Simple & Compound Interest Math Case Study for Class 8

Simple & Compound Interest Math Case Study for Class 8

Simple & Compound Interest Test Math Case Study for Class 8

The Simple & Compound Interest Test Math Case Study for Class 8 introduces students to important SI and CI concepts through practical examples. It helps them understand how interest grows over time, and it provides clear explanations to support better calculation skills.

Understanding SI and CI Concepts

This section explains the difference between simple interest and compound interest. Students learn to calculate interest step by step. Additionally, real-life examples make these concepts easier and more meaningful to understand.

Strengthening Analytical Skills

Case-based questions encourage logical thinking and structured problem-solving. Students explore situations involving savings, borrowing, and investment. Moreover, guided solutions allow them to correct mistakes quickly and improve confidence.

Case Study 5: Interest for Event Financing

**Riya** is designing a short-term corpus for an annual school festival where vendors and event teams will be paid from the interest earned on the corpus. She needs to balance modest growth with predictable liquidity. To do this she simulates realistic transactions: staggered deposits in accounts with different compounding conventions (**semi-annually, annually**), small personal loans to vendor partners, and conversions between nominal and effective rates to decide which bank offers better returns for the exact time horizon of the festival (between 2 and 4 years). She must also reconstruct missing data in some partner records (when maturity amount is recorded but principal or rate is missing) and compare nearly-equal offers to see whether the administrative difference materially affects cash flow for the event. All scenarios require careful rounding to two decimal places.

1. Riya places Rs. 3500 in an account that pays 7% per annum compounded semi-annually for 4 years. What is the maturity amount (rounded to two decimal places)?

Solution:
Semi-annual rate $r = \frac{7\%}{2} = 3.5\% = 0.035$. Number of periods $n = 4 \times 2 = 8$.
$A = P(1+r)^n = 3500(1.035)^8$
$A \approx 3500 \times 1.3168090 \approx 4608.83$
Correct answer is option **(a)**.

2. A small vendor record shows that Rs. 2500 earned Rs. 750 as simple interest in 5 years. What was the annual rate of simple interest (to two decimal places)?

Solution:
Using $SI=\frac{P\times R\times T}{100}$: $750=\frac{2500\times R\times 5}{100}$.
$R=\frac{750\times 100}{2500\times 5}=\frac{75000}{12500}=6$.
Thus, $R = 6.00\%$.
Correct answer is option **(b)**.

3. An advance of Rs. 9,200 becomes Rs. 11,529.60 in 3 years with annual compounding. What is the approximate annual interest rate (to two decimal places)?

Solution:
$\left(1+\frac{R}{100}\right)^3 = \frac{A}{P} = \frac{11529.60}{9200} \approx 1.25365217$
$1+\frac{R}{100} = 1.25365217^{1/3} \approx 1.07814077$
$R \approx 7.81\%$.
Correct answer is option **(b)**.

4. Riya compares two festival-bank offers for a 3-year deposit of Rs. 6000: Offer A is 6.5% p.a. compounded annually, Offer B is 6.4% p.a. compounded half-yearly. After 3 years, which offer gives a higher maturity and by approximately how much (rounded to two decimals)?

Solution:
**Offer A (Annual 6.5%):** $A_A=6000(1.065)^3 \approx 7247.70$
**Offer B (Half-yearly 6.4%, $r=3.2\%$, $n=6$):** $A_B=6000(1.032)^6 \approx 7248.19$
**Difference:** $A_B – A_A \approx 7248.19 – 7247.70 \approx 0.49$.
Offer **B** yields approximately **Rs. 0.49** more.
Correct answer is option **(b)**.

5. A partner borrowed Rs. 8200 and repaid Rs. 9842 after 3 years under simple interest. What annual simple interest rate was charged (rounded to two decimal places)?

Solution:
Total interest paid $SI = 9842 – 8200 = 1642$.
$R=\frac{SI \times 100}{P \times T} = \frac{1642 \times 100}{8200 \times 3} = \frac{164200}{24600} \approx 6.6748\%$.
Rounded to two decimals, $R \approx 6.67\%$.
Correct answer is option **(b)**.

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