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Number - Simple and compound interest

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Principal (P): The initial amount of money invested or borrowed.

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Interest Rate (r): The percentage charged or earned periodically, usually per annum (per year).

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Time (n or t): The duration for which the money is invested or borrowed.

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Simple Interest: Interest calculated only on the original principal amount throughout the period.

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Compound Interest: Interest calculated on the principal plus any interest accumulated from previous periods (interest on interest).

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Depreciation: The reduction in the value of an asset over time, calculated using the compound interest formula with a negative growth rate.

📐Formulae

Simple Interest: I=P×r×t100I = \frac{P \times r \times t}{100}

Total Amount (Simple Interest): A=P+IA = P + I

Compound Interest (Total Amount): A=P(1+r100)nA = P(1 + \frac{r}{100})^n

Compound Interest Earned: I=A−PI = A - P

Compound Depreciation: A=P(1−r100)nA = P(1 - \frac{r}{100})^n

💡Examples

Problem 1:

Calculate the simple interest earned on $3,500 invested at a rate of 4% per annum for 6 years.

Solution:

I=3500×4×6100=840I = \frac{3500 \times 4 \times 6}{100} = 840

Explanation:

Identify P=3500P = 3500, r=4r = 4, and t=6t = 6. Substitute these values into the simple interest formula I=Prt100I = \frac{Prt}{100} to find the interest alone.

Problem 2:

A bank offers a compound interest rate of 3.5% per year. If $8,000 is deposited for 4 years, calculate the total amount in the account at the end of the term.

Solution:

A=8000(1+3.5100)4=8000(1.035)4≈9180.18A = 8000(1 + \frac{3.5}{100})^4 = 8000(1.035)^4 \approx 9180.18

Explanation:

Use the compound interest formula A=P(1+r100)nA = P(1 + \frac{r}{100})^n. Here P=8000P = 8000, r=3.5r = 3.5, and n=4n = 4. Calculate the multiplier (1.035)(1.035) raised to the power of 4, then multiply by the principal.

Problem 3:

A laptop costs $1,200. Its value depreciates at a rate of 15% per year. What will be its value after 3 years?

Solution:

A=1200(1−15100)3=1200(0.85)3=736.95A = 1200(1 - \frac{15}{100})^3 = 1200(0.85)^3 = 736.95

Explanation:

Since the value is decreasing, use the depreciation formula A=P(1−r100)nA = P(1 - \frac{r}{100})^n. The multiplier is (1−0.15)=0.85(1 - 0.15) = 0.85. Apply the exponent for 3 years to find the final value.