Practice Maths

L12 — Simple and Compound Interest

Key Terms

Principal (P)
The initial sum of money invested or borrowed.
Interest rate (r)
The annual rate as a decimal — always divide the percentage by 100 before substituting.
Simple interest
Interest calculated only on the original principal each period; produces linear growth. I = Prn.
Compound interest
Interest calculated on the accumulated total (principal + prior interest); produces exponential growth. A = P(1 + r)n.
Compounding periods (k)
Number of times interest is applied per year. Period rate = r/k; total periods = kn. Formula: A = P(1 + r/k)kn.
Amount (A)
The total value after n years: A = principal + all interest earned.

Two Types of Interest

TypeFormulaGrowthKey feature
Simple Interest I = Prn  |  A = P(1 + rn) Linear Interest calculated on principal only
Compound Interest A = P(1 + r)n Exponential Interest earns interest
Compound (k per year) A = P(1 + r/k)kn Exponential k = compounding periods per year
  • P = principal (initial amount), r = annual interest rate as a decimal, n = number of years
  • I = interest earned, A = total amount (principal + interest)
  • Compound interest always overtakes simple interest over a long enough time period.
Simple Compound $1000 $1500 $2000 $2500 0 2 4 6 8 10 Years (n) Amount (A) P = $1000, r = 10% p.a.
Simple interest grows in a straight line; compound interest curves upward exponentially.
Hot Tip: Always convert the interest rate to a decimal before substituting: 6% → r = 0.06. Using r = 6 gives an answer 100 times too large — the most common error in financial maths.

Simple Interest

Simple interest is calculated only on the principal each period. The interest amount is the same every year, so the total grows in a straight line.

I = Prn    and    A = P(1 + rn) = P + I

Worked Example 1 — Simple Interest

Find the interest earned and total amount for $2500 at 4% p.a. simple interest over 6 years.

Step 1: Identify P = 2500, r = 0.04, n = 6.

Step 2: I = Prn = 2500 × 0.04 × 6 = $600

Step 3: A = P + I = 2500 + 600 = $3100

Alternatively: A = P(1 + rn) = 2500 × (1 + 0.04 × 6) = 2500 × 1.24 = $3100 ✓

Compound Interest

Compound interest is calculated on the total amount (principal + accumulated interest) each period. Interest earns interest, causing exponential growth.

A = P(1 + r)n

Worked Example 2 — Compound Interest

Find the total amount for $2500 at 4% p.a. compound interest over 6 years.

Step 1: P = 2500, r = 0.04, n = 6.

Step 2: A = 2500 × (1.04)6 = 2500 × 1.26532 ≈ $3163.30

Compare with simple interest: $3163.30 > $3100. Compound earns $63.30 more over 6 years.

Multiple Compounding Periods

When interest compounds more frequently than once per year:

A = P(1 + r/k)kn

where k = number of compounding periods per year.

CompoundingkPeriod rate
Annually1r
Semi-annually2r/2
Quarterly4r/4
Monthly12r/12
Daily365r/365

Worked Example 3 — Monthly Compounding

Find the total amount for $2500 at 6% p.a. compounded monthly for 3 years.

Step 1: P = 2500, r = 0.06, k = 12, n = 3.

Step 2: A = 2500 × (1 + 0.06/12)12×3 = 2500 × (1.005)36

Step 3: (1.005)36 ≈ 1.19668, so A ≈ 2500 × 1.19668 = $2991.70

Finding P, r, or n

Rearrange the simple interest formula to find any unknown:

  • Find P: P = I ÷ (rn)  or  P = A ÷ (1 + rn)
  • Find r: r = I ÷ (Pn)
  • Find n: n = I ÷ (Pr)

  1. Calculate simple interest. Fluency

    For each investment, find the interest earned (I) and the total amount (A).

    • (a) P = $500, r = 4% p.a., n = 3 years.
    • (b) P = $2000, r = 6% p.a., n = 5 years.
    • (c) P = $800, r = 2.5% p.a., n = 4 years.
    • (d) P = $1500, r = 8% p.a., n = 2.5 years.

  2. Calculate compound interest (annual compounding). Fluency

    Find the total amount A = P(1 + r)n for each investment.

    • (a) P = $1000, r = 5% p.a., n = 4 years.
    • (b) P = $3000, r = 3% p.a., n = 10 years.
    • (c) P = $500, r = 8% p.a., n = 6 years.
    • (d) P = $2000, r = 4% p.a., n = 3 years.

  3. Find the unknown (simple interest). Fluency

    • (a) I = $200, P = $1000, r = 5% p.a. Find n.
    • (b) I = $240, P = $1600, n = 3 years. Find r (as % p.a.).
    • (c) A = $1400, P = $1000, r = 5% p.a. Find n.
    • (d) A = $1500, r = 10% p.a., n = 5 years. Find P.

  4. Compound interest with multiple compounding periods. Fluency

    Use A = P(1 + r/k)kn. Give answers to the nearest cent.

    • (a) P = $1000, r = 6% p.a., n = 2 years, monthly compounding (k = 12).
    • (b) P = $5000, r = 4% p.a., n = 3 years, quarterly compounding (k = 4).
    • (c) P = $2000, r = 8% p.a., n = 5 years, semi-annual compounding (k = 2).
    • (d) P = $3000, r = 12% p.a., n = 1 year, monthly compounding (k = 12).

  5. Read from the graph: simple vs compound interest. Understanding

    The graph below shows two investments of $1000. Account A earns 12% p.a. simple interest (blue line). Account B earns 10% p.a. compound interest (orange curve).

    A B $1000 $1500 $2000 $2500 0 2 4 6 8 10 Years (n) Amount ($)
    • (a) At year 3, which account has more money? By approximately how much?
    • (b) Between which two years does Account B (compound) overtake Account A (simple)?
    • (c) Estimate the value of each account at year 7.
    • (d) Explain in one sentence why compound interest eventually wins despite a lower annual rate.

  6. Which account earns more? Understanding

    Zara has $4000 to invest. She compares two options for 5 years:

    • Account A: simple interest at 7% p.a.
    • Account B: compound interest at 5% p.a. (annual).
    • (a) Calculate the final value of Account A.
    • (b) Calculate the final value of Account B.
    • (c) Which account is better after 5 years?
    • (d) By trial and error (testing n = 10, 12, 14, 15), find approximately when Account B overtakes Account A.

  7. Interpret the compound interest formula. Understanding

    A bank offers 6% p.a. interest compounded monthly. Hamish invests $4000 for 2 years.

    • (a) What is the interest rate per month?
    • (b) How many compounding periods occur over 2 years?
    • (c) Write the compound interest formula A = P(1 + r/k)kn with all values substituted.
    • (d) Calculate the final amount.

  8. Find the principal. Understanding

    • (a) An investment earns simple interest at 5% p.a. The interest earned over 4 years is $600. Find P.
    • (b) A compound investment at 10% p.a. grows to $2662 after 3 years. Find P.
    • (c) How much must be invested at 5% p.a. simple interest to have $1300 after 6 years?
    • (d) An investment at 5% p.a. compound interest grows to $1276.28 after 5 years. Find P.

  9. Compare over multiple time periods. Problem Solving

    Lena has $5000 to invest. Option X: simple interest at 8% p.a. Option Y: compound interest at 6% p.a. (annual).

    • (a) Find the value of each option after 3 years.
    • (b) Find the value of each option after 8 years.
    • (c) Find the value of each option after 15 years.
    • (d) By testing n = 10 and n = 11, determine approximately when Option Y overtakes Option X.

  10. Doubling time. Problem Solving

    • (a) By testing successive values, find how many complete years it takes for an investment to double at 6% p.a. compound interest.
    • (b) Repeat for 8% p.a. compound interest. How many fewer years does it take?
    • (c) The Rule of 72 says the doubling time ≈ 72 ÷ rate%. Apply this to 6% and 8%. How accurate is it?
    • (d) What annual simple interest rate is needed to double a principal in exactly 10 years?
See Answers ➔