CA Foundation 2025 · Paper 3 · Quantitative AptitudeQ20 · 1 mark↻ Appears in 3 of 6 yearsOfficial key verified
Madhu invests ₹ 15,000 in a scheme and at the time of maturity the amount became ₹ 25,000. If CAGR for this investment is 8.88%, calculate the approximate number of years for which she has invested the amount. [Given that log⁡(1.667)=0.2219\log(1.667) = 0.2219 and log⁡(1.089)=0.037\log(1.089) = 0.037]
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n=log⁡(25000/15000)log⁡1.0888=0.22190.037≈6n = \frac{\log(25000/15000)}{\log 1.0888} = \frac{0.2219}{0.037} \approx 6 years.

Step 1 — Write the CAGR relation

(1+CAGR)n=Final valueInitial value(1 + \text{CAGR})^{n} = \frac{\text{Final value}}{\text{Initial value}}

2500015000=1.667,1+CAGR=1.0888\frac{25000}{15000} = 1.667,\qquad 1+\text{CAGR} = 1.0888

Step 2 — Take logarithms

n log⁡(1.0888)=log⁡(1.667)n\,\log(1.0888) = \log(1.667)

n=log⁡1.667log⁡1.089=0.22190.037n = \frac{\log 1.667}{\log 1.089} = \frac{0.2219}{0.037}

Step 3 — Compute

n=0.22190.037≈6 yearsn = \frac{0.2219}{0.037} \approx 6\text{ years}

Why the other options are wrong: (B) 7.7, (C) 5.5, (D) 7 do not satisfy 0.2219/0.0370.2219/0.037; they come from dividing by a wrong log value. …

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