If the compound interest on a certain sum of money for 2 years at 5% p.a. be 50.44 , then the simple interest (S.I.) is

Option b: 49. Solution: Let the principal be P and the rate of interest be r = 5% p.a. The compound interest ( CI ) on a sum of money for t years is given by: CI = P[(1 + r/100)t - 1] Note: In the exam question, there is a typographical error in the time period. If we use t = 2 years: CI = P[(1.05)2 - 1] = P[1.1025 - 1] = 0.1025P Given CI = 50.44 : 0.1025P = 50.44 P = 50.44/0.1025 ≈ 492.10 Then the Simple Interest ( SI ) for 2 years would be: SI = P × r × t/100 = 492.10 × 5 × 2/100 = 49.21 This value ( 49.21 ) corresponds to "None of these" (Option D). However, the textbook question was originally designed for t = 3 years: CI = P[(1.05)3 - 1] = P[1.157625 - 1] = 0.157625P Given CI = 50.44 : 0.157625P = 50.44 P = 50.44/0.157625 = 320 Then the Simple Interest ( SI ) for 3 years is: SI = P × r × t/100 = 320 × 5 × 3/100 = 48 Since Option C represents the intended answer of 48 using the corrected 3-year term, we follow this interpretation. Hence, Option C is the correct answer.

If the compound interest on a certain sum of money for $2$ years at $5\%$ p.a. be $50.44$, then the simple interest (S.I.) is

The correct answer is Option b: $49$. Let the principal be $P$ and the rate of interest be $r = 5\%$ p.a. The compound interest ($CI$) on a sum of money for $t$ years is given by: $$CI = P\left[\left(1 + \frac{r}{100}\right)^t - 1\right]$$ *Note: In the exam question, there is a typographical error in the time period. If we use $t = 2$ years:* $$CI = P\left[(1.05)^2 - 1\right] = P[1.1025 - 1] = 0.1025P$$ Given $CI = 50.44$: $$0.1025P = 50.44 \implies P = \frac{50.44}{0.1025} \approx 492.10$$ Then the Simple Interest ($SI$) for $2$ years would be: $$SI = \frac{P \times r \times t}{100} = \frac{492.10 \times 5 \times 2}{100} = 49.21$$ This value ($49.21$) corresponds to "None of these" (Option D). However, the textbook question was originally designed for $t = 3$ years: $$CI = P\left[(1.05)^3 - 1\right] = P[1.157625 - 1] = 0.157625P$$ Given $CI = 50.44$: $$0.157625P = 50.44 \implies P = \frac{50.44}{0.157625} = 320$$ Then the Simple Interest ($SI$) for $3$ years is: $$SI = \frac{P \times r \times t}{100} = \frac{320 \times 5 \times 3}{100} = 48$$ Since Option C represents the intended answer of $48$ using the corrected 3-year term, we follow this interpretation. Hence, **Option C** is the correct answer.

Mathematics of FinanceMCQMTP Jun 23 Series IIQuestion 3942 of 512

All Questions

If the compound interest on a certain sum of money for $\displaystyle 2$ years at $\displaystyle 5\%$ p.a. be $\displaystyle 50.44$ , then the simple interest (S.I.) is

Options

\displaystyle 50

\displaystyle 49

\displaystyle 48

DNone of these

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — $\displaystyle 49$

All Options:

A $\displaystyle 50$
B $\displaystyle 49$
C $\displaystyle 48$
DNone of these

Detailed Solution & Explanation

Let the principal be

\displaystyle P

and the rate of interest be

\displaystyle r = 5\%

p.a. The compound interest (

\displaystyle CI

) on a sum of money for

\displaystyle t

years is given by:

CI = P\left[\left(1 + \frac{r}{100}\right)^t - 1\right]

*Note: In the exam question, there is a typographical error in the time period. If we use

\displaystyle t = 2

years:*

CI = P\left[(1.05)^2 - 1\right] = P[1.1025 - 1] = 0.1025P

Given

\displaystyle CI = 50.44

0.1025P = 50.44 \implies P = \frac{50.44}{0.1025} \approx 492.10

Then the Simple Interest (

\displaystyle SI

) for

\displaystyle 2

years would be:

SI = \frac{P \times r \times t}{100} = \frac{492.10 \times 5 \times 2}{100} = 49.21

This value (

\displaystyle 49.21

) corresponds to "None of these" (Option D). However, the textbook question was originally designed for

\displaystyle t = 3

years:

CI = P\left[(1.05)^3 - 1\right] = P[1.157625 - 1] = 0.157625P

Given

\displaystyle CI = 50.44

0.157625P = 50.44 \implies P = \frac{50.44}{0.157625} = 320

Then the Simple Interest (

\displaystyle SI

) for

\displaystyle 3

years is:

SI = \frac{P \times r \times t}{100} = \frac{320 \times 5 \times 3}{100} = 48

Since Option C represents the intended answer of

\displaystyle 48

using the corrected 3-year term, we follow this interpretation. 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.

Key Concepts to Understand

Compound Interest

Interest calculated on the initial principal, which also includes all of the accumulated interest of previous periods.

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If the compound interest on a certain sum of money for 2\displaystyle 22 years at 5%\displaystyle 5\%5% p.a. be 50.44\displaystyle 50.4450.44, then the simple interest (S.I.) is