Question

The sum of digits of a two digit no. is 7 .If the no. obtained by reversing the digits is 11 more than the original no.,find the original no.

Answer 1

is it's answer 43 or 34 ???
if yes ..

Let a be the first digit and b be the second digit. The two digit number ( which looks like ab ) value is 10a+b

Also it is given that a+b=7a+b=7 -- ( 1 )

When the digits are interchanged ( it looks like baba ) the number's value becomes 10b+a10b+a

It is given that adding 9 to the original number makes it 10b+a10b+a ( interchanged )

So, 10a+b+9=10b+a10a+b+9=10b+a

Moving "a" and "b" terms to left side and moving numericals to right side,

10a+b−10b−a=−910a+b−10b−a=−9

9a−9b=−99a−9b=−9

9(a−b)=−99(a−b)=−9

a−b=−1a−b=−1

a=b−1a=b−1

Substituting for aa in ( 1 ), we get,

b−1+b=7b−1+b=7

2b=7+1=82b=7+1=8

b=82=4b=82=4

Hence a=b−1=4−1=3a=b−1=4−1=3

Hence original number is 34 and new number is 43.