Subject: General Aptitude · Chapter: Ratio and Proportion · Exam: 2006 · Marks: · Difficulty:
Suppose $y$ varies as the sum of two quantities of which one varies directly as $x$ and the other inversely as $x$. If $y=6$ when $x=4$ and $y=3 \frac{1}{3}$ when $x=3$, then the relation between $x$ and $y$ is
(a)$y=2 x-\frac{8}{x}$
(b)$y=x+\frac{4}{x}$
(c)$y=2 x+\frac{4}{x}$
(d)$y=2 x+\frac{8}{x}$
Answer
Answer (as printed): A
Explanation
$y \alpha\left(x+\frac{1}{x}\right) \Rightarrow y=k x+\frac{m}{x}$, where $k$ and $m$ are constants. Then, $4 k+\frac{m}{4}=6$ ...(i) and $$3 k+\frac{m}{3}=\frac{10}{3}$$ Multiplying (i) by 3 and (ii) by 4, we get : $$12 k+\frac{3 m}{4}=18$$ and $$12 k+\frac{4 m}{3}=\frac{40}{3}$$ Subtracting (iv) from (iii), we get : $$\frac{3 m}{4}-\frac{4 m}{3}=18-\frac{40}{3} \Rightarrow-\frac{7 m}{12}=\frac{14}{3} \Rightarrow m=-8$$ Putting $m=-8$ in (i), we get : $4 k+\frac{(-8)}{4}=6$ $$\Rightarrow 4 k=8 \Rightarrow k=2 \quad \therefore y=2 x-\frac{8}{x} .$$
Explanation as extracted from the printed page; notation may be imperfect.