A number consists of two digits whose sum is 9. The number formed by reversing the digits exceeds twice the original number by 18. Find the original number.

Let unit digit of the number be x and its ten’s digit be y.

Then the number will be 10×y + x = 10y + x

Sum of its digits is 9.

Therefore, the sum of x and y is 9

x + y = 9……(1)

On reversing the number, ten’s digit will become unit’s digit and unit’s digit will become ten’s digit.

So, after reversing the number,

Its ten’s digit = x and unit’s digit = y

Therefore, the new number formed = 10x + y

New number formed exceeds the twice the original number by 18

⇒ 10x + y = 2(10y + x) + 18

⇒ 10x + y = 20y + 2x + 18

⇒ 8x – 19y = 18…..(2)

From equation(1),

x + y = 9

⇒ x = 9 – y…..(3)

Substituting the value of x in eq(2)

8(9 – y) – 19y = 18

⇒ 72 – 8y – 19y = 18

⇒ – 27y = 18 – 72

⇒ – 27y = – 54

⇒ y = 2

Substituting the value of y in eq(3)

x = 9 – 2

⇒ x = 7

Hence, the original number is 10y + x = 10×2 + 7 = 27