RBSE Solutions for Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.7

RBSE Solutions for Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.7 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 6 Vedic Mathematics Exercise 6.7.

Rajasthan Board RBSE Class 7 Maths Chapter 6 Vedic Mathematics Ex 6.7

Question 1
Find the square root by (RBSESolutions.com) using ‘Viiokanam’ method :
(1) 169
(2) 324
(3) 576
(4) 2025
(5) 3025
Solution:
(1) 169
∵ Unit place of 169 is 9
∴ 169 is a perfect square number
Sum of digits = 1 + 6 + 9 = 16
Extreme digit (charam ank) of square root (RBSESolutions.com) will be either 1 or 3 and tens place digit will be 1.
Thus hence square root of 169 = 13

(2) 324
(i) Unit place of 324 is 4. So it is a perfect square number.
(ii) Sum of digits = 3 + 2 + 4 = 9 which is perfect square.
(iii) This number will have two digits in square root.
(iv) Making pair of 2-2 in 324 from right to left, second pair has only one digit 3.
∴ Tens place in square root of 324 will be 1.
(v) ∵ Extreme digit of number 324 is 3.
∴ Extreme digit of square root will be either 1 or 8.
Hence, the (RBSESolutions.com) square root of 324 is 18.

(3) 576
(i) Unit place digit in 576 is 6. So the given number is a perfect square.
(ii) Sum of digits = 5 + 7 + 6 = 18 which is not perfect square.
(iii) Square root of this number will have 2 digits.
(iv) On making the pairs of 2 – 2 from right to left, second pair has only one digit 5, therefore tens place digit of square root will be 2.
(v) Extreme digits of number will be 2.
So the extreme digit of its square root will be either 2 or 8 and extreme digit for tens place is 5 which is in between 4 – 8, therefore tens place digits in (RBSESolutions.com) square root will be 2.
∴ Square root of 576 may be either 24 or 28.
Perfect square root will be 24.
Hence square root = 24

(4) 2025
(i) Making pair from right to left first pair is 25 and another is 20.
(ii) Extreme digit of first pair is 5 therefore extreme digit of its square root will be 5.
(iii) The greatest square root within the maximum limit of 20 is 4.
Therefore possible (RBSESolutions.com) square root will be 45 Hence square root of 2025 = 45

(5) 3025
(i) Making pair from right to left, first pair is 25 and another pair is 30.
(ii) Extreme digit of first pair is 5 therefore extreme digit of its square roots will be 5.
(iii) The greatest square root within the maximum limit of 30 is 5. Therefore possible square root will be 55.
Hence square root of 3025 = 55.

(6) 9025
(i) Making (RBSESolutions.com) pairs from right to left, first pair is 25 and another pair is 90.
(ii) Extreme digit of first pair = 5 therefore extreme digit of possible square root will be 5.
(iii) Greatest square root within the maximum limit 90 is 9.
Hence square root of 9025 = 95

(7) 1024
(i) Making pairs from right to left, first pair = 24
(ii) Extreme digit of first pair = 4
∴ Extreme digit in possible square root may be either 2 or 8.
(iii) Greatest square root (RBSESolutions.com) within the maximum limit 10 = 3.
∴ Possible square root may be either 32 or 38.
Product of 3 and 8 = 3 x 8 = 24
(iv) ∵ 10 < 24
∴ Smallest number will be square root.
Hence square root = 32

(8) 441
(i) Unit place digit of 441 is 1. Therefore 441 may be a perfect square number.
(ii) Sum of (RBSESolutions.com) digit of 441 = 4 + 4 + 1 = 9 which is perfect square.
(iii) There may be two digits in square root of 441.
(iv) Making pair of 2 – 2 digits from right to left, second pair has only one digit 4. Therefore tens place digits in square root will be 2.
(v) Extreme digit of number is 1.
Therefore extreme digit in square root will be either 1 or 9 and for tens place is 2 which belongs a group 1 – 3 so the tens place will be 1.
(vi) Thus square root of 441 may be either 21 or 29.
(vii) On multiplying 2 with (2+1=3)
2 x 3 = 6
Here 4 digit of (RBSESolutions.com) second pair < Product 6
∴ Smaller number out of 21 or 29 will be taken as square root
Hence square root of 441 = 21

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