At what rate (in percentage) per annum will a sum of money dou... | Mathematics MCQ

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At what rate (in percentage) per annum will a sum of money double itself in 16 years on simple interest?

12.5%

5.6%

6.25%

Correct Answer

📐 MA • Math Direct FormulaMathematicsPrevious Year Paper 2024

Option C

6.25%

Quick Summary:

For a sum to become n times in T years, the rate is simply $R = \frac{100(n-1)}{T}$ . Here, n=2, so $R = \frac{100(2-1)}{16} = \frac{100}{16} = 6.25\%$ .

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📐MAMath SolutionDirect Formula

📋 Given

Time (T) = 16 years. The sum of money doubles itself (Amount A = 2 × Principal P).

🔢 Formula Used

R=100×(n−1)TR = \frac{100 \times (n-1)}{T}R=T100×(n−1)​

⚡ Exam Hall Shortcut / Speed Trick

For a sum to become n times in T years, the rate is simply $R = \frac{100(n-1)}{T}$ . Here, n=2, so $R = \frac{100(2-1)}{16} = \frac{100}{16} = 6.25\%$ .

⚠️ Common Student Trap / Pitfall

Confusing the amount formula with simple interest; remember that if a sum doubles, the Simple Interest (SI) gained is equal to the Principal ( $SI = P$ ).

🔢 Step-by-Step Solution

Identify the given variables

Let the principal be $P$ . If the amount doubles, the final amount $A = 2P$ . The interest earned is $SI = A - P = 2P - P = P$ . Time $T = 16$ years.

SI=P,T=16SI = P, T = 16SI=P,T=16

Apply Simple Interest Formula

The formula for simple interest is $SI = \frac{P \times R \times T}{100}$ . Substituting the known values:

P=P×R×16100P = \frac{P \times R \times 16}{100}P=100P×R×16​

Solve for Rate (R)

Divide both sides by $P$ and isolate $R$ :

1=R×16100  ⟹  R=10016=6.25%1 = \frac{R \times 16}{100} \implies R = \frac{100}{16} = 6.25\%1=100R×16​⟹R=16100​=6.25%

✅

C is correct because the rate required for the principal to double in 16 years is 6.25%.

Core Concepts Used

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Simple Interest Principal and Amount Relationship Rate of Interest

💡 EXAM TIP

This concept is a subset of simple interest problems involving 'times of itself' (e.g., tripling or quadrupling), which can be solved quickly using the $(n-1)$ multiplier on the numerator.

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