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: Set up the difference.
Let $D(n) = (1 + 0.12n) - (1.10)^n$, the extra amount Plan B gives over Plan A after $n$ years. Plan B is the better choice as long as $D(n) > 0$.

Step 2: Track $D(n)$ year by year.
$n=1$: $D = 1.12 - 1.10 = 0.02 > 0$.
$n=2$: $D = 1.24 - 1.21 = 0.03 > 0$.
$n=3$: $D = 1.36 - 1.331 = 0.029 > 0$.
$n=4$: $D = 1.48 - 1.4641 = 0.0159 > 0$, still positive but shrinking fast.
$n=5$: $D = 1.60 - 1.61051 = -0.01051 < 0$, Plan B has now fallen behind.

Step 3: Read off the crossover.
$D(n)$ stays positive through $n=4$ and turns negative at $n=5$. Compound growth is slow at first but eventually beats a fixed 12% add on every year, because compounding acts on an ever growing base while simple interest always adds the same fixed 12% of the original Re. 1.

Step 4: Final answer.
Plan B stays ahead for 4 full years before Plan A's compounding catches up.
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