Question

Mungeri Lal has two investment plans, A and B, to choose from. Plan A offers interest of 10% compounded annually, while plan B offers simple interest of 12% per annum. Till how many years is plan B a better investment?

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Compare \((1.10)^n\) with \(1+0.12n\) year by year until the compound value overtakes the simple value.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Rs. 1 invested in Plan A grows at compound interest of 10% a year, so after \(n\) years it becomes \((1.10)^n\). The same Rs. 1 in Plan B grows at simple interest of 12% a year, so after \(n\) years it becomes \(1 + 0.12n\). We need the largest \(n\) for which Plan B's amount is still bigger than Plan A's amount.

Step 2: Key Formula or Approach:
Compound interest amount: \(A_{comp} = (1.10)^n\).
Simple interest amount: \(A_{simp} = 1 + 0.12n\).
We check year by year until \(A_{comp}\) overtakes \(A_{simp}\).

Step 3: Detailed Explanation:
For \(n = 1\): \(A_{comp} = 1.10\), \(A_{simp} = 1.12\). Plan B is ahead.
For \(n = 2\): \(A_{comp} = 1.21\), \(A_{simp} = 1.24\). Plan B is still ahead.
For \(n = 3\): \(A_{comp} = 1.331\), \(A_{simp} = 1.36\). Plan B is still ahead.
For \(n = 4\): \(A_{comp} = 1.4641\), \(A_{simp} = 1.48\). Plan B is still ahead, by a small margin.
For \(n = 5\): \(A_{comp} = 1.61051\), \(A_{simp} = 1.60\). Now Plan A has overtaken Plan B.
So the crossover happens between year 4 and year 5, Plan B stays ahead through year 4 and loses the lead from year 5 onward.

Step 4: Final Answer:
Plan B remains the better investment up to 4 years.
\[ \boxed{4 \text{ years}} \]