ABC26GN1757 · Problems on Numbers
Subject: General Aptitude · Chapter: Problems on Numbers · Exam: 2006 · Marks: · Difficulty:
If the difference between the reciprocal of a positive proper fraction and the fraction itself be $\frac{9}{20}$, then the fraction is
(a)$\frac{3}{5}$
(b)$\frac{4}{5}$
(c)$\frac{5}{4}$
(d)$\frac{3}{10}$
Answer
Explanation
Let the fraction be $\frac{a}{1}$. Then, $\frac{1}{a}-a=\frac{9}{20} \Leftrightarrow \frac{1-a^{2}}{a}=\frac{9}{20}$ $$\begin{aligned} & \Leftrightarrow 20-20 a^{2}=9 a \\ & \Leftrightarrow 20 a^{2}+9 a-20=0 \\ & \Leftrightarrow 20 a^{2}+25 a-16 a-20=0 \\ & \Leftrightarrow 5 a(4 a+5)-4(4 a+5)=0 \\ & \Leftrightarrow(4 a+5)(5 a-4)=0 \Leftrightarrow a=\frac{4}{5} . \quad\left[\because a \neq-\frac{5}{4}\right] . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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