Question

In a two digit number, the unit digit is 5 less than ten's digit. Sum of digits one-eighth of the number find the number?​

Answer 1

Answer:

72

Step-by-step explanation:

Let the unit digit be b.  As this is 5 less than the ten's digit, the ten's digit is b+5.  Thus the number itself is

                             10(b+5)+b = 11b+50.

As the sum of the digits is one-eighth of the number itself, eight times the sum of the digits is equal to the number.  Expressed algebraically, this is

                         8 ( (b+5) + b ) = 11b + 50

                  =>    8 ( 2b + 5 ) = 11b + 50

                  =>    16b + 40 = 11b + 50

                  =>    5b = 10

                  =>    b = 2

Hence the units digit is 2 and the ten's digit is b+5 = 7, so the number is 72.

       

Answer 2

Answer:

72

Step-by-step explanation:

let's consider unit digit be x.As it is less than tens digit,tens digit will be b+5

10(b+5)+b=11b +50

a sum of digits is one eighth of the number itself. eight times the sum of digit is equal to the number

8[(b+5)+b]=11b+50

by solving we get,b=2

hence the unit digit is 2 and tens digit is b + 5 is equal to 7 then the number will be 72