UPSC CSE Prelims
Ratio & Proportion Previous Year Questions (PYQs)
Practice solved questions for Ratio & Proportion with detailed step-by-step solutions, key insights, and trend analysis for UPSC CSE PRELIMS.
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Suppose $x$, $y$ and $z$ are variables taking positive real numbers as their possible values. It is given that $y$ is directly proportional to $x^2$ and $x$ is inversely proportional to $z$. For $z = \frac{7}{25}$, the values of $x$ and $y$ are $5$ and $50$, respectively. If $y = 98$, what is $z$ equal to?
Detailed Explanation:
Given: $y \propto x^2 \Rightarrow y = kx^2$ and $x \propto \frac{1}{z} \Rightarrow xz = c$ (constant).
Using $z = \frac{7}{25}, x = 5, y = 50$: From $y = kx^2$, we get $50 = k(25) \Rightarrow k = 2$. From $xz = c$, we get $c = 5 \times \frac{7}{25} = \frac{7}{5}$. So $y = 2x^2$ and $xz = \frac{7}{5}$.
When $y = 98$: $2x^2 = 98 \Rightarrow x^2 = 49 \Rightarrow x = 7$. Thus $z = \frac{c}{x} = \frac{7/5}{7} = \frac{1}{5}$.
Therefore, the value of z is $\frac{1}{5}$, making Option 2 the correct answer.
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