Ratio, Proportion, Indices, LogarithmMCQPYQ Nov 19Question 877 of 302
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Value of [9n+14⋅3⋅3n3⋅3−n]1n\displaystyle \left[9^{n+14\displaystyle \frac{1}{4}} ⋅\displaystyle \cdot \frac{3⋅3n\displaystyle \sqrt{3 \cdot 3^n}}{3 ⋅\displaystyle \cdot 3−n\displaystyle \sqrt{3^{-n}}}\right]^{1n\displaystyle \frac{1}{n}}

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

✅ Option b — $27$

All Options:

  • A9\displaystyle 9
  • B$27$
  • C$81$
  • D3\displaystyle 3

Detailed Solution & Explanation

We want to simplify the expression:
[9n+14⋅3⋅3n3⋅3−n]1n\left[ 9^{n+14\displaystyle \frac{1}{4}} ⋅\displaystyle \cdot \frac{3⋅3n\displaystyle \sqrt{3 \cdot 3^n}}{3 ⋅\displaystyle \cdot 3−n\displaystyle \sqrt{3^{-n}}} \right]^{1n\displaystyle \frac{1}{n}}
Let us express each term inside the bracket in base 3\displaystyle 3:
1. 9n+14=(32)n+14=32(n+14)=32n+12\displaystyle 9^{n+14\displaystyle \frac{1}{4}} = \left(3^2\right)^{n+14\displaystyle \frac{1}{4}} = 3^{2\left(n+14\displaystyle \frac{1}{4}\right)} = 3^{2n + 12\displaystyle \frac{1}{2}}
2. 3⋅3n=3n+1=3n+12\displaystyle \displaystyle \sqrt{3 \cdot 3^n} = \sqrt{3^{n+1}} = 3^{n+12\displaystyle \frac{n+1}{2}}
3. 3⋅3−n=31⋅3−n2=31−n2\displaystyle 3 ⋅\displaystyle \cdot 3−n=31\displaystyle \sqrt{3^{-n}} = 3^1 ⋅\displaystyle \cdot 3^{-n2\displaystyle \frac{n}{2}} = 3^{1 - n2\displaystyle \frac{n}{2}}
Now substitute these simplified terms back into the fraction:
3⋅3n3⋅3−n=3n+1231−n2=3n+12−(1−n2)\frac{3⋅3n\displaystyle \sqrt{3 \cdot 3^n}}{3 ⋅\displaystyle \cdot 3−n\displaystyle \sqrt{3^{-n}}} = \frac{3^{n+12\displaystyle \frac{n+1}{2}}}{3^{1 - n2\displaystyle \frac{n}{2}}} = 3^{\displaystyle \frac{n+1}{2} - \left(1 - n2\displaystyle \frac{n}{2}\right)}
Simplify the exponent:
n+12−1+n2=n2+12−1+n2=n−12n+12−1+n2=n2+12−1+n2=n−12\displaystyle \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 3n−12\displaystyle 3^{n - 12\displaystyle \frac{1}{2}}. Multiply this with the first term:
32n+12⋅3n−12=3(2n+12)+(n−12)=33n3^{2n + 12\displaystyle \frac{1}{2}} ⋅\displaystyle \cdot 3^{n - 12\displaystyle \frac{1}{2}} = 3^{\left(2n + 12\displaystyle \frac{1}{2}\right) + \left(n - 12\displaystyle \frac{1}{2}\right)} = 3^{3n}
Now we raise this entire expression to the power of 1n\displaystyle 1n\displaystyle \frac{1}{n}:
(33n)1n=33n⋅1n=33=27\left(3^{3n}\right)^{1n\displaystyle \frac{1}{n}} = 3^{3n ⋅\displaystyle \cdot 1n\displaystyle \frac{1}{n}} = 3^3 = 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.

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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