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

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.