A number consists of two digit whose sum is $9$. If the digits are reversed, the new number is $\frac{3}{8}$ of the number. Find the number.

Given: A number consists of two digit whose sum is $9$. If the digits are reversed, the new number is $\frac{3}{8}$ of the number.

To do: To find the number.

Solution:

Let $x$ and $y$ be the two digit of the number.

As given, sum of the two digit $x+y=9\ .......\ ( i)$

Previous number$=xy$

Value of previous number$=10x+y$

While we reverse the digits, we get $yx$.

Value of new number $=10y+x$

As given, $10x+y=\frac{3}{8}( 10y+x)$

$\Rightarrow 8( 10x+y)=3( 10y+x)$

$\Rightarrow 80x+8y=30y+3x$

$\Rightarrow 80x-3x-30y+8y=0$

$\Rightarrow 77x-22y=0$

$\Rightarrow 11( 7x-2y)=0$

$\Rightarrow 7x-2y=0\ .........\ ( ii)$

On multiplying $( i)$ by $2$

$2x+2y=18\ ......\ ( iii)$

Add $( ii)$ and $( iii)$

$2x+2y+7x-2y=18+0$

$\Rightarrow 9x=18$

$\Rightarrow x=\frac{18}{9}$

$\Rightarrow x=2$

On putting $x=2$ in $( i)$

$2+y=9$

$\Rightarrow y=9-2=7$

The number is $"xy"=27$

Thus, the number is $27$.