The value of (243)0.13 × (243)0.07(7)0.25 × (49)0.075 × (343)0.2 is:

Option a: 3/7. Solution: We are given the expression: E = (243)0.13 × (243)0.07(7)0.25 × (49)0.075 × (343)0.2 Let us simplify the numerator and the denominator separately: 1) Numerator: Num = (243)0.13 × (243)0.07 = (243)0.13 + 0.07 = (243)0.2 Since 243 = 35 , we can write: Num = ( 35 )0.2 = 35 × 0.2 = 31 = 3 2) Denominator: Den = (7)0.25 × (49)0.075 × (343)0.2 Expressing all bases with base 7 (since 49 = 72 and 343 = 73 ): Den = 70.25 × ( 72 )0.075 × ( 73 )0.2 Den = 70.25 × 72 × 0.075 × 73 × 0.2 Den = 70.25 × 70.15 × 70.6 Adding the exponents: Den = 70.25 + 0.15 + 0.6 = 71.0 = 7 3) Combine the numerator and denominator: E = 3/7 Hence, Option A is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQMTP Sep 24 Series IIQuestion 927 of 305

All Questions

The value of $\displaystyle \frac{(243)^{0.13} \times (243)^{0.07}}{(7)^{0.25} \times (49)^{0.075} \times (343)^{0.2}}$ is:

Options

\displaystyle \frac{3}{7}

\displaystyle \frac{2}{7}

\displaystyle \frac{1}{7}

\displaystyle \frac{4}{7}

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Correct Answer

✅ Option a — $\displaystyle \frac{3}{7}$

All Options:

A $\displaystyle \frac{3}{7}$
B $\displaystyle \frac{2}{7}$
C $\displaystyle \frac{1}{7}$
D $\displaystyle \frac{4}{7}$

Detailed Solution & Explanation

We are given the expression:

E = \frac{(243)^{0.13} \times (243)^{0.07}}{(7)^{0.25} \times (49)^{0.075} \times (343)^{0.2}}

Let us simplify the numerator and the denominator separately:

1) Numerator:

\text{Num} = (243)^{0.13} \times (243)^{0.07} = (243)^{0.13 + 0.07} = (243)^{0.2}

Since

\displaystyle 243 = 3^5

, we can write:

\text{Num} = \left( 3^5 \right)^{0.2} = 3^{5 \times 0.2} = 3^1 = 3

2) Denominator:

\text{Den} = (7)^{0.25} \times (49)^{0.075} \times (343)^{0.2}

Expressing all bases with base

\displaystyle 7

(since

\displaystyle 49 = 7^2

and

\displaystyle 343 = 7^3

\text{Den} = 7^{0.25} \times \left( 7^2 \right)^{0.075} \times \left( 7^3 \right)^{0.2}

\text{Den} = 7^{0.25} \times 7^{2 \times 0.075} \times 7^{3 \times 0.2}

\text{Den} = 7^{0.25} \times 7^{0.15} \times 7^{0.6}

Adding the exponents:

\text{Den} = 7^{0.25 + 0.15 + 0.6} = 7^{1.0} = 7

3) Combine the numerator and denominator:

E = \frac{3}{7}

Hence, **Option A** 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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The value of (243)0.13×(243)0.07(7)0.25×(49)0.075×(343)0.2\displaystyle \frac{(243)^{0.13} \times (243)^{0.07}}{(7)^{0.25} \times (49)^{0.075} \times (343)^{0.2}}(7)0.25×(49)0.075×(343)0.2(243)0.13×(243)0.07​ is: