RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Additional Questions.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 6 |
| Subject | Maths |
| Chapter | Chapter 7 |
| Chapter Name | Vedic Mathematics |
| Exercise | Additional Questions |
| Number of Questions | 17 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Additional Questions
Multiple Choiee Questions
Question 1.
Meaning of (RBSESolutions.com) ekadhiken is :
(i) one less
(ii) one more
(iii) equal
(iv) zero.
Question 2.
Ekadhiken number of 7 will be :
(i) 6
(ii) 7
(iii) 8
(iv) 9
Question 3.
In 289, ekadhiken of 2 will be :
(i) 389
(ii) 489
(iii) 189
(iv) 589
Question 4.
In 46, ekadhiken (RBSESolutions.com) poorven of 4 will be :
(i) 56
(ii) 36
(iii) 47
(iv) 146
Question 5.
Eknunen of 8 will be :
(i) 7
(ii) 8
(iii) 9
(iv) 0
Question 6.
Sign of vinkulum is :
(i) +
(ii) –
(iii) ×
(iv) +
Question 7.
In 46, eknunen (RBSESolutions.com) poorven of 6 will be :
(i) 26
(ii) 46
(iii) 76
(iv) 36
Question 8.
Parammitra digit of 4 will be :
(i) 2
(ii) 3
(iii) 6
(iv) 7
Answers
1. (ii)
2. (iii)
3. (i)
4. (iv)
5. (i)
6. (ii)
7. (iv)
8. (iii)
Fill in the blanks
(i) Meaning of eknunen is ……………………
(ii) Ekadhiken (RBSESolutions.com) of 0 is ……………………
(iii) Value lesser than base is called …………………… deviation.
(iv) In vedic mathematics, we let base as …………………… or multiple of …………………… or power of ……………………
(v) Denoting negative numbers in positive form is called ……………………
Solutions.
(i) one less
(ii) 1
(iii) negative
(iv) 10, 10, 10,
(v) vinkulum.
Very Short Answer Type Questions
Question 1.
Write ekadhiken (RBSESolutions.com) poorven of digit 1 in 125.
Solution.
Ekadhiken poorven of digit 1 in 125 = \(\overset { . }{ 0 } \)125 = 1125
Question 2.
Write eknunen poorven of digit 6 in 2675.
Solution.
Eknunen poorven of digit 6 in 2675 \(\underset { . }{ 2 } 675\) = 1675
Question 3.
What will be the deviation of number 7 and 93 ?
Solution.
Base for number 7 = 10
∴ Deviation = 10 – 7 = 3
Base for number 93 = 100
∴ Deviation = 100 – 93 = 7
Question 4.
Write complementary of parammitra (RBSESolutions.com) digit of 2.
Solution.
Those two digits whose sum is 10, are called parammitra of each other.
∴ Parammitra digit of 2 = 10 – 2 = 8
Short/Long Answer Type Questions
Question 1.
Find the sum of 65 + 68, using the formula of ekadhiken poorven.
Solution.
Hints :
(a) Addition of digits of units place 5 + 8 = 13. Thus, we will mark one more (RBSESolutions.com) sign on poorven digit of 6 and we will put 3 below the sum (at the place of unit).
(b) In the addition of tens place 6 + \(\overset { . }{ 6 } \) = 13 (where \(\overset { . }{ 6 } \) = 7).
(c) Therefore, mark one more, sign on poorven digit of 6 i.e., 0.
(d) Write the remainder 3 at the place of addition (tens place).
(e) \(\overset { . }{ 0 } \) = 1 shall be written of hundreds place.
(f) Thus, 65 + 68 = 133
Question 2.
Find the subtraction 74 – 69, using the formula of eknunen poorven.
Solution.
Hint :
(a) 9 cannot be subtracted from 4. Therefore, we will add the (RBSESolutions.com) complementary digit of 9 i.e., 1 to 4, 4 + 1 = 5, write below the sum.
(b) Put a sign of less than on the poorven digit of 4 ie., 7, such as 7 = 6.
(c) Subtract 6 from \(\underset { . }{ 7 } \) (6 – 6 = 0), and write remainder 0 below.
(d) Thus, 74 – 69 = 05 or 5
Question 3.
Convert number 7 into vinkulum number.
Solution.
Hint :
(a) Vinkulum line on the parammitra digit of 7 i.e., 3.
(b) One more sign on (RBSESolutions.com) poorven digit of 7 i.e., 0.
(c) Write \(\overset { . }{ 0 } =1\)
(d) Thus, vinkulum of 7 is 1\(\overline { 3 } \) .
Question 4.
Solve 5 × 8 by formula of Nikhilam.
Solution.
5 × 8
Hint :
(a) Multiple number 5 = 10 – 5 which is 5 less than 10, and 8 = 10 – 2, which is 2 less than 10, write (- 5) and (- 2) as in the form of deviation.
(b) Write the numbers up and down and their (RBSESolutions.com) deviation in front of them.
(c) Product of deviation (- 5) × (- 2) = 10, would be written on right side of diagonal line.
(d) Write on left side (5 – 2) or (8 – 5) = 3.
(e) These will be only one number on the right side because in base 10, there is one zero.
(f) In the product of deviation (10) unit digit 0, on R.H.S. and add 1 on left side (in form of base 10).
(g) On L.H.S. 3 + 1 = 4.
(h) On removing slanted line, product is 40.
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