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Simple Interest for NSW Years 9–12: Formula, Calculator & HSC Practice

Simple interest is the extra money earned or paid on the original amount of money.

Formula: Simple Interest = (Principal * Rate * Time) / 100
Where:
- P = Principal amount
- R = Rate of interest per year (The rate is the whole number as shown. Eg. For 5%, use R = 5..)
- T = Time in years
Simple Interest Questions with detailed Explanations

Table of Contents

NSW Maths • Simple Interest

Simple Interest Explorer

Change the values and see the formula, working and graph update. You can find the interest, principal, rate, time or total amount.

What do you want to find?

Answer

How it works

See how the money grows

The graph updates after each calculation.

Enter your values and press Calculate to see how simple interest grows over time.

Remember: If the rate is 5%, use R = 5. If time is given in months, the tool converts it to years before calculating.
Interactive maths practice

Simple Interest Practice Test

Build confidence from Year 9 foundations to Year 12 HSC Standard applications. Practise simple interest, total amount, time conversions, rearranging the formula, daily rates, linear models and multi-step financial problems.

  • 8 progressive levels
  • 50 practice questions
  • Instant answer checks
  • Hints and full working
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Level 1 Simple Interest Basics 10 questions · Year 9 foundation 0/10

Tip: Use SI = (P × R × T) ÷ 100. Enter the interest only unless the question asks for the total amount.
  1. Calculate the simple interest on $1,000 at 5% p.a. for 2 years.

    View hint and full working

    Hint: Substitute P = 1000, R = 5 and T = 2 into the simple interest formula.

    Working
    SI = (P × R × T) ÷ 100
    SI = (1000 × 5 × 2) ÷ 100
    Answer: $100
  2. Calculate the simple interest on $2,500 at 4% p.a. for 3 years.

    View hint and full working

    Hint: Use the original principal for the whole calculation.

    Working
    SI = (2500 × 4 × 3) ÷ 100
    Answer: $300
  3. Find the simple interest on $1,800 at 6% p.a. for 4 years.

    View hint and full working

    Hint: Multiply the principal, rate and time, then divide by 100.

    Working
    SI = (1800 × 6 × 4) ÷ 100
    Answer: $432
  4. Find the simple interest on $3,200 at 3.5% p.a. for 2 years.

    View hint and full working

    Hint: The rate can include a decimal. Use R = 3.5.

    Working
    SI = (3200 × 3.5 × 2) ÷ 100
    Answer: $224
  5. Calculate the simple interest on $750 at 8% p.a. for 5 years.

    View hint and full working

    Hint: Use P = 750, R = 8 and T = 5.

    Working
    SI = (750 × 8 × 5) ÷ 100
    Answer: $300
  6. Find the simple interest on $1,250 at 2.4% p.a. for 3 years.

    View hint and full working

    Hint: Use R = 2.4, not 0.024, with this version of the formula.

    Working
    SI = (1250 × 2.4 × 3) ÷ 100
    Answer: $90
  7. Calculate the simple interest on $4,800 at 7.5% p.a. for 1 year.

    View hint and full working

    Hint: For one year, T = 1.

    Working
    SI = (4800 × 7.5 × 1) ÷ 100
    Answer: $360
  8. Find the simple interest on $950 at 4.2% p.a. for 2 years.

    View hint and full working

    Hint: Keep the decimal rate as 4.2.

    Working
    SI = (950 × 4.2 × 2) ÷ 100
    Answer: $79.80
  9. Calculate the simple interest on $6,000 at 3.25% p.a. for 4 years.

    View hint and full working

    Hint: Use R = 3.25 in the formula.

    Working
    SI = (6000 × 3.25 × 4) ÷ 100
    Answer: $780
  10. Find the simple interest on $2,750 at 5.6% p.a. for 1.5 years.

    View hint and full working

    Hint: A time of 1.5 years can be used directly as T = 1.5.

    Working
    SI = (2750 × 5.6 × 1.5) ÷ 100
    Answer: $231

Level 2 Total Amount and Time Conversions 10 questions · Years 9–10 core skills 0/10

Tip: For an annual rate, convert months to years first. Total amount = Principal + Simple Interest.
  1. An investment of $2,000 earns simple interest at 5% p.a. for 3 years. What is the total amount at the end?

    View hint and full working

    Hint: Find the interest first, then add it to the principal.

    Working
    SI = (2000 × 5 × 3) ÷ 100 = $300
    Total amount = 2000 + 300
    Answer: $2,300
  2. A principal of $1,500 is invested at 4% p.a. simple interest for 6 months. What is the total amount?

    View hint and full working

    Hint: Convert 6 months to 0.5 years before calculating interest.

    Working
    6 months = 6 ÷ 12 = 0.5 years
    SI = (1500 × 4 × 0.5) ÷ 100 = $30
    Answer: $1,530
  3. $3,200 is invested at 6.5% p.a. simple interest for 9 months. Find the total amount.

    View hint and full working

    Hint: 9 months = 9/12 = 0.75 years.

    Working
    9 months = 0.75 years
    SI = (3200 × 6.5 × 0.75) ÷ 100 = $156
    Answer: $3,356
  4. $5,000 earns simple interest at 3.2% p.a. for 18 months. What is the total amount?

    View hint and full working

    Hint: 18 months = 1.5 years.

    Working
    18 months = 18 ÷ 12 = 1.5 years
    SI = (5000 × 3.2 × 1.5) ÷ 100 = $240
    Answer: $5,240
  5. $840 is invested at 7.5% p.a. simple interest for 8 months. Find the total amount.

    View hint and full working

    Hint: Convert 8 months to 8/12 years.

    Working
    8 months = 8/12 = 2/3 year
    SI = (840 × 7.5 × 2/3) ÷ 100 = $42
    Answer: $882
  6. $12,000 is invested at 4.8% p.a. simple interest for 15 months. Find the total amount.

    View hint and full working

    Hint: 15 months = 1.25 years.

    Working
    15 months = 15 ÷ 12 = 1.25 years
    SI = (12000 × 4.8 × 1.25) ÷ 100 = $720
    Answer: $12,720
  7. $3,600 earns simple interest at 5.5% p.a. for 2.5 years. Find the total amount.

    View hint and full working

    Hint: Find the interest and then add it to $3,600.

    Working
    SI = (3600 × 5.5 × 2.5) ÷ 100 = $495
    Answer: $4,095
  8. $9,500 is invested at 2.8% p.a. simple interest for 30 months. Find the total amount.

    View hint and full working

    Hint: 30 months = 2.5 years.

    Working
    30 months = 30 ÷ 12 = 2.5 years
    SI = (9500 × 2.8 × 2.5) ÷ 100 = $665
    Answer: $10,165
  9. $4,200 earns simple interest at 6% p.a. for 1 year 3 months. Find the total amount.

    View hint and full working

    Hint: 1 year 3 months = 1.25 years.

    Working
    1 year 3 months = 1 + 3/12 = 1.25 years
    SI = (4200 × 6 × 1.25) ÷ 100 = $315
    Answer: $4,515
  10. $6,800 is invested at 3.75% p.a. simple interest for 20 months. Find the total amount.

    View hint and full working

    Hint: 20 months = 20/12 = 5/3 years.

    Working
    20 months = 20/12 = 5/3 years
    SI = (6800 × 3.75 × 5/3) ÷ 100 = $425
    Answer: $7,225

Level 3 Rearranging the Simple Interest Formula 5 questions · Year 10 core skills 0/5

Tip: Work backwards by rearranging SI = (P × R × T) ÷ 100 to find the missing value.
  1. The simple interest is $360 at 6% p.a. for 3 years. Find the principal.

    View hint and full working

    Hint: Use P = (SI × 100) ÷ (R × T).

    Working
    P = (360 × 100) ÷ (6 × 3)
    P = 36000 ÷ 18
    Answer: $2,000
  2. $3,500 earns $525 simple interest in 3 years. Find the annual interest rate.

    View hint and full working

    Hint: Use R = (SI × 100) ÷ (P × T).

    Working
    R = (525 × 100) ÷ (3500 × 3)
    R = 52500 ÷ 10500
    Answer: 5% p.a.
  3. $2,400 earns $480 simple interest at 5% p.a. Find the time.

    View hint and full working

    Hint: Use T = (SI × 100) ÷ (P × R).

    Working
    T = (480 × 100) ÷ (2400 × 5)
    T = 48000 ÷ 12000
    Answer: 4 years
  4. An investment earns $168 simple interest at 4% p.a. over 2.5 years. Find the principal.

    View hint and full working

    Hint: Rearrange the formula to make P the subject.

    Working
    P = (168 × 100) ÷ (4 × 2.5)
    P = 16800 ÷ 10
    Answer: $1,680
  5. $3,200 earns $288 simple interest over 1.5 years. Find the annual interest rate.

    View hint and full working

    Hint: Rearrange to R = (SI × 100) ÷ (P × T).

    Working
    R = (288 × 100) ÷ (3200 × 1.5)
    R = 28800 ÷ 4800
    Answer: 6% p.a.

Level 4 Simple Interest Word Problems 5 questions · Years 10–11 applications 0/5

Tip: Read carefully: decide whether the question wants interest, total amount, principal, rate or time.
  1. You borrow $4,500 at 7% p.a. simple interest for 2 years. What total amount must be repaid?

    View hint and full working

    Hint: Calculate the interest first, then add it to the amount borrowed.

    Working
    SI = (4500 × 7 × 2) ÷ 100 = $630
    Total repayment = 4500 + 630
    Answer: $5,130
  2. A $6,000 investment earns 4.5% p.a. simple interest for 18 months. How much interest is earned?

    View hint and full working

    Hint: Convert 18 months to 1.5 years.

    Working
    18 months = 1.5 years
    SI = (6000 × 4.5 × 1.5) ÷ 100
    Answer: $405
  3. An investment of $3,000 earns $270 simple interest in 18 months. Find the annual interest rate.

    View hint and full working

    Hint: Convert 18 months to 1.5 years, then solve for R.

    Working
    T = 18/12 = 1.5 years
    R = (270 × 100) ÷ (3000 × 1.5)
    Answer: 6% p.a.
  4. How much must be invested at 3% p.a. simple interest for 5 years to earn $225 interest?

    View hint and full working

    Hint: The principal is unknown, so use P = (SI × 100) ÷ (R × T).

    Working
    P = (225 × 100) ÷ (3 × 5)
    P = 22500 ÷ 15
    Answer: $1,500
  5. How long will $4,000 need to be invested at 4% p.a. simple interest to earn $640?

    View hint and full working

    Hint: Solve for T.

    Working
    T = (640 × 100) ÷ (4000 × 4)
    T = 64000 ÷ 16000
    Answer: 4 years

Level 5 Daily Rates and Mixed Time Periods 5 questions · Year 11 financial maths 0/5

Tip: Match the rate and time units. A daily rate can be used with days; an annual rate needs time in years.
  1. A balance of $2,500 is charged simple interest at 0.04% per day for 30 days. Find the interest charged.

    View hint and full working

    Hint: Because the rate is per day, use T = 30 days directly with R = 0.04.

    Working
    SI = (2500 × 0.04 × 30) ÷ 100
    Answer: $30
  2. A debt of $1,800 is charged simple interest at 0.03% per day for 45 days. Find the interest charged.

    View hint and full working

    Hint: Use the daily rate with the number of days.

    Working
    SI = (1800 × 0.03 × 45) ÷ 100
    Answer: $24.30
  3. $7,200 is invested at 6% p.a. simple interest for 75 days. Assume 365 days in a year. Find the interest, to the nearest cent.

    View hint and full working

    Hint: Convert 75 days to 75/365 of a year.

    Working
    T = 75/365 years
    SI = (7200 × 6 × 75/365) ÷ 100
    Answer: $88.77
  4. A lender charges simple interest at 0.05% per day. What is the equivalent simple annual rate if a year has 365 days?

    View hint and full working

    Hint: Multiply the daily percentage rate by 365.

    Working
    Annual simple rate = 0.05% × 365
    Answer: 18.25%
  5. $4,000 earns $120 simple interest over 120 days. Assuming 365 days in a year, find the annual simple interest rate.

    View hint and full working

    Hint: Convert 120 days to years, then rearrange for R.

    Working
    T = 120/365 years
    R = (120 × 100) ÷ (4000 × 120/365)
    Answer: 9.125% p.a.

Level 6 Simple Interest Graphs and Linear Models 5 questions · Years 11–12 0/5

Tip: Simple interest grows at a constant rate, so its graph is linear. The gradient represents the interest added per time period.
  1. An investment of $2,000 earns 5% p.a. simple interest. What is the yearly increase in the total amount?

    View hint and full working

    Hint: The yearly increase is 5% of the original principal.

    Working
    Yearly interest = (2000 × 5 × 1) ÷ 100
    Answer: $100 per year
  2. The total value of an investment is modelled by A = 3500 + 210t, where t is in years. What simple interest rate is being earned?

    View hint and full working

    Hint: The gradient 210 is the interest earned each year. Compare it with the principal 3500.

    Working
    Annual interest = $210
    R = (210 ÷ 3500) × 100
    Answer: 6% p.a.
  3. The simple interest earned is modelled by I = 144t. If the principal is $2,400, find the annual interest rate.

    View hint and full working

    Hint: The gradient 144 is one year of interest.

    Working
    Annual interest = $144
    R = (144 ÷ 2400) × 100
    Answer: 6% p.a.
  4. A simple-interest investment follows A = 5000 + 225t. Find the total amount after 7 years.

    View hint and full working

    Hint: Substitute t = 7 into the linear model.

    Working
    A = 5000 + 225(7)
    A = 5000 + 1575
    Answer: $6,575
  5. $8,000 is invested at 4.5% p.a. simple interest. What total amount would the graph show at t = 6 years?

    View hint and full working

    Hint: Find six years of simple interest and add it to the principal.

    Working
    Annual interest = (8000 × 4.5) ÷ 100 = $360
    Interest after 6 years = 360 × 6 = $2,160
    Answer: $10,160

Level 7 HSC Standard Financial Applications 5 questions · Year 12 HSC Standard 0/5

Tip: HSC-style questions often add an extra step such as a fee, tax, comparison, daily rate or linear model.
  1. A $12,000 loan is charged simple interest at 7.2% p.a. for 3 years. There is also a $250 establishment fee. What is the total cost of borrowing, excluding repayment of the principal?

    View hint and full working

    Hint: Add the simple interest and the establishment fee.

    Working
    SI = (12000 × 7.2 × 3) ÷ 100 = $2,592
    Cost of borrowing = 2592 + 250
    Answer: $2,842
  2. $15,000 is invested at 5.4% p.a. simple interest for 4 years. If 10% of the interest earned is paid as tax, how much interest remains after tax?

    View hint and full working

    Hint: Find the interest first, then keep 90% of it.

    Working
    SI = (15000 × 5.4 × 4) ÷ 100 = $3,240
    Tax = 10% of 3240 = $324
    Answer: $2,916
  3. An investment of $10,000 must grow to $12,500 in 5 years using simple interest. What annual interest rate is required?

    View hint and full working

    Hint: The interest earned must be $2,500, then solve for R.

    Working
    SI = 12500 − 10000 = $2,500
    R = (2500 × 100) ÷ (10000 × 5)
    Answer: 5% p.a.
  4. A credit balance of $2,400 is charged simple interest at 0.06% per day for 25 days. Find the total amount owing.

    View hint and full working

    Hint: Calculate the daily simple interest first, then add it to the balance.

    Working
    SI = (2400 × 0.06 × 25) ÷ 100 = $36
    Total owing = 2400 + 36
    Answer: $2,436
  5. The value of a simple-interest investment is modelled by A = 18000 + 990t. Find the annual simple interest rate.

    View hint and full working

    Hint: The coefficient of t is the annual interest.

    Working
    Annual interest = $990
    R = (990 ÷ 18000) × 100
    Answer: 5.5% p.a.

Level 8 HSC Standard Challenge Problems 5 questions · Year 12 challenge 0/5

Tip: Break multi-step questions into stages. Identify the unknown, keep units consistent and state the final financial meaning.
  1. How long will it take $20,000 invested at 5% p.a. simple interest to reach a total value of $25,000?

    View hint and full working

    Hint: The required interest is $5,000. Then solve for time.

    Working
    SI = 25000 − 20000 = $5,000
    T = (5000 × 100) ÷ (20000 × 5)
    Answer: 5 years
  2. An investment earns simple interest equal to 12% of its principal over 30 months. Find the annual simple interest rate.

    View hint and full working

    Hint: 30 months = 2.5 years. The total interest is 12% of P.

    Working
    T = 30/12 = 2.5 years
    0.12P = (P × R × 2.5) ÷ 100
    12 = 2.5R
    Answer: 4.8% p.a.
  3. Investment A places $10,000 at 5.2% p.a. simple interest for 4 years. Investment B places $10,000 at 5.5% p.a. for the first 2 years and 4.5% p.a. for the next 2 years, with simple interest always calculated on the original principal. By how much does the better investment exceed the other after 4 years?

    View hint and full working

    Hint: Calculate the total simple interest for each investment separately.

    Working
    A: SI = (10000 × 5.2 × 4) ÷ 100 = $2,080, so A = $12,080
    B: interest = $1,100 + $900 = $2,000, so B = $12,000
    Answer: $80
  4. A $18,500 loan is charged 8.4% p.a. simple interest for 15 months. The total amount owing is repaid in 15 equal monthly payments. Find each monthly payment, to the nearest cent.

    View hint and full working

    Hint: Convert 15 months to 1.25 years, find the total owing, then divide by 15.

    Working
    T = 15/12 = 1.25 years
    SI = (18500 × 8.4 × 1.25) ÷ 100 = $1,942.50
    Total owing = 18500 + 1942.50 = $20,442.50
    Answer: $1,362.83
  5. An investment is worth $13,640 after 4 years at 4% p.a. simple interest. Find the original principal, to the nearest cent.

    View hint and full working

    Hint: The final amount is P plus 16% of P, because 4% is earned for 4 years.

    Working
    A = P + SI
    13640 = P + (P × 4 × 4) ÷ 100
    13640 = 1.16P
    Answer: $11,758.62

Simple interest is an important financial mathematics skill from the junior years through to HSC Mathematics Standard. The aim is not just to memorise one formula. You should be able to identify the principal, rate and time, rearrange the formula, convert time units and explain what your answer means.

What level should I practise?

Years 9–10

Focus on the formula, simple interest, total amount and converting months to years.

Year 11

Build confidence with rearranging formulas, financial applications, rates, units and linear relationships.

HSC Mathematics Standard 1

Practise investment, interest, loans, credit-card situations and making decisions from financial information.

HSC Mathematics Standard 2

Practise multi-step financial modelling, comparisons, borrowing, investments and interpreting results.

Best way to use this page: learn one idea, follow the worked example, try the Simple Interest Explorer, then test yourself in the interactive practice questions.

If percentages are still difficult, revise how to calculate percentages first. For broader Stage 5 revision, use the Year 9 NSW Maths guide or the Year 10 NSW Maths guide.

Simple Interest Formula

Simple interest is calculated on the original principal. If the principal and rate stay the same, the same amount of interest is added during each equal time period.

When the rate is entered as a percentage such as 5: SI = (P × R × T) ÷ 100
SymbolMeaningExample
SISimple interest earned or charged$300
PPrincipal — the original amount invested or borrowed$2,000
RInterest rate per year5% means use R = 5
TTime in years when the rate is per annum3 years
Another formula you may see: I = Prt, where the interest rate is written as a decimal. For example, 5% becomes 0.05. Both forms are equivalent. This page mainly uses SI = PRT ÷ 100, so 5% is entered as 5.

The official NSW Mathematics K–10 glossary also defines simple interest using the principal, rate and number of time periods.

How to Calculate Simple Interest

Example: $2,000 is invested at 5% p.a. for 3 years. Find the simple interest.

Step 1: Identify the values.

P = 2000, R = 5, T = 3

Step 2: Substitute into the formula.

SI = (2000 × 5 × 3) ÷ 100

Step 3: Calculate.

SI = 300

Answer: The simple interest is $300.

Interest is not always the final amount

If the question asks for the total amount, add the interest to the original principal.

A = P + SI

In the example above, the total amount is $2,000 + $300 = $2,300.

Common mistake: $300 is the interest, not the final balance. Always check whether the question asks for interest earned or the total amount.

How to Find the Principal, Rate or Time

In harder questions, the simple interest may already be given and another value is missing. Rearranging the formula first usually makes the working clearer. If algebraic rearranging is unfamiliar, revise Algebra Made Easy.

FindFormula
Simple InterestSI = (P × R × T) ÷ 100
PrincipalP = (SI × 100) ÷ (R × T)
RateR = (SI × 100) ÷ (P × T)
TimeT = (SI × 100) ÷ (P × R)

Example: Find the interest rate.

Dana invests $2,000 and earns $400 simple interest over 4 years. Find the annual interest rate.

R = (SI × 100) ÷ (P × T)
R = (400 × 100) ÷ (2000 × 4)
R = 5

Answer: 5% p.a.

Example: Find the principal.

An investment earns $360 at 6% p.a. simple interest over 3 years. Find the original principal.

P = (360 × 100) ÷ (6 × 3)
P = 2000

Answer: The principal is $2,000.

Simple Interest With Months and Days

The rate and the time must refer to the same time period. If the rate is per annum, time should normally be written in years before you substitute into the formula.

Given timeConvert to years
3 months3 ÷ 12 = 0.25 years
6 months6 ÷ 12 = 0.5 years
9 months9 ÷ 12 = 0.75 years
18 months18 ÷ 12 = 1.5 years

Example: Interest for 9 months.

$4,000 is invested at 6% p.a. simple interest for 9 months.

T = 9 ÷ 12 = 0.75 years
SI = (4000 × 6 × 0.75) ÷ 100
SI = 180

Answer: $180 simple interest.

What if the rate is given per day?

If a question gives a daily rate, the number of days can be used directly because the units already match. For example, a debt of $1,800 charged at 0.05% per day for 30 days gives:

SI = (1800 × 0.05 × 30) ÷ 100 = $27

Senior students can explore official examples in the NSW Department of Education's Year 11 Mathematics Standard financial mathematics resources.

What Does a Simple Interest Graph Show?

Simple interest creates a straight-line relationship when the principal and rate remain constant. This happens because the same amount of interest is added during every equal time period.

Example: $2,000 at 5% p.a.

Interest each year = (2000 × 5 × 1) ÷ 100 = $100
Total amount after t years: A = 2000 + 100t

In A = 2000 + 100t, the value $2,000 is the starting amount and $100 is the increase each year. On a graph, $100 is the gradient. This links simple interest to linear relationships.

If gradient or straight-line equations need revision, use our Number Plane and Straight Lines guide. You can also use the interactive graph in the Simple Interest Explorer above.

Simple Interest for HSC Mathematics Standard 1 and Standard 2

Simple interest is not only a Year 9 topic. In senior Mathematics Standard, students use financial mathematics to solve practical problems and make decisions about investments, loans and borrowing.

Mathematics Standard 1

Financial mathematics includes investment, depreciation and loans. Students need to select formulas, calculate accurately and interpret the result.

Mathematics Standard 2

Financial mathematics extends into investment and loans, annuities and more detailed financial modelling and comparisons.

Under the NSW Mathematics Standard 11–12 Syllabus (2024), Standard 1 includes solving problems involving simple and compound interest to make decisions about financial situations, while Standard 2 includes modelling financial situations involving interest, depreciation and borrowing money. See the official Mathematics Standard outcomes and course structure.

NSW syllabus transition: Year 11 began the new Mathematics Standard 11–12 Syllabus (2024) in 2026. Year 12 continues the 2017 syllabus during 2026, with the new Year 12 syllabus starting in Term 4 2026 and the first HSC examination under the new syllabus in 2027. See the official NESA implementation information.

What changes in an HSC-style question?

The arithmetic may still be simple, but the question often requires more than direct substitution. You may need to:

  • decide which values represent principal, interest, rate and time
  • convert months, days or rates into matching units
  • find a missing value by rearranging a formula
  • calculate a final balance rather than only the interest
  • compare two financial options
  • explain which option is better and why
  • interpret a table, graph or spreadsheet result

For additional official senior resources, see the NSW Department of Education pages for Year 12 Mathematics Standard 1 financial mathematics and Year 12 Mathematics Standard 2 financial mathematics.

Common Simple Interest Mistakes

1. Using 0.05 when the formula already divides by 100

For SI = PRT ÷ 100, use R = 5 for 5%. If you use I = Prt, use r = 0.05.

2. Mixing months and years

If the rate is per annum, 6 months should usually be written as 6/12 = 0.5 years.

3. Confusing interest with total amount

SI is only the interest. If the question asks for the final amount, use A = P + SI.

4. Using the wrong value as the principal

The principal is the original amount invested or borrowed, not the final balance.

5. Giving a number without units

Money answers need dollars, rate answers need %, and time answers need a unit such as years or months.

HSC habit: after calculating, read the question again. A correct calculation can still be incomplete if the question asked for a comparison, total amount or recommendation and you only give the interest.

Simple Interest vs Compound Interest

The main difference is what the interest is calculated on. With simple interest, interest is calculated on the original principal. With compound interest, interest is added to the balance and future interest can then be calculated on that larger amount.

Simple InterestCompound Interest
Interest calculated onOriginal principalGrowing balance
Typical growthLinearExponential
GraphStraight lineCurved growth
Common formulaSI = PRT ÷ 100A = P(1 + r)n

HSC-style comparison: $10,000 at 5% p.a. for 3 years.

Simple interest:

SI = (10000 × 5 × 3) ÷ 100 = $1,500
Final amount = $11,500

Compound interest, compounded annually:

A = 10000(1.05)³ = $11,576.25

The compound-interest investment gives $76.25 more after 3 years.

NSW Student Study Pathway

Simple interest connects to several other topics. If a question feels difficult, the problem may actually be a missing skill in percentages, algebra, rates or linear relationships.

Ready to practise?
Go directly to the 50-question Simple Interest Practice Test. Start with the easier levels and move towards Year 11 and HSC Standard questions.

Simple Interest FAQs

What is simple interest in one sentence?

Simple interest is interest calculated on the original principal rather than on previously earned interest.

What is the simple interest formula?

When the rate is entered as a percentage, use SI = (P × R × T) ÷ 100.

How do I find the total amount?

Calculate the simple interest first, then use A = P + SI.

How do I use 5% in the formula?

For SI = PRT ÷ 100, use R = 5. For I = Prt, use r = 0.05.

How do I calculate simple interest for 6 months?

If the rate is per annum, convert the time to years: 6 ÷ 12 = 0.5 years.

How do I find the interest rate?

Rearrange the formula to R = (SI × 100) ÷ (P × T).

How do I find the principal?

Use P = (SI × 100) ÷ (R × T).

How do I find the time?

Use T = (SI × 100) ÷ (P × R).

Why is the simple-interest graph a straight line?

Because the same amount of interest is added in each equal time period when the principal and rate remain constant.

Is simple interest useful for HSC Mathematics Standard?

Yes. Financial mathematics is part of Mathematics Standard, and senior questions can involve interest, investments, loans, comparisons and financial decision-making.

Trusted NSW Mathematics Links

Curriculum note: This learning guide is designed to support NSW students from Stage 5 through Mathematics Standard 1 and Standard 2. Schools may teach content in different sequences, so students should also follow their school's assessment notification and scope-and-sequence documents.

Next step: use the interactive practice test below to move from direct formula questions to multi-step HSC-style applications.

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Choosing the right selective high school in NSW can shape your child’s academic journey. This guide breaks down the six types of selective high schools in NSW — fully selective, partially selective, agricultural, Aurora College, Conservatorium, and boarding schools—with pros, cons, and examples of each. Learn how to match the right school type to your child’s strengths, passions, and family situation so you can make confident application choices.

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