Komal invested a sum of ₹5000 at 20% per annum compound intere... | General Science MCQ | Examoogle

General SciencePrevious Year Paper 2024

Komal invested a sum of ₹5000 at 20% per annum compound interest, componded annually. If she received an amount of ₹8640 after n years, the value of n is:

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral SciencePrevious Year Paper 2024

Option C

Quick Summary:

Compound interest calculates interest on both the initial principal and the accumulated interest from previous periods. Using the compound interest formula, the amount after $n$ years is given by $A = P(1 + \frac{R}{100})^n$ , where $A$ is the final amount, $P$ is the principal, $R$ is the annual rate of interest, and $n$ is the time period in years.

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🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of compound interest like a snowball rolling down a hill; as it gathers snow (interest), it grows larger and collects even more snow with each turn, growing exponentially faster over time.

🧠 Memory Mnemonic / Shortcut Aid

Compound Interest Formula: 'A = P (1 + r/t)^nt' — Amount Equals Principal times One plus Rate over 100 to the power of time.

🔢 Key Formulas

$A = P \times (1 + \frac{R}{100})^n$ — Compound Amount formula

$C.I. = A - P$ — Compound Interest earned

⚙️ Working Principle

Substituting the given values into the compound interest equation: $8640 = 5000(1 + \frac{20}{100})^n$ . Simplifying the term inside the parenthesis gives $\frac{8640}{5000} = (1.2)^n$ . Dividing both sides by 1000 and simplifying the fraction $\frac{864}{500} = \frac{216}{125}$ , we get $(1.2)^n = \frac{216}{125}$ . Since $\frac{216}{125} = (\frac{6}{5})^3 = (1.2)^3$ , comparing both sides yields $n = 3$ .

📌 Key Points

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Compound interest grows exponentially with time $n$ .
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The base ratio $1 + \frac{R}{100}$ simplifies to $1.2$ for a $20\%$ interest rate.
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Equating powers of identical bases allows direct solving for the exponent $n$ .

✅ Advantages

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Yields higher returns over longer investment periods compared to simple interest.
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Accurately reflects financial growth in banking and market investments.

❌ Disadvantages / Limitations

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Increases liability burdens quickly if applied to loans over long durations.

🛠️ Applications / Uses

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Banking fixed deposits and recurring deposits
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Loan amortizations and mortgage calculations

📄 Additional Information

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Options A, B, and D yield incorrect amounts when evaluated at fractional or alternative integer powers of 1.2.
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For $n = 3$ , $5000 \times (1.2)^3 = 5000 \times 1.728 = 8640$ , matching the given amount perfectly.

📊 Diagram / Illustration

✅

C is correct — The value of $n$ is 3 years as verified by solving the compound interest equation $5000(1 + 0.20)^n = 8640$ .

Core Concepts Used

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Compound Interest Exponential Growth Exponents and Powers

💡 EXAM TIP

When solving exponential equations in compound interest, always attempt to express both sides of the equation with the same base to easily equate exponents.

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