Question

Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two-digit number,

four times the sum of its two digits is 21 less than the reversed two-digit number.
(Hint: You can use variables to represent the digits of a number. If a two-digit number has the digit x in tens place and yin one's place, the
number will be 10x + y. Reversing the order of the digits will change their place value and the reversed number will 10y + x.)
The difference of the original two-digit number and the number with reversed digits is

Answer 1

Answer: 9

Explanation:

let the two-digit number has the digit 'x' in ten's place and 'y' in one's place.

So, The Original Number will be 10x+y and the Reversed Number will be 10y+x.

Now, According to the given information :

⇒ 5(x+y) + 13 = 10x + y

⇒ 4y + 13 = 5x ......(equation 1)

and,

⇒ 4(x+y) + 21 = 10y + x

⇒ 6y - 21 = 3x ......(equation 2)

By solving equation 1 and 2, we get

y = 8 and x = 9

Hence,

The Original Number = 10x + y = 10×9 + 8 = 98

The Reversed number = 10y + x = 10×8 + 9 = 89

and,

The difference of the original two-digit number and the number with reversed digits is ( 98 - 89 ) = 9.