Question

The middle digit of a three digit number is 0 and the sum of the other two digits is 11 . If the number obtained by reversing the digits exceeds the original number by 495,find the number.

Answer 1

Dear Student ,

Let the number in 100s digit be x, and the number in 1s digit be y,

Then,the number would be x0y : 100x+y

Given ,
Sum of the digits : x+y=11.....(i)

If it is reversed ,it would be y0x : 100y+x,

Given,
The difference between them: 100x+y -(100y+x) = -495

100x+y-100y-x =-495

99x-99y = -495 (divided by 99)

x-y=-5.....(ii)

Adding (i) and (ii), we get,

(x+y)+(x-y)=11+(-5)

x+y+x-y=11-5

2x=6

∴ x=3

Substituting value of x in (i),

x+y=11

3+y=11

y=8

∴The number is 308

Regards