If 9n × 35 × (27)5/3 × (81)4 = 27 , then the value of n is

Option b: 0. Solution: Given the equation: 9n × 35 × (27)5/3 × (81)4 = 27 Let us express all terms with base 3 using the laws of exponents: - 9n = (32)n = 32n - 275 = (33)5 = 315 - 814 = (34)4 = 316 - 27 = 33 Substitute these expressions back into the original equation: 32n × 35 × 3153 × 316 = 33 Simplify the numerator using xa × xb = xa+b : Numerator = 32n + 5 + 15 = 32n + 20 Simplify the denominator: Denominator = 31 × 316 = 317 Now divide the numerator by the denominator: 32n+20317 = 32n + 20 - 17 = 32n+3 Equate this to the right-hand side ( 33 ): 32n+3 = 33 Since the bases are identical, we equate the exponents: 2n + 3 = 3 2n = 0 n = 0 This matches Option B. Hence, Option B is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQPYQ Dec 23Question 891 of 305

All Questions

If $\displaystyle \frac{9^n \times 3^5 \times (27)^5}{3 \times (81)^4} = 27$ , then the value of $\displaystyle n$ is

Options

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Correct Answer

✅ Option b — 0

All Options:

Detailed Solution & Explanation

Given the equation:

\frac{9^n \times 3^5 \times (27)^5}{3 \times (81)^4} = 27

Let us express all terms with base

\displaystyle 3

using the laws of exponents:
-

\displaystyle 9^n = (3^2)^n = 3^{2n}

\displaystyle 27^5 = (3^3)^5 = 3^{15}

\displaystyle 81^4 = (3^4)^4 = 3^{16}

\displaystyle 27 = 3^3

Substitute these expressions back into the original equation:

\frac{3^{2n} \times 3^5 \times 3^{15}}{3 \times 3^{16}} = 3^3

Simplify the numerator using

\displaystyle x^a \cdot x^b = x^{a+b}

\text{Numerator} = 3^{2n + 5 + 15} = 3^{2n + 20}

Simplify the denominator:

\text{Denominator} = 3^1 \times 3^{16} = 3^{17}

Now divide the numerator by the denominator:

\frac{3^{2n+20}}{3^{17}} = 3^{2n + 20 - 17} = 3^{2n+3}

Equate this to the right-hand side (

\displaystyle 3^3

3^{2n+3} = 3^3

Since the bases are identical, we equate the exponents:

2n + 3 = 3

2n = 0 \implies n = 0

This matches **Option B**.
Hence, **Option B** 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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If 9n×35×(27)53×(81)4=27\displaystyle \frac{9^n \times 3^5 \times (27)^5}{3 \times (81)^4} = 273×(81)49n×35×(27)5​=27, then the value of n\displaystyle nn is