NCERT Solutions Class 6 Mathematics Chapter 2 Whole Numbers

Whole Numbers

Q.1) Write the next three natural numbers after 10999.
Sol.1) We can find next natural number to 10,999 by adding 1 to it which would be 10,999 + 1 = 11,000
We can find next natural number to 11,000 by adding 1 to it which would be 11,000 + 1 = 11,001
We can find next natural number to 11,001 by adding 1 to it which would be 11,001 + 1 = 11,002
Therefore, next three natural numbers after 10,999 are 11,000, 11,001 and 11,002.

Q.2) Write the three whole numbers occurring just before 10001.
Sol.2) We can find whole number occurring just before 10,001 by subtracting 1 from 10,001 which would be 10,001 − 1 = 10,000
We can find whole number occurring just before 10,000 by subtracting 1 from 10,000 which would be 10,000 − 1 = 9,999
We can find whole number occurring just before 9,999 by subtracting 1 from 9,999 which would be 9,999 − 1 = 9,998
Therefore, three whole numbers occurring just before 10,001 are 10,000, 9,999 and 9,998.

Q.3) Which is the smallest whole number?
Sol.3) The number 0 is the first and the smallest whole nos.

Q.4) How many whole numbers are there between 32 and 53?
Sol.4) We need to find number of whole numbers between 32 and 53 which means that we need to count number of steps it would take us to reach 52 starting from 32.
Let us write the whole numbers.
32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53
Now count the numbers between 32 and 53:
32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53
Number of steps = 52 − 32 = 20
Therefore, there are 21 whole numbers between 32 and 53.
There are 20 whole numbers between 32 and 53.

Q.5) Write the successor of:
(a) 2440701 (b) 100199
(c) 1099999 (d) 2345670
Sol.5) (a) Successor of 2440701 is 2440701 + 1 = 2440702
(b) Successor of 100199 is 100199 + 1 = 100200
(c) Successor of 1099999 is 1099999 + 1 = 1100000
(d) Successor of 2345670 is 2345670 + 1 = 2345671

Q.6) Write the predecessor of:
(a) 94 (b) 10000
(c) 208090 (d) 7654321
Sol. 6) By subtracting 1 from a number, we get its predecessor.
(a) 94
Predecessor of 94 is 94 -1 = 93
(b) 10000
Predecessor of 10000 – 1 = 9,999
(c) 208090
Predecessor of 208090 is 208089
(d) 7654321
Predecessor of 7654321 is 7654320

Q.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
Sol.7) Draw a line, mark a point on it and label it 0. Mark out points to the right of 0, at equal intervals. Label them as 1, 2, 3,... Thus, we have a number line with the whole numbers represented on it.

Observe that, out of any two whole numbers, the number on the right of the other number is the greater number. [3 is to the right of 2. So 3 is the greater number and 2 is smaller]
The number line can be used to represent larger numbers.
a) 503 is on the left of 530 on the number line because 503 is smaller than 530.
Therefore, 530 > 503.
b) 307 is on the left of 370 on the number line because 307 is smaller than 370.
Therefore, 370 > 307.
c) 56,789 is on the left of 98,765 on the number line because 56,789 is smaller than 98,765.
Therefore, 98,765 > 56,789.
d) 98,30,415 is on the left of 1,00,23,001 on the number line because 98,30,415 is smaller than 1,00,23,001.
Therefore, 98,30,415 < 1,00,23,001.

Q.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.
Sol.8) a) Zero is the smallest natural number.
False, zero is not a natural number. It is a whole number.
b) 400 is the predecessor of 399.
False, 400 is the successor of 399, (399 − 1 = 398) is the predecessor.
c) Zero is the smallest whole number. 
True
d) 600 is the successor of 599. (599 + 1 = 600)
True
e) All natural numbers are whole numbers.
Whole no.s = 0 & natural no.s = 0,1,2,3,4
True
f) All whole numbers are natural numbers.
False, every whole number except 0 is a natural number.
g) The predecessor of a two-digit number is never a single digit number.
False, the predecessor of 10 is 9. (10 − 1 = 9)
h) 1 is the smallest whole number.
False, 0 is the smallest whole number.
i) The natural number 1 has no predecessor.
True
j) The whole number 1 has no predecessor.
False, 0 is a predecessor of the whole number 1 (1 − 1 = 0)
k) The whole number 13 lies between 11 and 12.
False, 13 is the successor of 12. (11 + 1 = 12, 12 + 1 = 13)
l) The whole number 0 has no predecessor.
True
m) The successor of a two-digit number is always a two-digit number.
False, it can be a three-digit number.(99 + 1 = 100)

Exercise 2.2

Q.1) Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647
Sol.1) a) Rearrange the sum 837 + 208 + 363 as (837 + 363) + 208.
(837 + 363) + 208 = 1,200 + 208 = 1408
b) Rearrange the sum 1962 + 453 + 1538 + 647 as (1962 + 1538) + (453 + 647)
(1962 + 1538) + (453 + 647) = 3500 + 1100 = 4,600

Q.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
Sol. 2) a) 2 × 1768 × 50
= (2 × 50) × 1768
= 100 × 1768
= 1,76,800
(b) 4 × 166 × 25
= (4 × 25) × 166
= 100 × 166
= 16,600
(c) 8 × 291 × 125
= (8 × 125) × 291
= 1000 × 291
= 2,91,000
(d) 625 × 279 × 16
= (625 × 16) × 279
= 10,000 × 279
= 2,790,000
(e) 285 × 5 × 60
= 285 × (5 × 60)
= 285 × 300
= 85,500
(f) 125 × 40 × 8 × 25
= (125 × 40) × (8 × 25)
= 5,000 × 200
= 10,00,000

Q.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
Sol.3) (a) 297 × 17 + 297 × 3
= 297 × ( 17 + 3 )
= 297 × (20)
= 297 × 10 × 2
= 2970 × 2
= 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
= 54279 × (100)
= 54279 × 100
= 5427900
(c) 81265 × 169 – 81265 × 69
= 81265 × ( 169 – 69)
= 81265 × (100)
= 81265 × 100
= 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 769 × 5 × 5 × 782 + 769 × 25 × 218
= 769 × (5 × 5 × 782 + 25 × 218)
= 769 × (25 × 782 + 25 × 218)
= 769 × 25(782 + 218)
= 769 × 25(1000)
= 769 × 25 × 1000
= 19225000

Q.4) Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102
(c) 258 × 1008 (d) 1005 × 168
Sol.4) (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
                       = 85400 + 1708 = 87108 (𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
(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`

Q.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?
Sol.5) Method 2: Find the total litres of petrol filled for both days and multiply by the cost per litre.
A taxi driver filled his car petrol tank on Monday= 40 𝑙𝑖𝑡𝑟𝑒𝑠
A taxi driver filled his car petrol tank on next day= 50 𝑙𝑖𝑡𝑟𝑒𝑠
The total litres of petrol he filled his car petrol tank= 40 + 50 = 90 𝑙𝑖𝑡𝑟𝑒𝑠
The price of 1 𝑙𝑖𝑡𝑟𝑒 of petrol= 𝑅𝑠 44
∴ the price of 90 litres of petrol= 𝑅𝑠. 90 × 44 = 𝑅𝑠. 3960
Method 2: find the amount spent for each day and then add.
Amount spent for petrol on Monday = 40 × 44 = 𝑅𝑠. 1760
Amount spent for petrol on Tuesday = 50 × 44 = 𝑅𝑠. 2200
Total amount spent in all on petrol = 𝑅𝑠. 1760 + 𝑅𝑠. 2200 = 𝑅𝑠. 3960

Q.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?
Sol.6) Milk supply by vendor in the morning = 32 𝑙𝑖𝑡𝑟𝑒𝑠
Milk supply by vendor in the evening = 68 𝑙𝑖𝑡𝑟𝑒𝑠
Cost of milk per litre = 𝑅𝑠. 15
Total cost of supplied milk  = 𝑅𝑠. 15 × (32 + 68)
                                           = 𝑅𝑠. 15 × 100
                                           = 𝑅𝑠. 1500

Q.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.
Sol.7) i) 425 × 136 = 425 × (6 + 30 + 100)
We see that the number 136 is rewritten as 6 + 30 + 100
Distributivity of multiplication over addition (c)
ii) 2 × 49 × 50 = 2 × 50 × 49
We see that the numbers are rearranged under the multiplication operation.
Commutativity under multiplication (a)
iii) 80 + 2005 + 20 = 80 + 20 + 2005
The numbers are rearranged under the addition operation.
Commutativity under addition (b)

Exercise 2.3

Q.1) Which of the following will not represent zero:
(a) 1 + 0 (b) 0 × 0 (c) 0/2 (d) 10−10/2
Sol.1) Option (a) will not represent zero.
a) does not represent 0, as 1 + 0 is 1.
b) 0 × 0 is 0, as any number multiplied by 0 becomes zero.
c) 0/2 is 0, as 0 cannot be divided.
d) 10−10/2 is 0, as 10 – 10 is 0 and 0 cannot be divided.

Q.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.2) Yes, if the product of two numbers is 0, then one or both of them will be zero.
Let us consider two numbers, for example 12 and 15.
12 × 15 = 180
The numbers are non-zero and so is the product.
Similarly, if we consider any two non-zero numbers, the product is a non-zero number.
Let us now consider numbers 12 and 0 OR 0 and 135.
We know that any number multiplied by 0 becomes 0, and hence the product of both 12 and 0 AND 0 and 135 are 0.
Also 0 × 0 = 0.
Hence, if the product of any two whole numbers is zero, then one or both of them will be zero.

Q.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.3) If the product of two whole numbers is 1, then both the numbers should be 1.
Let us consider two numbers, for example 20 and 1.
20 × 1 = 20
We see that the product is not 1.
But, 1 × 1 = 1.
Hence, if the product of any two whole numbers is 1, then both of them should be 1.

Q.4) Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001 (c) 824 × 25 (d) 4275 × 125
(e) 504 × 35
Sol.4) a) 728 × 101 = 728 × (100 + 1)
= 728 × 100 + 728
= 72800 + 728
= 73,528
b) 5437 × 1001 = 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437
= 54,42,437
c) 824 × 25 = 824 × (20 + 5)
= 824 × 20 + 824 × 5
= 16480 + 4120
= 20,600
d) 4275 × 125 = 4275 × (100 + 20 + 5)
= 4275 × 100 + 4275 × 20 + 4275 × 5
= 427500 + 85,500 + 21375
= 5,34,375
e) 504 × 35 = (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17,640

Q.5) Study the pattern :
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Sol.5) Let see how the pattern works.
1 × 8 + 1 = 9
12 × 8 + 2 = 98
𝑖. 𝑒. (11 + 1) × 8 + 2 = 11 × 8 + 1 × 8 + 2 = 88 + 8 + 2 = 98
123 × 8 + 3 = 987 ;
𝑖. 𝑒. (111 + 11 + 1) × 8 + 3 = 111 × 8 + 11 × 8 + 1 × 8 + 2
= 888 + 88 + 8 + 3 = 987
1234 × 8 + 4 = 9876 ;
𝑖. 𝑒. (1111 + 111 + 11 + 1) × 8 + 4 = 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 + 4
= 8888 + 888 + 88 + 8 + 4 = 9876
12345 × 8 + 5 = 98765
𝑖. 𝑒. (11111 + 1111 + 111 + 11 + 1) × 8 + 5
= 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 + 5
= 88888 + 8888 + 888 + 88 + 8 + 5 = 98765
Extending the pattern, we have
123456 × 8 + 6
= (111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 5
= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 + 5
= 888888 + 88888 + 8888 + 888 + 88 + 8 + 6 = 987654
1234567 × 8 + 7
= (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 7
= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8 + 7
= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8 + 7 = 9876543