Answer
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Hint: First assume the two digits as $x$ and $y$. Then do as the question says and make the reverse of the digits accordingly. Then write the value in terms of $x$ or $y$ and substitute. You will get the answer.
Complete step-by-step answer:
It is given that there are two digits. So let the two digits be $x$ and $y$.
So the original number becomes $10x+y$.
It is given in question that the number obtained by reversing the order of the digits, so the reversing order means the tens digit becomes ones digit and the ones digit becomes tens digit, i.e., $10x+y$ will become $10y+x$.
It is mentioned in question that the sum of two digit numbers and the number obtained by reversing the order of its digit is 99.
So, it becomes,
$10x+y+10y+x=99$
Simplifying in simple manner, we get,
$10x+x+10y+y=99$
$11x+11y=99$
Now dividing the above equation by 11, we get the equation as,
$\dfrac{11x+11y}{11}=\dfrac{99}{11}$
Simplifying in simple manner we get,
$\begin{align}
& x+y=\dfrac{11\times 9}{11} \\
& x+y=9......(i) \\
\end{align}$
Now it is mentioned in question that the digits differ by 3 i.e., the difference between the
digit is 3, so the equation becomes,
$x-y=3$
So, simplifying the above equation, i.e., writing the equation in terms of $y$we get,
$y=x-3$………………… (ii)
So substituting equation (ii) in equation (1), we get,
$\begin{align}
& x+x-3=9 \\
& 2x-3=9 \\
& \Rightarrow 2x=12 \\
\end{align}$
Dividing the above equation by 2, we get,
$\begin{align}
& \dfrac{2x}{2}=\dfrac{12}{2} \\
& \Rightarrow x=6 \\
\end{align}$
Now substituting value of $x$ in equation (ii), we get,
$y=6-3=3$
So, the two digits i.e. $x$ and $y$ are 6 and 3 respectively.
The original number is 63.
Note: Read the question thoroughly. So assume the two digits. Don’t jumble while solving or reversing the digits. So it is given in question that the number obtained in reversing the order, so the reversing order means reverse of $10x+y=10(6)+3=63$ which is $10y+x=10(3)+6=36$ don’t confuse yourself here. Use the proper substitution. As we have taken substitution in terms of $y$, you can also take the substitution in terms of $x$.
Complete step-by-step answer:
It is given that there are two digits. So let the two digits be $x$ and $y$.
So the original number becomes $10x+y$.
It is given in question that the number obtained by reversing the order of the digits, so the reversing order means the tens digit becomes ones digit and the ones digit becomes tens digit, i.e., $10x+y$ will become $10y+x$.
It is mentioned in question that the sum of two digit numbers and the number obtained by reversing the order of its digit is 99.
So, it becomes,
$10x+y+10y+x=99$
Simplifying in simple manner, we get,
$10x+x+10y+y=99$
$11x+11y=99$
Now dividing the above equation by 11, we get the equation as,
$\dfrac{11x+11y}{11}=\dfrac{99}{11}$
Simplifying in simple manner we get,
$\begin{align}
& x+y=\dfrac{11\times 9}{11} \\
& x+y=9......(i) \\
\end{align}$
Now it is mentioned in question that the digits differ by 3 i.e., the difference between the
digit is 3, so the equation becomes,
$x-y=3$
So, simplifying the above equation, i.e., writing the equation in terms of $y$we get,
$y=x-3$………………… (ii)
So substituting equation (ii) in equation (1), we get,
$\begin{align}
& x+x-3=9 \\
& 2x-3=9 \\
& \Rightarrow 2x=12 \\
\end{align}$
Dividing the above equation by 2, we get,
$\begin{align}
& \dfrac{2x}{2}=\dfrac{12}{2} \\
& \Rightarrow x=6 \\
\end{align}$
Now substituting value of $x$ in equation (ii), we get,
$y=6-3=3$
So, the two digits i.e. $x$ and $y$ are 6 and 3 respectively.
The original number is 63.
Note: Read the question thoroughly. So assume the two digits. Don’t jumble while solving or reversing the digits. So it is given in question that the number obtained in reversing the order, so the reversing order means reverse of $10x+y=10(6)+3=63$ which is $10y+x=10(3)+6=36$ don’t confuse yourself here. Use the proper substitution. As we have taken substitution in terms of $y$, you can also take the substitution in terms of $x$.
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