Subject: General Aptitude · Chapter: Problems on Numbers · Exam: 2008 · Marks: · Difficulty:
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?
(a)15.6
(b)24
(c)37.2
(d)39
(e)None of these
Answer
Answer (as printed): A
Explanation
Let ten's digit $=x$. Then, unit's digit $=x^{2}$. Then, number $=10 x+x^{2}$. Clearly, since $x^{2}>x$, so the number formed by interchanging the digits is greater than the original number. $$\therefore \quad \begin{aligned} & \left(10 x^{2}+x\right)-\left(10 x+x^{2}\right) \\ & =54 \Leftrightarrow 9 x^{2}-9 x=54 \Leftrightarrow x^{2}-x \\ & =6 \Leftrightarrow x^{2}-x-6=0 \end{aligned}$$ $\Leftrightarrow \quad x^{2}-3 x+2 x-6=0$ $\Leftrightarrow \quad x(x-3)+2(x-3)=0$ $\Leftrightarrow(x-3)(x+2)=0$ $\Leftrightarrow \quad x=3$. So, ten's digit = 3, unit's digit = $3^{2}=9$. $\therefore \quad$ Original number = 39. Required result $=40 \%$ of $39=15.6$.
Explanation as extracted from the printed page; notation may be imperfect.