RBSE Solutions for Class 8 Maths Chapter 5 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 5 Vedic Mathematics Additional Questions.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 5 |
| Chapter Name | Vedic Mathematics |
| Exercise | Additional Questions |
| Number of Questions | 13 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 5 Vedic Mathematics Additional Questions
I. Objective type Questions
Question 1.
How many groups makes when two digit number is multiplied by two digit number
(a) 1
(b) 2
(c) 3
(d) 4
Question 2.
How many groups makes when three digit number is multiplied by three digit number
(a) 3
(b) 5
(c) 6
(d) 9
Question 3.
If base is 10 and sub-base is 30 then what sub-base digit will be
(a) 3
(b) 2
(c) 1
(d) 43
Question 4.
Sub-base digit is calculated as
(a) Base ÷ Sub-base
(b) Sub-base – Base
(c) Sub-base ÷ Base
(d) Base – Sub-base
Question 5.
In multiplication process by Urdhva(RBSESolutions.com)Tiryagbhyam, the number of groups is known by
(a) (2n – 1)
(b) (2n + 2)
(c) (2n + 1)
(d) (2n – 2)
Question 6.
If the number is 22 and the base is 20, then the deviation is
(a) 10
(b) 20
(c) 2
(d) – 2
Answer
1. (c)
2. (b)
3. (a)
4. (c)
5. (a)
6. (c).
II. Short Answer Type Questions
Question 1.
Find the product of following
(i) 98 x 98
(ii) 102 x 102
Solution
(i) 98 x 98
Base = 10
Sub-base = 9 x 10 = 90
Sub-base digit = 90 ÷ 10 = 9
Deviation(RBSESolutions.com)from Sub-base = 98 – 90 = + 8
98 – 90 = + 8
Therefore, the required product of 98 x 98 will be 9604
(ii) 102 x 102
Base = 10
Sub-base = 10 x 10 = 100
Sub-base digit = 100 ÷ 10 = 10
Deviation from Sub-base = 102 – 100 = + 2
102 – 100 = + 2
Therefore, the required product of 102 x 102 will be 10404.
Question 2.
Find the cube of 103.
Solution
Base = 100, Deviation = + 03
(103)3 = 103 + 2 x 03 / 3 x (03)2 / (03)3
= 103 + 6 / 27 / 27
= 1092727
Question 3.
Find the cube of 96.
Solution
Base = 100, Deviation = (- 04)
(96)3 = 96 + 2 x (- 04)/3 x (- 04)2/(- 04)3
= 96 – 08 / 3 x 16 / – 64
= 88 / 48 / – 64
= 88 / 48 /1 – 64
= 88 / 47 / 100 – 64
= 88 / 47 / 36
= 884736
Question 4.
Find the cube of 15.
Solution
Base = 10, Deviation = + 5
Then,
(15)3 = 15 + 2 x 5 / 3 x 52 / 53
= 15 + 10 / 75 / 125
= 25 / 75 / 125
= 25 /8 7 / 5
= 33/7/5
= 3375
Question 5.
Find the value of 101 x 102 x 103 by applying ‘nikhilam’ formula.
Solution
Product(RBSESolutions.com)by ‘nikhilam’ formula Base = 100
= 106 / 11 / 06
= 1061106
Question 6.
Multiply by Urdhwtirygbhyaam method
426 x 351
Solution
Hence the required product 426 x 351
= 149526
Question 7.
Multiply by(RBSESolutions.com)using Nikhilam formula :
22 x 23
Solution
Base = 10
Sub-base = 2 x 10 = 20
Sub-base digit = 20 ÷ 10 = 2
Deviation from sub-base = 22 – 20 = 2,
23 – 20 = 3
Hence the required product of 22 x 23 = 506
Question 8.
Multiply by using Nikhilam formula
103 x 103 x 103
Solution
Base = 100
= 103 + 3 + 3 / 9 + 9 + 9 / 27
= 109 / 27 / 27
= 1092727
Question 9.
Divide by using Dhwajank Method 18542 ÷ 52
Solution
∴Quotient = 356 and Remainder = 30
Question 10.
Solve 415 x 132 using ‘Urdhavtirgbhyaam’ sutra.
Solution
415 x 132
Step 1. 415 is multiplicand and 132 is multi-plier. Writing in the form of multiplication.
Step 2. Number of groups will be 5 in three digit numbers.
= 54780
Question 11.
Solve 3732 ÷ 42 by ‘Dhwajank’ Sutra.
Solution
3732 ÷ 42
Hint—(i) 37 ÷ 4, first digit of quotient = 8, remainder = 5
(ii) New dividend = 53
Modified dividend = 53 – 8 x 2 = 37
(iii) 37 ÷ 4, second digit of quotient = 8, remainder = 5
(iv) New dividend = 52
Modified(RBSESolutions.com)dividend = 52 – 8 x 2 = 36
= Final remainder
∴ Quotient = 88, Remainder = 36.
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