## NCERT Solutions for Class 6th Maths: Chapter 2 – Whole Numbers

Class 6th Maths Chapter 2 Whole Numbers NCERT Solution is given below.
Exercise – 2.1
Question 1:
Write the next three natural numbers after 10999.
Answer:
Next three natural numbers after 10999 are 11000, 11001, 11002
Question 2:
Write the three whole numbers occurring just before 10001.
Answer:
3 whole numbers just before 10001 are 10000, 9999, 9998
Question 3:
Which is the smallest whole number?
Answer:
The smallest whole number is 0.
Question 4:
How many whole numbers are there between 32 and 53?
Answer:
Whole numbers between 32 and 53 = 20 (53 − 32 − 1 = 20)
(33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52)
Question 5:
Write the successor of:
(a) 2440701 (b) 100199
(c) 1099999 (d) 2345670
Answer:
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671
Question 6:
Write the predecessor of:
(a) 94 (b) 10000
(c) 208090 (d) 7654321
Answer:
(a) 94 − 1 = 93
(b) 10000 − 1 = 9999
(c) 208090 − 1 =208089
(d) 7654321 − 1 = 7654320
Question 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
As 530 > 503,
503 is on the left side of 530 on the number line.
(b) 370, 307
As 370 > 307,
307 is on the left side of 370 on the number line.
(c) 98765, 56789
As 98765 > 56789,
56789 is on the left side of 98765 on the number line.
(d) 9830415, 10023001
Since 98, 30, 415 < 1, 00, 23, 001, 98,30,415 is on the left side of 1,00,23,001 on the number line.
Question 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) False, 0 is not a natural number.
(b) False, as predecessor of 399 is 398 (399 − 1 = 398).
(c) True
(d) True, as 599 + 1 = 600
(e) True
(f) False, as 0 is a whole number but it is not a natural number.
(g) False, as predecessor of 10 is 9.
(h) False, 0 is the smallest whole number.
(i) True, as 0 is the predecessor of 1 but it is not a natural number.
(j) False, as 0 is the predecessor of 1 and it is a whole number.
(k) False, 13 does not lie in between 11 and 12.
(l) True, predecessor of 0 is −1, which is not a whole number.
(m) False, as successor of 99 is 100.
Exercise 2.2
Question 1:
Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647
Answer:
(a) 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408
(b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600
Question 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 = 2 × 50 × 1768 = 100 × 1768 = 176800
(b) 4 × 166 × 25 = 4 × 25 × 166 = 100 × 166 = 16600
(c) 8 × 291 × 125 = 8 × 125 × 291 = 1000 × 291 = 291000
(d) 625 × 279 × 16 = 625 × 16 × 279 = 10000 × 279 = 2790000
(e) 285 × 5 × 60 = 285 × 300 = 85500
(f) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25 = 1000 × 1000 = 1000000
Question 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
Answer:
(a) 297 × 17 + 297 × 3 = 297 × (17 + 3) = 297 × 20 = 5940
(b) 54279 × 92 + 8 × 54279 = 54279 × 92 + 54279 × 8
= 54279 × (92 + 8)
= 54279 × 100 = 5427900
(c) 81265 × 169 − 81265 × 69 = 81265 × (169 − 69) = 81265 × 100 = 8126500
(d) 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
Question 4:
Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102
(c) 258 × 1008 (d) 1005 × 168
Answer:
(a) 738 × 103 = 738 × (100 + 3)
= 738 × 100 + 738 × 3 (Distributive property)
= 73800 + 2214
= 76014
(b) 854 × 102 = 854 × (100 + 2)
= 854 × 100 + 854 × 2 (Distributive property)
= 85400 + 1708 = 87108
(c) 258 × 1008 = 258 × (1000 + 8)
= 258 × 1000 + 258 × 8 (Distributive property)
= 258000 + 2064 = 260064
(d) 1005 × 168 = (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (Distributive property)
= 168000 + 840 = 168840
Question 5:
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 Rs 44 per litre, how much did he spend in all on petrol?
Answer:
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
Question 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 15 per litre, how much money is due to the vendor per day?
Answer:
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
Question 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) [Distributivity of multiplication over addition] Hence, (c)
(ii) 2 × 49 × 50 = 2 × 50 × 49 [Commutativity under multiplication] Hence, (a)
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 [Commutativity under addition] Hence, (b)
Exercise 2.3
Question 1:
Which of the following will not represent zero?
(a) 1 + 0 (b) 0 × 0
Question 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.
Answer:
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.)
Question 3:
If the product of two whole numbers is 1, can we say that one of both of them will be 1? Justify through examples.
Answer:
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.
Question 4:
Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001
(c) 824 × 25 (d) 4275 × 125
(e) 504 × 35
Answer:
(a) 728 × 101= 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 72800 + 728 = 73528
(b) 5437 × 1001 = 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437 = 5442437
(c) 824 × 25 = (800 + 24) × 25
= (800 + 25 − 1) × 25
= 800 × 25 + 25 × 25 − 1 × 25
= 20000 + 625 − 25
= 20000 + 600 = 20600
(d) 4275 × 125 = (4000 + 200 + 100 − 25) × 125
= 4000 × 125 + 200 × 125 + 100 × 125 − 25 × 125
= 500000 + 25000 + 12500 − 3125
= 534375
(e) 504 × 35 = (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140 = 17640
Question 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