Question

Let D be a two digit number. If the half of D exceeds one third of D by the sum of the digits in D. Then the sum of the digits in D is?
(a) 6
(b) 8
(c) 9
(d) 15

Answer 1

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 a linear equation in two variables. Find the possible conditions for the sum to exist and hence find the sum.

Complete step-by-step solution -
Let us assume that the two digit number is ‘$xy$’. It is given that, half of D exceeds one third of D by the sum of the digits in D. Therefore, mathematically,
$\dfrac{xy}{2}=\dfrac{xy}{3}+(x+y)....................(i)$
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 (i) in this form, we get,
$\begin{align}
  & \dfrac{10x+y}{2}=\dfrac{10x+y}{3}+\left( x+y \right) \\
 & \dfrac{10x+y}{2}-\dfrac{10x+y}{3}=\left( x+y \right) \\
\end{align}$
Taking L.C.M we get,
$\begin{align}
  & \dfrac{(30x+3y)-(20x+2y)}{6}=\left( x+y \right) \\
 & \dfrac{10x+y}{6}=\left( x+y \right) \\
\end{align}$
Cross-multiplying the terms we get,
$\begin{align}
  & 10x+y=6x+6y \\
 & 10x-6x=6y-y \\
 & 4x=5y \\
 & x=\dfrac{5y}{4} \\
\end{align}$
Now, the sum of the digits of D is $(x+y)$. Substituting the value of $x$ we get,
Sum of digits $=\dfrac{5y}{4}+y=\dfrac{9y}{4}$. Now, since D is a two digit number, therefore it must be an integer. Hence, the sum of digits of D must be an integer. Therefore, $\dfrac{9y}{4}$ must be an integer.
Now, for $\dfrac{9y}{4}$ to be an integer, $y$ must be 4. Therefore, the sum of digits of D is 9.
Hence, option (c) is the correct answer.

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.