RBSE Solutions for Class 8 Maths Chapter 5 Vedic Mathematics In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 5 Vedic Mathematics In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 5 |
| Chapter Name | Vedic Mathematics |
| Exercise | In Text Exercise |
| Number of Questions | 4 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 5 Vedic Mathematics In Text Exercise
Page No: 58
Question 1.
Multiplying the following by Urdhwtirgbhyaam Method
(i) 15 × 12
(ii) 60 × 18
(iii) 71 × 8
(iv) 122 × 4
(v) 706 × 56
(vi) 497 × 173
Solution
(i) 15 × 12
Step 1.
Step 2.
Step 3.
1×1 / 1×2+5×1 / 5×2
Step 4.
1 / 2+5 / 10
Step 5.
1 / 7 /10
Step 6.
Step 7.
Therefore, the required(RBSESolutions.com)product we get
15 × 12 = 180
(ii) 60 × 18
Step 1.
Step 2.
Step 3.
6×1 / 6×8+0×1 / 0 ×8
Step 4.
6 / 48+0 / 0
Step 5.
6 / 48 / 0
Step 6.
Step 7.
Therefore, the required(RBSESolutions.com)product we get
60 × 18 = 1080
(iii) 71 × 8
Step 1.
Step 2.
Step 3.
7×0 / 7×8+1×0 / 1×8
Step 4.
0 / 56 + 0 / 8
Step 5.
0 / 56 / 8
Step 6.
Step 7.
Therefore, the required product we get 71 × 8 = 568
(iv) 122 x 4
Step 1.
Step 2.
Step 3.
1×0 / 1×0+2×0 / 1×4+2×0+2×0 / 2×4+2×0 / 2×4
Step 4.
0 / 0 + 0/ 4 + 0 + 0 / 8 + 0/ 8
Step 5. 0 / 0 / 4 / 8 / 8
Step 6.
Step 7.
Therefore, the required product we get 122 × 4 = 488
(v) 706 × 56
Step 1.
Step 2.
Step 3.
7×0 / 7×5+0+0 /7×6+6×0+0×5 / 0×6+6×5 / 6×6
Step 4.
0 / 35 + 0 / 42 + 0 + 0 / 0 + 30 / 36
Step 5.
0 / 35 / 42 / 30 / 36
Step 6.
Step 7.
Therefore, the required product we get 706 × 56 = 39536
(vi) 497 × 173
Step 1.
Step 2.
Step 3.
4×1 / 4×7+9×1 / 4×3+x 1+9×7 / 9×3+7×7 / 7×3
Step 4.
4 / 28 + 9 / 12 + 7 × 63 / 27 + 49 / 21
Step 5.
4/37/82/76/21
Step 6.
Step 7.
Therefore, the required product we get 497 × 173 = 85981
Page No: 60
Question 2.
Find the product of the following
(i) 11 x 15
(ii) 12 x 18
(iii) 19 x 17
(iv) 28 x 22
(v) 51 x 49
(vi) 99 x 96
Solution
(i) 11 x 15
Base = 10
Sub-base = 1 x 10 = 10
Sub-base digit = 10 ÷ 10 = 1
Deviation(RBSESolutions.com)from Sub-base = 11 – 10 = + 1
15 – 10 = + 5
Therefore, the required product of 11 x 15
we get 165
(ii) 12 x 18
Base = 10
Sub-base = 1 x 10 = 10
Sub-base digit = 10 ÷ 10 = 1
Deviation from Sub-base = 12 – 10 = + 2
18 – 10 = + 8
Therefore, the required(RBSESolutions.com)product of 12 x 18
we get 216
(iii) 19 x 17
Base = 10
Sub-base = 1 x 10
Sub-base digit = 10 ÷ 10 = 1
Deviation from Sub-base = 19 – 10 = + 9
17 – 10 = + 7
Therefore, the required product of 19 x 17
we get 323
(iv) 28 x 22
Base = 10
Sub-base = 2 x 10 = 20
Sub-base digit = 20 ÷ 10 = 2
Deviation from Sub-base = 28 – 20 = + 8
22 – 20 = + 2
Therefore, the required(RBSESolutions.com)product of 28 x 22
we get 616
(v) 51 x 49
Base = 10
Sub-base = 5 x 10 = 50
Sub-base digit = 50 ÷ 10 = 5
Deviation from Sub-base = 51 – 50 = + 1
49 – 50 = – 1
= 2499
Therefore, the required product of 51 x 49 we get 2499.
(vi) 99 x 96
Base = 10
Sub-base = 9 x 10 = 90
Sub-base digit = 90 ÷ 10 = 9
Deviation from Sub-base = 99 – 90 = + 9
96 – 90 = + 6
Therefore, the required(RBSESolutions.com)product of 99 x 96 we get 9504
Page No: 63
Question 3.
Multiply the following three numbers by formula Nikhilam
(i) 11 x 12 x 13
(ii) 8 x 9 x 10
(iii) 6 x 7 x 8
(iv) 27 x 28 x 29
(v) 98 x 99 x 99
(vi) 51 x 52 x 53
Solution
(i) 11 x 12 x 13
Step 1. Base of 11, 12, 13 = 10
Deviation = 1, 2, 3
Therefore, the required product of Number
11 x 12 x 13 = 1716
(ii) 8 x 9 x 10
Step 1. Base of 8, 9, 10 = 10
Deviation = – 2, -1, 0
Therefore, the required Product of Number
8 x 9 x 10 = 720
(iii) 6 x 7 x 8
Step 1. Base of 6, 7, 8 = 10
Deviation = – 4, – 3, – 2
Therefore, the required Product of Number
6 x 7 x 8 = 336
(iv) 27 x 28 x 29
Step 1. Base of 27, 28, 29 = 20
Deviation = 7, 8, 9
Therefore, the required Product of Number 27 x 28 x 29 = 21924
(v) 98 x 99 x 99
Therefore, the required Product(RBSESolutions.com)of Number 98 x 99 x 99 = 960498
(vi) 51 x 52 x 53
Therefore, the required Product of Number 51 x 52 x 53 = 140556
Page No: 68
Question 4.
Division operation using Dhwajank method
(1) 1737 ÷ 21
(2) 37941 ÷ 47
(3) 23754 ÷ 74
(4) 3257 ÷ 74
(5) 7453 ÷ 79
(6) 59241 ÷ 82
Solution.
(1) 1737 ÷ 21
Quotient = 82
Remainder = 15
(2) 37941 ÷ 47
Quotient = 807
Remainder = 12
(3) 23754 ÷ 74
Quotient = 321
Remainder = 0
(4) 3257 ÷ 74
Quotient = 44
Remainder = 1
(5) 7453 ÷ 79
Quotient = 94
Remainder = 27
(6) 59241 ÷ 82
Quotient = 722
Remainder = 37
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