Question

the sum of the digit of a two digit number is 12. the number obtained by interchanging its digit exceed the given number by 18. find the number .

Answer 1

Let us assume x and y are the 2digit of the number

.: 2 digit no. is = 10x + y  & the reversed number = 10y + x

Given

x + y = 12

y = 12 - x   ----------------------(i)

Also ,

10y +x - 10x -y = 18

9y -9x = 18

Taking Common [9]

9[ y - x ] = 18

y - x = 2  ---------------(ii)

Substituting The value of y from eq 1 & 2

12 -x - x = 2

12 - 2x = 2

12 - 2 = 2x

10 = 2x

x = 5

.: y = 12 - x

y = 12 - 5

y = 7

.: Two digit Number will be : 10x + y = 10*5 + 7

50 + 7

= 57

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Answer 2

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