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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