Mathematics for FinanceMTP Sep 24 Series IIQuestion 1428 of 512
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6,400\displaystyle 6,400 amounts to 7840\displaystyle 7840 in two years at simple interest. How much will a sum of 84\displaystyle 84 invested at the same rate of simple interest amount in four years?

Options

A11.20\displaystyle 11.20
B112.20\displaystyle 112.20
C121.80\displaystyle 121.80
D121.80\displaystyle 121.80
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Correct Answer

✅ Option d — 121.80\displaystyle 121.80

All Options:

  • A11.20\displaystyle 11.20
  • B112.20\displaystyle 112.20
  • C121.80\displaystyle 121.80
  • D121.80\displaystyle 121.80

Detailed Solution & Explanation

First, let's find the simple interest rate of the first investment. Given parameters: * Initial Principal (P1\displaystyle P_1) = Rs. 6,400\displaystyle \text{Rs. }6,400 * Final Amount (A1\displaystyle A_1) = Rs. 7,840\displaystyle \text{Rs. }7,840 * Time (t1\displaystyle t_1) = 2\displaystyle 2 years The Simple Interest (SI1\displaystyle SI_1) earned is: SI1=A1−P1=7,840−6,400=1,440SI_1 = A_1 - P_1 = 7,840 - 6,400 = 1,440 The formula for Simple Interest is: SI1=P1×r×t1100SI_1 = \frac{P_1 \times r \times t_1}{100} 1,440=6,400×r×21001,440 = \frac{6,400 \times r \times 2}{100} 1,440=128r1,440 = 128 r Solving for r\displaystyle r: r=1,440128=11.25% p.a.r = \frac{1,440}{128} = 11.25\% \text{ p.a.} Now, we calculate the amount (A2\displaystyle A_2) for the second investment: * Principal (P2\displaystyle P_2) = Rs. 84\displaystyle \text{Rs. }84 * Rate (r\displaystyle r) = 11.25%\displaystyle 11.25\% p.a. * Time (t2\displaystyle t_2) = 4\displaystyle 4 years The interest (SI2\displaystyle SI_2) earned is: SI2=P2×r×t2100=84×11.25×4100=84×45100=37.80SI_2 = \frac{P_2 \times r \times t_2}{100} = \frac{84 \times 11.25 \times 4}{100} = \frac{84 \times 45}{100} = 37.80 The final amount (A2\displaystyle A_2) is: A2=P2+SI2=84+37.80=121.80A_2 = P_2 + SI_2 = 84 + 37.80 = 121.80 Thus, the amount is Rs. 121.80\displaystyle \text{Rs. }121.80. Hence, **Option D** is the correct answer.

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