The sum of the digits of a two digit number is 12, if the new number formed by reversing the digits is greater than the original number by 54, find the original number.
Answers
a+b=12 ........(1)
54+(10a+b)=(10b+a)
54+10a+b=10b+a
9a-9b=-54
a-b=-6........(2)
adding (1) and (2)
a+b=12
a-b=-6
----------
2a=6
a=3
so b=9
39+54=93 (verified)
orig no.=39 and No. when digits reversed=93
Given : The sum of the digits of a two digit number is 12, the new number formed by reversing the digits is greater than the original number by 54 .
To find : the original number
Solution :
Let one of those no. be x
Since , their sum → 12
Therefore , the other no. → 12 - x
The two digit number
formed → 10 ( 12 - x) + x
→ 120 - 10x + x
→ 120 - 9x
The other two digit no. formed by reversing the above
number gives → 10 × x + 12 - x
→ 10x - x + 12
→ 9x + 12
ATQ ,
↝ [ 9x + 12 ] - [ 120 - 9x ] = 54
↝ 9x + 12 - 120 + 9x = 54
↝ 18x - 108 = 54
↝ 18x = 54 + 108
↝ x = 162 / 18
↝ x = 9
The two digit number formed → 12 - x
→ 12 - 9 = 3
→ 39 Ans !!