In a two-digit positive number, the digit in the unit's place is equal to the square of the digit in ten's place, and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number ?

Answer: Option A

Let ten's digit = x
Then, unit's digit = x²
Then, number = 10x + x²
Clearly, since x² > x,
So, the number formed by interchanging the digits is greater than the original number.
$$\begin{array}{rl}& \therefore \left(10x^2+x\right)-\left(10x+x^2\right)=54\\ & \iff 9x^2-9x=54\\ & \iff x^2-x=6\\ & \iff x^2-x-6=0\\ & \iff x^2-3x+2x-6=0\\ & \iff \left(x-3\right)\left(x+2\right)=0\\ & \iff x=3\end{array}$$
So. ten's digit = 3, unit's digit = 3² = 9
∴ Original number = 39
Required result :
= 40% of 39
= 15.6