Solutions

NCERT Class 6 Maths Solutions for Chapter 2 Whole Numbers

Exercise 2.1 (Page 31)

1. Write the next three natural numbers after 10999.

The next three natural numbers after 10999 are 11000, 11001, and 11002.

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

The three whole numbers occurring just before 10001 are 10000, 9999, and 9998.

3. Which is the smallest whole number?

The smallest whole number is 0.

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

The whole numbers between 32 and 53 are as follows:

(33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52)

Hence, there are 20 whole numbers between 32 and 53.

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

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

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

(a) 94 – 1 = 93

(b) 10000 – 1 = 9999

(c) 208090 – 1 = 208089

(d) 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

(a) 530 > 503

Hence, 503 is on the left side of 530 on the number line.

(b) 370 > 307

Hence, 307 is on the left side of 370 on the number line.

(c) 98765 > 56789

Hence, 56789 is on the left side of 98765 on the number line.

(d) 9830415 < 10023001

Hence, 9830415 is on the left side of 10023001 on the number 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.

(a) False – 0 is not a natural number.

(b) False – The predecessor of 399 is 398 because (399 – 1 = 398).

(c) True – Zero is the smallest whole number.

(d) True – Since (599 + 1 = 600).

(e) True – All natural numbers are whole numbers.

(f) False – 0 is a whole number but is not a natural number.

(g) False – For example, the predecessor of 10 is 9.

(h) False – 0 is the smallest whole number.

(i) True – The predecessor of 1 is 0, but it is not a natural number.

(j) False – 0 is the predecessor of 1 and is a whole number.

(k) False – 13 does not lie between 11 and 12.

(l) True – The predecessor of 0 is -1 and is not a whole number.

(m) False – For example, the successor of 99 is 100.

Exercise 2.2 (Page 40)

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

(a) Given 837 + 208 + 363

= (837 + 363) + 208

= 1200 + 208

= 1408

(b) Given 1962 + 453 + 1538 + 647

= (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

(a) Given 2 × 1768 × 50

= 2 × 50 × 1768

= 100 × 1768

= 176800

(b) Given 4 × 166 × 25

= 4 × 25 × 166

= 100 × 166

= 16600

(c) Given 8 × 291 × 125

= 8 × 125 × 291

= 1000 × 291

= 291000

(d) Given 625 × 279 × 16

= 625 × 16 × 279

= 10000 × 279

= 2790000

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 1000000

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

(a) Given 297 × 17 + 297 × 3

= 297 × (17 + 3)

= 297 × 20

= 5940

(b) Given 54279 × 92 + 8 × 54279

= 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

(c) Given 81265 × 169 – 81265 × 69

= 81265 × (169 – 69)

= 81265 × 100

= 8126500

(d) Given 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

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

(a) Given 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (using distributive property)

= 168000 + 840

= 168840

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 ₹ 44 per litre, how much did he spend in all on petrol?

Petrol quantity filled on Monday = 40 litres

Petrol quantity filled on Tuesday = 50 litres

Total petrol quantity filled = (40 + 50) litre

Cost of petrol per litre = ₹ 44

Total money spent = 44 × (40 + 50)

= 44 × 90

= ₹ 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 ₹ 45 per litre, how much money is due to the vendor per day?

Milk supplied in the morning = 32 litres

Milk supplied in the evening = 68 litres

Total milk supplied = (32 + 68) litres

Cost of milk per litre = ₹ 45

Total money due to the vendor = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

Exercise 2.3 (Page 42)

1. 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 (not zero)

(b) 0 × 0 = 0

(c) 0 ÷ 2 = 0

(d) (10 – 10) ÷ 2 = 0 ÷ 2 = 0

Therefore, option (a) will not 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.

Yes, if the product of two whole numbers is zero, then one or both of them will definitely be zero.

For example:

0 × 4 = 0 (one number is zero)

0 × 0 = 0 (both numbers are zero)

This shows that if the product is zero, at least one of the numbers must be zero.

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.

Yes, if the product of two whole numbers is 1, then one or both of them must be 1.

For example:

1 × 1 = 1 (both numbers are 1)

This shows that if the product is 1, both numbers must be 1.

4. Find the using distributive property:
  • (a) 728 × 101
  • (b) 5437 × 1001
  • (c) 824 × 25
  • (d) 4275 × 125

(a) Given 728 × 101

= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728

= 73528

(b) Given 5437 × 1001

= 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c) Given 824 × 25

= 824 × (20 + 5)

= 824 × 20 + 824 × 5

= 16480 + 4120

= 20600

(d) Given 4275 × 125

= 4275 × (100 + 25)

= 4275 × 100 + 4275 × 25

= 427500 + 106875

= 534375

5. Study the pattern:
  • 4 × 6 = 24
  • 4 × 66 = 264
  • 4 × 666 = 2664
  • 4 × 6666 = 26664
  • 4 × 66666 = 266664

Observe the number of sixes and number of digits in the product.

6. Using distributive property, solve the following:
  • (a) 258 × 1008
  • (b) 1005 × 168

(a) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8

= 258000 + 2064

= 260064

(b) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168

= 168000 + 840

= 168840

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