Question

the two digits of a two digit number differ by 5 . If the digits are reversed , the sum of the two numbers is 99 . Find the numbers

Answer 1

Hi there !!
Here's your answer

Let the digit in the one's place be x
Digit in tens place = x + 5

The original number formed will be
10(x+5) + x
= 10x + 50 + x
= 11x + 50 ______(i)

Given that,
when the numbers are reversed , the sum of original and new number is 99

By interchanging the digits , we have,
Digit in units place = x + 5
Digit in tens place = x

The new number formed will be :
10(x) + x + 5
= 11x + 5 ______(ii)

So,
adding (i) and (ii),
we have,

11x + 50 + 11x + 5 = 99
22x + 55 = 99
22x = 99 - 55 = 44
x = 44/22
x = 2

So,
digit in the units place = x = 2
Digit in tens place = x + 5 = 2 + 5 = 7

Thus,
the original and the new numbers are 27 and 72 respectively

Note : If you take x as the digit in tens place , then the original and the new numbers will be 72 and 27 respectively.


Verification [ if required ]
Original number = 27
New number = 72

Their sum = 99

Thus, our answer is correct !

____________________

Hope it helps :D

Answer 2

Answer:

Original no. 27

Interchange no. 72