Value of [9n+1/4 × (3 × 3n)3 × sqrt(3-n)]1/n

Option b: 27. Solution: We want to simplify the expression: [ 9n+1/4 × (3 × 3n)3 × sqrt(3-n) ]1/n Let us express each term inside the bracket in base 3 : 1. 9n+1/4 = (32)n+1/4 = 32(n+1/4) = 32n + 1/2 2. sqrt(3 × 3n) = sqrt(3n+1) = 3n+1/2 3. 3 × sqrt(3-n) = 31 × 3-n/2 = 31 - n/2 Now substitute these simplified terms back into the fraction: (3 × 3n)3 × sqrt(3-n) = 3+1/231 - n/2 = 3n+1/2 - (1 - n/2) Simplify the exponent: n+1/2 - 1 + n/2 = n/2 + 1/2 - 1 + n/2 = n - 1/2 So the fraction term is 3n - 1/2 . Multiply this with the first term: 32n + 1/2 × 3n - 1/2 = 3(2n + 1/2) + (n - 1/2) = 33n Now we raise this entire expression to the power of 1/n : (33n)1/n = 33n × 1/n = 33 = 27 This matches Option B. Note: The textbook answer key incorrectly specifies Option C ( 81 ) as correct. However, our rigorous mathematical simplification clearly proves the value is 27 (Option B). Hence, Option B is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQPYQ Nov 19Question 877 of 305

All Questions

Value of $\displaystyle \left[9^{n+\frac{1}{4}} \cdot \frac{\sqrt{3 \cdot 3^n}}{3 \cdot \sqrt{3^{-n}}}\right]^{\frac{1}{n}}$

Options

\displaystyle 9

\displaystyle 27

\displaystyle 81

\displaystyle 3

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

Correct Answer

✅ Option b — $\displaystyle 27$

All Options:

A $\displaystyle 9$
B $\displaystyle 27$
C $\displaystyle 81$
D $\displaystyle 3$

Detailed Solution & Explanation

We want to simplify the expression:

\left[ 9^{n+\frac{1}{4}} \cdot \frac{\sqrt{3 \cdot 3^n}}{3 \cdot \sqrt{3^{-n}}} \right]^{\frac{1}{n}}

Let us express each term inside the bracket in base

\displaystyle 3

:
1.

\displaystyle 9^{n+\frac{1}{4}} = \left(3^2\right)^{n+\frac{1}{4}} = 3^{2\left(n+\frac{1}{4}\right)} = 3^{2n + \frac{1}{2}}

\displaystyle \sqrt{3 \cdot 3^n} = \sqrt{3^{n+1}} = 3^{\frac{n+1}{2}}

\displaystyle 3 \cdot \sqrt{3^{-n}} = 3^1 \cdot 3^{-\frac{n}{2}} = 3^{1 - \frac{n}{2}}

Now substitute these simplified terms back into the fraction:

\frac{\sqrt{3 \cdot 3^n}}{3 \cdot \sqrt{3^{-n}}} = \frac{3^{\frac{n+1}{2}}}{3^{1 - \frac{n}{2}}} = 3^{\frac{n+1}{2} - \left(1 - \frac{n}{2}\right)}

Simplify the exponent:

\frac{n+1}{2} - 1 + \frac{n}{2} = \frac{n}{2} + \frac{1}{2} - 1 + \frac{n}{2} = n - \frac{1}{2}

So the fraction term is

\displaystyle 3^{n - \frac{1}{2}}

. Multiply this with the first term:

3^{2n + \frac{1}{2}} \cdot 3^{n - \frac{1}{2}} = 3^{\left(2n + \frac{1}{2}\right) + \left(n - \frac{1}{2}\right)} = 3^{3n}

Now we raise this entire expression to the power of

\displaystyle \frac{1}{n}

\left(3^{3n}\right)^{\frac{1}{n}} = 3^{3n \cdot \frac{1}{n}} = 3^3 = 27

This matches **Option B**.
Note: The textbook answer key incorrectly specifies **Option C** (

\displaystyle 81

) as correct. However, our rigorous mathematical simplification clearly proves the value is

\displaystyle 27

(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.

Ready to Master Ratio, Proportion, Indices, Logarithm?

Practice all 305 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

Value of [9n+14⋅3⋅3n3⋅3−n]1n\displaystyle \left[9^{n+\frac{1}{4}} \cdot \frac{\sqrt{3 \cdot 3^n}}{3 \cdot \sqrt{3^{-n}}}\right]^{\frac{1}{n}}[9n+41​⋅3⋅3−n​3⋅3n​​]n1​