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YEAR 11 · INVESTING AND BORROWING · 20 MIN

Simple interest

Use and rearrange the simple-interest model with consistent time units.

In this lesson

Before you start
  • Annual rates
  • Rearranging equations
01 / LEARN

Understand the idea.

Identify principal and time

With simple interest, each period's interest is based on the original principal, not the growing balance. In I=P⁢r⁢t, use principal P, decimal annual rate r and time t in years. Add the principal only when the question asks for the final balance.

Match the time unit to the rate. An annual rate requires years, so nine months is 912=0.75 year. For a calculation involving days, the question must supply the day-count convention rather than leaving you to choose one silently.

Find an unknown

Rearrange I=P⁢r⁢t to isolate the unknown quantity. For a rate, divide the interest by principal times time; for a time, divide by principal times rate. The amount substituted for I must be interest alone, rather than the final balance.

Simple interest adds the same dollar amount each year when the principal and rate stay fixed. This makes the balance grow linearly. Check the answer by substituting it back into the original relationship, and express a rate as a percentage only after solving in decimals.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

Find interest and final balance on AUD 3200 at 4.5% simple interest for 9 months.

Worked solution

  1. 1

    Convert the rate and time to compatible forms.

    r=0.045,t=912=0.75year

    A decimal annual rate must be paired with a time measured in years.

  2. 2

    Calculate the interest.

    I=3200⁢(0.045)⁢(0.75)=AUD 108

    Simple interest uses the original AUD 3200 throughout the nine months.

  3. 3

    Add the interest to the principal.

    A=3200+108=AUD 3308

    The final balance contains both the original deposit and the interest earned.

Answer

AUD 108 interest; AUD 3308 final

EXAMPLE 02

Worked example

AUD 2500 earns AUD 300 simple interest in 3 years. Find the annual rate.

Worked solution

  1. 1

    Substitute into the simple-interest relationship.

    300=2500⁢(r)⁢(3)

    The AUD 300 is interest earned, while AUD 2500 is the original principal.

  2. 2

    Isolate the annual decimal rate.

    r=3002500⁢(3)=0.04

    Divide both sides by the principal and the three-year duration.

  3. 3

    Express and check the annual percentage.

    r=4⁢% p.a.,2500⁢(0.04)⁢(3)=300

    The recovered rate reproduces the supplied total interest over three years.

Answer

4% p.a.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find simple interest on AUD 1500 at 6% p.a. for 4 months.

Show answer & reasoning

AUD 30.00

Four months is 412 year. Therefore I=1500⁢(0.06)412=AUD 30.00. The interest for four months is one third of the annual simple interest.

Keep practising

6 worksheet questions · 12 marks

Read the questions online
  1. Interest on AUD 4800 at 3.5% simple interest for 2 years?
    2 marks
  2. Final amount on AUD 1800 at 5% simple interest for 18 months?
    2 marks
  3. Find the principal earning AUD 144 in 2 years at 4.5% simple interest.
    2 marks
  4. Time for AUD 2000 to earn AUD 350 at 7% simple interest?
    2 marks
  5. Annual simple rate when AUD 3600 earns AUD 81 in 9 months?
    2 marks
  6. Using 365 days/year, find interest on AUD 7300 at 5% for 40 days.
    2 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Interest on AUD 4800 at 3.5% simple interest for 2 years?
    2 marks

    4800×0.035×2=336.

    AUD 336.00
  2. Final amount on AUD 1800 at 5% simple interest for 18 months?
    2 marks

    I=1800×0.05×1.5=135.

    A=1935.

    AUD 1,935.00
  3. Find the principal earning AUD 144 in 2 years at 4.5% simple interest.
    2 marks

    P=144(0.045×2)=1600.

    AUD 1,600.00
  4. Time for AUD 2000 to earn AUD 350 at 7% simple interest?
    2 marks

    t=350(2000×0.07)=2.5.

    2.5 years
  5. Annual simple rate when AUD 3600 earns AUD 81 in 9 months?
    2 marks

    r=81(3600×0.75)=0.03.

    3%
  6. Using 365 days/year, find interest on AUD 7300 at 5% for 40 days.
    2 marks

    I=7300×0.05×40365=40.

    AUD 40.00
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: investing and borrowing. Supplied educational models cover interest, investment comparisons, shares and credit; they are not recommendations of real products or statements of personal tax liability.

Official curriculum source ↗

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