RBSE Solutions for Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.2

RBSE Solutions for Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 6 Vedic Mathematics Exercise 6.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 6
Chapter Name	Vedic Mathematics
Exercise	Ex 6.2
Number of Questions	2
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.2

Question 1
Subtract using (RBSESolutions.com) Nikhilam formula

Solution:
(i) Count from starting from right to left,
write 9 in place of every 0 and 10 on the place of last 0.
The digit just left to 0 will be become lesser 1.

(ii) (a) Here 2 < 7
∴ 7 – 2 = 5 take complement of 5.
Here (RBSESolutions.com) complement of 5 which will be kept on unit place.

(b) 7 < 8 ∴ 8 – 7 = 1
on subtracting 1 from 9 we get 8.
complement of 1 = 9

(c) 8 < 9 ∴ 9 – 8 = 1
Again on subtracting 1 from its complement 9 gives 9 – 1 = 8
(d) on (RBSESolutions.com) subtracting 2 from 5 – 1 = 4 remainder is 4 – 2 = 2

(iii) (a) Here 7 < 8
∴ Take complement of 8 – 7 = 1
∴ On unit place 10 – 1 = 9

(b) 8 < 9 ∴ 9 – 8 = 1,
On subtracting 1 from its complement 9, 9 – 1 = 8
(c) ∴ 9 > 8 ∴ 9 – 8 = 1
On (RBSESolutions.com) subtracting 9 (complement of 1) form 9 = 9 – 1 = 0
(d) 4 – 1 = 3

Question 2
Multiply using appropriate formula
(i) 87 x 10
(ii) 53 x 100
(iii) 432 x 1000
(iv) 64 x 5
(v) 72 x 50
(vi) 81 x 99
(vii) 99 x 999
(viii) 99 x 9
Solution:
(i) 87 x 10 = 870
(ii) 53 x 100 = 5300
(iii) 432 x 1000 = 432000
(iv) 64 x 5 = 64 x \(\frac { 10 }{ 2 }\) = \(\frac { 64 }{ 2 }\) x 10
= 32 x 10 = 320
(v) 72 x 50 = 72 x \(\frac { 100 }{ 2 }\) = \(\frac { 72 }{ 2 }\) x 100
= 36 x 100 = 3600
(vi) 81 x 99 =

(vii)

(viii)

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