NCERT Solutions | Class 6 Maths Chapter 2 | Whole Numbers

CBSE Solutions | Maths Class 6

Check the below NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Pdf free download. NCERT Solutions Class 6 Maths were prepared based on the latest exam pattern. We have Provided Whole Numbers Class 6 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 6 Maths

NCERT Solutions Class 6 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	6th
Subject:	Maths
Chapter:	2
Chapters Name:	Whole Numbers
Medium:	English

Whole Numbers | Class 6 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 6 Maths Chapter 2 Whole Numbers to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.1

Ex 2.1 Class 6 Maths Question 1.

Write the next three natural numbers after 10999.

Solution:

The next three natural numbers after 10999 are 10999 +1,10999 + 2 and 10999 + 3
i.e.,11000, 11001 and 11002.

Ex 2.1 Class 6 Maths Question 2.

Write the three whole numbers occurring just before 10001.

Solution:

The three whole numbers occurring just before 10001 are 10001 – 1, 10001 – 2 and 10001 – 3
i.e., 10000, 9999 and 9998.

Ex 2.1 Class 6 Maths Question 3.

Which is the smallest whole number?

Solution:

The smallest whole number is 0.

Ex 2.1 Class 6 Maths Question 4.

How many whole numbers are there between 32 and 53?

Solution:

There are (53 – 32) -1 = 21 – 1= 20 whole numbers between 32 and 53.

Ex 2.1 Class 6 Maths Question 5.

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

Solution:

(a) Successor of 2440701 = 2440701 + 1 = 2440702
(b) Successor of 100199 = 100199 + 1 = 100200
(c) Successor of 1099999 =1099999 +1 = 1100000
(d) Successor of 2345670 = 2345670 + 1 = 2345671

Ex 2.1 Class 6 Maths Question 6.

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

Solution:

(a) Predecessor of 94 = 94 – 1 = 93
(b) Predecessor of 10000 = 10000 – 1 = 9999
(c) Predecessor of 208090 = 208090 – 1 = 208089
(d) Predecessor of 7654321 =7654321 – 1 = 7654320

Ex 2.1 Class 6 Maths 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

Solution:

(a) Here 503 lies on the left of 530 on the number line.
∴ 530 > 503
(b) Here 307 lies on the left of 370 on the number line.
∴ 370 > 307
(c) Here 56789.lies on the left of 98765 on the number line.
∴ 98765 > 56789
(d) Here 9830415 lies on the left of 10023001
∴ 9830415 > 10023001

Ex 2.1 Class 6 Maths 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.

Solution:

(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

Ex 2.2 Class 6 Maths Question 1.

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

Solution:

Ex 2.2 Class 6 Maths Question 2.

Find the product by suitable rearrangement:
(a) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25

Solution:

(a) 2 x 1768 x 50 = (2 x 50) x 1768
= 100 x 1768 = 176800
(b) 4 x 166 x 25 = (4 x 25) x 166
= 100 x 166 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291
= 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279
= 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60)
= 285 x 300 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25)
= 1000 x 1000
= 1000000

Ex 2.2 Class 6 Maths Question 3.

Find the value of the following:
(a) 297 x 17 + 297 x 3
(b) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218

Solution:

(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)
= 297 x 20
= 5949
(b) 54279 x 92 + 8 x 54279
= 54279 x (92 + 8)
= 54279 x 100 = 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 -69)
= 81265 x 100
= 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845x5x782 + 3845x5x218
= (3845 x 5) x (782+ 218)
=19225 x 1000
=19225000

Ex 2.2 Class 6 Maths Question 4.

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

Solution:

(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 = 73800 + 2214 = 76014
(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 = 85400 + 1708 = 87108
(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 = 258000 + 2064 = 260064
(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840 = 168840

Ex 2.2 Class 6 Maths Question 5.

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

Solution:

Petrol filled on Monday and Tuesday
= 40 litres + 50 litres
= 90 litres
Cost of petrol @ ₹ 44 per litre
= 90 x ₹ 44 = ₹ 3960

Ex 2.2 Class 6 Maths 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 ? 15 per litre, how much money is due to the vendor per day?

Solution:

Milk supplied to a hotel in the morning and evening
= 32 litres + 68 litres = 100 litres
Money due to vendor per day.
= 100 x ₹ 15 = ₹ 1500

Ex 2.2 Class 6 Maths Question 7.

Match the following:
(i) 425 x 136 = 425 x (6 + 30 +100) (a) Commutativity under multiplication
(ii) 2 x 49 x 50 = 2 x 50 x 49 (b) Commutativity under addition
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition

Solution:

Matching is as under :
(i) → (c)
(ii) → (a)
(iii) → (b)

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.3

Ex 2.3 Class 6 Maths Question 1.

Which of the following will not represent zero :
(a) a + 0
(b) 0 x 0
(c) \(\frac { 0 }{ 2 } \)
(d) \(\frac { 10-10 }{ 2 } \)

Solution:

(a) 1 + 0 = 1 ≠ 0
(b) 0 x 0 = 0
(c)\(\frac { 0 }{ 2 } =0\times \frac { 1 }{ 2 } =0\)
(d) \(\frac { 10-10 }{ 2 } =\frac { 0 }{ 2 } =0\times \frac { 1 }{ 2 } =0\)
Thus, only (a) does not represent zero.

Ex 2.3 Class 6 Maths 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.

Solution:

We know that the product of any whole number and zero is always zero i. e., a x 0 = 0, where a is any whole number.
NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers 2

Ex 2.3 Class 6 Maths Question 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.

Solution:

We know that if a is any whole number, then
a x l = a -1 x a
∴ Clearly a must be = 1
Thus, both the numbers should be 1 i. e., 1 x 1 = 1

Ex 2.3 Class 6 Maths Question 4.

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

Solution:

Ex 2.3 Class 6 Maths Question 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?

Solution:

Write the next two steps :
123456 x 8 + 6 = 987654
1234567×8 + 7 = 9876543
How the pattern works?
1 x8 + 1= 9
(11 + 1) x 8 + 2 = 12 x 8 + 2 = 98
(111 + 11 + 1) x 8 + 3 = 123 x 8 + 3 = 987 .
(1111 + 111 + 11 + 1) x 8 + 4 =1234 x 8 + 4 =9876
(11111 + 1111 + 111 + 11 + 1) x 8 + 5
= 12345 x 8 + 5=98765
(111111 + 11111 + 1111 + 111 + 11 + 1) x 8+ 6
= 123456 x 8 + 6 = 987654
and (1111111 + 111111 + 11111 +1111 +111 +11 +1) x 8 + 7
= 1234567 x 8 + 7 = 9876543