BETAThis platform is under active development; bugs, missing features, and risk of data loss are present. Thank you for your support!

Simple And Compound Interest Growth And Decay

Learn it by playing

Answer these questions to earn energy, then fish and explore. No account needed.

For teachers: ready-to-use lesson slides, revision notes, diagrams for Simple And Compound Interest Growth And Decay (Maths [CIE], Extended) — use them in your lesson, or run the topic as a live class game.

Notes

Simple Interest

  • **Simple interest** is calculated only on the original amount (principal), so each interest payment is the same.
  • Total interest = principal ×(rate/100)×\times (rate/100) \times time (years).
  • Final amount = principal + total interest.
  • Example: $250 at 4% for 6 years → interest per year = $10, total interest = $60, final amount = $310.
  • To find the rate: divide total interest by (principal×time)(principal \times time).

Compound Interest

  • **Compound interest** is calculated on the running total, so interest is earned on interest.
  • Final amount = principal ×(1+r100)n\times (1 + \frac{r}{100})^{n}, where r is the annual percentage rate and n is the number of years.
  • The multiplier is (1 + r/100); e.g., 5% → multiplier 1.05.
  • For different rates over time, apply each multiplier sequentially.
  • Reverse problems: original amount = final amount ÷(1+r100)n\div (1 + \frac{r}{100})^{n}.

Depreciation

  • **Depreciation** is a percentage decrease in value over time (e.g., cars, phones).
  • Final value = principal ×(1r100)n\times (1 - \frac{r}{100})^{n}, where r is the annual depreciation rate.
  • The multiplier is (1 - r/100); e.g., 15% depreciation → multiplier 0.85.
  • Amount lost = original value - final value.

Exponential Growth & Decay

  • **Exponential growth**: quantity increases by a fixed percentage each period; multiplier k>1k > 1.
  • **Exponential decay**: quantity decreases by a fixed percentage each period; multiplier 0<k<10 < k < 1.
  • General model: B=A×knB = A \times k^{n}, where A is initial amount, k is multiplier, n is number of periods.
  • Examples: population growth, bacterial growth (growth); cooling, radioactive decay (decay).
  • To find k:k=(BA)(1n)k: k = (\frac{B}{A})^(\frac{1}{n}). To find n: use trial and improvement.

Key Differences: Simple vs Compound

  • Simple interest: interest only on principal; linear growth.
  • Compound interest: interest on principal + accumulated interest; exponential growth.
  • For the same rate and time, compound interest yields a higher final amount than simple interest.
  • Always check whether the question asks for interest earned or final amount.

Common Exam Tips

  • Identify whether the problem involves simple interest, compound interest, depreciation, or exponential growth/decay.
  • Use the correct multiplier: growth =1+(rate/100)= 1 + (rate/100), decay =1(rate/100)= 1 - (rate/100).
  • Round answers as instructed (e.g., nearest hundred, nearest dollar).
  • For reverse problems, rearrange the formula to solve for the unknown.

Simple Interest vs Compound Interest Growth

Simple vs Compound Interest Growth01234100110120130Time (years)Amount ($)Simple InterestCompound Interest

Exponential Growth and Decay Curves

Exponential Growth & Decay0123450100150200TimeAmountGrowth (k>1)Decay (0<k<1)

Practice questions

Free preview — 8 of 40 questions. Sign up to see them all.

  1. 1.What is simple interest?

    Easy
    • AInterest calculated only on the original amount
    • BInterest calculated on the running total each year
    • CInterest that increases each year
    • DInterest that decreases each year
  2. 2.Which multiplier represents a 4% increase?

    Easy
    • A1.4
    • B0.96
    • C1.04
    • D0.04
  3. 3.A car depreciates by 15% each year. What is the multiplier for one year?

    Easy
    • A1.15
    • B0.15
    • C0.85
    • D1.85
  4. 4.Paula invests $600 at a rate of r% per year simple interest. At the end of 10 years, the total interest earned is $90. Find r.

    Medium
    • A1.5
    • B15
    • C0.15
    • D1.5%
  5. 5.Jan invests $800 at a rate of 3% per year simple interest. Calculate the value of her investment at the end of 4 years.

    Medium
    • A$896
    • B$824
    • C$96
    • D$800
  6. 6.Toby invested £7500 for 2 years in a savings account. He was paid 4% per annum compound interest. How much money did Toby have at the end of 2 years?

    Medium
    • A£8112
    • B£7800
    • C£8100
    • D£8112.00
  7. 7.Eric invests an amount in a bank that pays compound interest at a rate of 2.16% per year. At the end of 5 years, the value of his investment is $6999.31. Calculate the amount Eric invests.

    Hard
    • A$6300
    • B$6500
    • C$6200
    • D$6000
  8. 8.The population of a town decreases exponentially at a rate of 1.7% per year. The population now is 250 000. Calculate the population at the end of 5 years, correct to the nearest hundred.

    Hard
    • A229 500
    • B229 400
    • C230 000
    • D228 800

Unlock all 40 questions, slides & more

Create a free account to see every question, the slides, flashcards and revision notes for this topic.

Past papers

Past-paper practice for this topic is coming soon.

🗂️ Coming soon