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Number and Algebra - Simple and compound interest

Grade 11IB_AI

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

🔑Concepts

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Simple Interest: Interest calculated only on the initial amount (principal) borrowed or invested. The interest II remains constant for each time period.

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Compound Interest: Interest calculated on the principal plus any interest accumulated from previous periods. In the IB AI course, we often use the GDC (Graphic Display Calculator) Finance solver for these calculations.

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Compounding Frequency (kk): The number of times interest is added per year. Common values: Annually (k=1k=1), Semi-annually (k=2k=2), Quarterly (k=4k=4), Monthly (k=12k=12), and Daily (k=365k=365).

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Depreciation (Reducing Balance): A decrease in the value of an asset over time at a fixed percentage rate. It follows the same mathematical structure as compound interest but with a negative growth rate.

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GDC Finance Solver Variables: NN (number of years or total periods), I%I\% (annual interest rate), PVPV (present value), PMTPMT (payment per period, usually 0 for simple growth), FVFV (future value), P/YP/Y (payments per year), C/YC/Y (compounding periods per year).

📐Formulae

I=P×r×n100I = \frac{P \times r \times n}{100}

FV=PV×(1+r100k)knFV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}

FV=PV×(1−r100)nFV = PV \times \left(1 - \frac{r}{100}\right)^{n}

I=FV−PVI = FV - PV

💡Examples

Problem 1:

A student invests $4000 in a savings account that pays a simple interest rate of 3%3\% per annum. Calculate the total amount in the account after 55 years.

Solution:

  1. Identify variables: P=4000P = 4000, r=3r = 3, n=5n = 5.
  2. Calculate Interest: I=4000×3×5100=600I = \frac{4000 \times 3 \times 5}{100} = 600.
  3. Total Amount: A=P+I=4000+600=4600A = P + I = 4000 + 600 = 4600.

Explanation:

Simple interest is calculated using the formula I=Prn100I = \frac{Prn}{100}. The total amount is the sum of the principal and the interest earned.

Problem 2:

An investment of $8000 earns interest at a rate of 5%5\% per annum, compounded monthly. Find the value of the investment after 22 years.

Solution:

  1. Identify variables: PV=8000PV = 8000, r=5r = 5, k=12k = 12 (monthly), n=2n = 2.
  2. Use the compound interest formula: FV=8000×(1+5100×12)12×2FV = 8000 \times \left(1 + \frac{5}{100 \times 12}\right)^{12 \times 2}.
  3. FV=8000×(1+51200)24≈8000×(1.004167)24≈8839.54FV = 8000 \times \left(1 + \frac{5}{1200}\right)^{24} \approx 8000 \times (1.004167)^{24} \approx 8839.54.

Explanation:

Since interest is compounded monthly, we divide the annual rate by 1212 and multiply the number of years by 1212 to get the total number of periods.

Problem 3:

A car was purchased for $25000 and depreciates at a rate of 12%12\% per year. Calculate the value of the car after 44 years and find the total decrease in value.

Solution:

  1. Calculate Future Value: FV=25000×(1−0.12)4=25000×(0.88)4≈14992.38FV = 25000 \times (1 - 0.12)^{4} = 25000 \times (0.88)^{4} \approx 14992.38.
  2. Calculate total decrease (Interest lost) using vertical subtraction: 25000.00−14992.3810007.62\begin{array}{r} 25000.00 \\ -14992.38 \\ \hline 10007.62 \end{array}

Explanation:

Depreciation uses the formula FV=PV(1−r)nFV = PV(1 - r)^n. The total loss in value is the difference between the initial price and the value after 44 years.