Question

A two-digit number is such that the product of its digit is $12$. When $9$ is added to the number, the digits interchange their places, find the number:
A. $62$
B. $34$
C. $26$
D. $43$

Answer 1

Hint: We will first assume the two digits as $x$ which is at one’s place and $y$ which is at tens place. Now the number formed by these digits is $10y+x$ and the number that is formed when the two digits are interchanged is $10x+y$. In the problem, they have mentioned that ‘When $9$ is added to the number, the digits interchange their places’ from this we can obtain an equation in $x$ and $y$. In the problem, they have also mentioned that the product of the digits i.e. product of $x$ and $y$ is $12$. From this, we can obtain another equation in terms of $x$ and $y$. Now to get the values of $x$ and $y$ we need to solve the obtained two equations.

Complete step-by-step solution
Let the two digits of the number are $x$ which is at one’s place and $y$ which is at tens place. Now the number is $10y+x$.
The number when the digits are interchanged is $10x+y$
In the problem we have given that When $9$ is added to the number, the digits interchange their places, mathematically
$10y+x+9=10x+y$
Rearranging the terms in the above equation so that all the variables are at one side and constants are another side.
$\begin{align}
  & \Rightarrow 10y+x-10x-y=-9 \\
 & \Rightarrow 9y-9x=-9 \\
\end{align}$
Dividing the above equation with $-9$ on both sides, then we will get
$x-y=1...\left( \text{i} \right)$.
In the problem they have also mentioned that the product of the digits is equal to $12$. i.e. $xy=12$
We have an algebraic equation i.e. ${{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy$
Substituting the all the values we have in the above equation, then we will get
$\begin{align}
  & {{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy \\
 & \Rightarrow {{\left( x+y \right)}^{2}}={{\left( 1 \right)}^{2}}+4\left( 12 \right) \\
 & \Rightarrow {{\left( x+y \right)}^{2}}=1+48 \\
 & \Rightarrow {{\left( x+y \right)}^{2}}=49 \\
 & \Rightarrow x+y=\sqrt{49} \\
 & \Rightarrow x+y=7....\left( \text{ii} \right) \\
\end{align}$
Up to now we have the equations $x-y=1$, $x+y=7$. To get the values of $x$ and $y$ we need to solve the above equations. Now the value of $y$ from $x-y=1$ is $y=x-1$, substituting this value in $x+y=7$, then we will get
$\begin{align}
  & x+y=7 \\
 & \Rightarrow x+\left( x-1 \right)=7 \\
 & \Rightarrow x+x-1=7 \\
 & \Rightarrow 2x=7+1 \\
 & \Rightarrow 2x=8 \\
 & \Rightarrow x=4 \\
\end{align}$
Now the value of $y$ is $y=x-1=4-1=3$
$\therefore $ The number is $10y+x=10\times 3+4=34$ and the number when the digits are interchanged is $43$.

Note: We can solve the equation $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ by adding both of them
$\begin{align}
  & \Rightarrow \left( x-y \right)+\left( x+y \right)=1+7 \\
 & \Rightarrow x-y+x+y=8 \\
 & \Rightarrow 2x=8 \\
 & \Rightarrow x=4 \\
\end{align}$
From equation $\left( \text{i} \right)$ $x-y=1\Rightarrow y=4-1=3$.
From both the methods we got the same result.