Practice Maths

Simple Interest

Key Terms

Simple interest
is calculated only on the original principal — it does not compound.
P
= Principal (original amount invested or borrowed).
r
= interest rate per year (as a decimal, e.g. 5% p.a. → r = 0.05).
t
= time in years. Convert months to years: divide by 12. Convert days to years: divide by 365.
I
= interest earned or charged.
A
= total amount (principal + interest).
To find P, r, or t: rearrange I = Prt. e.g. P = I ÷ (rt), r = I ÷ (Pt), t = I ÷ (Pr).
Simple interest grows linearly — equal amounts of interest are added each year.
Key Formulas
• I = Prt
• A = P + I = P(1 + rt)
• P = I ÷ (rt)     r = I ÷ (Pt)     t = I ÷ (Pr)
• Time conversions: t (years) = months ÷ 12 = days ÷ 365
Variable Meaning Rearranged formula
IInterestI = Prt
PPrincipalP = I ÷ (rt)
rAnnual rate (decimal)r = I ÷ (Pt)
tTime (years)t = I ÷ (Pr)
ATotal amountA = P + I

Simple interest: A vs t graph for P = $1 000, r = 8% p.a. (linear growth)

t (yrs) A ($) 1000 1080 1160 1240 1320 1400 1 2 3 4 5 A = 1000 + 80t
Hot Tip Always convert the interest rate to a decimal and the time to years before substituting into I = Prt. A common mistake is to use r = 5 instead of r = 0.05. Also, if time is given in months, divide by 12 before substituting. For example, 18 months = 18/12 = 1.5 years.

Worked Example 1 — Finding simple interest and total amount

Question: Calculate the simple interest and total amount for $3 500 invested at 6% p.a. for 4 years.

Identify: P = $3 500, r = 0.06, t = 4

Interest: I = Prt = 3 500 × 0.06 × 4 = $840

Total amount: A = P + I = $3 500 + $840 = $4 340

Worked Example 2 — Finding the principal

Question: How much must be invested at 4.5% p.a. simple interest for 30 months to earn $270 in interest?

Convert time: t = 30 ÷ 12 = 2.5 years

Rearrange: P = I ÷ (rt) = 270 ÷ (0.045 × 2.5) = 270 ÷ 0.1125 = $2 400

Where Does the Formula Come From?

Simple interest is built on one very intuitive idea: interest is directly proportional to both the amount borrowed and the length of time it is borrowed for. If you borrow twice as much money, you should pay twice as much interest. If you borrow it for twice as long, you should also pay twice as much interest. These two proportional relationships give us:

I ∝ P   and   I ∝ t

Combining them, and including the interest rate r as the constant of proportionality (the "price" per dollar per year), we get I = Prt. This is why the formula looks the way it does — it is not arbitrary, it is a precise description of proportional growth.

Because I grows at a constant rate with respect to t, the graph of the total amount A against time t is a straight line. The gradient of that line equals P × r — the annual interest earned. This is why simple interest is sometimes called linear growth.

Interest Rate: The Decimal Trap

The most common — and most catastrophic — error in simple interest problems is substituting the percentage directly without converting it to a decimal. The rate r in the formula I = Prt must always be expressed as a decimal fraction, never as a percentage number.

For example, an interest rate of 5% per annum means the bank charges 5 cents for every $1 for every year. As a decimal, that is r = 0.05. If you accidentally substitute r = 5, you will calculate 100 times too much interest — a catastrophic error that is unfortunately very easy to make under exam pressure.

The fix is simple: whenever you see a percentage, divide it by 100 before writing it into the formula. If the rate is given as 4.5%, write r = 0.045. If it is 12%, write r = 0.12. Make this a reflex.

Common Mistake — Using r = 5 Instead of r = 0.05: A student calculates interest on $2 000 at 5% p.a. for 3 years as I = 2000 × 5 × 3 = $30 000. The correct answer is I = 2000 × 0.05 × 3 = $300. The error is a factor of 100. Always convert: 5% → 0.05, not 5.

Rearranging the Formula — Finding P, r, or t

You will not always be asked to find the interest. Sometimes you are given the interest earned and need to find the principal, the rate, or the time. All three rearrangements come from the same formula — just divide both sides by the variables you know.

Finding the Principal (P): You know I, r, and t. Divide both sides of I = Prt by rt:

I = Prt   →   P = I ÷ (rt)

Example: I = $450, r = 6% p.a. = 0.06, t = 5 years → P = 450 ÷ (0.06 × 5) = 450 ÷ 0.30 = $1 500

Finding the Rate (r): You know I, P, and t. Divide both sides by Pt:

r = I ÷ (Pt)   → then multiply by 100 to express as a percentage

Example: I = $600, P = $4 000, t = 3 years → r = 600 ÷ (4000 × 3) = 600 ÷ 12000 = 0.05 = 5% p.a.

Finding the Time (t): You know I, P, and r. Divide both sides by Pr:

t = I ÷ (Pr)   → answer is in years

Example: I = $320, P = $2 000, r = 0.08 → t = 320 ÷ (2000 × 0.08) = 320 ÷ 160 = 2 years

Time Must Be in Years

The rate r is an annual rate — it tells you how much interest is earned per year. This means the time t must also be expressed in years for the formula to be consistent. If the time is given in months, divide by 12. If it is given in days, divide by 365.

For example, "18 months" becomes t = 18/12 = 1.5 years. "73 days" becomes t = 73/365 = 0.2 years. Many exam questions deliberately give time in months or days to test whether you remember to convert.

Exam Tip — Time Conversions: Always write your time conversion as a fraction before substituting. Write t = 18/12 = 1.5, not just t = 1.5 out of nowhere. This makes it clear to the marker that you have correctly converted, and it protects you from arithmetic slips.

Simple Interest in the Real World

Simple interest is used in situations where the interest does not need to compound — typically for short-term lending or fixed-term products. Common real-world examples include:

  • Personal loans: Some car loans and small personal loans use a flat (simple) interest structure, making the total cost easy to calculate upfront.
  • Term deposits (short-term): A bank may offer a 6-month term deposit at a fixed simple interest rate, guaranteeing a known return.
  • Hire purchase agreements: The total interest on a purchase paid off over time is sometimes calculated as simple interest on the full purchase price.

For longer-term products like mortgages, superannuation, and savings accounts, banks use compound interest — which is why understanding the difference matters. Simple interest keeps the maths transparent; compound interest better reflects how money actually grows over time.

Mastery Practice

See Answers ➔
  1. Fluency

    Calculate the simple interest for each investment.

    1. (a) P = $2 000, r = 5% p.a., t = 3 years
    2. (b) P = $5 500, r = 4% p.a., t = 2 years
    3. (c) P = $800, r = 7.5% p.a., t = 18 months
    4. (d) P = $12 000, r = 3.2% p.a., t = 6 months
  2. Fluency

    Calculate the total amount (A) after simple interest.

    1. (a) P = $4 000, r = 6% p.a., t = 5 years
    2. (b) P = $750, r = 8% p.a., t = 3 years
    3. (c) P = $1 600, r = 4.5% p.a., t = 2.5 years
  3. Fluency

    Find the missing variable in each case.

    1. (a) Find P if I = $360, r = 6% p.a., t = 3 years
    2. (b) Find r if I = $875, P = $2 500, t = 7 years
    3. (c) Find t if I = $480, P = $3 200, r = 5% p.a.
  4. Fluency

    Time conversions. Express each time in years, then calculate the interest.

    1. (a) P = $900, r = 6% p.a., t = 8 months
    2. (b) P = $2 400, r = 5% p.a., t = 146 days
    3. (c) P = $6 000, r = 3% p.a., t = 2 years 6 months
  5. Understanding

    Working backwards from total amount.

    Note: If you are given A and need I, use I = A − P. If given A and need P, rearrange A = P(1 + rt).
    1. (a) An investment grows to $3 920 after 4 years at 5% p.a. simple interest. Find the principal.
    2. (b) A loan of $5 000 accrues to a total of $6 250 under simple interest over 5 years. Find the annual interest rate.
    3. (c) An amount of $2 000 grows to $2 300 at 6% p.a. simple interest. How long does this take?
  6. Understanding

    Comparing investments.

    1. (a) Bank A offers 5.5% p.a. simple interest. Bank B offers 5% p.a. simple interest. For a $4 000 investment over 3 years, how much more interest does Bank A earn?
    2. (b) Over what time period does $1 000 at 4% p.a. simple interest equal $1 000 at 6% p.a. for 2 years?
  7. Understanding

    Reading the graph.

    Refer to the linear graph in the Key Ideas section (P = $1 000, r = 8% p.a.).
    1. (a) What is the total amount after 3 years? Read from the graph and verify using A = P(1 + rt).
    2. (b) At what time has the total amount doubled?
    3. (c) What is the gradient of the A vs t graph? What does it represent?
  8. Understanding

    Simple interest on a loan.

    1. (a) Marcus borrows $8 500 for a car at 9% p.a. simple interest over 3 years. Find the total interest charged and the total amount repaid.
    2. (b) What are his monthly repayments (assuming equal repayments over 36 months)?
    3. (c) If Marcus could find a lender charging only 6% p.a. simple interest, how much would he save in total interest?
  9. Problem Solving

    Finding rate given repayments.

    Challenge. A personal loan of $6 000 is repaid over 2 years with total repayments of $7 440.
    1. (a) Calculate the total interest paid.
    2. (b) Find the annual simple interest rate.
    3. (c) What would the total repayments be if the rate were 15% p.a. simple interest over the same period?
    4. (d) The loan can be repaid in 18 months instead if monthly repayments are increased to $460. Under these conditions, find the effective annual rate.
  10. Problem Solving

    Investment planning.

    Challenge. Sophie wants to save $20 000 for a house deposit. She has $14 000 to invest now.
    1. (a) At 6% p.a. simple interest, how many years will it take to reach $20 000?
    2. (b) If she needs the money in 4 years, what annual simple interest rate does she need?
    3. (c) Suppose instead she invests $14 000 at 6% p.a. simple interest for 2 years, then reinvests the total amount at 8% p.a. simple interest for a further 2 years. What is the final total? (Treat the total after year 2 as the new principal.)