Question

a number consists of two digits and the digit in the ten's place exceeds that in the unit's place by 5. if 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed. then the sum of digits of the number is?

Answer 1

sum of the digits is (x+y),i.e.,7+2=9

Answer 2

Hi there !!
Here's your answer

Given that ,
the digit in the ten's place exceeds that in the unit's place by 5

So,
Let the digit in units be x
Digit in tens place = x + 5

The number is 10(x+5) + x

= 10x + 50 + x = 11x + 50 ____(i)

Given,
if 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed

So,
5 times the sum of digits = 5(x + x + 5 )

= 5(2x + 5) = 10x + 25

Also,
If this sum is subtracted from the number, the digits are reversed
So,

By interchanging the digits,we have,

Digit in units place = x + 5
Digit in tens place = x

The new number is 10(x) + x + 5

= 10x + x + 5 = 11x + 5_____(i)

Now,
Forming a balanced equation, we have,

11x + 50 - (10x + 25) = 11x + 5

11x + 50 - 10x - 25 = 11x + 5

x + 25 = 11x + 5

25 - 5 = 11x - x

20 = 10x


$x = \frac{20}{10}$
x = 2

Thus,

Digit in units place = x = 2
Digit in tens place = x + 5 = 2 + 5 = 7

So,
the sum of digits is 2 + 7 = 9

The original number is 72