RBSE Solutions for Class 6 Maths Chapter 3 Whole Numbers Ex 3.2

RBSE Solutions for Class 6 Maths Chapter 3 Whole Numbers Ex 3.2 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 3 Whole Numbers Exercise 3.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Whole Numbers
Exercise	Ex 3.2
Number of Questions	10
Category	RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 3 Whole Numbers Ex 3.2

Question 1.
Add the following by (RBSESolutions.com) arranging in proper order.
(i) 85 + 186 + 15
(ii) 175 + 96 + 25
(iii) 65 + 75 + 35
(iv) 55 + 86 + 45
Solution.
Add under associativity –
(i) 85 + 186 + 15
(85 + 186) + 15 = 85 + (186 + 15)
271 + 15 = 85 + 201 ⇒ 286 = 286

(ii) 175 + 96 + 25
(175 + 96) + 25 = 175 + (96 + 25)
271 + 25 = 175 + 121 ⇒ 296 = 296

(iii) 65 + 75 + 35
(65 + 75) + 35 = 65 + (75 + 35)
140 + 35 = 65 + 110 ⇒ 175 = 175

(iv) 55 + 86 + 45
(55 + 86) + 45 = 55 + (86 + 45)
141 + 45 = 55 + 131 ⇒ 186 = 186

Question 2.
Find out the (RBSESolutions.com) multiplication by proper order.
(i) 4 × 1225 × 25
(ii) 4 × 158 × 125
(iii) 4 × 85 × 25
(iv) 8 × 20 × 125
Solution.
Multiply under associativity –
(i) (4 × 1225) × 25 = 1225 × (4 × 25)
4900 × 25 = 1225 × 100 = 122500
(ii) (4 × 158) × 125 = 158 × (4 × 125)
632 × 125 = 158 × 500 = 79000
(iii) (4 × 85) × 25 = 85 × (4 × 25)
340 × 25 = 85 × 100 = 8500
(iv) (8 × 20) × 125 = 20 × (8 × 125)
160 × 125 = 20 × 1000 = 20000

Question 3.
Find out the value of each of the following by (RBSESolutions.com) distributive property.
(i) 185 × 25 + 185 × 75
(ii) 4 × 18 + 4 × 12
(iii) 54279 × 92 + 8 × 54279
(iv) 12 × 8 + 12 × 2
Solution.
Finding each value by distributive law –
(i) 185 × 25 + 185 × 75 = 185 × (26 + 75) – 185 × 100 = 18500
(ii) 4 × 18 + 4 × 12 = 4 × (18 + 12) = 4 × 30 = 120
(iii) 54279 × 92 + 8 × 54279 = 54279 × (92 + 8) = 54279 × 100 = 5427900
(iv) 12 × 8 + 12 × 2 = 12 × (8 + 2) = 12 × 10 = 120

Question 4.
Find out the multiplication by using proper property.
(i) 185 × 106
(ii) 208 × 185
(iii) 54 × 102
(iv) 158 × 1008
Solution.
Solving by (RBSESolutions.com) distribution method of multiplication on addition—
(i) 185 × 106 = 185 × (100 + 6)
= 185 × 100 +185 × 6 = 18500 +1110 = 19610
(ii) 208 × 185 = 185 × 208 = 185 × (200 + 8)
= 185 × 200 + 185 × 8 = 37000 + 1480 = 38480
(iii) 54 × 102 = 54 × (100 + 2)
= 54 × 100 + 54 × 2 = 5400 + 108 = 5508
(iv) 158 × 1008 = 158 × (1000 + 8)
= 158 × 1000 + 158 × 8 = 158000 + 1264 = 159264

Question 5.
Match the following

Solution.
(i) (b)
(ii) (a)
(iii) (d)
(iv) (c)

Question 6.
If the multiplication of any two whole (RBSESolutions.com) numbers is zero, can we say that one or both of the numbers must be zero? Give an example to prove it.
Solution.
Yes, because when zero is multiplied by any number, result is zero
e.g.- 0 × 7 = 0, 8 × 0 = 0, 0 × 0 = 0

Question 7.
If the multiplication of two whole numbers is 1, then can we say that one or both of the numbers are equal to 1? Prove your answer with example.
Solution.
If multiplication of two whole numbers is 1, then definitely one or both the numbers are equal to 1 or should be reciprocal to each other.
eg.- 1 × 1 = 1 and 5 × \(\frac { 1 }{ 5 } \) =1

Question 8.
Find out the following (RBSESolutions.com) by distributive method.
(i) 138 × 101
(ii) 125 × 400
(iii) 608 × 35
Solution.
Solving by distribution method of multiplication on addition –
(i) 138 × 101 = 138 × (100 + 1)
= 138 × 100 + 138 × 1 = 13800 + 138 = 13938
(ii) 125 × 400 = 125 × (300 + 100)
= 125 × 300 + 125 × 100 = 37500 + 12500 = 50000
(iii) 608 × 35 = 608 × (30 + 5)
= 608 × 30 + 608 × 5 = 18240 + 3040 = 21280

Question 9.
Which of the following will not result in zero ?
(i) 1 + 0
(ii) 0 × 0
(iii) \(\frac { 0 }{ 2 } \)
(iv) 10 – \(\frac { 10 }{ 2 } \)
Solution.
Solving each
(i) 1 + 0 = 1 will not result in zero
(ii) 0 × 0 = 0 will result in zero
(iii) \(\frac { 0 }{ 2 } \) = 0 will result in zero
(iv) 10 – \(\frac { 10 }{ 2 } \) = 10 – 5 = 5 will not result in zero

Question 10.
Choose a, b, c…. and write (RBSESolutions.com) in the bracket.
(i) Which of the following has the commutative property of addition?
(a) 5 × 8 = 8 × 5
(b) (2 × 3) × 5 = 2 × (3 × 5)
(c) (12 + 8) + 10 = (2 + 8) + 10
(d) 15 + 8 = 8 + 15
(ii) Which of the following has commutative property of multiplication.
(a) 10 × 20 = 20 × 10
(b) 10 × 10 = 20 × 20
(c) (10 × 20) = 10 × 1
(d) 10 + 20 = 10 × 20
Solution.
(i) (d)
(ii) (a).

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