Question

The sum of two digit number and the number formed by reversing the digits is 65. Find the number, if one of the digits is one more than the other?

Answer 1

Answer:

13 is the answer of the ques5ion

Answer 2

Answer:

Step-by-step explanation:Let the 2 digit number as(10a+b)

Number obtain by reversing the digit=(10b+a)

Given,

( a+b=11)~equation-1

also,

=(10a+b)-(10b+a)=9

=10a+b-10b-a=9

=9a-9b=9

=9(a-b)=9

=(a-b)=9–9

=a-b=1

=a+b+a-b=11+1

=2a=12

=a=6

Putting the value of a in equation 1, we get

=6+b=11

=b=11–6

=b=5

Original number=10×6+5=60+5=65

So the require number is 65