Question

The sum of the digits of $2$- digit number is $11$. The number obtained by adding $4$ to this number is $41$ less than the reversed number. Find the original number.

Answer 1

Hint: For finding the original number, we will assume the two digit number as $10x+y$. Then, we will make an equation according to the given conditions in the question. After that we will solve the equations to get the value of variables. Now, we will substitute the value of variables in the assumed digit and will get the final answer.

Complete step by step answer:
Let’s suppose that the two digit number is $10x+y$, where $x$ and $y$ are positive integers and digits of the original number.
The first condition given in the question is that the sum of the digits of a two digit number is $11$. So, according to the condition:
$\Rightarrow x+y=11$ … $\left( i \right)$
The second condition is the number obtained by adding $4$ to this number is $41$ less than the reversed number. So, according to this condition:
$\Rightarrow 10x+y+4=10y+x-41$
Now, we will make equal like terms and variable one side as:
$\Rightarrow 10x-x+y-10y=-41-4$
Here, we will do necessary calculations accordingly as:
\[\Rightarrow 9x-9y=-45\]
We will divide by $9$ in the above step as:
\[\Rightarrow \dfrac{\left( 9x-9y \right)}{9}=\dfrac{-45}{9}\]
After solving above step we will have:
$\Rightarrow x-y=-5$ … $\left( ii \right)$
Now, we will add equation $\left( i \right)$ and equation $\left( ii \right)$:
$\Rightarrow x+y+x-y=11+\left( -5 \right)$
Here, we will do required mathematical operations to solve the above equation as:
$\begin{align}
  & \Rightarrow 2x=11-5 \\
 & \Rightarrow 2x=6 \\
\end{align}$
Now, we will divide by $2$each side as:
$\dfrac{2x}{2}=\dfrac{6}{2}$
Here, we will cancel out the equal like term and use division as:
$x=3$
Since, we got the value of $x$. We will substitute $3$ for $x$ in equation $\left( i \right)$:
$\Rightarrow 3+y=11$
Here, we will subtract $3$ both sides as:
$\Rightarrow 3+y-3=11-3$
Now, we will cancel out equal like term and will do necessary calculation as:
$\Rightarrow y=8$
Since, we got both the digits. We will substitute it in the assumed number to get the original number as:
$\begin{align}
  & \Rightarrow 10x+y \\
 & \Rightarrow 10\times 3+8 \\
\end{align}$
After required calculation, we will have:
$\begin{align}
  & \Rightarrow 30+8 \\
 & \Rightarrow 38 \\
\end{align}$
Hence, the original number is $38$.

Note: we will check that our solution is correct or not with the help of the condition given in the question that the number obtained by adding $4$ to this number is $41$ less than the reversed number. So, according to this condition:
$\Rightarrow 10x+y+4=10y+x-41$
Now, we will substitute $3$ for $x$ and $8$ for \[y\] in the above equation as:
$\Rightarrow 10\times 3+8+4=10\times 8+3-41$
Here, we will complete the multiplication of numbers as:
$\Rightarrow 30+8+4=80+3-41$
Now, we will do the addition and subtraction accordingly as:
$\begin{align}
  & \Rightarrow 38+4=83-41 \\
 & \Rightarrow 42=42 \\
\end{align}$
Since, $L.H.S.=R.H.S$.
Hence, the solution is correct.