Ratio, Proportion, Indices, LogarithmMCQPYQ Nov 18Question 874 of 305
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Value of 2m+1×32m−n+3×5n+m+4×62n+m62m+n×10n+1×15m+3\displaystyle \frac{2^{m+1} \times 3^{2m-n+3} \times 5^{n+m+4} \times 6^{2n+m}}{6^{2m+n} \times 10^{n+1} \times 15^{m+3}}

Options

A32m−2n\displaystyle 3^{2m-2n}
B6\displaystyle 6
C1\displaystyle 1
DNone of these
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Correct Answer

✅ Option c — 1\displaystyle 1

All Options:

  • A32m−2n\displaystyle 3^{2m-2n}
  • B6\displaystyle 6
  • C1\displaystyle 1
  • DNone of these

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Detailed Solution & Explanation

Let us simplify the given fraction by expressing all composite bases (6\displaystyle 6, 10\displaystyle 10, and 15\displaystyle 15) in terms of their prime factors (2\displaystyle 2, 3\displaystyle 3, and 5\displaystyle 5):
62n+m=(2×3)2n+m=22n+m×32n+m6^{2n+m} = (2 \times 3)^{2n+m} = 2^{2n+m} \times 3^{2n+m}
62m+n=(2×3)2m+n=22m+n×32m+n6^{2m+n} = (2 \times 3)^{2m+n} = 2^{2m+n} \times 3^{2m+n}
10n+1=(2×5)n+1=2n+1×5n+110^{n+1} = (2 \times 5)^{n+1} = 2^{n+1} \times 5^{n+1}
15m+3=(3×5)m+3=3m+3×5m+315^{m+3} = (3 \times 5)^{m+3} = 3^{m+3} \times 5^{m+3}

Now, substitute these prime factor representations into the numerator and denominator of the given expression:

**Numerator (N\displaystyle N):**
N=2m+1×32m−n+3×5n+m+4×22n+m×32n+mN = 2^{m+1} \times 3^{2m-n+3} \times 5^{n+m+4} \times 2^{2n+m} \times 3^{2n+m}
Group the powers of 2\displaystyle 2, 3\displaystyle 3, and 5\displaystyle 5 using the laws of exponents (xa⋅xb=xa+b\displaystyle x^a \cdot x^b = x^{a+b}):
- Exponent of 2\displaystyle 2: (m+1)+(2n+m)=2m+2n+1\displaystyle (m+1) + (2n+m) = 2m+2n+1
- Exponent of 3\displaystyle 3: (2m−n+3)+(2n+m)=3m+n+3\displaystyle (2m-n+3) + (2n+m) = 3m+n+3
- Exponent of 5\displaystyle 5: n+m+4\displaystyle n+m+4
So:
N=22m+2n+1×33m+n+3×5n+m+4N = 2^{2m+2n+1} \times 3^{3m+n+3} \times 5^{n+m+4}

**Denominator (D\displaystyle D):**
D=(22m+n×32m+n)×(2n+1×5n+1)×(3m+3×5m+3)D = (2^{2m+n} \times 3^{2m+n}) \times (2^{n+1} \times 5^{n+1}) \times (3^{m+3} \times 5^{m+3})
Group the powers of 2\displaystyle 2, 3\displaystyle 3, and 5\displaystyle 5 in the denominator:
- Exponent of 2\displaystyle 2: (2m+n)+(n+1)=2m+2n+1\displaystyle (2m+n) + (n+1) = 2m+2n+1
- Exponent of 3\displaystyle 3: (2m+n)+(m+3)=3m+n+3\displaystyle (2m+n) + (m+3) = 3m+n+3
- Exponent of 5\displaystyle 5: (n+1)+(m+3)=n+m+4\displaystyle (n+1) + (m+3) = n+m+4
So:
D=22m+2n+1×33m+n+3×5n+m+4D = 2^{2m+2n+1} \times 3^{3m+n+3} \times 5^{n+m+4}

Comparing the simplified expressions of the numerator and the denominator, we see that they are exactly identical:
N=DN = D
Thus, the value of the fraction is:
ND=1\frac{N}{D} = 1
This matches **Option C**.
Note: The textbook answer key incorrectly designates **Option A** as correct. However, our rigorous prime factorization proves that the numerator and denominator are identical, so the value must be exactly 1\displaystyle 1 (Option C).
Hence, **Option C** 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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