Ratio, Proportion, Indices, LogarithmMCQMTP Dec 22 Series IIQuestion 966 of 305

$\displaystyle \frac{p^2 - q^2}{pr - qr} + \log \frac{r^2}{pq} =$

\displaystyle pqr

\displaystyle \frac{1}{pqr}

\displaystyle 1

\displaystyle 0

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Correct Answer

✅ Option d — $\displaystyle 0$

Re-reading the question: this is likely

\displaystyle \log \frac{p^2}{qr} + \log \frac{q^2}{pr} + \log \frac{r^2}{pq}

(same pattern as Q155).

= \log\left(\frac{p^2}{qr} \cdot \frac{q^2}{pr} \cdot \frac{r^2}{pq}\right)

= \log\left(\frac{p^2 q^2 r^2}{p^2 q^2 r^2}\right) = \log 1 = 0

Alternatively, if the expression is exactly as written with the fraction:

\frac{p^2 - q^2}{pr - qr} = \frac{(p-q)(p+q)}{r(p-q)} = \frac{p+q}{r}

Then

\displaystyle \frac{p+q}{r} + \log\frac{r^2}{pq}

— this doesn't simplify to 0 in general. So the intended reading is the logarithm version.

**The answer is (d) 0.**

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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