
A certain number between $10$ and $100$ is 8 times the sum of digits. If 45 is subtracted from it the digits will be reversed. Find the number.
Answer
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Hint: First assume the digits of one’s place and ten’s place to be $x\And y$ respectively. So, we will get the number as $10y+x$. Now, by using the information that the number is 8 times the sum of digits and if 45 be subtracted from it the digits will be reversed, frame two equations. By solving these equations we get the desired answer.
Complete step-by-step answer:
We have been given that a number between $10$ and $100$ is 8 times the sum of digits. Also, if 45 is subtracted from it the digits will be reversed.
We have to find the number.
Let us assume that the digits of one’s place is $x$ and ten’s place is $y$.
Now, the number will be $10y+x$ .
Now, given in the question that number is 8 times the sum of digits,
So, mathematically we have
$10y+x=8\left( x+y \right)$
Simplifying further we get
$10y+x=8x+8y$
$\begin{align}
& \Rightarrow 10y-8y=8x-x \\
& \Rightarrow 2y=7x \\
& y=\dfrac{7x}{2}..........(i) \\
\end{align}$
Now, it is given in the question that if 45 be subtracted from the number digits will be reversed.
So, we have $10y+x-45=10x+y$
Simplifying further, we get
$\begin{align}
& \Rightarrow 10y-y-45=10x-x \\
& \Rightarrow 9y=9x+45 \\
\end{align}$
Dividing the whole equation by $9$, we get
$\Rightarrow y=x+5$
Now, substituting the value of $y$ from equation (i) and simplifying further, we have
$\begin{align}
& \Rightarrow \dfrac{7x}{2}=x+5 \\
& \Rightarrow 7x=2\left( x+5 \right) \\
& \Rightarrow 7x=2x+10 \\
& \Rightarrow 7x-2x=10 \\
& \Rightarrow 5x=10 \\
& \Rightarrow x=\dfrac{10}{5} \\
& x=2 \\
\end{align}$
Now, the value of $y$ will be
$\begin{align}
& \Rightarrow y=x+5 \\
& y=2+5 \\
& y=7 \\
\end{align}$
So, we have digits of one’s place is $2$ and ten’s place is $7$.
Now, the number will be
$\begin{align}
& \Rightarrow 10\times 7+2 \\
& \Rightarrow 70+2 \\
& =72 \\
\end{align}$
So, the number is $72$.
Note: The possibility of mistake while considering the number is that if it is calculated as $xy$ or $x+y$ instead of $10y+x$. While writing the value of a number keep in mind that every digit is multiplied by its place value.
Complete step-by-step answer:
We have been given that a number between $10$ and $100$ is 8 times the sum of digits. Also, if 45 is subtracted from it the digits will be reversed.
We have to find the number.
Let us assume that the digits of one’s place is $x$ and ten’s place is $y$.
Now, the number will be $10y+x$ .
Now, given in the question that number is 8 times the sum of digits,
So, mathematically we have
$10y+x=8\left( x+y \right)$
Simplifying further we get
$10y+x=8x+8y$
$\begin{align}
& \Rightarrow 10y-8y=8x-x \\
& \Rightarrow 2y=7x \\
& y=\dfrac{7x}{2}..........(i) \\
\end{align}$
Now, it is given in the question that if 45 be subtracted from the number digits will be reversed.
So, we have $10y+x-45=10x+y$
Simplifying further, we get
$\begin{align}
& \Rightarrow 10y-y-45=10x-x \\
& \Rightarrow 9y=9x+45 \\
\end{align}$
Dividing the whole equation by $9$, we get
$\Rightarrow y=x+5$
Now, substituting the value of $y$ from equation (i) and simplifying further, we have
$\begin{align}
& \Rightarrow \dfrac{7x}{2}=x+5 \\
& \Rightarrow 7x=2\left( x+5 \right) \\
& \Rightarrow 7x=2x+10 \\
& \Rightarrow 7x-2x=10 \\
& \Rightarrow 5x=10 \\
& \Rightarrow x=\dfrac{10}{5} \\
& x=2 \\
\end{align}$
Now, the value of $y$ will be
$\begin{align}
& \Rightarrow y=x+5 \\
& y=2+5 \\
& y=7 \\
\end{align}$
So, we have digits of one’s place is $2$ and ten’s place is $7$.
Now, the number will be
$\begin{align}
& \Rightarrow 10\times 7+2 \\
& \Rightarrow 70+2 \\
& =72 \\
\end{align}$
So, the number is $72$.
Note: The possibility of mistake while considering the number is that if it is calculated as $xy$ or $x+y$ instead of $10y+x$. While writing the value of a number keep in mind that every digit is multiplied by its place value.
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