Question

A number lies between 300 and 400. If the number is added to the number formed by reversing its digits, the sum is 888 and if the unit’s digit and the ten’s digit change places, then the new number exceeds the original number by 9. Then the number is:
(a) 354
(b) 341
(c) 348
(d) 345

Answer 1

Hint: Assume that the required number is ‘3xy’ with ‘3’ as its hundred’s place digit, ‘x’ as its ten’s place digit and ‘y’ as its unit’s place digit. Therefore, the number formed by reversing its digits is ‘yx3’. Also, the number formed by reversing the unit’s place and ten’s place digit is ‘3yx’. Write the number ‘3xy’ in the form: $300+10x+y$, ‘yx3’ in the form: $100y+10x+3$ and ‘3yx’ in the form: $300+10y+x$. 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 required number is ‘3xy’, as the number lies between 300 and 400, so its hundred’s place digit should be 3. Therefore, the number formed by reversing its digits is ‘yx3’. Also, the number formed by reversing the unit’s place and ten’s place digit is ‘3yx’.
Now, a number of the form ‘3yz’ is written as: $300+10x+y$, because 3, x and y are simply the face values of a number but the place value of ‘3’ is hundred, ‘x’ is ten and ‘y’ is unit. Similarly, ‘yx3’ is written as: $100y+10x+3$ and ‘3yx’ is written as: $300+10y+x$.
It is given that, the sum of the number obtained by adding the assumed number and the reverse of the number is 888. Therefore, mathematically,
$\begin{align}
  & 300+10x+y+100y+10x+3=888 \\
 & \Rightarrow 303+20x+101y=888 \\
 & \Rightarrow 20x+101y=888-303 \\
 & \Rightarrow 20x+101y=585..................(i) \\
\end{align}$
Also, we have been provided with the information that, when the unit’s digit and the ten’s digit change places, then the new number exceeds the original number by 9. Therefore, mathematically,
$\begin{align}
  & 300+10y+x=300+10x+y+9 \\
 & \Rightarrow 10y+x=10x+y+9 \\
 & \Rightarrow 10y-y+x-10x=9 \\
 & \Rightarrow 9y-9x=9 \\
 & \Rightarrow y-x=1 \\
 & \Rightarrow y=x+1..................(ii) \\
\end{align}$
Substituting the value of y from equation (ii) in equation (i), we have,
\[\begin{align}
  & 20x+101\left( x+1 \right)=585 \\
 & \Rightarrow 20x+101x+101=585 \\
 & \Rightarrow 121x=585-101 \\
 & \Rightarrow 121x=484 \\
 & \Rightarrow x=\dfrac{484}{121} \\
 & \Rightarrow x=4 \\
\end{align}\]
Now, substituting the value of x in equation (ii), we get,
$\begin{align}
  & y=x+1 \\
 & \Rightarrow y=4+1 \\
 & \Rightarrow y=5 \\
\end{align}$
Therefore, the required number is ‘3xy’ = 345. Hence, option (d) 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 ‘3xy’ and, 3, x and y are simply the face values of hundred’s place digit, ten’s place digit and one’s place digit respectively. Now, we have assumed the number as ‘3xy’, because it is lying between 300 and 400, as provided in the question. So, the hundred’s place must be 3.