Question

SUM OF THE DIGIT OF A TWO DIGIT NO. IS 11 . WHEN WE INTERCHANGED THE DIGIT IT IS FOUND THAT THE RESULTING NEW NO. IS GREATER THAN THE ORIGINAL NO. BY 63 . FIND THE TWO DIGIT NO.

Answer 1

Let the tens digit be x and the ones digit be y.
x + y = 11 … (1)
Original number = 10x + y
Reverse number = 10y + x
Given, Reverse number = Original number + 63
∴10y + x = 10x + y + 63
⇒ 9y – 9x = 63
⇒ y – x = 7 … (2)
Adding (1) and (2),

x+y = 11
-x+y=7
2y=18
⇒ y = 9
When y = 9,
x + 9 = 11 (Using (1))
x = 11 – 9 = 2
∴ Original number = 10x + y = 20 + 9 = 29

Answer 2

Let the tens digit be a and the ones digit be b.

a+ b = 11  … (1)

Original number = 10a + b
Reverse number = 10b +a
 Reverse number = Original number + 63
∴10b + a = 10a+ b + 63

⇒ 9b– 9a = 63

⇒ b – a =  7 … (2)

Adding (1) and (2),

 a+b = 11
-a+b=7

2b=18

⇒ b = 9

When b = 9,

a + 9 = 11  (Using (1))

a = 11 – 9 = 2

∴ Original number = 10a + b = 20 + 9 = 29