## RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A
- RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B
- RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C
- RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D

**Question 1.**

**Solution:**

Let tens digit = x

Units digit = 3

∴Number = 3 + 10x

According to the condition,

7 (x + 3) = 3 + 10x

7x + 21 = 3 + 10x

21 – 3 = 10x – 7x

=> 3x = 18

x = \(\\ \frac { 18 }{ 3 } \)

∴Number = 3 + 10x

= 3 + 10 x 6 = 3 + 60 = 63

**Question 2.**

**Solution:**

Let ten’s digit = x

Then units digit = 2x

and number = 10x + 2x = 12x

According to the condition,

12x = x + 2x + 18

12x – x – 2x = 18

=> 9x = 18

x = \(\\ \frac { 18 }{ 9 } \) = 2

∴Number = 12x = 2 x 12 = 24

**Question 3.**

**Solution:**

Let units digit = x

and tens digit = y

Number = x + 10y

Now x + 10y = 4 (x + y) + 3

=> x + 10y = 4x + 4y + 3

10y – 4y – 4x + x = 3

=> 6y – 3x = 3

2y – x = 1 ….(i)

∴Number by reversing the order of digits = y + 10x

=>x + 10y + 18 = y + 10x

=>10x – x + y – 10y = 18

=> 9x – 9y = 18

x – y = 2 ….(ii)

∴Adding (i) and (ii)

=> 2y – y = 3

y = 3

x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5

∴Number = x + 10y = 5 + 3 x 10

= 5 + 30 = 35

**Question 4.**

**Solution:**

Sum of two digits of a number =15

Let units digit = x

Then tens digit = 15 – x

∴Number = 10 (15 – x) + x

= 150 – 10x + x = 150 – 9x

By interchanging the digits, the new number will be

= 10x + 15 – x = 9x + 15

According to the condition,

9x + 15 = 9 + 150 – 9x

9x + 9x = 159-15 = 144

18x = 144

=>x = \(\\ \frac { 144 }{ 18 } \) = 8

∴Number = 150 – 9x = 150 – 9 x 8

= 150 – 72 = 78

**Question 5.**

**Solution:**

Let units place digit = x

and tens place digit = y

Then number = x + 10y

By interchanging the positions of the digits then

Units digits = y

and tens digit = x

∴Number = y + 10x

(x + 10y) – (y + 10x) = 63

=> x + 10y – y – 10x = 63

9y – 9x = 63

=> 9(y – x) = 63

y – x = \(\\ \frac { 63 }{ 9 } \) = 7

∴Hence, difference of its digits = 7 Ans.

**Question 6.**

**Solution:**

Sum of three digits of a number = 16

Let units digit of a three-digit number = x

Then tens digit = 3x

and hundreds digit = 4x

∴Number = x + 10 x 3x + 100 x 4x

= x + 30x + 400x = 431x

But x + 3x + 4x = 16 => 8x = 16

∴x = \(\\ \frac { 16 }{ 8 } \) = 2

∴Number = 431 x 862

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A are helpful to complete your math homework.

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