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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