Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2
1. Find the sum by suitable arrangement of terms:
Question (a)
837 + 208 + 363
Solution:
837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
Question (b)
1962 + 453 + 1538 + 647.
Solution:
1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
2. Find the product by suitable arrangement of terms:
Question (a)
2 × 1497 × 50
Solution:
= (2 × 50) × 1497
= 100 × 1497
= 149700
Question (b)
4 × 263 × 25
Solution:
= (4 × 25) × 263
= 100 × 263
= 26300
Question (c)
8 × 163 × 125
Solution:
= (8 × 125) × 163
= 1000 × 163
= 163000
Question (d)
963 × 16 × 25
Solution:
= 963 × (16 × 25)
= 963 × 400
= 385200
Question (e)
5 × 171 × 60
Solution:
= (5 × 60) × 171
= 300 × 171
= 51300
Question (f)
125 × 40 × 8 × 25
Solution:
= (125 × 40) × (8 × 25)
= 5000 × 200
= 1000000
Question (g)
30921 × 25 × 40 × 2
Solution:
= 30921 × (25 × 40) × 2
= 30921 × 1000 × 2
= 61842000
Question (h)
4 × 2 × 1932 × 125
Solution:
4 × 2 × 1932 × 125
= 1932 × (4 × 2 × 125)
= 1932 × 1000
= 1932000
Question (i)
5462 × 25 × 4 × 2.
Solution:
= 5462 × 2 × 25 × 4
= 10924 × 100
= 1092400
3. Find the value of each of the following using distributive property:
Question (a)
(649 × 8) + (649 × 2)
Solution:
(649 × 8) + (649 × 2)
= 649 × (8 + 2)
= 649 × 10
= 6490
Question (b)
(6524 × 69) + (6524 × 31)
Solution:
(6524 × 69) + (6524 × 31)
= 6524 × (69 + 31)
= 6524 × 100
= 652400
Question (c)
(2986 × 35) + (2986 × 65)
Solution:
(2986 × 35) + (2986 × 65)
= 2986 × (35 + 65)
= 2986 × 100
= 298600
Question (d)
(6001 × 172) – (6001 × 72).
Solution:
(6001 × 172) – (6001 × 72)
= 6001 × (172 – 72)
= 6001 × 100
= 600100
4. Find the value of the following:
Question (a)
493 × 8 + 493 × 2
Solution:
(a) 493 × 8 + 493 × 2
= 493 × (8 + 2) = 493 × 10
= 4930
Question (b)
24579 × 93 + 7 × 24579
Solution:
24579 × 93 + 7 × 24579
= 24579 × (93 + 7)
= 24579 × 100
= 2457900
Question (c)
3845 × 5 × 782 + 769 × 25 × 218
Solution:
3845 × 5 × 782 + 769 × 25 × 218
= 769 × 5 × 5 × 782 + 769 × 5 × 5 × 218
= 769 × 5 × 5 × (782 + 218)
= 769 × 25 × 1000
= 19225 × 1000
= 19225000
Question (d)
3297 × 999 + 3297.
Solution:
3297 × 999 + 3297
= (3297) × (999 + 1)
= 3297 × 1000
= 3297000
5. Find the product, using suitable properties:
Question (a)
738 × 103
Solution:
= 738 × (100 + 3)
= (738 × 100) + (738 × 3)
[Using a × (b + c) = (a × b) + (a × c)]
= 73800 + 2214
= 76014
Question (b)
854 × 102
Solution:
= 854 × (100 + 2)
= (854 × 100) + (854 × 2)
[Using a × (b + c) = (a × b) + (a × c)]
= 85400 + 1708
= 87108
Question (c)
258 × 1008
Solution:
= 258 × (1000 + 8)
= (258 × 1000) + (258 x 8)
[Using a × (b + c) = (a × b) + (a × c)]
= 258000 + 2064
= 260064
Question (d)
736 × 93
Solution:
= 736 × (100 – 7)
= 736 × 100 = 736 × 7
[Using a × (b – c) = (a × b) – (a × c)]
= 73600 – 5152
= 68448
Question (e)
816 × 745
Solution:
= (800 + 16) × 745
= 800 × 745 + 16 × 745
[Using a × (b + c) = (a × b) + (a × c)]
= 596000 + 11920
= 607920
Question (f)
2032 × 613
Solution:
= 2032 × (600 + 13)
= 2032 × 600 + 2032 × 13
[Using a × (b + c) = (a × b) + (a × c)]
= 1219200 + 26416
= 1245616
6. A taxi driver 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 ₹ 78 per litre, how much he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 Litres
Petrol filled on Tuesday = 50 Litres
Total Petrol filled = 40 + 50 = 90 Litres
Cost per litre = ₹ 78
Total Cost = 90 × ₹ 78
= ₹ 7020
7. 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 ₹ 35 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 Litres
Milk supplied in the evening = 68 Litres
Total milk supplied = 32 + 68
= 100 Litres
Cost Per litre = ₹ 35
∴ Total cost of milk per day = 100 × ₹ 35
= ₹ 3500
8. We know that 0 × 0 = 0. Is there any other whole number which when multiplied by itself gives the product equal to the number itself? Find out the number.
Solution:
Yes, there is a whole number 1.
Here 1 × 1 = 1.
9. Fill in the blanks:
Question (i)
(a) 15 × 0 = ………….. .
(b) 15 + 0 = ………….. .
(c) 15 – 0 = ………….. .
(d) 15 ÷ 0 = ………….. .
(e) 0 × 15 = ………….. .
(f) 0 + 15 = ………….. .
(g) 0 ÷ 15 = ………….. .
(h) 15 × 1 = ………….. .
(i) 15 ÷ 1 = ………….. .
(j) 1 ÷ 1 = ………….. .
Solution:
(a) 0,
(b) 15,
(c) 15,
(d) Not defined,
(e) 0,
(f) 15,
(g) 0,
(h) 15,
(i) 15,
(j) 1.
Question 10.
The product of two Whole numbers is zero. What do you conclude. Explain with example.
Solution:
We conclude that one number must be zero such 25 × 0 = 0.
Question 11.
Match the following:
| 1. 537 × 106 = 537 ×100 + 537 × 6 | (a) Commutativity under multiplication |
| 2. 4 × 47 × 25 = 4 × 25 × 47 | (b) Commutativity under addition |
| 3. 70 + 1923 + 30 = 70 + 30 + 1923 | (c) Distributivity of multiplication over addition. |
Solution:
| 1. 537 × 106 = 537 × 100 + 537 × 6 | (c) Distributivity of multiplication over addition. |
| 2. 4 × 47 × 25 = 4 × 25 × 47 | (a) Commutativity under multiplication |
| 3. 70 + 1923 + 30 = 70 + 30 + 1923 | (b) Commutativity under addition |