Question

The sum of the digits of a two-digit number is 12. the no obtained by interchanging the order of 2 digit exceeds the given no by 18. find the 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

here's the ans to your ques :)

hope it helps you ....

let the 1st no be x

2nd no be y

according to the ques

x+y=12 --------1

the original no is 10x+y

on interchanging the digits ,

10y+x=10x+y+18

10y-y+x-10x=18

9y-9x=18

9(y-x)=18

y-x=18/9

y-x=2

-x+y=2 ---------2

now we can solve the equations 1 and 2 by using elimination method

x+y=12

-(-x+y=2)

---------------

2x = 10

x=10/2

x=5

now we can substitute the value of x in eq 1

now

x+y=12

5+y=12

y=12-5

y=7

hence ,

1st no = x= 5

2nd no =y= 7

so the no is 57