Simple And Compound Interest Growth And Decay

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선생님을 위해: Simple And Compound Interest Growth And Decay(Maths [CIE], Extended)을(를) 위한 바로 쓸 수 있는 수업 슬라이드, 복습 노트, 다이어그램 — 수업에 사용하거나, 학생들이 실시간 게임으로 즐기는 인터랙티브 클래스 활동으로 진행하세요.

수업 노트

Simple Interest

Simple interest is calculated only on the original amount (principal), so each interest payment is the same.
Total interest = principal × (rate/100) × 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).

Compound Interest

Compound interest is calculated on the running total, so interest is earned on interest.
Final amount = principal × (1 + 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 + r/100)ⁿ.

Depreciation

Depreciation is a percentage decrease in value over time (e.g., cars, phones).
Final value = principal × (1 - 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 > 1.
Exponential decay: quantity decreases by a fixed percentage each period; multiplier 0 < k < 1.
General model: B = A × kⁿ, 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 = (B/A)^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), decay = 1 - (rate/100).
Round answers as instructed (e.g., nearest hundred, nearest dollar).
For reverse problems, rearrange the formula to solve for the unknown.

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연습 문제

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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.Which multiplier represents a 4% increase?
Easy
- A1.4
- B0.96
- C1.04
- D0.04
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.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.Jan invests $800 at a rate of 3% per year simple interest. Calculate the value of her investment at the end of 4 years.
Easy
- A$896
- B$824
- C$96
- D$800
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£8000
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, correct to the nearest $100.
Hard
- A$6300
- B$6500
- C$6200
- D$6000
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.
Medium
- A229 500
- B229 400
- C230 000
- D228 800

더 많은 문제

9.Dean invests $500 for 10 years at a rate of 1.7% per year simple interest. Calculate the total interest earned.
Easy
- A$85
- B$8.50
- C$850
- D$58.50
10.Gretal invests $5000 at a rate of 2% per year compound interest. Calculate the value of her investment at the end of 3 years.
Medium
- A$5306.04
- B$5300
- C$5306
- D$5306.40
11.The population of a city is increasing exponentially at a rate of 2% each year. The population now is 256 000. Calculate the population after 30 years, correct to the nearest thousand.
Medium
- A464 000
- B463 000
- C462 000
- D465 000
12.Marlene invests $550 at a rate of 1.9% per year compound interest. Calculate the value of the investment at the end of 10 years.
Medium
- A$663.90
- B$654.50
- C$651.52
- D$663.15
13.Which of the following is an example of exponential decay?
Easy
- APopulation increasing by 2% each year
- BCar value decreasing by 15% each year
- CInvestment earning 5% simple interest
- DTemperature rising by 1°C each hour
14.Roland invests $12 000 in an account that pays compound interest at a rate of 2.2% per year. Calculate the value of his investment at the end of 6 years, correct to the nearest dollar.
Medium
- A$13 674
- B$13 677
- C$13 670
- D$13 700
15.Sara invests $500 for 15 years at a rate of 2% per year simple interest. Calculate the total interest Sara receives.
Easy
- A$150
- B$15
- C$300
- D$165
16.Newton has a population of 23 000. The population decreases exponentially at a rate of 1.4% per year. Calculate the population of Newton after 5 years, correct to the nearest hundred.
Medium
- A21 400
- B21 500
- C21 300
- D21 600
17.A town has a population of 45 000. This population increases exponentially at a rate of 1.6% per year. Find the population of the town at the end of 5 years. Give your answer correct to the nearest hundred.
Medium
- A48 700
- B48 600
- C48 800
- D49 000
18.When Heidi was born, her grandfather invested some money in an account that paid compound interest. The graph shows exponential growth. Use the graph to find the original amount of money invested.
Medium
- A2000
- B2500
- C1500
- D3000
19.Use the graph to find the number of years it took for the original amount to double.
Medium
- A14
- B12
- C10
- D16
20.There are 30 000 lions in Africa. The number of lions in Africa decreases exponentially by 2% each year. Find the number of lions in Africa after 6 years. Give your answer correct to the nearest hundred.
Medium
- A26 600
- B26 500
- C26 700
- D26 400
21.Viv wants to invest £2000 for 2 years. Bank A offers 3% per year simple interest. Bank B offers 2.8% per year compound interest. Which bank gives more money at the end of 2 years?
Hard
- ABank A
- BBank B
- CBoth same
- DCannot determine
22.Jack has £15 000 to invest for 3 years. Account X pays 3.2% per year simple interest. Account Y pays 3% per year compound interest. Which account should he choose to have the most money?
Hard
- AAccount X
- BAccount Y
- CBoth equal
- DNeither
23.Ian invested an amount of money at 3% per annum compound interest. At the end of 2 years the value of the investment was £2652.25. Work out the amount of money Ian invested.
Medium
- A£2500
- B£2550
- C£2600
- D£2450
24.Liam invests £6200 for 3 years in a savings account. He gets 2.5% per annum compound interest. How much money will Liam have at the end of 3 years?
Medium
- A£6676.72
- B£6665.00
- C£6680.00
- D£6670.00
25.Katy invests £200 000 in a savings account for 4 years. The account pays compound interest at a rate of 1.5% per annum. Calculate the total amount of interest Katy will get at the end of 4 years.
Medium
- A£12 272
- B£12 000
- C£12 500
- D£12 272.71
26.Boris invests $550 at a rate of 2% per year simple interest. Calculate the value of the investment at the end of 10 years.
Easy
- A$660
- B$650
- C$670
- D$640
27.A car costs $20,000 and depreciates at a rate of 15% per year. Which expression gives the value of the car after 3 years?
Medium
- A20000 × 0.85³
- B20000 × 1.15³
- C20000 × 0.15³
- D20000 × 3 × 0.85
28.A population of bacteria doubles every hour. Initially there are 500 bacteria. How many bacteria are there after 4 hours?
Easy
- A8000
- B2000
- C4000
- D16000
29.A house is bought for $250,000. Its value increases exponentially at a rate of 3% per year. After how many complete years will its value first exceed $300,000?
Medium
- A6 years
- B7 years
- C5 years
- D8 years
30.What is the multiplier for a 12% increase?
Easy
- A1.12
- B0.12
- C1.012
- D0.88
31.A radioactive substance decays exponentially. After 10 years, its mass is half of the original. What is the annual decay rate?
Hard
- A6.7%
- B10%
- C5%
- D7.2%
32.A laptop worth $1200 depreciates by 20% each year. At the same time, a tablet worth $800 depreciates by 10% each year. After 2 years, which is worth more?
Medium
- AThe laptop
- BThe tablet
- CThey are equal
- DCannot determine
33.An investment of $5000 grows to $5512.50 after 2 years with simple interest. What is the annual interest rate?
Medium
- A5.125%
- B5%
- C5.25%
- D4.875%
34.A city's population decreases exponentially at a rate of 2% per year. The population is now 500,000. Which expression gives the population 5 years ago?
Medium
- A500000 / 0.98⁵
- B500000 × 0.98⁵
- C500000 / 1.02⁵
- D500000 × 1.02⁵
35.A bank offers 3% per year compound interest. How much would $2000 grow to in 3 years?
Medium
- A$2185.45
- B$2180.00
- C$2185.00
- D$2185.54
36.Which of the following represents exponential decay?
Easy
- Ay = 100 × 0.9^x
- By = 100 × 1.1^x
- Cy = 100 + 10x
- Dy = 100 - 10x
37.A $1000 investment earns 5% per year compound interest for 3 years, then 4% per year for the next 2 years. What is the final value?
Hard
- A$1252.09
- B$1293.67
- C$1276.28
- D$1300.00
38.A machine costs $8000 and depreciates by 12% per year. What is its value after 4 years, correct to the nearest dollar?
Medium
- A$4800.00
- B$4798.00
- C$4812.00
- D$4790.00
39.What is the multiplier for a 6% decrease?
Easy
- A0.94
- B0.06
- C1.06
- D0.6
40.Which of the following is an example of exponential growth?
Easy
- AA car losing value each year
- BA population increasing by a fixed number each year
- CA bank account earning compound interest
- DA radioactive substance decaying
41.Simple interest is calculated on the current balance, including previously earned interest.
Easy
True or false?
42.A house is bought for $250,000. Its value increases exponentially at a rate of 3% per year. Which expression gives its value after 5 years?
Medium
- A250000 × 1.03⁵
- B250000 × 0.97⁵
- C250000 × 5 × 0.03
- D250000 × (1 + 0.03 × 5)
43.Arrange the following investments in order of final value after 3 years, from smallest to largest. (Assume all start with $1000.)
Medium
- Simple interest at 5% per year
- Compound interest at 5% per year
- Depreciation at 5% per year
44.Match each scenario to the correct multiplier.
Easy
- 10% increase
- 10% decrease
- 10% simple interest for 1 year
- 1.10
- 0.90
45.Match each scenario to the correct multiplier for one year.
Easy
- A 5% increase
- A 5% decrease
- A 12% increase
- A 12% decrease
- 0.95
- 1.05
- 1.12
- 0.88
46.Arrange the following investments in order of final value after 3 years, from smallest to largest. (Assume all start with $1000 and each uses the given annual rate.)
Medium
- Simple interest at 5%
- Compound interest at 5%
- Compound interest at 4%
- Simple interest at 6%
47.Which of the following statements about simple and compound interest are true? (Select all that apply.)
Medium
- ASimple interest is calculated on the original amount only.
- BCompound interest is calculated on the current balance including previously earned interest.
- CFor the same principal, rate, and time, simple interest always gives a larger final amount than compound interest.
- DThe multiplier for a 10% increase is 0.10.
48.A 20% decrease followed by a 20% increase returns the original value.
Medium
True or false?
49.Fill in the blank.
Easy
The formula for compound interest is final amount = P × (1 + ____) ^ n, where r is the percentage rate and n is the number of years.
50.A population of 50,000 bacteria doubles every hour. How many bacteria are there after 4 hours?
Easy
51.A car costs $20,000 and depreciates at a rate of 15% per year. Find its value after 3 years. Give your answer to the nearest dollar.
Medium
52.State the multiplier for a 12% increase.
Easy
53.A radioactive substance decays exponentially. After 10 years, its mass is half of the original. Find the annual decay rate (as a percentage, correct to 1 decimal place).
Hard
54.A machine costs $8000 and depreciates at a rate of 12% per year. Find its value after 4 years. Give your answer to the nearest dollar.
Medium
55.Complete the following sentence about exponential growth.
Easy
If a quantity grows exponentially by 8% each year, the multiplier is ____. After 5 years, the original amount is multiplied by ____.
56.A population of 50,000 bacteria doubles every hour. How many bacteria are there after 4 hours?
Easy
- A880000
- B720000
- C800000
- D960000
57.A car costs $20,000 and depreciates at a rate of 15% per year. Find its value after 3 years. Give your answer to the nearest dollar.
Medium
- A11055 $
- B14739 $
- C13511 $
- D12283 $
58.A radioactive substance decays exponentially. After 10 years, its mass is half of the original. Find the annual decay rate (as a percentage, correct to 1 decimal place).
Hard
- A6.7 %
- B7.7 %
- C8.7 %
- D5.7 %

복습 노트

## Simple & Compound Interest, Growth & Decay ### Simple Interest - Interest calculated only on the **original amount** (principal). - Each interest payment is the same. - **Formula:** Total interest = Principal × (rate/100) × time. - Final amount = Principal + Total interest. - Example: $250 at 4% for 6 years → interest = 250 × 0.04 × 6 = $60; total = $310. ### Compound Interest - Interest calculated on the **running total** (principal + accumulated interest). - Use a **multiplier**: for r% increase, multiplier = 1 + r/100. - **Formula:** Final amount = P × (1 + r/100)^n, where P = principal, r = rate (%), n = number of years. - To find original amount (reverse): P = Final amount / (1 + r/100)^n. - Example: $1200 at 4% for 7 years → 1200 × 1.04^7 ≈ $1579. ### Depreciation - Value **decreases** by a fixed percentage each year. - Multiplier = 1 - r/100 (e.g., 15% decrease → 0.85). - **Formula:** Final value = P × (1 - r/100)^n. - Example: £20,000 car depreciates 15% p.a. for 3 years → 20000 × 0.85^3 = £12,282.50. ### Exponential Growth & Decay - General model: B = A × k^n, where A = initial amount, B = new amount, k = multiplier (k>1 for growth, 0<k<1 for decay), n = number of periods. - **Growth:** e.g., population increases 1.6% p.a. → k = 1.016. - **Decay:** e.g., population decreases 1.7% p.a. → k = 0.983. - To find k: k = (B/A)^(1/n). - To find n: use trial and improvement. - Round answers as required (e.g., nearest hundred, nearest thousand). ### Key Points - **Simple vs Compound:** Simple interest uses only original amount; compound uses running total. - **Multiplier method:** Works for both growth and decay. - **Reverse problems:** Divide final amount by multiplier^n to find original. - **Check question:** Are they asking for interest earned or total amount?

Simple Interest: Interest calculated only on the original principal amount; each interest payment is the same.
Compound Interest: Interest calculated on the running total (principal plus accumulated interest); interest earns interest.
Multiplier: A decimal factor used to increase or decrease an amount by a percentage; for growth: 1 + r/100, for decay: 1 - r/100.
Depreciation: The decrease in value of an asset over time, usually calculated as a percentage decrease each year.
Exponential Growth: A quantity increases by a fixed percentage each time period; modelled by B = A × k^n with k > 1.
Exponential Decay: A quantity decreases by a fixed percentage each time period; modelled by B = A × k^n with 0 < k < 1.
Principal: The original amount of money invested or borrowed.
Interest Rate: The percentage at which interest is calculated, usually per year.

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