How to Solve: Compound Interest

How to Solve: Compound Interest

A Complete Guide for Students & Teachers

Introduction

"If you want to know how much your savings will grow—or how much that loan will cost you—you need to master compound interest. It’s on every major exam, and it’s how banks, investments, and even credit cards work in real life."

What You Need To Know First

Before tackling compound interest, you must understand: 1. Simple Interest – Interest calculated only on the original amount (principal). 2. Exponents – How to calculate numbers raised to a power (e.g., (2^3 = 8)). 3. Percentage Increase/Decrease – How to apply a percentage change to a value.

If you’re shaky on any of these, review them first.

Key Vocabulary

Formulas To Know

1. Compound Interest Formula (Most Common)

Formula: [ A = P \left(1 + \frac{r}{n}\right)^{n \times t} ]

Variables: - (A) = Final amount (including interest) - (P) = Principal (starting amount) - (r) = Annual interest rate (in decimal form, e.g., 5% = 0.05) - (n) = Number of times interest is compounded per year - (t) = Time in years

MEMORISE THIS – It’s the most important formula for compound interest.

2. Compound Interest Only (Not Given on Exam Sheet)

Formula: [ \text{Compound Interest} = A - P ] (Subtract the principal from the final amount to find just the interest earned.)

Example: If (A = \$1,157.63) and (P = \$1,000), then: [ \text{Compound Interest} = 1,157.63 - 1,000 = \$157.63 ]

3. Continuous Compounding (Rare, but on some exams)

Formula: [ A = P \times e^{r \times t} ] (Where (e \approx 2.71828) – usually given on the exam sheet.)

When to use it: Only if the question says "compounded continuously."

Step-by-Step Method

Step 1: Identify the Given Values

Read the problem carefully and write down: - (P) (Principal) - (r) (Annual interest rate – convert to decimal if needed) - (n) (Compounding periods per year) - (t) (Time in years)

Example Problem: "You invest $2,000 at an annual interest rate of 6%, compounded quarterly, for 5 years. What is the final amount?"

Values: - (P = \$2,000) - (r = 6\% = 0.06) - (n = 4) (quarterly = 4 times per year) - (t = 5) years

Step 2: Plug Values into the Formula

Use: [ A = P \left(1 + \frac{r}{n}\right)^{n \times t} ]

Substitute: [ A = 2,000 \left(1 + \frac{0.06}{4}\right)^{4 \times 5} ]

Step 3: Simplify Inside the Parentheses

Calculate (\frac{r}{n}) first: [ \frac{0.06}{4} = 0.015 ]

Now add 1: [ 1 + 0.015 = 1.015 ]

Step 4: Calculate the Exponent

Multiply (n \times t): [ 4 \times 5 = 20 ]

Now the formula looks like: [ A = 2,000 \times (1.015)^{20} ]

Step 5: Compute the Exponent

Use a calculator to find ((1.015)^{20}): [ (1.015)^{20} \approx 1.346855 ]

Step 6: Multiply to Find the Final Amount

[ A = 2,000 \times 1.346855 \approx 2,693.71 ]

Answer: The final amount is \$2,693.71.

Step 7: Find the Interest Earned (If Needed)

[ \text{Compound Interest} = A - P = 2,693.71 - 2,000 = \$693.71 ]

Worked Examples

Example 1 – Basic (Annual Compounding)

Problem: "You deposit $5,000 in a bank at 4% interest compounded annually for 3 years. What is the final amount?"

Solution: 1. Identify values:
- (P = \$5,000)
- (r = 4\% = 0.04)
- (n = 1) (annually)
- (t = 3) years

1. Plug into formula:
   [ A = 5,000 \left(1 + \frac{0.04}{1}\right)^{1 \times 3} ]
   [ A = 5,000 (1.04)^3 ]

2. Calculate exponent:
   [ (1.04)^3 = 1.124864 ]

3. Multiply:
   [ A = 5,000 \times 1.124864 = 5,624.32 ]

Answer: The final amount is \$5,624.32.

What we did and why: We used the compound interest formula with annual compounding ((n=1)). The key was converting the percentage to a decimal and correctly calculating the exponent.

Example 2 – Medium (Quarterly Compounding)

Problem: "A loan of $10,000 is taken at 8% interest compounded quarterly for 2 years. How much interest is paid?"

Solution: 1. Identify values:
- (P = \$10,000)
- (r = 8\% = 0.08)
- (n = 4) (quarterly)
- (t = 2) years

1. Plug into formula:
   [ A = 10,000 \left(1 + \frac{0.08}{4}\right)^{4 \times 2} ]
   [ A = 10,000 (1.02)^8 ]

2. Calculate exponent:
   [ (1.02)^8 \approx 1.171659 ]

3. Multiply:
   [ A = 10,000 \times 1.171659 = 11,716.59 ]

4. Find interest:
   [ \text{Interest} = A - P = 11,716.59 - 10,000 = 1,716.59 ]

Answer: The interest paid is \$1,716.59.

What we did and why: We adjusted for quarterly compounding ((n=4)) and calculated the exponent carefully. Then, we subtracted the principal to find just the interest.

Example 3 – Exam Style (Disguised Problem)

Problem: "Sarah invests $3,000 in an account that earns 5% interest per year, compounded monthly. After how many years will her investment grow to $4,000?"

Solution: 1. Identify values:
- (P = \$3,000)
- (A = \$4,000)
- (r = 5\% = 0.05)
- (n = 12) (monthly)

1. Plug into formula:
   [ 4,000 = 3,000 \left(1 + \frac{0.05}{12}\right)^{12 \times t} ]

2. Simplify inside parentheses:
   [ 1 + \frac{0.05}{12} = 1.0041667 ]

3. Divide both sides by 3,000:
   [ \frac{4,000}{3,000} = (1.0041667)^{12t} ]
   [ 1.3333 = (1.0041667)^{12t} ]

4. Take the natural logarithm (ln) of both sides:
   [ \ln(1.3333) = 12t \times \ln(1.0041667) ]

5. Calculate logs:
   [ \ln(1.3333) \approx 0.287682 ]
   [ \ln(1.0041667) \approx 0.004158 ]

6. Solve for (t):
   [ 0.287682 = 12t \times 0.004158 ]
   [ t = \frac{0.287682}{12 \times 0.004158} \approx 5.76 \text{ years} ]

Answer: It will take approximately 5.76 years (or 5 years and 9 months).

What we did and why: This was a "solve for time" problem, which required logarithms. We isolated the exponent, took logs, and solved for (t). This is a common exam trick—don’t panic if the variable is in the exponent!

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in. Compound interest means interest on interest. The formula you must memorise is: [ A = P \left(1 + \frac{r}{n}\right)^{n \times t} ] Where: - (P) = starting money - (r) = annual rate (as a decimal) - (n) = how often it compounds per year - (t) = time in years

Steps: 1. Write down all given values. 2. Plug them into the formula. 3. Simplify inside the parentheses first. 4. Calculate the exponent. 5. Multiply to find (A). 6. Subtract (P) if you need just the interest.

Watch out for: - Forgetting to convert % to decimal (5% = 0.05). - Mixing up (n) and (t) (compounding periods vs. years). - Rounding too early (keep decimals until the end).

If the question asks for time, use logarithms. If it says ‘compounded continuously,’ switch to (A = Pe^{rt}).

You’ve got this. Now go practice—exams love compound interest!