Mathematics for FinanceMTP Sep 24 Series IIQuestion 1427 of 512
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At what time a certain sum of money amounts to 400\displaystyle 400 at 10%\displaystyle 10\% p.a. S.I. and to 200\displaystyle 200 at 4%\displaystyle 4\% p.a. S.I.

Options

A10\displaystyle 10 Yrs.
B30\displaystyle 30 Yrs.
C50\displaystyle 50 Yrs.
DNone of these
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Correct Answer

✅ Option c — 50\displaystyle 50 Yrs.

All Options:

  • A10\displaystyle 10 Yrs.
  • B30\displaystyle 30 Yrs.
  • C50\displaystyle 50 Yrs.
  • DNone of these

Detailed Solution & Explanation

Let the principal amount be P\displaystyle P and the time period be T\displaystyle T years. The formula for Simple Interest amount (A\displaystyle A) is: A=P(1+r×T100)A = P\left(1 + \frac{r \times T}{100}\right) We are given two cases: 1. At 10%\displaystyle 10\% p.a. S.I., the amount is Rs. 400\displaystyle \text{Rs. }400: 400=P(1+0.10T)— (Equation 1)400 = P(1 + 0.10 T) \quad \text{--- (Equation 1)} 2. At 4%\displaystyle 4\% p.a. S.I., the amount is Rs. 200\displaystyle \text{Rs. }200: 200=P(1+0.04T)— (Equation 2)200 = P(1 + 0.04 T) \quad \text{--- (Equation 2)} Dividing Equation 1 by Equation 2: 400200=P(1+0.10T)P(1+0.04T)\frac{400}{200} = \frac{P(1 + 0.10 T)}{P(1 + 0.04 T)} 2=1+0.10T1+0.04T2 = \frac{1 + 0.10 T}{1 + 0.04 T} Cross-multiplying: 2(1+0.04T)=1+0.10T2(1 + 0.04 T) = 1 + 0.10 T 2+0.08T=1+0.10T2 + 0.08 T = 1 + 0.10 T 2−1=0.10T−0.08T2 - 1 = 0.10 T - 0.08 T 1=0.02T1 = 0.02 T Solving for T\displaystyle T: T=10.02=50 yearsT = \frac{1}{0.02} = 50 \text{ years} Thus, the time period is 50\displaystyle 50 years. Hence, **Option C** is the correct answer.

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