RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.3

RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.3 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.3.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 6
Subject	Maths
Chapter	Chapter 7
Chapter Name	Vedic Mathematics
Exercise	Ex 7.3
Number of Questions	1
Category	RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.3

Question 1.
Convert the general (RBSESolutions.com) numbers into its vinkulum.
(i) 8
(ii) 27
(iii) 82
(iv) 78
(v) 96
Sol.
(i) 8
Hints :
(a) Put vinkulum line on the parammitra digit of 8 i.e.,2 = \(\overline { 2 }\)
(b) Put one more sign on poorven digit of 8 i.e., 0 = \(\dot { 0 } \)\(\overline { 2 }\)
(c) Write \(\dot { 0 } \) = 1.
(d) Thus, we got the (RBSESolutions.com) vinkulum of 8 = \(\overline { 2 }\)

(ii) 27
Hints :
= \(\overline { 8 }\) 7
(a) Digit 7 would be as it is and
= \(\dot { 0 } \)\(\overline { 8 }\)7 vinkulum line would be on param
= 187 mitra digit of 2 i.e., 8.
(b) One more sign on the poorven of 8 i.e., 0.
(c) Write \(\dot { 0 } \) = 1,
(d) Thus, we got vinkulum of 27

(iii) 82
Hint :
= \(\overline { 2 }\)2 (a) Digit 2 would be as it is and vinkulum line would be on
= \(\dot { 0 } \)\(\overline { 2 }\)2 parammitra digit of 8 i.e., 2.
= 1\(\overline { 2 }\)2 (b) One more sign (RBSESolutions.com) on the poorven digit of 8 ie., 0.
(c) Write \(\dot { 0 } \) = 1
(d) Thus, we get vinkulum of 82.

(iv) 78
Hint :
= \(\overline { 3 }\)8 (a) Digit 8 would be as it is and vinkulum line would be on
= \(\dot { 0 } \)\(\overline { 3 }\)8 parammitra digit of 7 i.e., 3.
= 1\(\overline { 3 }\)8 (b) One more (RBSESolutions.com) sign on the poorven digit of 7 ie., 0.
(c) Write \(\dot { 0 } \) = 1.
(d) Thus, we got vinkulum of 78.

(v) 96
Hint :
= \(\overline { 1 }\)6 (a) Digit 6 would be as it is and
= \(\dot { 0 } \)\(\overline { 1 }\)6 vinkulum line would be on parammitra digit of 9 i.e., 1.
= 1\(\overline { 1 }\)6 (b) One more sign (RBSESolutions.com) on the poorven digit of 9 Le., 0.
(c) Write \(\dot { 0 } \) = 1.
(d) Thus, we got vinkulum of 96

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