Ratio, Proportion, Indices, LogarithmMCQPYQ June 19Question 875 of 305
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If 2x2=3y2=12z2\displaystyle 2^{x^2} = 3^{y^2} = 12^{z^2} then

Options

A1x2+1y2=1z2\displaystyle \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}
B1x2+2y2=1z2\displaystyle \frac{1}{x^2} + \frac{2}{y^2} = \frac{1}{z^2}
C2x2+1y2=1z2\displaystyle \frac{2}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}
DNone of these
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Correct Answer

✅ Option c — 2x2+1y2=1z2\displaystyle \frac{2}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}

All Options:

  • A1x2+1y2=1z2\displaystyle \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}
  • B1x2+2y2=1z2\displaystyle \frac{1}{x^2} + \frac{2}{y^2} = \frac{1}{z^2}
  • C2x2+1y2=1z2\displaystyle \frac{2}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}
  • DNone of these

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Detailed Solution & Explanation

Let us set the given equal expressions to a constant k\displaystyle k (where k>1\displaystyle k > 1):
2x2=3y2=12z2=k2^{x^2} = 3^{y^2} = 12^{z^2} = k
This allows us to express the base numbers 2\displaystyle 2, 3\displaystyle 3, and 12\displaystyle 12 in terms of k\displaystyle k:
2=k1x22 = k^{\frac{1}{x^2}}
3=k1y23 = k^{\frac{1}{y^2}}
12=k1z212 = k^{\frac{1}{z^2}}
We know the mathematical relation between the numbers 12\displaystyle 12, 2\displaystyle 2, and 3\displaystyle 3 is:
12=4×3=22×312 = 4 \times 3 = 2^2 \times 3
Substitute the expressions of these numbers in terms of k\displaystyle k into this equation:
k1z2=(k1x2)2×k1y2k^{\frac{1}{z^2}} = \left(k^{\frac{1}{x^2}}\right)^2 \times k^{\frac{1}{y^2}}
Using the laws of exponents (xa⋅xb=xa+b\displaystyle x^a \cdot x^b = x^{a+b} and (xa)b=xab\displaystyle (x^a)^b = x^{ab}):
k1z2=k2x2×k1y2k^{\frac{1}{z^2}} = k^{\frac{2}{x^2}} \times k^{\frac{1}{y^2}}
k1z2=k2x2+1y2k^{\frac{1}{z^2}} = k^{\frac{2}{x^2} + \frac{1}{y^2}}
Equating the exponents since the bases are identical:
2x2+1y2=1z2\frac{2}{x^2} + \frac{1}{y^2} = \frac{1}{z^2}
This corresponds to **Option C**.
Note: The textbook designates **Option B** as correct. This discrepancy arises because of a typographical swap of bases in the question text. If the question had been written as 3x2=2y2=12z2\displaystyle 3^{x^2} = 2^{y^2} = 12^{z^2}, then we would have 3=k1/x2\displaystyle 3 = k^{1/x^2} and 2=k1/y2\displaystyle 2 = k^{1/y^2}, which yields 1x2+2y2=1z2\displaystyle \frac{1}{x^2} + \frac{2}{y^2} = \frac{1}{z^2} (Option B). However, for the given question text, the correct mathematical derivation leads to **Option C**.
Hence, **Option C** is the correct answer.

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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