
In a two digit number, the ten’s digit is three times the unit’s digit. When the number is decreased by 54, the digits are reversed. Find the number.
Answer
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Hint: Assume that the required two digit number is ‘$xy$’ with $x$ as its ten’s place digit and $y$ as its unit’s place digit. Use the given information to form two linear equations in two variables. Solve the obtained equations to get the number.
Complete step-by-step answer:
Let us assume that the two digit number is ‘$xy$’. It is given that the ten’s digit is three times the unit’s digit. Therefore, Mathematically,
$x=3y.................................(i)$
This is assumed as equation (i).
Now, we have been provided the information that, when 54 is subtracted from ‘$xy$’ then digits are reversed, that is, the number becomes ‘$yx$’. Therefore, mathematically,
$\begin{align}
& xy-54=yx \\
& xy-yx=54....................(ii) \\
\end{align}$
We know that any number has its place value and face value. Place value states the position of a digit in a given number, whereas face value describes the value of the digit. For example: we take a number 74. In this number tens place digit is 7, so 10 is the place value and digit 7 is the face value. Similarly, at ones place 4 is present, so place value is 1 and face value is 4. So, 74 can be written as $74=7\times 10+4\times 1$.
Similarly, writing equation (ii) in this form, we get,
$\begin{align}
& (10x+y)-(10y+x)=54 \\
& 10x+y-10y-x=54 \\
& 9x-9y=54 \\
& \\
\end{align}$
Cancelling 9 from both sides, we get,
$x-y=6....................(iii)$
Now, substituting the value of $x$ from equation (i) in equation (iii), we get,
$\begin{align}
& 3y-y=6 \\
& 2y=6 \\
& \therefore y=3 \\
\end{align}$
Substituting the value of $y$ in equation (i), we get,
$\begin{align}
& x=3\times 3 \\
& \therefore x=9 \\
\end{align}$
Hence, the two digit number is $xy=93$.
Note: One should not get confused in place value and face value of any number. The number we have assumed is ‘$xy$’ and, $x\text{ and }y$ are simply the face values of tens place digit and ones place digit respectively.
Complete step-by-step answer:
Let us assume that the two digit number is ‘$xy$’. It is given that the ten’s digit is three times the unit’s digit. Therefore, Mathematically,
$x=3y.................................(i)$
This is assumed as equation (i).
Now, we have been provided the information that, when 54 is subtracted from ‘$xy$’ then digits are reversed, that is, the number becomes ‘$yx$’. Therefore, mathematically,
$\begin{align}
& xy-54=yx \\
& xy-yx=54....................(ii) \\
\end{align}$
We know that any number has its place value and face value. Place value states the position of a digit in a given number, whereas face value describes the value of the digit. For example: we take a number 74. In this number tens place digit is 7, so 10 is the place value and digit 7 is the face value. Similarly, at ones place 4 is present, so place value is 1 and face value is 4. So, 74 can be written as $74=7\times 10+4\times 1$.
Similarly, writing equation (ii) in this form, we get,
$\begin{align}
& (10x+y)-(10y+x)=54 \\
& 10x+y-10y-x=54 \\
& 9x-9y=54 \\
& \\
\end{align}$
Cancelling 9 from both sides, we get,
$x-y=6....................(iii)$
Now, substituting the value of $x$ from equation (i) in equation (iii), we get,
$\begin{align}
& 3y-y=6 \\
& 2y=6 \\
& \therefore y=3 \\
\end{align}$
Substituting the value of $y$ in equation (i), we get,
$\begin{align}
& x=3\times 3 \\
& \therefore x=9 \\
\end{align}$
Hence, the two digit number is $xy=93$.
Note: One should not get confused in place value and face value of any number. The number we have assumed is ‘$xy$’ and, $x\text{ and }y$ are simply the face values of tens place digit and ones place digit respectively.
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