NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

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Updated at 2021-05-07


NCERT solutions for class 6 Maths Chapter 2 Whole Numbers Exercise 2.1

Q.1 Write the next three natural numbers after 10999.



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

11000, 11001 and 11002

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

10000, 9999 and 9998

Q.3. Which is the smallest whole number?



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

0

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

There are 20 whole numbers between 32 and 53 which are given below:
(33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52)

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) 2440702
(b) 100200
(c) 1100000
(d) 2345671

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) 93
(b) 9999
(c) 208089
(d) 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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) \(\because \)530 > 503
\(\therefore \) 503 is on the left side of 530 on the number line
(b) \(\because \) 370 > 307
\(\therefore \) 307 is on the left side of 370 on the number line
(c) \(\because \) 98765 > 56789
\(\therefore \). 56789 is on the left side of 98765 on the number line
(d) \(\because \) 9830415 < 10023001
\(\therefore \) 9830415 is on the left side of 10023001 on the number line

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.



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) False
(b) False
(c) True
(d) True
(e) True
(f) False
(g) False
(h) False
(i) True
(J) False
(k) False
(l) True
(m) False

NCERT solutions for class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) Given 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) Given 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(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

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(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

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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) Given 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3
= 73800 + 2214
= 76014
(b) Given 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2
= 85400 + 1708
= 87108
(c) Given 258 × 1008
= 258 × (1000 + 8)
= 258 × 1000 + 258 × 8
= 258000 + 2064
= 260064
(d) Given 1005 × 168
= (1000 + 5) × 168
= 1000 × 168 + 5 × 168
= 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 ? 44 per litre, how much did he spend in all on petrol?



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

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 = Rs. 44
Total money spent = 44 × (40 + 50)
= 44 × 90
= 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 Ra. 45 per litre, how much money is due to the vendor per day?



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

Solutions: Milk quantity supplied in the morning = 32 litres
Milk quantity supplied in the evening = 68 litres
Cost of milk per litre = Rs. 45
Total cost of milk per day = 45 × (32 + 68)
= 45 × 100
= Rs. 4500
Hence, the money is due to the vendor per day is Rs. 4500

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.



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(i) 425 × 136 = 425 × (6 + 30 + 100) (c) Distributivity of multiplication over addition. Hence (c) is the correct answer
(ii) 2 × 49 × 50 = 2 × 50 × 49 (a) Commutativity under multiplication Hence, (a) is the correct answer
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (b) Commutativity under addition Hence, (b) is the correct answer

NCERT solutions for class 6 Maths Chapter 2 Whole Numbers 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



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(a) 1 + 0 = 1
Hence, it does not represent zero
(b) 0 × 0 = 0
Hence, it represents zero
(c) 0 / 2 = 0
Hence, it represents zero
(d) (10 – 10) / 2 = 0 / 2 = 0
Hence, it represents zero

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.



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

Yes, if the product of two whole number is zero we can say that one or both of them will be zero
Examples: 0 × 81 = 0, 1 × 0 = 0 and 0×0 = 0

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.



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

No, the product of two whole number is 1, if both number will be 1.
Example: 1×1 = 1 but 1×8= 8

Q.4 Find using distributive property:
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

(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
= (800 + 24) × 25
= (800 + 25 – 1) × 25
= 800 × 25 + 25 × 25 – 1 × 25
= 20000 + 625 – 25
= 20000 + 600
= 20600
(d) Given 4275 × 125
= 4275 × (100+20 + 5 )
= 427500 + 85500 + 21375
= 534375
(e) Given 504 × 35
= (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140= 17640

Q.5 Study the pattern:
1 x 8 + 1= 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?



NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers


Answer :

Step I: 123456 x 8 + 6 = 987654
Step II: 1234567 x 8 + 7 = 9876543
Working pattern:
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765



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