# Rd Sharma Solutions Class 8 Chapter 5

## Rd Sharma Solutions Class 8 Chapter 5 Playing with Numbers

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 5, Playing with Numbers. Here students can easily find Exercise wise solution for chapter 5, Playing with Numbers. Students will find proper solutions for Exercise 5.1, 5.2 and 5.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Playing with Numbers Exercise 5.2 Solution :

Question no – (1)

Solution :

Since 35a64 is multiple of 3.

3 + 5 + a + 6 + 4 is a multiple of 3

= a + 6 is also multiple of 3

a + 6 = 0, 3, 6, 9, 12, 15 ……….(i)

But a is a digit of the number 35a64

∴ a can take values 0,1,2,3, …..9

= a + 6 can take values 6,7, 8, 9, 10, …15 …..(ii)

From (i) and (ii) we get,

a + 6 = 6 or 9, or 12 or 12 or 15

= a = 0, a = 3, a = 6, a = 9

Hence, x = 3 , 6,

Therefore, the possible values of a will be 0,3, 6, 9

Question no – (2)

Solution :

Since 18×71 is multiple of 3

1 + 8 + x + 7 + 1 is x multiple of 3

= 8 + x is also multiple of 3

(8+x) = 1, 4, 7, ……(i)

But, x is a digit of the number 18×71

x can take values 1,2,3, ……9

= (8+x) = can take values 6,7,8,9,10 …15 ….(ii)

From, (i) and (ii) we get

8 + x = 9 or 12 or 15

= x = 1, x = 4, x =7,

Hence x = 1, 4, 7,

Therefore, the possible values of x will be 1, 4, 7

Question no – (3)

Solution :

Since, 66784x is multiple of 9

6 + 6 + 7 + 8 + 4 + x is multiple of 9

= 4 + x is also multiple of 9

4 + x = 5, 14,………

= x = 5 or 14

Therefore, the possible values of x will be 5 or 14.

Question no – (4)

Solution :

67 × 19 is Divisible by 11.

(6 + x + 9) – (7+1) is a multiple of 11

= (15 + x) – 8 is a multiple of 11

(15 + x) 8 is a multiple of 11

(15 + x) = 0 or, 22, or 33

When x = 7 … then (15 + 7) = 2 multiple by 11.

Question no – (5)

Solution :

3 × 2 is multiple of 11.

We have,

Sum of the digits in odd places = 3 + 2 = 5

Sum of the digits in even places = y

Sum of the digits in even places – sum of digit in odd places = y – 5

If 3×2 is multiple by 11, then

y – 5 must be a multiple of 11.

= y – 5 = 0 or 11 or 22 or 33…….

y = 5 or 16 or 27……

Question no – (6)

Solution :

98215 × 2 is divisible by 4.

We have,

x² = 10x + 2 where x = 1 ,2, 3, 4, 5, 6, 7, 8, 9,

x² = 12, 22, 32, 42, 52, 62, 72, 82, 92,…..

out of those values of a8 only 12, 32, 52, 72, 92, are divisible by 4.

x = 1, 3, 5, 7, 9,………

Therefore, all the possible values of x will be 1, 3, 5, 7, 9

Question no – (7)

Solution :

67×19 is Divisible by 11.

(6 + x + 9) – (7 + 1) is a multiple of 11

= (15 + x) – 8 is a multiple of 11

(15 + x) 8 is a multiple of 11

(15 + x) = 0 or, 22, or 33

When x = 7 … then (15+7) = 2 multiple by 11.

Question no – (12)

Solution :

6 is divisible by 2 not by 4.

Question no – (13)

Solution :

(i) If a number is divisible by 3 it must be divisible by 9

= False.

(ii) If a number is divisible by 9, it must be divisible by 3

= Yes, It is True that, if a number is divisible by 9, it must be divisible by 3.

(iii) If number is divisible by 4, it must be divisible by 8

= False.

(iv) If a number is divisible by 8, it must be divisible by 4

= True.

(v) A number is divisible by 18, if it is divisible by both 3 and 6

= False
Because, a number is divisible by 18, if it is divisible by both 2 and 9.

(vi) If a number is divisible by both 9 and 10, it must be divisible by 90

= True.

(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately

= False.

(viii) If a number divides three numbers exactly, it must divide their sum exactly

= True.

(ix) If two numbers are co-prime, at least one of them must be a prime number

= False.

(x) The sum of two consecutive odd numbers is always divisible by 4

= True.

Playing with Numbers Exercise 5.3 Solution :

Question no – (1)

Solution :

3 7

+ A B
———————-
9 A

A = 5, B = 8

Question no – (2)

Solution :

A B

+ 7 3
———————-
9 A

A = 2, B = 5,

Question no – (3)

Solution :

A 1

+ 1 B
———————-
B 0

A = 7, B = 9

Question no – (4)

Solution :

2 A B

+A B 1
———————-
B 1 8

A = 4, B = 7

Question no – (5)

Solution :

1 2 A

+ 6 A B
———————-
A 0 9

A = 8, B = 1

Question no – (6)

Solution :

A B 7

+ 7 A B
———————-
9 8 A

A = 2, B = 5

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Updated: June 13, 2023 — 2:36 pm