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 = x 10

= 32 x 10 = 320

(v) 72 x 50 = 72 x = x 100

= 36 x 100 = 3600

(vi) 81 x 99 =

(vii)

(viii)

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