Question

The digit in ten's place of a two-digit number is three times that in one's place. if the digits are reversed the new number will be 36 less than the original number. Find the number.

• A. $64$
• B. $93$
• C. $62$
• D. $42$

Answer 1

The correct option is C

$62$

Step 1: Consider the given details.

Let the digit in ten's place be $a$ and the digit in one's place be $b$.

$\therefore \text{Number}=10a+b$

Also, the reverse number$=a+10b$.

Here, the digit in the ten's place of a two-digit number is three times that in the one's place.

$\Rightarrow a=3b...\left(1\right)$

Also, if the digits are reversed, the new number will be 36 less than the original number.

$\begin{array}{rcl}& \Rightarrow & \left(10a+b\right)-36=a+10b\\ & \Rightarrow & 10a-a+b-10b=36\\ & \Rightarrow & 9a-9b=36\\ & \Rightarrow & a-b=4....\left(2\right)\end{array}$

Step 2: Apply the substitution method.

Now, substitute the value of equation (1) in equation (2):

$\begin{array}{rcl}& \Rightarrow & 3b-b=4\\ & \Rightarrow & 2b=4\\ & \Rightarrow & b=2\end{array}$

And, $a=3\left(2\right)=6$.

Thus, the required number is:

$\begin{array}{rcl}\text{Number}& =& 10a+b\\ & =& 10\left(6\right)+2\\ & =& 62\end{array}$

Hence option (c) is correct.