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.

(i) By Sutra Ekadhikena Purvena, (RBSESolutions.com) add the following:

(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) 12^{3} = 12 + 2 x (2) / 3(2)^{2} / (2)^{3}

= 16 / 12 / 8 = 1728

(b) 105^{3}

= 105 + 2 x (05) / 3 x (05)^{2} / (05)^{3} = 115 / 75 /_{1} 25 = 1157625

(ii) (a) 35^{3}

= 3^{2} (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) 497^{3}

= 5^{2} {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) 53214^{2}

= 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

= 5^{2} / 2 x 5 x 3 / 2 x (2 x 5) + 3^{2} / 2 x (5 x 1) + 2 (3 x 2) / 2 (5 x 4) + 2 (3 x 1) + 2^{2} / 2 (3 x 4) + 2 (2 x 1) / 2 x (2 x 4) + 1^{2} / 2 x 1 x 4 / 4^{2}

= 25 / 30 / 29 / 22 / 50 / 28 / 17 / 8 / 16

= 2831729796

(b) 31321^{2}

= 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

= 3^{2} / 2 x 1 x 3 / 2 (1 x 2) + 3^{2} / 2 x (3 x 2) + 2 (1 x 3) / 2 (3 x 1) + 2 (1 x 2) + 3^{2} / 2 (1 x 1) + 2 (3 x 2) / 2 x (3 x 1) + 2^{2} / 2 x 2 x 1 / 1^{2}

= 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 – 4^{2} = 0, New dividend = 09, Divisor = 4 x 2 = 8.

(ii) 9 ÷ 8, Quotient digit = 1, Remainder = 1.

(iii) New dividend = 17,

Adjusted dividend = 17 – 1^{2} = 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 – 3^{2} = 1, New dividend = 13, Divisor = 3 x 2 = 6.

(ii) 13 ÷ 6, Quotient digit = 2, Remainder = 1.

(iii) New dividend = 12,

Adjusted dividend = 12 – 2^{2} = 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 + 1^{2}) = 0

(ix) 9 – 2 x 1 x 4 = 1

(x) 16 – 4^{2} = 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

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