Simple and Compound Interest

CSEC Mathematics: Consumer Arithmetic

Essential Understanding: Interest is the cost of borrowing money or the reward for saving it. Understanding how simple and compound interest work is crucial for making smart financial decisions about loans, savings, and investments.

Key Skill: Calculating Interest
Exam Focus: Formula Application
Real World: Loans & Savings

Key Terminology

P

Principal (P)

The original amount of money borrowed or invested before any interest is added.

Example: If you deposit $5,000 in a bank, the principal is $5,000.

R

Rate (R)

The percentage charged or earned per time period (usually per year/annum).

Example: A rate of 5% per annum means 5% interest each year.

T

Time (T)

The duration for which the money is borrowed or invested (usually in years).

Example: A 3-year loan has T = 3.

Simple Interest

Simple interest is calculated only on the original principal. The interest earned or paid remains the same each year.

Simple Interest Formula

\[ I = \frac{P \times R \times T}{100} \]

Where:

  • \( I \) = Interest earned/paid
  • \( P \) = Principal (original amount)
  • \( R \) = Rate (percentage per annum)
  • \( T \) = Time (in years)

Total Amount (Simple Interest)

\[ A = P + I \quad \text{or} \quad A = P\left(1 + \frac{RT}{100}\right) \]

\( A \) = Final amount (Principal + Interest)

1

Worked Example: Simple Interest

Problem: Maria invests $8,000 at a simple interest rate of 6% per annum for 4 years. Calculate:

(a) The simple interest earned

(b) The total amount after 4 years

1
Identify the values: \( P = \$8000 \), \( R = 6\% \), \( T = 4 \) years
2
(a) Calculate Interest: \[ I = \frac{P \times R \times T}{100} = \frac{8000 \times 6 \times 4}{100} = \frac{192000}{100} = \$1920 \]
3
(b) Calculate Total Amount: \[ A = P + I = 8000 + 1920 = \$9920 \]

Answer: (a) Interest = $1,920   (b) Total Amount = $9,920

Compound Interest

Compound interest is calculated on the principal plus any previously accumulated interest. This means you earn "interest on interest," making your money grow faster over time.

Compound Interest Formula

\[ A = P\left(1 + \frac{R}{100}\right)^n \]

Where:

  • \( A \) = Final amount
  • \( P \) = Principal
  • \( R \) = Rate per period (%)
  • \( n \) = Number of compounding periods

Compound Interest Earned

\[ \text{Compound Interest} = A - P = P\left(1 + \frac{R}{100}\right)^n - P \]
2

Worked Example: Compound Interest

Problem: John deposits $5,000 in a savings account that pays 8% per annum compound interest. Calculate the amount in the account after 3 years.

1
Identify the values: \( P = \$5000 \), \( R = 8\% \), \( n = 3 \) years
2
Apply the formula: \[ A = P\left(1 + \frac{R}{100}\right)^n = 5000\left(1 + \frac{8}{100}\right)^3 \]
3
Simplify: \[ A = 5000(1.08)^3 = 5000 \times 1.259712 = \$6298.56 \]

Answer: Amount after 3 years = $6,298.56

Compound Interest earned: \( \$6298.56 - \$5000 = \$1298.56 \)

Year-by-Year Method (Alternative Approach)

For CSEC exams, you can also calculate compound interest year by year:

3

Year-by-Year Calculation

Problem: Calculate compound interest on $5,000 at 8% for 3 years (same as Example 2).

Year 1

Principal: $5,000

Interest: \( 5000 \times 0.08 = \$400 \)

Amount at end of Year 1: $5,400

Year 2

Principal: $5,400

Interest: \( 5400 \times 0.08 = \$432 \)

Amount at end of Year 2: $5,832

Year 3

Principal: $5,832

Interest: \( 5832 \times 0.08 = \$466.56 \)

Amount at end of Year 3: $6,298.56

Total Compound Interest: $400 + $432 + $466.56 = $1,298.56

Growth Comparison Visualizer

Adjust the sliders to see how Compound Interest pulls away from Simple Interest over time.

Simple Interest: $0.00
Simple Total: $0.00
Compound Interest: $0.00
Compound Total: $0.00
Difference: $0.00
Aspect Simple Interest Compound Interest
Calculation Base Original principal only Principal + accumulated interest
Interest Each Period Same amount every period Increases each period
Growth Pattern Linear (straight line) Exponential (curved)
Total Interest Lower over time Higher over time
Formula \( I = \frac{PRT}{100} \) \( A = P(1 + \frac{R}{100})^n \)
Best For Short-term borrowing Long-term saving

Interactive Interest Calculator

Calculate Your Interest

Results:

Simple Interest: $150.00

Simple Interest Total: $1,150.00


Compound Interest: $157.63

Compound Interest Total: $1,157.63


Difference (Compound - Simple): $7.63

Appreciation and Depreciation

The compound interest formula can also be used for:

+

Appreciation

When the value of an asset increases over time (e.g., property, antiques).

\[ A = P\left(1 + \frac{R}{100}\right)^n \]

Same formula as compound interest!

-

Depreciation

When the value of an asset decreases over time (e.g., cars, electronics).

\[ A = P\left(1 - \frac{R}{100}\right)^n \]

Note the minus sign!

4

Worked Example: Depreciation

Problem: A car was purchased for $25,000. If it depreciates at 15% per annum, find its value after 3 years.

1
Identify values: \( P = \$25000 \), \( R = 15\% \), \( n = 3 \) years
2
Apply depreciation formula: \[ A = P\left(1 - \frac{R}{100}\right)^n = 25000\left(1 - \frac{15}{100}\right)^3 \]
3
Calculate: \[ A = 25000(0.85)^3 = 25000 \times 0.614125 = \$15,353.13 \]

Answer: The car is worth $15,353.13 after 3 years.

Total Depreciation: $25,000 - $15,353.13 = $9,646.87

Remember the Signs!

Appreciation (value goes UP): Use + in the formula: \( (1 + \frac{R}{100}) \)

Depreciation (value goes DOWN): Use - in the formula: \( (1 - \frac{R}{100}) \)

Memory Tip: "De-preciation" has "De-" meaning decrease, so use the minus sign!

Past Paper Style Questions

CSEC-Style Question 1

A woman deposited $12,000 in a bank which offers 5% per annum simple interest.

(a) Calculate the interest earned after 4 years. [2 marks]

(b) Calculate the total amount in her account after 4 years. [1 mark]

(c) How long would it take for the interest to equal the principal? [2 marks]

Solutions:

(a) \( I = \frac{PRT}{100} = \frac{12000 \times 5 \times 4}{100} = \frac{240000}{100} = \$2400 \)

(b) \( A = P + I = 12000 + 2400 = \$14,400 \)

(c) For interest to equal principal: \( I = P = \$12000 \)

\( 12000 = \frac{12000 \times 5 \times T}{100} \)

\( 12000 = 600T \)

\( T = 20 \) years

CSEC-Style Question 2

Kevin invested $50,000 at a rate of 6% per annum compound interest.

(a) Calculate the value of the investment at the end of the first year. [2 marks]

(b) Calculate the value of the investment at the end of 3 years. [2 marks]

(c) Calculate the compound interest earned over the 3 years. [1 mark]

Solutions:

(a) After 1 year: \( A = 50000 \times 1.06 = \$53,000 \)

(b) After 3 years: \( A = 50000(1.06)^3 = 50000 \times 1.191016 = \$59,550.80 \)

(c) Compound Interest: \( 59550.80 - 50000 = \$9,550.80 \)

CSEC-Style Question 3

A machine costing $80,000 depreciates at 12% per annum.

(a) Calculate the value of the machine after 2 years. [3 marks]

(b) Calculate the total depreciation over the 2 years. [1 mark]

Solutions:

(a) Using depreciation formula:

\( A = P(1 - \frac{R}{100})^n = 80000(1 - 0.12)^2 = 80000(0.88)^2 \)

\( A = 80000 \times 0.7744 = \$61,952 \)

(b) Total depreciation: \( 80000 - 61952 = \$18,048 \)

CSEC Practice Arena

Test Your Understanding

1
Calculate the simple interest on $2,000 at 4% per annum for 3 years.
$200
$240
$24
$2,400
Solution: \( I = \frac{PRT}{100} = \frac{2000 \times 4 \times 3}{100} = \frac{24000}{100} = \$240 \)
2
$10,000 is invested at 10% per annum compound interest. What is the amount after 2 years?
$12,000
$12,100
$11,000
$21,000
Solution: \( A = P(1 + \frac{R}{100})^n = 10000(1.10)^2 = 10000 \times 1.21 = \$12,100 \)
3
Which formula is used to calculate depreciation?
\( A = P(1 + \frac{R}{100})^n \)
\( A = P(1 - \frac{R}{100})^n \)
\( I = \frac{PRT}{100} \)
\( A = P + I \)
Explanation: Depreciation means the value decreases, so we subtract the rate. The formula is \( A = P(1 - \frac{R}{100})^n \).
4
A laptop worth $1,500 depreciates by 20% in its first year. What is its value after one year?
$300
$1,800
$1,200
$1,500
Solution: \( A = 1500 \times (1 - 0.20) = 1500 \times 0.80 = \$1,200 \)
5
If $5,000 is invested for 2 years at 8% per annum, what is the difference between compound interest and simple interest?
$800
$64
$32
$0
Solution:
Simple Interest: \( I = \frac{5000 \times 8 \times 2}{100} = \$800 \)
Compound Interest: \( A = 5000(1.08)^2 = 5000 \times 1.1664 = \$5832 \)
CI = $5832 - $5000 = $832
Difference: $832 - $800 = $32
Target

CSEC Examination Tips

  • Read carefully: Check if the question asks for simple or compound interest - they give different answers!
  • Show your formula: Always write down the formula you're using before substituting values.
  • Units matter: Make sure time is in years. Convert months to years if needed (e.g., 6 months = 0.5 years).
  • Calculator tip: For compound interest, use the power button (^ or \( y^x \)) on your calculator.
  • Check reasonableness: Compound interest should always be slightly more than simple interest for the same conditions.
  • Depreciation: Don't forget to use subtraction in the formula!

Summary: Key Formulas

Simple Interest

  • \( I = \frac{PRT}{100} \)
  • \( A = P + I \)
  • Interest is constant each year

Compound Interest

  • \( A = P(1 + \frac{R}{100})^n \)
  • Interest on interest
  • Growth is exponential

Depreciation

  • \( A = P(1 - \frac{R}{100})^n \)
  • Value decreases over time
  • Note the minus sign!
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