RBSE Solutions for Class 5 Maths Chapter 4 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 4 Vedic Mathematics Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 5 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Vedic Mathematics |

Exercise |
Additional Questions |

Number of Questions |
23 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 5 Maths Chapter 4 Vedic Mathematics Additional Questions

**Multiple Choice Questions**

Question 1.

Ekadhik of 6 i.e. value of 6 is-

(a) 5

(b) 7

(c) 6

(d) 8

Question 2.

Ekanyun of 7 or meaning of 7 is-

(a) 8

(b) 7

(c) 5

(d) 6

Question 3.

Param mitra digit of 8 is called-

(a) 2

(b) 3

(c) 4

(d) 5

Question 4.

Ekadhik of 18 is-

(a) 17

(b) 16

(c) 19

(d) 20

Question 5.

In 63, Ekadhikena purvena number of digit 6 is-

(a) 62

(b) 64

(c) 0

(d) 163

Question 6.

In 523, Ekanyuna purvena number of digit 2 is-

(a) 524

(b) 423

(c) 422

(d) 623

Question 7.

Param mitra digit of a digit 1 is-

(a) 9

(b) 0

(C) 8

(d) 3

Question 8.

Extreme digit in(RBSESolutions.com)number 623 is-

(a) 6

(b) 2

(c) 3

(d) 0

Answers:

1. (b)

2. (d)

3. (a)

4. (c)

5. (d)

6. (b)

7. (a)

8. (c)

**Fill in the blanks in the following**

1. Extreme digit of any number is its digit at ………………….

2. Ekadhik of 5 is represented as ………………….

3. Ekanyunof 8 is ………………….

4. Ekadhiken(RBSESolutions.com)purvena of digit 3 in number 23 is ………………….

5. In vedic mathematics …………………. is one of the method of multiplication.

Answers

1. Unit place

2. 5

3. 7

4. 33

5. Nikhilam method (sutra).

**Very Short Answer Type Questions**

Question 1.

Which are param mitra digits ?

Solution:

If sum of two digits is(RBSESolutions.com)equal to 10, then they are mutually called param mitra digits.

Question 2.

Write Ekanyun purvena of digit 0 in 2710.

Solution:

2710 = \(27\underset { \bullet }{ 1 } 0\) = 2700

Question 3.

Which are extreme digits ?

Solution:

Extreme digit of any number is its digit at unit’s place.

Question 4.

Write deviation in(RBSESolutions.com)number 13 on base 10.

Solution:

Deviation = +3

Question 5.

Which digits are called Nikhilam digits?

Solution:

All digits of a number are called Nikhilam digits.

Question 6.

Write deviation in(RBSESolutions.com)number 8 on base 10.

Solution:

Deviation = -2.

**Short Answer Type Questions**

Question 1.

Find the value of 11 × 16 by using Nikhilam formula with base 10.

Solution:

Value of 11 × 16 by Nikhilam sutra on base 10 is-

**Directions-**

(i) Closest base is 10, so deviations are +1 and +6.

(ii) Multiplic ation of deviations = (+1) × (+6) = 6.

(iii) On the left(RBSESolutions.com)hand side, write 11 + 6 = 17 or 16 + 1 = 17.

(iv) Required solution is 176.

Question 2.

Multiply by Nikhilam Sutra- 14 × 13.

Solution:

Question 3.

Multiply 11 × 15 by vedic(RBSESolutions.com)mathematics method.

Solution:

**Directions –**

(i) Closest base = 10 so deviations are + 1 and + 5.

(ii) Multiplication of deviations =1 × 5 – 5.

(iii) On the left(RBSESolutions.com)hand side, wirte 11 + 5 = 16 or 15 + 1 = 16.

(iv) Required solution is 165.

Question 4.

Multiply 9 × 11

Solution:

**Directions –**

(i) Closest base = 10, so deviations are -1 and 1.

(ii) Multiplication of deviations = -1 × 1 = -1.

(iii) On the left hand side, write 9 + 1 = 10 or 11 – 1 = 10.

(iv) Negative(RBSESolutions.com)number can not be put on right side. So borrow one carry from left side and make it positive.

(v) On the left and side, write 10 – 1 = 9.

(vi) Required solution is 99.

Question 5.

Multiply 14 × 17

Solution:

**Directions –**

(i) Closest base = 10 so deviations are +4 and +7.

(ii) Multiplication of deviations = 7 × 4 = 28.

(iii) On the left hand side, write 14 + 7 = 21 or 17 + 4 = 21.

(iv) On left(RBSESolutions.com)side there are 2 digits, so we have to shift one digit from right side (because base 10 has only one zero, therefore only one digit will remain on right side).

(v) On left side we will write 21 + 2 = 23.

(v) Required solution is 238.

Question 6.

Do subtraction- 753 – 584

Solution:

**Directions –**

(i) 4 can not be subtracted from 3, so add param mitra digit of 4 which is 6 in 3 and write the sum = 9 below in answer.

(ii) Put Ekanyun mark on 5 which is purvena digit of 3. i.e. \(\underset { . }{ 5 } \).

(iii) \(\underset { \bullet }{ 5 } =4\), 8 can(RBSESolutions.com)not be subtracted from 4, so add param mitra digit of 8 which is 2 in 4 and write the sum = 6 below in answer.

(iv) Put Ekanyun mark on 7 which is purvena digit of \(\underset { \bullet }{ 5 } \) i.e., \(\underset { \bullet }{ 7 } \)

(v) \(\underset { \bullet }{ 7 } \) = 6 – 5 = 1

Question 7.

Do subtraction- 8321 – 7654

Solution:

**Directions –**

(i) 4 can not be subtracted from 1 so add, param mitra digit of 4 which is 6 in 1 and write the sum = 7 below in answer.

(ii) Put Ekanyun mark on 2 which is purvena digit of 1 i.e. \(\underset { \bullet }{ 2 } \)

(iii) \(\underset { \bullet }{ 2 } =1\),5 can not be subtracted from 1 so add param mitra digit of 5 which is 5 in 1 and write the sum = 6 below in answer.

(iv) Put Ekanyun mark on 3(RBSESolutions.com)which is purvena digit of 2 i.e., 3.

(v) \(\underset { \bullet }{ 3 } =2\), 6 can not be subtracted from 2 so add param mitra digit of 6 which is 4 in 2 and write the sum = 6 below in answer.

(vi) Put Ekanyun mark on 8 which is purvena digit of 3 i.e. \(\underset { \cdot }{ 8 } \).

(vii) \(\underset { \bullet }{ 8 } \) = 7, write 7 – 7 = 0.

Question 8.

Subtract- 700 – 432

Solution:

**Directions –**

(i) 2 can not be subtracted from 0, so add param mitra digit of 2 which is 8 in 0 and write the sum = 8 below in answer and put Ekadhik mark on purvena digit of 2 in below number.

(ii) \(\overset { . }{ 3 } =4\),4 can not be subtracted from 0, so add param mitra digit of 4 which is 6 in 0 and write the sum = 6 below in answer.

(iii) Put Ekadhik mark on 4 (RBSESolutions.com)which is purvena digit of \overset { \bullet }{ 3 }.

(iv) \(\overset { \bullet }{ 4 } =5\), subtract 5 from 7 and write (7 – 5) = 2.

We hope the RBSE Solutions for Class 5 Maths Chapter 4 Vedic Mathematics Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 4 Vedic Mathematics Additional Questions, drop a comment below and we will get back to you at the earliest.

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