# RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions

RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions.

 Board RBSE Class Class 9 Subject Maths Chapter Chapter 1 Chapter Name Vedic Mathematics Exercise Additional Questions Number of Questions Solved 12 Category RBSE Solutions

## RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions

Question 1. (ii) By vedic method, subtract (RBSESolutions.com) the following:  (iii) By Sutra Ekadhikena Purvena, multiply (RBSESolutions.com) the following:
(a) 586 x 514
(b) 3993 x 3007
Solution  Question 2.
(i) Multiply with the (RBSESolutions.com) help of Sutra Ekanyunena Purvena
(a) 8567 x 9999
(b) 512 x 99
(ii) By vedic method, multiply the following:
(a) $$11\frac { 1 }{ 3 }$$ x $$11\frac { 2 }{ 3 }$$
(b) 1.15 x 1.85
Solution
(i) (a) L.H.S = 8567 – 1 = 8566
R.H.S = 9999 – 8566 = 1433
∴8567 x 9999
= 8567 – 1 / 9999 – 8566
= 85661433
(b) L.H.S = 512 – 1 = 511
R.H.S = 99 – 511
∴512 x 99
= 511 / 99 – 511
= 51199 – 511
= 50688 Question 3.
(i) Change the following (RBSESolutions.com) into vinculum number:
(a) 898
(b) 18469
(ii) Change the following into ordinary number: Solution   Question 4.
(i) Multiply the following (RBSESolutions.com) by Sutra Nikhilam (base method):
(a) 92 x 87
(b) 1007 x 1012
(c) 103 x 105 x 106
(d) 12 x 13 x 15
(ii) Multiply the following by Sutra Nikhilam (sub-base method):
(a) 54 x 56
(b) 206 x 212
(c) 21 x 24 x 25
(d) 502 x 503 x 504
Solution   Question 5.
(i) Find the square of the (RBSESolutions.com) following by Sutra Nikhilam (base method):
(a) 17
(b) 104
(ii) Find the square of the following by Sutra Nikhilam (sub-base method):
(a) 64
(b) 308
Solution
(i) (a) 17² = 17 + 7 / 7²
= 24 /4 9 = 289
(b) 104² = 104 + 04 / (04)²
= 108 / 16
= 10816
(ii) (a) 64² = 6 (64 + 4) / 4²
= 408 /1 6
= 4096
(b) 308² = 3 (308 + 08) / (08)²
= 948 / 64
= 94864

Question 6.
(i) Find the cube of the (RBSESolutions.com) following by Sutra Nikhilam (base method):
(a) 12
(b) 105
(ii) Find the cube of the following by Sutra Nikhilam (sub-base method):
(a) 35
(b) 497
Solution
(i) (a) 123 = 12 + 2 x (2) / 3(2)2 / (2)3
= 16 / 12 / 8 = 1728
(b) 1053
= 105 + 2 x (05) / 3 x (05)2 / (05)3 = 115 / 75 /1 25 = 1157625
(ii) (a) 353
= 32 (35 + 2 x 5) / 3 x 3 x (5)2 / (5)3 = 9 x 45 / 9 x 25 / 125 = 405 /22 5 /12 5 = 42875
(b) 4973
= 52 {497 + 2 x (- 03)} / 5 x 3 x (- 03)2 / (- 03)3
= 25 x 491 / 5 x 27 / (- 27)
= 12275 /1 35 /(- 27)
= 12276 / 34 / 100 – 27 = 122763473 Question 7.
Divide the following (RBSESolutions.com) by using Nikhilam Sutra:
(a) 243 ÷ 9
(b) 1356 ÷ 96
Solution Question 8.
Find the product by using (RBSESolutions.com) Sutra Urdhva tiryagbhyam or by Sutra vertically and crosswise.
(a) 447 x 028
(b) 108 x 112
Solution  Question 9.
Find the square of the following (RBSESolutions.com) numbers by Dvanda yoga method or Duplex process:
(a) 53214
(b) 31321
Solution
(a) 532142
= Dvanda of 5 / Dvanda of 53 / Dvanda of 532 / Dvanda of 5321 / Dvanda of 53214 / Dvanda of 3214 / Dvanda of 214 / Dvanda of 14 / Dvanda of 4
= 52 / 2 x 5 x 3 / 2 x (2 x 5) + 32 / 2 x (5 x 1) + 2 (3 x 2) / 2 (5 x 4) + 2 (3 x 1) + 22 / 2 (3 x 4) + 2 (2 x 1) / 2 x (2 x 4) + 12 / 2 x 1 x 4 / 42
= 25 / 30 / 29 / 22 / 50 / 28 / 17 / 8 / 16
= 2831729796
(b) 313212
= Dvanda of 3 / Dvanda of 13 / Dvanda of 132 / Dvanda of 3132 / Dvanda of 31321 / Dvanda of 1321 / Dvanda of 321 / Dvanda of 21 / Dvanda of 1
= 32 / 2 x 1 x 3 / 2 (1 x 2) + 32 / 2 x (3 x 2) + 2 (1 x 3) / 2 (3 x 1) + 2 (1 x 2) + 32 / 2 (1 x 1) + 2 (3 x 2) / 2 x (3 x 1) + 22 / 2 x 2 x 1 / 12
= 9/6/13/18/19/14/10/4/1
= 981005041 Question 10.
Calculate square root of the following (RBSESolutions.com) numbers by Dvanda yoga method or Duplex process:
(a) 169744
(b) 10329796
Solution Steps:
(i) 16 – 42 = 0, New dividend = 09, Divisor = 4 x 2 = 8.
(ii) 9 ÷ 8, Quotient digit = 1, Remainder = 1.
(iii) New dividend = 17,
Adjusted dividend = 17 – 12 = 16.
(iv) 16 ÷ 8, Quotient digit = 2, Remainder = 0.
(v) Final remainder = 0.4 – 2 x 1 x 2
= 0
Therefore, (RBSESolutions.com) square root of 169744 is 412. Steps:
(i) 10 – 32 = 1, New dividend = 13, Divisor = 3 x 2 = 6.
(ii) 13 ÷ 6, Quotient digit = 2, Remainder = 1.
(iii) New dividend = 12,
Adjusted dividend = 12 – 22 = 8.
(iv) 8 ÷ 6, Quotient digit = 1, Remainder = 2.
(v) New dividend = 29,
Adjusted dividend = 29 – 2 x 1 x 2 = 25
(vi) 25 ÷ 6, Quotient digit = 4, Remainder = 1
(vii) New dividend = 17
Now find final (RBSESolutions.com) remainder as follows:
(viii) 17 – (2 x 4 x 2 + 12) = 0
(ix) 9 – 2 x 1 x 4 = 1
(x) 16 – 42 = 0

Question 11.
Divide the following by Dhwajanka method:
(a) 12474 ÷ 54
(b) 21112 ÷ 812
Solution Steps:
(i) 12 ÷ 5, Quotient digit = 4 is written below (RBSESolutions.com) the horizontal line and write remainder = 2 before 4.
(ii) New dividend = 24 Adjusted dividend
= New dividend – Quotient digit x Dhwajanka
= 24 – 2 x 4 = 16
(iii) 16 ÷ 5, Quotient digit = 3, Remainder = 1
(iv) Adjusted divisor = 16 – 3 x 4 = 4
(v) 4 ÷ 4, Quotient digit 1,
Remainder = 0
(vi) Final remainder = 04 – 1 x 4 = 0
Quotient = 231, Remainder = 0 Steps:
(i) 21 ÷ 8, Quotient digit = 2, Remainder = 5
(ii) New dividend = 51,
Corrected (RBSESolutions.com) dividend = 51 – 2 x 1
= 49
(iii) 49 ÷ 8, Quotient digit = 6, Remainder = 1
(iv) Final remainder
= 112 – (6 x 1 + 2 x 2) 10 – 6 x 2 = 112 – 100 – 12 = 0
Quotient = 26, Remainder = 0.

Question 12.
Divide the following by (RBSESolutions.com) Paravartya yojayet method:
(a) 6534 ÷ 123
(b) 13999 ÷ 1112
Solution  We hope the given RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 1 Vedic Mathematics Additional Questions, drop a comment below and we will get back to you at the earliest.