RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.8

RBSE Solutions for Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.8 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.8.

Rajasthan Board RBSE Class 6 Maths Chapter 7 Vedic Mathematics Ex 7.8

Question 1.
Divide by formula (RBSESolutions.com) of Nikhilam :
(i) 124 ÷ 89
(ii) 406 ÷ 9
(iii) 298 ÷ 96
(iv) 1358 ÷ 113
(v) 1234 ÷ 112
(vi) 306 ÷ 8
Solution.
On dividing by formula of Nikhilam :
(i) 124 ÷ 89
Divisor = 89
Complementary number = 100 – 89 = 11
On dividing in three section :

Taking base 100, there are two zeroes in (RBSESolutions.com) base number. Therefore remainder should also be of maximum two digits. For this we draw a straight line from right side leaving 2 digits. We draw one more straight line on the left side. Now write divisor 89 on the left side of this line and write complementary number 11 below this. Process of solution is like this :

First of all write divisor, then digit 1 under middle section. Then, after multiplying this by complementary number, write below the divisor. Now, add the digits on right side. Middle section is quotient and third section is remainder. Hence, quotient = 1 and remainder = 35.

(ii) 406 ÷ 9
Divisor = nearest base of 9 = 10
Complementary number = 1
Here is base 10, there is only a single 0. Therefore, write 6 of divisor (RBSESolutions.com) in the third section. In the middle section digit of divisible is 40.

Hint :
Number 10 greater than 9 by 1 number, then we write 1 below 4 in middle section and 9 below 10 in third section. We will write parammitra of 9 below (RBSESolutions.com) in sum. Hence, quotient = 45 and remainder = 1.

(iii) 298 ÷ 96
Divisor = 96
Complementary number = 100 – 96 = 04

Hint :
(a) Base is 100. Therefore, there are written two digits on right side.
(b) Write 2 and multiplied it by complementary number and write in third section.
(c) Remainder was 106 but there shall be two digits in the third section (since, base = 100). Therefore, we will 02 from 102 in third section and add 1 in middle section and multiply 1 with the complementary number and write it in third section.
(d) Add the numbers in (RBSESolutions.com) third section and middle section.
(e) Hence, quotient and remainder are 3 and 10 respectively.

(iv) 1358 ÷ 113

Hint :
(i) Divisor = 113, base = 100, deviation = 13
(ii) Converted digits = – 1, – 3
(iii) On writing 58 in third section and 13 in middle section.
(iv) Write 13 below 1, in place of sum, product 1 × -1 -3 = -1 -3
(v) Write -1 below 3 and 5 below -3.
(vi) Writing sum 3 – 1 = 2 in place of sum.
(vii) Again (RBSESolutions.com) product = 2 × -1 -3 = -2 -6, write below last numbers.
(viii) While adding, quotient = 12 and remainder = 02

(v) 1234 ÷ 112

Hint :
(i) Divisor = 112, base = 100, deviation = 12
(ii) Converted digits = -1, -2
(iii) On putting 34 in third section and 12 in middle section.
(iv) Write 1 below 12 in place of sum, product 1 × -1 -2 = -1 -2
(v) Write -1 below 2 and -2 below 3.
(vi) Write sum 2 -1 = 1 in place of sum.
(vii) Again product = 1 × -1 -2 = -1 -2, write (RBSESolutions.com) below last digits.
(viii) While adding, quotient = 11 and remainder = 02 ,

(vi) 306 ÷ 8
Divisor = 8
Complementary nunber = 10 – 8 = 2

Hint :
(a) In middle section, write 3 below in place of sum.
(b) Digit 3 × complementary number = 3 × 2 = 6, write it below 0 and leave third section empty.
(c) Write sum 0 + 6 = 6, below in place of sum.
(d) Again 6 x complementary number = 6 × 2 = 12, write it in third section below 6.
(e) Write sum 6 + 6 = 12, below in place of sum.
(f) Remainder is 12, but on 1 digit can be put in third section, because base is 10. Thus, add 2 from 12 in third section and add 1 in middle section after (RBSESolutions.com) multiplying it by complementary number.
(g) Add the numbers in middle section and third section.
(h) Hence, quotient = 38 and remainder = 2.

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