ncert class 6 maths
6th maths book
6th standard maths
class 6 maths solutions

EXERCISE 2.1
1. Write the next three natural numbers after 10999.
Answer – The next three natural numbers after 10999 are – 
10999 + 1 = 11000
11000 + 1 = 11001
11001 + 1 = 11002

2. Write the three whole numbers occurring just before 10001.
Answer – The three whole numbers occurring just before 10001 are – 
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998

3. Which is the smallest whole number?
Answer – The smallest whole number 0

4. How many whole numbers are there between 32 and 53?
Answer – 20 whole numbers are there between 32 and 53.

5. Write the successor of :

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

Answer – 
(a) 2440701
2440701 + 1 = 2440702

(b) 100199
100199 + 1 = 100200

(c) 1099999
1099999 + 1 = 1100000

(d) 2345670
2345670 + 2345671

6. Write the predecessor of :

(a) 94

(b) 10000

(c) 208090

(d) 7654321

Answer – 
(a) 94
94 − 1 = 93

(b) 10000
10000 − 1 = 9999

(c) 208090
208090 − 1 = 208089

(d) 7654321
7654321 − 1 = 7654320

7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.

(a) 530, 503

(b) 370, 307

(c) 98765, 56789

(d) 9830415, 10023001

Answer – 
(a) 530, 503
Sol – 530 > 503
So, 503 is right side of 530 on numbers line.

(b) 370, 307
Sol – 370 > 307
So, 307 is right side of 370 on numbers line.

(c) 98765, 56789
Sol – 98765 > 56789
So, 56789 is right side of 98765 on numbers line.

(d) 9830415, 10023001
Sol – 10023001 < 9830415
So, 9830415 is right side of 10023001 on numbers line.

8. Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f ) All whole numbers are natural numbers.
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
Answer – 

(a) Zero is the smallest natural number.	(F)
(b) 400 is the predecessor of 399.	(F)
(c) Zero is the smallest whole number.	(T)
(d) 600 is the successor of 599.	(T)
(e) All natural numbers are whole numbers.	(T)
(f ) All whole numbers are natural numbers.	(F)
(g) The predecessor of a two digit number is never a single digit number.	(F)
(h) 1 is the smallest whole number.	(F)
(i) The natural number 1 has no predecessor.	(T)
(k) The whole number 13 lies between 11 and 12.	(F)
(l) The whole number 0 has no predecessor.	(T)
(m) The successor of a two digit number is always a two digit number.	(F)

EXERCISE 2.2

1. Find the sum by suitable rearrangement :

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

Answer – 
(a) 837 + 208 + 363
Sol – (837 + 363) + 208
= 1200 + 208
= 1408

(b) 1962 + 453 + 1538 + 647 
Sol – (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

Answer – 
(a) 2 × 1768 × 50
Sol – 2 × 50 × 1768
= 100 × 1768
= 176800

(b) 4 × 166 × 25
Sol – 4 × 25 × 166
= 100 × 166
= 16600

(c) 8 × 291 × 125
Sol – 8 × 125 × 291
= 1000 × 291
= 291000

(d) 625 × 279 × 16
Sol – 625 × 16 × 279
= 10000 × 279
= 2790000

(e) 285 × 5 × 60
Sol – 285 × 300
= 85500

(f) 125 × 40 × 8 × 25
Sol – 125 × 8 × 40 × 25
= 1000 × 1000
= 1000000

3. Find the value of the following :

(a) 297 × 17 + 297 × 3	(b) 54279 × 92 + 8 ×
(c) 81265 × 169 – 81265 × 69	(d) 3845 × 5 × 782 + 769 × 25 × 218

Answer – 
(a) 297 × 17 + 297 × 3
Sol – 297 × (17 + 3)
= 297 × 20
= 5940

(b) 54279 × 92 + 8 × 54279
Sol – 54279 × (92 + 8)
= 54279 × 100
= 5427900

(c) 81265 × 169 – 81265 × 69
Sol – 81265 × (169 − 69)
= 81265 × 100
= 8126500

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

4. Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

1005 × 168

Answer – 
(a) 738 × 103
Sol – 738 × (100 + 3) 3 (Distributive property)
= 738 × 100 + 738 × 3
= 73800 +  2214
= 76014

(b) 854 × 102
Sol – 854 × 100 + 854 × 2 (Distributive property)
= 85400 + 1708
= 87108

(c) 258 × 1008
Sol – 258 × (1000 + 8)
= 258 × 1000 + 258 × 8 (Distributive property)
= 258000 + 2064
= 260064

(d) 1005 × 168
Sol – (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (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 rs 44 per litre, how much did he spend in all on petrol?
Sol – Quantity of petrol filled on Monday = 40 l
Quantity of petrol filled on Tuesday = 50 l
Total quantity filled = (40 + 50) l
Cost of petrol (per l) = Rs 44
Total money spent = 44 × (40 + 50)
= 44 × 90 = Rs 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 rs 45 per litre, how much money is due to the vendor per day?
Sol – Quantity of milk supplied in the morning = 32 l
Quantity of milk supplied in the evening = 68 l
Total of milk per litre = (32 + 68) l
Cost of milk per litre = Rs 15
Total cost per day = 15 × (32 + 68)
= 15 × 100 = Rs 1500

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.

Answer – 

(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.

EXERCISE 2.3

1. Which of the following will not represent zero :

(a) 1 + 0

(b) 0 × 0

(c) \(\displaystyle \frac{0}{2}\)

(d) \(\displaystyle \frac{10-10}{2}\)

Answer – 
(a) 1 + 0
Sol – 1 + 0
= 1
No, it is not represent zero.

(b) 0 × 0
Sol – 0
Yes, it is represent zero.

(c) \(\displaystyle \frac{0}{2}\)
Sol – 0
Yes, it is represent zero.

(d) \(\displaystyle \frac{10-10}{2}\)
Sol – \(\displaystyle \frac{0}{2}\)
Yes, it is represent zero.

2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Sol – If the product of 2 whole numbers is zero, then one of them is definitely zero.
For example, 0 × 2 = 0 and 17 × 0 = 0
If the product of 2 whole numbers is zero, then both of them may be zero.
0 × 0 = 0
However, 2 × 3 = 6
(Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.)

3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Sol – If the product of 2 numbers is 1, then both the numbers have to be equal to 1.
For example, 1 × 1 = 1
However, 1 × 6 = 6
Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.

4. Find using distributive property :

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

Answer – 
(a) 728 × 101
Sol – 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 72800 + 728
= 73528

(b) 5437 × 1001
Sol – 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437
= 5442437

(c) 824 × 25
Sol – (800 + 24) × 25
= (800 + 25 − 1) × 25
= 800 × 25 + 25 × 25 − 1 × 25
= 20000 + 625 − 25
= 20000 + 600
= 20600

(d) 4275 × 125
Sol – (4000 + 200 + 100 − 25) × 125
= 4000 × 125 + 200 × 125 + 100 × 125 − 25 × 125
= 500000 + 25000 + 12500 − 3125
= 534375

(e) 504 × 35
Sol – (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17640

5. Study the pattern :

1 × 8 + 1 = 9	1234 × 8 + 4 = 9876
12 × 8 + 2 = 98	12345 × 8 + 5 = 98765
123 × 8 + 3 = 987

Write the next two steps. Can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).
Answer – 
123456 × 8 + 6 = 987648 + 6 = 987654
1234567 × 8 + 7 = 9876536 + 7 = 9876543
Yes, the pattern works.
As 123456 = 111111 + 11111 + 1111 + 111 + 11 + 1,
123456 × 8 = (111111 + 11111 + 1111 + 111 + 11 + 1) × 8
= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8
= 888888 + 88888 + 8888 + 888 + 88 + 8 = 987648
123456 × 8 + 6 = 987648 + 6 = 987654