Growth, decay and compound interest

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Notas de aula

Simple vs Compound Interest

Simple interest is calculated only on the original principal each period.
Compound interest is calculated on the principal plus any interest already added.
Compound interest grows faster than simple interest over time because interest earns interest.
The extra growth from compounding is called anatocism in older texts.

The Compound Interest Formula

The final amount after compound interest is given by A = P\left(1 + \frac{r}{n}\right)^nt.
In this formula: A = final amount, P = principal, r = annual interest rate (as a decimal), n = number of compounding periods per year, t = time in years.
The total compound interest earned is I = A - P.
The multiplier for one period is 1 + \frac{r}{n}; raising it to the power nt applies it repeatedly.

Compounding Frequency

Compounding frequency is how many times per year interest is added to the principal.
Common frequencies: annually (n = 1), half-yearly (n = 2), quarterly (n = 4), monthly (n = 12), weekly (n = 52), daily (n = 365).
More frequent compounding gives a slightly higher final amount for the same annual rate.
For monthly compounding at an annual rate r, the monthly rate is \frac{r}{12}.

Growth and Decay Multipliers

For a percentage increase of p%, the multiplier is 1 + \frac{p}{100}.
For a percentage decrease of p%, the multiplier is 1 - \frac{p}{100}.
Repeated growth or decay over t years uses the multiplier raised to the power t.
Example: a 5% annual increase for 3 years gives a total multiplier of 1.05³.

Depreciation

Depreciation is a repeated percentage decrease in value.
If an item loses p% of its value each year, its value after t years is V = P(1 - \frac{p}{100})^t.
Depreciation is a form of exponential decay.
Example: a car worth £12000 losing 15% per year is worth 12000 \times 0.85^t after t years.

Population Growth and Decay

Populations can grow or decay by a fixed percentage each year.
For a population growing by r% per year, after t years the population is P(1 + \frac{r}{100})^t.
For a population decaying by r% per year, after t years the population is P(1 - \frac{r}{100})^t.
These are examples of exponential growth and exponential decay.

Solving Problems with Iteration

To find how many years it takes to reach a target value, substitute different values of t until the amount passes the target.
This trial-and-improvement method is called iteration.
For growth, the amount increases with t; for decay, it decreases with t.
Example: to find when £500 at 4% compound interest exceeds £600, test t = 1, 2, 3, … until A > 600.

The Rule of 72

The Rule of 72 estimates the number of years for an investment to double.
Years to double ≈ \frac{72}{\text{interest rate as a percentage}}.
Example: at 6% compound interest, money doubles in about 72 ÷ 6 = 12 years.
This rule was described by Luca Pacioli in 1494.

Annual Equivalent Rate (AER)

The annual equivalent rate (AER) shows the effective annual interest when compounding happens more than once a year.
AER allows fair comparison between accounts with different compounding frequencies.
AER is the total interest earned in one year divided by the principal.
Other names include effective annual rate, effective interest rate, and annual percentage yield.

Slides

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Questões de prática

Prévia grátis — 8 de 67 perguntas. Cadastre-se para ver todas.

1.In the compound interest formula A = P\left(1 + \frac{r}{n}\right)^nt, what does n represent?
Easy
- AThe number of times interest is compounded per year
- BThe number of years the money is invested
- CThe annual interest rate as a decimal
- DThe final amount of money
2.Which multiplier should be used to increase an amount by 6% each year for 3 years?
Medium
- A1.06³
- B1.6³
- C1.06 × 3
- D0.06³
3.A car is worth £12,000. It depreciates by 15% each year. Which calculation gives its value after 2 years?
Easy
- A12000 × 0.85²
- B12000 × 0.15²
- C12000 × 1.15²
- D12000 × (1 − 0.15 × 2)
4.£500 is invested at 4% per year compound interest. What is the total amount after 3 years, to the nearest penny?
Medium
- A£562.43
- B£560.00
- C£562.00
- D£624.32
5.£800 is invested at a nominal annual rate of 6% compounded monthly. What is the amount after 2 years, to the nearest penny?
Medium
- A£901.60
- B£898.88
- C£896.00
- D£904.32
6.Which of the following statements about compound interest are true? (Select all that apply.)
Medium
- ACompound interest is interest earned on both the principal and previously accumulated interest.
- BSimple interest is always greater than compound interest for the same rate and time.
- CThe annual equivalent rate (AER) allows fair comparison of different compound interest products.
- DIncreasing the compounding frequency always decreases the final amount.
- ECompound interest can be calculated using a multiplier raised to a power.
7.Compound interest is calculated only on the original principal sum, not on any interest already earned.
Easy
True or false?
8.Match each term with its correct meaning.
Medium
- Principal
- Compounding frequency
- Annual equivalent rate (AER)
- The original sum of money invested or borrowed
- The number of times per year that interest is added
- The effective annual rate allowing fair comparison of products

Mais perguntas

9.Match each compounding frequency with its correct value of n in the compound interest formula.
Easy
- Annually
- Quarterly
- Monthly
- Weekly
- n = 1
- n = 4
- n = 12
- n = 52
10.Order these steps to calculate the final amount when £600 is invested at 3% per year compound interest for 5 years.
Medium
- Write the multiplier as 1.03
- Raise the multiplier to the power 5
- Multiply the principal £600 by the result
- Round the final amount to the nearest penny
11.Which of the following statements about compound interest and simple interest are true? (Select all that apply.)
Medium
- ACompound interest is calculated on the principal and on previously earned interest.
- BSimple interest is calculated only on the original principal.
- CFor the same rate and time, compound interest always gives a larger final amount than simple interest.
- DCompound interest can be calculated using a multiplier raised to a power.
- ESimple interest uses a different formula that involves a power.
12.A house is bought for £250,000. Its value increases by 4% each year. Which calculation gives its value after 3 years?
Medium
- A250000 × (1.04)³
- B250000 × (1.4)³
- C250000 × (1 + 0.04 × 3)
- D250000 × (0.96)³
13.Depreciation means the value of an item decreases by a fixed percentage each year.
Easy
True or false?
14.£2000 is invested at a nominal annual rate of 5% compounded quarterly. What is the amount after 1 year, to the nearest penny?
Medium
- A£2100.00
- B£2101.89
- C£2102.50
- D£2155.06
15.A car worth £15,000 depreciates by 12% in the first year and then by 8% in the second year. Which of the following calculations correctly give its value after 2 years? (Select all that apply.)
Hard
- A15000 × 0.88 × 0.92
- B15000 × (0.88 × 0.92)
- C15000 × (0.88 + 0.92)
- D15000 × (0.88)² × (0.92)²
- E15000 × 0.88 × 0.92 × 1
16.£5000 is invested at 4% per year compound interest. How many complete years will it take for the investment to first exceed £6000?
Medium
- A5 years
- B6 years
- C7 years
- D8 years
17.Order these multipliers from smallest to largest.
Medium
- 0.95
- 1.02
- 1.05
- 1.10
18.A company's profits are £40,000 in year 1 and increase by 10% each year. What is the profit in year 3?
Medium
- A£48,400
- B£48,000
- C£44,000
- D£52,000
19.A radioactive substance has a mass of 80 g and decays by 25% each hour. What is its mass after 2 hours?
Medium
- A45 g
- B30 g
- C40 g
- D20 g
20.A jacket costs £80. Its price is reduced by 20% in a sale. What is the sale price?
Easy
- A£64
- B£60
- C£76
- D£100
21.A tree is 2 m tall and grows by 15% each year. What is its height after 1 year?
Easy
- A2.30 m
- B2.15 m
- C2.45 m
- D2.60 m
22.Which expression gives the value of an item after 3 years if it starts at £2000 and depreciates by 10% each year?
Medium
- A£2000 × (0.90)³
- B£2000 × (1.10)³
- C£2000 × (1 - 0.10 × 3)
- D£2000 × (0.10)³
23.A sum of £500 is invested at 5% per year compound interest. What is the amount after 2 years?
Medium
- A£551.25
- B£550.00
- C£525.00
- D£552.50
24.A car worth £8000 depreciates by 15% in the first year and then by 10% in the second year. What is its value after 2 years?
Medium
- A£6120
- B£6000
- C£6800
- D£7200
25.The multiplier for a 7% increase is 1.07.
Easy
True or false?
26.Depreciation means the value of an item increases by a fixed percentage each year.
Easy
True or false?
27.Match each percentage change with its correct multiplier.
Easy
- Increase by 12%
- Decrease by 12%
- Increase by 5%
- Decrease by 5%
- 1.12
- 0.88
- 1.05
- 0.95
28.A rectangular garden is 20 m by 15 m. Its length increases by 10% and its width decreases by 10%. What is the percentage change in the area of the garden?
Hard
- A1% decrease
- B1% increase
- Cno change
- D2% decrease
29.A company's profits are £50,000 in year 1. They increase by 10% in year 2 and by 5% in year 3. Which of the following statements are true? (Select all that apply.)
Hard
- AThe profit in year 3 is £57,750.
- BThe overall percentage increase from year 1 to year 3 is 15.5%.
- CThe profit in year 3 is £57,500.
- DThe overall percentage increase from year 1 to year 3 is 15%.
- EThe profit in year 3 is £55,000.
30.If a quantity increases by 10% and then decreases by 10%, the final value is the same as the original.
Easy
True or false?
31.Match each situation with the correct multiplier for one year.
Medium
- Increase by 5%
- Decrease by 5%
- Increase by 10%
- Decrease by 10%
- 1.05
- 0.95
- 1.10
- 0.90
32.Order these steps to solve: A town's population is 15,000 and increases by 4% each year. Find the population after 3 years, to the nearest whole number.
Hard
- Identify P = 15000, r = 0.04, t = 3
- Calculate the multiplier 1 + r = 1.04
- Raise the multiplier to the power t: 1.04³
- Multiply P by the result: 15000 × 1.04³
- Round to the nearest whole number
33.A cube has side length 5 cm. Each side is increased by 20%. What is the percentage increase in the volume of the cube?
Hard
- A72.8%
- B60%
- C20%
- D80%
34.In the formula A = P\left(1 + \frac{r}{n}\right)^nt, which letter represents the original principal sum?
Easy
- AP
- BA
- Cr
- Dt
35.Which value of n should be used in the compound interest formula when interest is compounded weekly?
Easy
- A52
- B12
- C7
- D365
36.Compound interest is interest accumulated from a principal sum and previously accumulated interest.
Easy
True or false?
37.In the compound interest formula, the compounding frequency n is the number of times interest is added each year.
Easy
True or false?
38.Which of the following are valid compounding frequencies for compound interest? (Select all that apply.)
Medium
- AYearly
- BHalf-yearly
- CQuarterly
- DMonthly
- ENever
39.Order these values of n from smallest to largest for the compound interest formula.
Medium
- 12 (monthly)
- 1 (annually)
- 365 (daily)
- 4 (quarterly)
- 52 (weekly)
40.Which of the following statements about simple interest and compound interest are true? (Select all that apply.)
Medium
- ASimple interest is calculated only on the principal sum.
- BCompound interest is calculated on the principal sum plus previously accumulated interest.
- CSimple interest always earns more than compound interest over the same time period.
- DCompound interest depends on the compounding frequency.
- ESimple interest depends on the compounding frequency.
41.A population of 2000 increases by 5% each year. What is the population after 2 years?
Medium
42.£1500 is invested at 3% per year compound interest. How much interest is earned after 4 years? Give your answer to the nearest penny.
Medium
43.Complete the sentence about compound interest.
Easy
In the formula A = P(1 + r)^t, the letter ____ represents the original principal sum.
44.A town's population is 8000. It decreases by 2% each year. What is the population after 3 years? Give your answer to the nearest whole number.
Medium
45.Complete the sentence about the compound interest formula.
Easy
In the formula A = P(1 + \frac{r}{n})^nt, the letter ____ represents the original principal sum, and the letter ____ represents the nominal annual interest rate.
46.A town has a population of 5000. It increases by 3% each year. What is the population after 1 year?
Medium
47.£1200 is invested at 4% per year compound interest. What is the amount after 2 years? Give your answer to the nearest penny.
Medium
48.£900 is invested at 5% per year compound interest. What is the amount after 2 years? Give your answer to the nearest penny.
Medium
49.A car is bought for £18,000 and depreciates by 20% each year. What is its value after 2 years? Give your answer to the nearest pound.
Medium
50.£700 is invested at a nominal annual rate of 4% compounded semi-annually. What is the amount after 3 years? Give your answer to the nearest penny.
Medium
51.A town's population is 12,000 and decreases by 3% each year. What is the population after 4 years? Give your answer to the nearest whole number.
Medium
52.A radioactive substance has a mass of 200 g and decays by 20% each hour. What is its mass after 3 hours? Give your answer to the nearest gram.
Medium
53.Explain the difference between simple interest and compound interest.
Hard
54.Explain why compound interest gives a larger final amount than simple interest over the same time period at the same rate.
Hard
55.Explain why a 10% increase followed by a 10% decrease does not return an amount to its original value.
Hard
56.Explain how you would use iteration to find the number of years for an investment of £4000 at 3% per year compound interest to first exceed £5000.
Hard
57.A rectangular field measures 40 m by 30 m. The length is increased by 10% and the width is decreased by 10%. Explain how the area changes and calculate the new area.
Hard
58.A population of 2000 increases by 5% each year. What is the population after 2 years?
Medium
- A2647
- B2205
- C1984
- D2426
59.£1500 is invested at 3% per year compound interest. How much interest is earned after 4 years? Give your answer to the nearest penny.
Medium
- A169.43 £
- B225.92 £
- C188.26 £
- D207.09 £
60.A town's population is 8000. It decreases by 2% each year. What is the population after 3 years? Give your answer to the nearest whole number.
Medium
- A6776 people
- B9035 people
- C7529 people
- D8282 people
61.A town has a population of 5000. It increases by 3% each year. What is the population after 1 year?
Medium
- A5150 people
- B4635 people
- C5665 people
- D6180 people
62.£1200 is invested at 4% per year compound interest. What is the amount after 2 years? Give your answer to the nearest penny.
Medium
- A1168.13 £
- B1557.5 £
- C1297.92 £
- D1427.71 £
63.£900 is invested at 5% per year compound interest. What is the amount after 2 years? Give your answer to the nearest penny.
Medium
- A1190.71 £
- B1091.48 £
- C992.25 £
- D893.02 £
64.A car is bought for £18,000 and depreciates by 20% each year. What is its value after 2 years? Give your answer to the nearest pound.
Medium
- A13824 £
- B12672 £
- C10368 £
- D11520 £
65.£700 is invested at a nominal annual rate of 4% compounded semi-annually. What is the amount after 3 years? Give your answer to the nearest penny.
Medium
- A709.24 £
- B788.05 £
- C866.86 £
- D945.67 £
66.A town's population is 12,000 and decreases by 3% each year. What is the population after 4 years? Give your answer to the nearest whole number.
Medium
- A11692 people
- B12755 people
- C9566 people
- D10629 people
67.A radioactive substance has a mass of 200 g and decays by 20% each hour. What is its mass after 3 hours? Give your answer to the nearest gram.
Medium
- A102 g
- B112 g
- C92 g
- D122 g

Resumos de revisão

## Growth, decay and compound interest **Compound interest** is interest earned on both the original **principal** and on interest already added. **Simple interest** pays interest only on the original principal each period. ### The multiplier method For a percentage change of r% per period, the **multiplier** is: - increase: 1 + r/100 - decrease (decay/depreciation): 1 − r/100 Repeated over t periods, raise the multiplier to the power t: \(A = P\left(1 + \frac{r}{n}\right)^{tn}\) where A = final amount, P = original principal sum, r = nominal annual interest rate, n = compounding frequency (1: annually, 12: monthly, 52: weekly, 365: daily), t = overall length of time (same units as r, usually years). Compound interest generated: I = A − P. ### Simple vs compound Simple interest adds the same amount each period; compound interest grows faster because each period's interest is added to the principal. ### Depreciation and decay For a value falling by r% per year, use multiplier (1 − r/100)^t. This models depreciation, population decay and similar decreases. ### Growth For a population or value rising by r% per year, use multiplier (1 + r/100)^t. ### Compounding frequency - Yearly: n = 1 - Half-yearly: n = 2 - Quarterly: n = 4 - Monthly: n = 12 - Weekly: n = 52 - Daily: n = 365 More frequent compounding gives a higher final amount for the same nominal rate. ### Annual equivalent rate (AER) The effective annual rate is the total accumulated interest payable by the end of one year divided by the principal sum. It lets you compare products fairly. ### Finding how many years (iteration) To find how long it takes to reach a target value, try values of t one at a time until the amount passes the target. This is **iteration**. ### Rule of 72 To estimate the number of years for an investment to double, divide 72 by the interest rate (as a number, e.g. 6 for 6%). ### Quick checks - Interest rate as a decimal: r/100. - Multiplier for 5% increase: 1.05. - Multiplier for 5% decrease: 0.95. - After t years at 5% compound: multiply by 1.05^t.

Compound interest: Interest accumulated from a principal sum and previously accumulated interest; the result of reinvesting or retaining interest that would otherwise be paid out.
Simple interest: Interest where previously accumulated interest is not added to the principal amount of the current period.
Principal: The original sum of money (P) on which interest is calculated.
Compounding frequency: The number of times per given unit of time the accumulated interest is capitalized, on a regular basis (n).
Annual equivalent rate (AER): The total accumulated interest that would be payable up to the end of one year, divided by the principal sum; used to compare financial products fairly.
Depreciation: A decrease in value over time, modelled with a multiplier less than 1, e.g. (1 − r/100)^t.
Rule of 72: To find the number of years for an investment at compound interest to double, divide the interest rate into 72.
Iteration: Repeating a calculation with different values (e.g. trying different numbers of years) to find how many years it takes to reach a target value.

Flashcards

What is compound interest?

Interest calculated on the original principal plus any interest already added. Each period's interest is added to the amount before the next period's interest is worked out.

How does simple interest differ from compound interest?

With simple interest, previously accumulated interest is never added to the principal. With compound interest, accumulated interest is added to the principal each period, so later interest is earned on it too.

What does the compounding frequency tell you?

It is the number of times per unit of time that accumulated interest is added to the principal. It can be yearly, half-yearly, quarterly, monthly, weekly or daily.

For monthly compounding with an annual rate, what is the compounding frequency?

n = 12, and time is measured in months.

State the compound interest formula and name each symbol.

A = P\left(1 + \frac{r}{n}\right)^tn where A = final amount, P = original principal, r = nominal annual interest rate, n = compounding frequency, t = time in years.

How do you find the total compound interest earned?

Subtract the principal from the final amount: I = A − P.

What is the multiplier for a 5% increase?

1.05 (since 100% + 5% = 105%, and 105 ÷ 100 = 1.05).

What is the multiplier for a 12% decrease?

0.88 (since 100% − 12% = 88%, and 88 ÷ 100 = 0.88).

How do you calculate repeated percentage change over several years?

Raise the multiplier to the power of the number of years, then multiply by the starting amount. For example, after 3 years at 5%: amount × 1.05³.

What is depreciation, and how is it calculated?

Depreciation is a repeated percentage decrease in value. Use a multiplier below 1 raised to the number of years, e.g. a 15% yearly fall gives value × 0.85^t.

How is compound interest used to model population growth?

Treat the yearly percentage increase as a multiplier above 1 and raise it to the number of years: population × (1 + rate)^t.

How is exponential decay modelled with a multiplier?

Use a multiplier between 0 and 1 raised to the power of the number of periods, e.g. a 20% yearly decrease gives amount × 0.8^t.

How can iteration find how many years it takes to reach a target value?

Apply the multiplier year by year, recording the amount after each year, until the amount first reaches or passes the target. The number of steps is the number of years.

What is the annual equivalent rate (AER)?

The total interest payable by the end of one year divided by the principal, expressed as an annual rate. It lets you compare products with different compounding frequencies fairly.

Why is the yearly compounded rate higher than a disclosed rate paid twice yearly?

Interest paid after six months is added to the principal, so the second half-year's interest is earned on a larger amount.

State the Rule of 72.

To estimate the number of years for an investment at compound interest to double, divide the interest rate into 72.

Which mathematician discovered the constant e while studying compound interest, and when?

Jacob Bernoulli, in 1683.

Who wrote Arithmeticall Questions (1613), a landmark book devoted to compound interest?

Richard Witt, a London mathematical practitioner. The book contained 124 worked examples and tables based on 10% interest.

What is continuous compounding?

Compounding at ever-increasing frequency until a limit is approached, so interest is added continuously rather than at set intervals. It is often used in valuing financial derivatives.

An amount of £500 grows by 4% each year. Write the expression for its value after 6 years.

500 × 1.04⁶.

A car worth £12000 depreciates by 18% each year. Write the expression for its value after 4 years.

12000 × 0.82⁴.

A population of 2500 increases by 3% per year. Write the expression for the population after 10 years.

2500 × 1.03¹⁰.

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