Mathematics of FinancePYQ May 25Question 4010 of 507
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A sum of ₹ 725 is lent in the beginning of a year at a certain rate of simple interest. After 8 months, a sum of ₹ 362.50 more is lent but at the rate twice the former. At the end of the year, ₹ 33.50 is earned as interest from both the loans. What was the original rate of interest?

Options

A3.6%
B4.54%
C3.46%
D4.12%
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Correct Answer

✅ Option c — 3.46%

All Options:

  • A3.6%
  • B4.54%
  • C3.46%
  • D4.12%

Detailed Solution & Explanation

Let the original rate of interest be R%\displaystyle R\% per annum.
1. **First loan interest calculation**:
Principal amount (P1\displaystyle P_1) = ₹725\displaystyle ₹ 725
Time duration (T1\displaystyle T_1) = 1 year\displaystyle 1 \text{ year} (since it is lent in the beginning of the year)
Interest from the first loan (I1\displaystyle I_1):
I1=P1×R×T1100=725×R×1100=7.25RI_1 = \frac{P_1 \times R \times T_1}{100} = \frac{725 \times R \times 1}{100} = 7.25R

2. **Second loan interest calculation**:
Principal amount (P2\displaystyle P_2) = ₹362.50\displaystyle ₹ 362.50
This amount is lent after 8 months, so it earns interest for the remaining 4 months of the year:
T2=12−8=4 months=412=13 yearT_2 = 12 - 8 = 4 \text{ months} = \frac{4}{12} = \frac{1}{3} \text{ year}
The interest rate is twice the original rate, so Rate (R2\displaystyle R_2) = 2R%\displaystyle 2R\% per annum.
Interest from the second loan (I2\displaystyle I_2):
I2=P2×(2R)×T2100=362.50×2R×13100=725R300I_2 = \frac{P_2 \times (2R) \times T_2}{100} = \frac{362.50 \times 2R \times \frac{1}{3}}{100} = \frac{725R}{300}

3. **Total interest calculation**:
The total interest earned from both loans at the end of the year is ₹33.50\displaystyle ₹ 33.50:
I1+I2=33.50I_1 + I_2 = 33.50
7.25R+725R300=33.507.25R + \frac{725R}{300} = 33.50
725R100+725R300=33.50\frac{725R}{100} + \frac{725R}{300} = 33.50
Multiply the entire equation by 300\displaystyle 300 to clear the denominators:
3(725R)+725R=33.50×3003(725R) + 725R = 33.50 \times 300
2175R+725R=100502175R + 725R = 10050
2900R=100502900R = 10050
R=100502900=20158≈3.4655%R = \frac{10050}{2900} = \frac{201}{58} \approx 3.4655\%
The original rate of interest is approximately 3.46%\displaystyle 3.46\%.
Hence, **Option C** is the correct answer.

About This Chapter: Mathematics of Finance

Paper

Paper 3: Quantitative Aptitude

Weightage

12-16 Marks

Key Topics

Simple & Compound Interest, Annuity, Perpetuity

The most important mathematical chapter in the entire syllabus. It covers Simple Interest (SI), Compound Interest (CI), Nominal vs Effective rates, Present and Future Value, Annuities (Ordinary and Due), Sinking Funds, and Perpetuities. The concepts learned here are applied heavily in CA Intermediate and Final.

View Official ICAI Syllabus

Exam Strategy Tip

Guaranteed 12-16 marks. Master your calculator! Learn the 'GT' and compound interest M+/M- tricks to solve annuity questions in 10 seconds without writing long formulas.

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