Question

the sum of digits of a two digit number is 11.if the digit at tens place is increased by 5 and the digit at unit place is decreased by 5 the digit of a number are found to be reversed. find the original number

Answer 1

let two digit number be 10x + y and it's reverse number be 10y + x

sum of digits of the 10x + y two digit number is 11
so, x+y = 11...........(1)

the digit at tens place is increased by 5 and the digit at unit's place is increased by 5 the digit of a number are found to be reversed
so,
10(x + 5) + y - 5 = 10y + x
10x + 50 + y - 5 = 10y + x
10x + y + 45 = 10y + x
10x - x + 45 = 10y - y
9x + 45 = 9y
divide by 9 on both sides
x + 5 = y
y - x = 5.........(2)
sum (1) and (2)
x + y - x + y = 11 + 5
2y = 16
y = 8 substitute in (1)
x + 8 = 11
x = 11 - 8
x = 3
so the original number is (10x + y) = 10*3 + 8 = 30+8 = 38