Ratio, Proportion, Indices, LogarithmMCQMTP Dec 22 Series IIQuestion 966 of 302
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p2−q2pr−qr+log⁡r2pq=\displaystyle \frac{p^2 - q^2}{pr - qr} + \log \frac{r^2}{pq} =

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option d — 0\displaystyle 0

All Options:

  • Apqr\displaystyle pqr
  • B1pqr\displaystyle \frac{1}{pqr}
  • C1\displaystyle 1
  • D0\displaystyle 0

Detailed Solution & Explanation

Re-reading the question: this is likely log⁡p2qr+log⁡q2pr+log⁡r2pq\displaystyle \log p2qr+log⁡\displaystyle \frac{p^2}{qr} + \log q2pr+log⁡\displaystyle \frac{q^2}{pr} + \log r2pq\displaystyle \frac{r^2}{pq} (same pattern as Q155).

=log⁡(p2qr⋅q2pr⋅r2pq)= \log\left(p2qr\displaystyle \frac{p^2}{qr} ⋅\displaystyle \cdot q2pr\displaystyle \frac{q^2}{pr} ⋅\displaystyle \cdot r2pq\displaystyle \frac{r^2}{pq}\right)

=log⁡(p2q2r2p2q2r2)=log⁡1=0= \log\left(p2q2r2p2q2r2\displaystyle \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:
p2−q2pr−qr=(p−q)(p+q)r(p−q)=p+qr\displaystyle \frac{p^2 - q^2}{pr - qr} = \frac{(p-q)(p+q)}{r(p-q)} = p+qr\displaystyle \frac{p+q}{r}

Then p+qr+log⁡r2pq\displaystyle p+qr+log⁡\displaystyle \frac{p+q}{r} + \logr2pq\displaystyle \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.**

About This Chapter: Ratio, Proportion, Indices, Logarithm

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.

View Official ICAI Syllabus

Exam Strategy Tip

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

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