Question

The sum of the digits of a two digit number is 8. If the digits are reversed, the number is decreased by 54. The number is

Answer 1

Answer:

given a+b=8-->1

10b+a=a+10a-54(- bcoz reverse of digit dec by 54)-->2

from a+b=8

a=8-b sub this in eq 2

then we get 18b=18

b=1

a=7

so original numb is 71

Answer 2

Answer:

The number is 71.

Step-by-step explanation:

Given the sum of the digits of a two digit number is 8

        Let the digits be $a$ and $b$

        and hence the number is $10a+b$

        Then the sum of the digits is $a+b=8$                          ...(1)

and given, if the digits are reversed, the number is decreased by 54

                    $(10a+b)-(a+10b)=54$

                                  $\implies 9a-9b=54\implies a-b=\frac{54}{9}$

                                  $\implies a-b=6$                                        ...(2)

Adding equations (1) and (2),

                   $2a=14\implies a=\frac{14}{2} =7$

Substituting $a$ in equation (1),

                  $7+b=8\implies b=8-7=1$

Then the number is $10a+b=10\times 7+1=71$

The number is 71.