Question

the sum of a two digit no and the no obtained by reversing it's digits is 121. find the no if it's units place digit is greater than the tens place digit by 7.​

Answer 1

let the 2 digit number be 10x+y.

now the number obtained by reversing the digits is 10y+x.

according to the question,

(10x+y)+(10y+x)=121-----(1)

y=x+7-----(2)

(1)---

10x+x+10y+y=121

11x+11y=121

11(x+y)=121

x+y=11-----(3)

substituting the value of y from (2) into (3)

x+(x+7)=11

2x=11-7

2x=4

so, x= 2

now substitute the value of x in (2)

y=2+7

so y=9

therefore the number 10x+y is 10(2)+9=29