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.

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

The correct answer is Option a: 3/7. 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 302

All Questions

The value of $\displaystyle$ is:

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

For any discrepancies in this question, email contact@cadada.in

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 =

Let us simplify the numerator and the denominator separately:

1) Numerator:

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

Since $243 = 3^5

\displaystyle , we can write:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mtext>Num</mtext><mo>=</mo><msup><mrow><mo fence="true">(</mo><msup><mn>3</mn><mn>5</mn></msup><mo fence="true">)</mo></mrow><mn>0.2</mn></msup><mo>=</mo><msup><mn>3</mn><mrow><mn>5</mn><mo>×</mo><mn>0.2</mn></mrow></msup><mo>=</mo><msup><mn>3</mn><mn>1</mn></msup><mo>=</mo><mn>3</mn></mrow><annotation encoding="application/x-tex">

\displaystyle (since &#36;49 = 7^2

and $343 = 7^3$):

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

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

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

Adding the exponents:

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

3) Combine the numerator and denominator:

E =

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.

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Exam Strategy Tip

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

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Correct Answer

All Options:

Detailed Solution & Explanation

About This Chapter: Ratio, Proportion, Indices, Logarithm

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