Class VI - Mathematics

# NCERT Solutions: Chapter - 1 Whole Numbers

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Page 31

Exercise 2.1

Q1. Write the next three natural numbers after 10999.

Add 1 to the given number to get the next number

1. 10999 + 1 = 11000
2. 11000 + 1 = 11001
3. 11001 + 1 = 11002

Q2. Write the three whole numbers occurring just before 10001.

Subtract 1 from the given number to get the number occurring before it

1. 10001 – 1 = 10000
2. 10000 – 1 = 9999
3. 9999 – 1 = 9998

Q3. Which is the smallest whole number?

The natural numbers along with zero form the collection of whole numbers.
Therefore, the smallest whole number is zero (0).

Q4. How many whole numbers are there between 32 and 53?

The number of whole numbers lying between 32 and 53 is 20 (53 – 32 – 1 = 20).
They are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52.

Q5. Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670

Add 1 to the given number to get the next number i.e. its successor
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671

Q6. Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321

Subtract 1 from the given number to get the number occurring before it i.e. its predecessor
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320

Q7. 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

In number line, the smaller number is on the left and the greater number falls on the right.
(a) 530, 503
530 > 503
Since 503 is smaller than 530, it lies on the left of 530 on the number line.

(b) 370, 307
370 > 307
Since 307 is smaller than 370, it lies on the left of 370 on the number line.

(c) 98765, 56789
98765 > 56789
Since 56789 is smaller than 9876, it lies on the left of 98765 on the number line.

(d) 9830415, 10023001
9830415 < 10023001
Since 9830415 is smaller than 10023001, it lies on the left of 10023001 on the number line.

Q8. 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.

(a) False because zero (0) is not a natural number.
(b) False because 399 - 1 = 398 therefore the predecessor of 399 is 398 and NOT 400.
(c) True because the natural numbers along with zero form the collection of whole numbers.
(d) True because 599 + 1 = 600.
(e) True because the natural numbers along with zero form the collection of whole numbers.
(f) False because the natural numbers along with zero form the collection of whole numbers and zero is not a natural number.
(g) False, 10 is a 2 digit number but its predecessor, which is 10 – 1 = 9, is a single digit number.
(h) False because the natural numbers along with zero form the collection of whole numbers so smallest whole number is zero.
(i) True because 1 is the smallest natural number.
(j) False because 0 is a whole number and is a predecessor of 1, 1 – 1 = 0.
(k) False because 12 + 1 = 13 so 13 is a successor of 12 and will come after 12.
(l) True because 0 is the smallest whole number.
(m) False, 999 which is a 3 digit number but its successor, which is 999 + 1 = 1000, is a four digit number.

Page 40

Exercise 2.2

Q1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647

a) (837 + 363) + 208 = 1325
b) (1962 + 1538) + (647 + 453) = 3480 + 1086 = 4566

Q2. 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

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 = 3750 × 279 = 10,46,250

e) 285 × 5 × 60
= 285 × (5 × 60) = 285 × 300 = 85,500

f) 125 × 40 × 8 × 25
= (125 × 8) × (40 × 25) = 1000 × 1000 = 10,00,000

Q3. 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

a) 297 × 17 + 297 × 3
Since 297 is repeated, we can take it as a common term
297 × (17 + 3) = 297 × 20 = 5940

b) 54279 × 92 + 8 × 54279
Since 54279 is repeated, we can take it as a common term
54279 × (92 + 8) = 54279 × 100 = 54,27,900

c) 81265 × 169 – 81265 × 69
Since 81265 is repeated, we can take it as a common term
81265 × (169 - 69) = 81265 × 100 = 81,26,500

d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218
= 3845 × 5 × 782 + (769 × 5) × 5 × 218
= 3845 × 5 × 782 + 3845 × 5 × 218
Since 3845 and 5 are repeated, we can take them as common term
= 3845 × 5 × (782 + 218)
= 19,225 × 1000
= 1,92,25,000

Q4. Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168

a) 738 × 103
= 738 × (100 + 3)
By distributive law
A × (B + C) = A × B + A × C
= 738 × 100 + 738 × 3
= 73800 + 2214
= 76,014

b) 854 × 102
= 854 × (100 + 2)
By distributive law,
A × (B + C) = A × B + A × C
= 854 × 100 + 854 × 2
= 85400 + 1708
= 87,108

c) 258 × 1008
= 258 × (1000 + 8)
By distributive law
A × (B + C) = A × B + A × C
= 258 × 1000 + 258 × 8
= 258000 + 2064
= 260,064

d) 1005 × 168
= (1000 + 5) × 168
By distributive law
A × (B + C) = A × B + A × C
= 1000 × 168 + 5 × 168
= 168000 + 840
= 1,68,840

Q5. 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?

The quantity of diesel filled on Monday = 40L
The quantity of diesel filled on Tuesday = 50L
The total quantity filled = (40 + 50)L
Cost of diesel (per L) = Rs. 44
The total money spent on diesel = 44 × (40 + 50)
= 44 × 90
= Rs. 3960
The total amount spent on diesel = Rs. 3960

Q6. 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?

The quantity of milk supplied in the morning = 32 litres
The quantity of milk supplied in the evening = 68 litres
The total quantity filled = (32 + 68) litres
Cost of milk (per litre) = Rs. 15
The total cost of milk = 15 × (32 + 68)
= 15 × 100
= Rs. 1500
The total amount due to the vendor per day = Rs. 1500

Q7. Match the following:

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

(ii) 2 × 49 × 50 = 2 × 50 × 49
This is commutative property under multiplication

(iii) 80 + 2005 + 20 = 80 + 20 + 2005
This is commutative property under addition

Page 43

Exercise 2.3

Q1. Which of the following will not represent zero:
(a) 1 + 0
(b) 0 × 0
(c) 0/2
(d) (10 – 10)/2

a) 1 + 0 = 1
It does not represent zero, since the answer is 1.

b) 0 × 0 = 0
It represents zero, since zero multiplied with anything is zero.

c) 0/2= 0
It represents zero, since zero divided with anything is zero.

d) (10 - 10) / 2 = 0
It represents zero.

Q2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

1. The product of 2 whole numbers will be zero only when one of the numbers is multiplied with zero.
For example
2 × 0 = 0
120 × 0 = 0

2. The product of 2 whole numbers will be zero when both the numbers are zero.
0 × 0 = 0

Q3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

1. The product of 2 whole numbers will be one only if both the numbers are 1.
For example
1 × 1 = 1

2. If only one of the numbers is one the product will not be one.
For example
1 × 2 = 2
1 × 100 = 100

Q4. Find using distributive property:
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35

By distributive law,
A × (B + C) = A × B + A × C

(a) 728 × 101
= 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 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) + (85500) + (21375)
= 5,34,375

(e) 504 × 35
= 504 × (30 + 5)
= (504 × 30) + (504 × 5)
= 15120 + 2520
= 17,640

Q5. 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?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

We know that
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

So, (123456 × 8) + 6 = 987648 + 6 = 987654

Similarly,
(1234567 × 8) + 7 = 9876536 + 7 = 9876543

Hence, the next two steps will be
123456 x 8 + 6 = 987654 and
1234567 x 8 + 7 = 9876543

Pattern of the working is
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543