Question

the sum of the digits of the two digit number is 12 if the new number formed by reversing the digit is greater than the original number by 54 find the original number

Answer 1

let a and b the digit
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

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

Digit(Unit place) = t

Digit(Tens place) = p

Now,

Number = 10p + t

When number Reversed = 10t + p

A/Q we have,

First situation :-

⇒ p + t = 12

⇒ t = 12 - p .... (1)

Second Situation :-

⇒ 10t + p = 10p + t + 54

⇒ 10t - t - 10p + p = 54

⇒ 9t - 9p = 54

Putting value of t from (1) we have,

⇒ 9(12 - p) - 9p = 54

⇒ 108 - 9p - 9p = 54

⇒ -18p = 54 - 108

⇒ -18p = -54

⇒ p = 54/18

⇒ p = 3

Substitute the value of p in (1) :-

⇒ t = 12 - p

⇒ t = 12 - 3

⇒ t = 9

So,

Number = 10p + t

= 10(3) + 9

= 39

Hence,

Required number = 39