Question

the sum of the digit of a two digit number is 12 the number obtained by reversing the order of the digit of the given number exceeds the given number by 18 find the 2 digit number

Answer 1

Let us assume x and y are the two digits of the number

Therefore, two-digit number is = 10x + y and the reversed number = 10y + x

Given:

x + y = 12

y = 12 – x -----------1

Also given:

10y + x - 10x – y = 18

9y – 9x = 18

y – x = 2 -------------2

Substitute the value of y from eqn 1 in eqn 2

12 – x – x = 2

12 – 2x = 2

2x = 10

x = 5

Therefore, y = 12 – x = 12 – 5 = 7

Therefore, the two-digit number is 10x + y = (10*5) + 7 = 57

Answer 2

One digit number is X
sum of the digits =12
tens digit = 12-x
original no.= 10(12-x)+X
New no. after reserving the digits=10×x+(12-x)
original no.+18= New no.
10(12-x)+X+18=10x+12-x
10×12-10×X+x+18=9x+12
120-9x+18=9x+12
-9x-9x=12-120-18
-18x=12-138
-18x=-126
X=126÷18
X=7
One digit no. is 7
tens digit= 12-7
=5
So the no.is 57 or 75