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Number - Percentages and Interest (Simple and Compound)

Grade 12IGCSE

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

🔑Concepts

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Percentage Multipliers: Representing a percentage increase as (1 + r/100) and a decrease as (1 - r/100).

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Reverse Percentages: Finding the original value after a percentage change has occurred (e.g., finding the pre-tax price).

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Simple Interest: Interest calculated only on the initial principal amount throughout the duration.

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Compound Interest: Interest calculated on the initial principal and also on the accumulated interest of previous periods.

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

📐Formulae

Percentage Change=New Value−Original ValueOriginal Value×100\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100

Simple Interest (I)=P×R×T100\text{Simple Interest (I)} = \frac{P \times R \times T}{100}

Total Amount (Simple Interest)=P+I\text{Total Amount (Simple Interest)} = P + I

Compound Interest Amount (A)=P(1+r100)n\text{Compound Interest Amount (A)} = P \left(1 + \frac{r}{100}\right)^n

Depreciation (Value after n years)=P(1−r100)n\text{Depreciation (Value after } n \text{ years)} = P \left(1 - \frac{r}{100}\right)^n

\text{Original Value (Reverse %)} = \frac{\text{Final Value}}{\text{Multiplier}}

💡Examples

Problem 1:

A laptop is sold for $864 after a 20% discount. Find the original price of the laptop.

Solution:

Original Price = 864/0.80=864 / 0.80 = 1,080

Explanation:

To find the original value before a 20% decrease, we identify the multiplier: 100%−20%=80%100\% - 20\% = 80\% or 0.800.80. We divide the final price by this multiplier to reverse the change.

Problem 2:

Calculate the total amount in a savings account after 5 years if $4,000 is invested at a compound interest rate of 3.5% per annum.

Solution:

A=4000×(1+0.035)5=4000×(1.187686)≈A = 4000 \times (1 + 0.035)^5 = 4000 \times (1.187686) \approx 4,750.74$

Explanation:

Using the compound interest formula A=P(1+r/100)nA = P(1 + r/100)^n, where P=4000P=4000, r=3.5r=3.5, and n=5n=5. The multiplier is 1.0351.035 raised to the power of the number of years.

Problem 3:

A car depreciates in value by 12% each year. If the car is worth $25,000 now, what will it be worth in 3 years? Give your answer to the nearest dollar.

Solution:

Value = 25000×(1−0.12)3=25000×(0.88)3≈25000 \times (1 - 0.12)^3 = 25000 \times (0.88)^3 \approx 17,037$

Explanation:

This is a negative growth problem. We use the depreciation formula with a multiplier of 0.880.88 (since 100%−12%=88%100\% - 12\% = 88\%) and apply it for n=3n=3 years.