres
Petrol quantity filled on Tuesday = 50 litres
Total petrol quantity filled = (40 + 50) litre
Cost of petrol per litre = ₹ 44
Total money spent = 44 × (40 + 50)
= 44 × 90
= ₹ 3960

6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 45 per litre, how much money is due to the vendor per day?

Solutions:
Milk quantity supplied in the morning = 32 litres
Milk quantity supplied in the evening = 68 litres
Cost of milk per litre = ₹ 45
Total cost of milk per day = 45 × (32 + 68)
= 45 × 100
= ₹ 4500
Hence, the money is due to the vendor per day is ₹ 4500

7. Match the following:

(i) 425 × 136 = 425 × (6 + 30 + 100)  (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49  (b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005  (c) Distributivity of multiplication over addition.

Solutions:

(i) 425 × 136 = 425 × (6 + 30 + 100)  (c) Distributivity of multiplication over addition.
(ii) 2 × 49 × 50 = 2 × 50 × 49  (a) Commutativity under multiplication
(iii) 80 + 2005 + 20 = 80 + 20 + 2005  (b) Commutativity under addition
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