Lesson 6: Representing Simple Interest with Tables and Graphs

Prepared by Yueyi Wang

Year level: Year 10 (middle ability set)Duration: 60 minutesFocal concept/topic: Initial value, annual rate of change and connections among formula, table and graph

Lesson aim:

Interpret the graph constructed in Lesson 5 and investigate how a change in the annual rate changes a simple-interest amount rule. Students explain the initial value and gradient in context and connect matching representations.

Learning intention:

We are learning to interpret simple-interest tables and graphs, explain their initial value and gradient, and predict how changing the annual rate changes the graph.

Success criteria:

□ I can identify the principal as the amount graph’s initial value.
□ I can calculate and interpret the gradient as dollars earned per year.
□ I can match a rule, a table and a graph using their starting amount and yearly increase.
□ I can predict, check and explain what changes when the annual rate changes and the principal stays fixed.

Syllabus outcomes and content:

NSW Education Standards Authority (NESA, 2022a).

MA5-FIN-C-01: solves financial problems involving simple interest, earning money and spending money
MAO-WM-01: develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently and clearly
Establish and use the formula I = Prn to find simple interest where I = simple interest, P = principal, r = interest rate per time period and n = number of time periods
Apply the simple interest formula to solve problems related to investing money at simple interest rates, both algebraically and graphically

Working Mathematically indicators:

Teacher-written indicators, linked to activities:

Connect the coefficient of n with equal table differences and the graph’s gradient (Interpret the graph).
Predict and test the effect of changing r while keeping P fixed (ICT investigation).
Use initial value and rate of change to justify matching representations (Match and explain).
Communicate the meaning of a rule, graph and units independently (Checkpoint).

ICT:

Pairs use the L6 Rates tab with P = $2000 and annual rates 4%, 6% and 8%. They record predictions before viewing the table and chart, verify a row by hand, and explain the graphs using Pr.
The chart is already set up with fixed axes and cell-linked series. Students focus on the relationship rather than learning chart controls during the mathematics lesson.
Paper fallback: the resource pack contains the same labelled graph, a blank values table and the prediction questions.

LIT:

Use initial value and gradient with labelled examples. Read the gradient as dollars per year, rather than saying only “steeper”.
Pairs use “The initial value stays … because …” and “The gradient changes from … to … because …”.
Students justify a match by naming two features, then write an independent explanation. Word meanings and visuals remain available (NESA, 2022b).

Prerequisite knowledge:

Lesson 5 rule A = 2000 + 120n, values table and completed graph.
Calculating Pr, reading coordinates and finding the change between two values.
One-step inverse operations from Lesson 5. The new term gradient describes a rate already present in the table.

Resources:

Student Resources: Lesson 6 pages, prediction table, representations task, independent checkpoint and exit ticket.
ICT Workbook (Excel): L6 Rates tab; one device per pair, projector and calculators.
Printed Lesson 5 graph and the prepared paper fallback graph.
Simple-interest assessment record and the checkpoint marking guide below.

Before the lesson:

Return corrected Lesson 5 graphs. Choose an opening question from the exit evidence; use the rate correction or axis correction only if needed.
Open L6 Rates and keep the table and chart out of view until pairs have recorded predictions. Set P = 2000 and the rate cells to 4%, 6% and 8%.
Check that all three series use the same time values and fixed axes. Do not let automatic rescaling turn a visual change into evidence about the rate.
Print the Lesson 6 resource pages. For the paper fallback, cover the graph until predictions are written. Prepare the paper fallback and the checkpoint. Finding graph intersections or comparing fees is outside this lesson’s core.

Labels used in this plan: WM Working Mathematically LIT Literacy ICT Technology AfL Assessment for learning AoL Assessment of learning

Key teacher question → Expected: suggested student response. Self-designed WM indicators are in italics.

Teaching sequence

Steps & timing	Teaching & learning activities
ORIENTATION0–6 minRetrieve the graph already madeWMAfL	Teacher Display the Lesson 5 graph and A = 2000 + 120n. R1: state P, the yearly interest and the total amount at 2 years. R2: calculate the increase from year 1 to year 2. Use simultaneous mini-whiteboard responses. Correct one selected algebra or axis error from Lesson 5 if needed, then show today’s learning intention. Key question: What does the number 120 describe? Expected: The fixed increase of $120 for each extra year; it is the annual simple interest. Students Use their own graph to answer and check the labels. Check / support The retrieval establishes that the graph and table are familiar. If a student still confuses I and A, return to the P + I bar before asking about gradient.
ENHANCING6–16 minInterpret initial value and gradientWMLITAfL	Teacher Annotate the same amount graph. Identify the initial value as A when n = 0, which is P = $2000. Model gradient using (1, 2120) and (2, 2240): change in amount divided by change in time. Link $120/year to Pr and the coefficient of n in the rule. We do W1: use (0, 2000) and (2, 2240) to find the same gradient over a two-year interval. Briefly connect the known interest rule I = 120n: at n = 0, I = 0, while A = 2000. The two graphs have the same annual increase but different starting values. Do not require a second hand-drawn graph. Model / board working Gradient = (2240 − 2120)/(2 − 1) = 120 dollars per year. W1: (2240 − 2000)/(2 − 0) = 240/2 = 120 dollars per year. Key question: Why is gradient not just the increase of $240 in W1? Expected: The $240 increase takes 2 years. 240 ÷ 2 = $120 per year. Key question: Why do I = 120n and A = 2000 + 120n have the same gradient? Expected: Adding the fixed principal changes the initial value, but not the amount added each year. Students Annotate the starting point and a rise/run triangle. Explain the units of gradient and complete W1. Check / support If students give $240 as the gradient for W1, ask how many years that change covers. “Steeper” is accepted as a comparison only when accompanied by the yearly dollar change.
ENHANCING16–34 minICT investigation: change the rateWMICTLITAfL	Teacher Give pairs 3 minutes to predict before showing the workbook: with P fixed at $2000, what happens to the initial value and annual increase when r is 4%, 6% or 8%? Pairs open L6 Rates. One student operates the device and the other records; swap roles after checking the first row. Students verify by hand the amounts at n = 1.5 for all three rates. Students compare the table and chart. Ask them to name the annual increase for each line, explain the common starting point and explain why all three lines are straight. Pairs briefly change the 6% rate to 5% only after predicting the new annual increase and amount at 2 years. Check, record and restore 6%. This tests a prediction with one changed parameter. Use the same time range and scale for all lines. Linking the rule, differences and graph is the purpose of the technology task (Ainsworth, 2006). Model / board working 4%: A = 2000 + 80n. 6%: A = 2000 + 120n. 8%: A = 2000 + 160n. Key question: What stays the same when only the annual rate changes? Expected: P stays $2000, so each amount graph starts at (0, 2000). Key question: Why does 8% give the greatest gradient? Expected: 2000 × 0.08 = $160/year, greater than $120/year and $80/year for the other rates. Students Record predictions, calculate one row by hand, inspect the chart, then write a claim with supporting numbers. Explain to a partner how the coefficient Pr changes. Use the paper version if devices are unavailable. Check / support Ask each pair to point to a formula cell and a graph feature that supports its claim. If a pair cannot explain the spreadsheet result, return to Pr before extending the investigation.
ENHANCING34–44 minMatch representations and explainWMLITAfL	Teacher Give the two formula cards F1: A = 1500 + 60n and F2: A = 1500 + 90n. Show table T1 with amounts 1500, 1590, 1680, 1770 and T2 with 1500, 1560, 1620, 1680 at n = 0, 1, 2, 3. Show graph G1 through (0, 1500) and (3, 1680), and graph G2 through (0, 1500) and (3, 1770), both on the same labelled axes. Students match each formula to a table and graph. Require two reasons: initial value and annual increase. Then ask which relationship has the larger gradient and what the coefficients mean. Extension E1: for F1, find the annual percentage rate and verify it. Explain why the number 60 in the rule is not a 60% rate. Key question: Why can’t we call 90 the percentage rate in F2? Expected: 90 is dollars earned per year. The annual rate is 90/1500 = 0.06 = 6% p.a. Students Match F1/F2 with T1/T2 and G1/G2. Write a two-feature justification before discussing it with a partner. Check / support If a student matches only by endpoint, ask them to calculate a yearly difference. Support with a matching frame: “F… matches T… and G… because both start at … and increase by … per year.”
ENHANCING44–54 minIndependent checkpointWMLITAoL	Teacher Allow 10 minutes of individual work, with calculator and the common formula card but no partner or teacher solution prompts. C1 A principal of $2500 earns $100 simple interest per year. Write a total-amount rule. (1 mark) C2 Complete a table at n = 0, 1, 3 and use it to plot those points on the supplied axes. (2 marks: correct table; correctly plotted and joined points) C3 State the initial value and gradient, including what each means. (2 marks) C4 A $2000 investment earns $120 simple interest over 18 months. Find its annual percentage rate and check by substitution. (2 marks) C5 For C1, explain why simple-interest growth is linear. (1 mark) Collect the checkpoint before feedback. Students Complete C1–C5 independently. Show setup and units. Indicate whether the optional rearranged-formula card was used. Check / support Use the 8-mark guide and criterion record below. Credit correct reasoning after a single arithmetic error where it remains consistent. Record the support used so numerical marks and independence can be interpreted together.
SYNTHESIZING54–60 minClose the saving sequenceWMAfL	Teacher Use 3 minutes to connect the three lessons: establish the formula, solve an unknown, and explain the relationship in a table or graph. Ask students to point to one representation that supports the claim “interest is added at a constant rate”. Use the final 3 minutes for exit X1: two rules are A = 1500 + 60n and A = 1500 + 90n. What stays the same, and which has the greater gradient? Give units. X2: why are both amount graphs straight lines? Collect the exit ticket. Explain that Lessons 7–9 use percentage calculations, clear time units and total amounts when examining spending and borrowing. Students Write a short comparison and explanation independently. Check / support Use the checkpoint for the end-of-sequence judgement. Use the exit reasoning to plan a short correction or feedback task before students move to spending contexts. After the lesson AoL: mark the 8-mark checkpoint against the guide. Record four criteria separately: amount rule/units, table/graph, contextual interpretation and inverse application. A total mark alone can hide a specific gap. AfL: use the exit explanation and checkpoint working to assign a brief follow-up: R decimal-to-percentage; T annual units; U inverse operation; G initial value/gradient; B fixed principal. Give a worked correction or a comparison task matched to that need. Share the criterion summary with the unit teacher before Lesson 7. The group’s Lesson 9 assessment focuses on spending and borrowing, so this checkpoint provides the direct evidence for the simple-interest sequence.

Worked solutions

Task / question	Worked solution
R1–R2	P = $2000. Annual interest is $120. At 2 years, A = 2000 + 120 × 2 = $2240. From year 1 to year 2 the increase is 2240 − 2120 = $120.
Gradient model and W1	(2240 − 2120)/(2 − 1) = 120 dollars per year. W1: (2240 − 2000)/(2 − 0) = 240/2 = 120 dollars per year. Both I = 120n and A = 2000 + 120n increase by $120 per year; their initial values are $0 and $2000 respectively.
ICT yearly increase and rules	At P = 2000: 4% gives Pr = $80/year and A = 2000 + 80n; 6% gives $120/year and A = 2000 + 120n; 8% gives $160/year and A = 2000 + 160n. Each starts at $2000. All are linear because P and r are fixed in each model.
ICT table: n = 0, 0.5, 1, 1.5, 2, 3	4% amounts: $2000, $2040, $2080, $2120, $2160, $2240. 6%: $2000, $2060, $2120, $2180, $2240, $2360. 8%: $2000, $2080, $2160, $2240, $2320, $2480. Interest is each amount minus $2000.
ICT hand check and 5% test	At 1.5 years: 4% I = 2000 × 0.04 × 1.5 = $120, A = $2120; 6% I = $180, A = $2180; 8% I = $240, A = $2240. Changing 6% to 5% gives $100/year, A = 2000 + 100n and A = $2200 at n = 2. The starting amount remains $2000.
Representations match	F1 matches T2 and G1: each starts at $1500 and adds $60/year. F2 matches T1 and G2: each starts at $1500 and adds $90/year. F2/G2 has the greater gradient. F2 rate = 90/1500 = 0.06 = 6% p.a.
E1	For F1, Pr = 60, so r = 60/1500 = 0.04 = 4% p.a. Check: 1500 × 0.04 = $60 in one year. The coefficient 60 is a dollar increase per year, not a percentage.
C1–C2: 3 marks	C1 A = 2500 + 100n. (1) C2 table: (0, 2500), (1, 2600), (3, 2800). (1 for table) Plot these three points on the supplied axes and join with a straight segment for 0 ≤ n ≤ 3. (1) See checkpoint graph.
C3: 2 marks	Initial value: $2500, the original principal when no time has elapsed. (1) Gradient: $100 per year, the fixed yearly increase in the total amount. (1)
C4: 2 marks	n = 18/12 = 1.5. 120 = 2000 × r × 1.5 = 3000r, so r = 120/3000 = 0.04 = 4% p.a. (1) Check: 2000 × 0.04 × 1.5 = $120. (1)
C5: 1 mark	It is linear because the amount increases by the same $100 for every extra year. Interest is based on the fixed original principal. Accept an equivalent explanation of constant change. (1)
Exit X1–X2	X1: both start at $1500. The second has the larger gradient: $90/year rather than $60/year. X2: each rule adds the same dollar amount for every extra year within that model. P and r stay fixed, so the rate of change is constant.

Worked solutions — C2 checkpoint graph

C2: (0, 2500), (1, 2600), (3, 2800). Join the points with a straight segment for 0 ≤ n ≤ 3. The total amount rises by $100 per year.

Adjustments and anticipated misconceptions

Anticipated misconception	Why students may think this	How this lesson addresses it
The amount graph must start at zero.	Students may recall I = 0 at time zero and transfer this to A.	Return to A = P + I and label the existing principal at n = 0.
Gradient is the total rise, regardless of time.	Students may subtract two amounts but omit division by the interval.	Compare one-year and two-year intervals on the same line. State dollars per year.
120 in A = 2000 + 120n means 120%.	A coefficient and percentage rate are both described as rates, but have different units.	Label 120 as dollars/year and connect it to Pr. If needed, calculate r = 120/2000 = 6%.
A larger annual rate changes the starting value or makes the graph curved.	Students may mix the principal with the increase, or assume all interest is compounded.	Hold P fixed, predict before changing r, and check the common point (0, 2000) and equal differences.
A larger final amount proves a larger percentage rate.	The initial principal may be overlooked when comparing graphs.	If this comparison arises, use the Lesson 4 extension to distinguish P, Pr and r. This is a comparison of mathematical relationships.

Adjustments

Students who need…	Adjustment
More support	Use the corrected Lesson 5 graph, prelabelled axes and a labelled rise/run triangle. Provide a partly completed table and sentence frame. Pair a recorder with a device operator, then swap. The paper fallback uses the same values and scale.
More challenge	E1 recovers the percentage rate from the rule and verifies it. Ask students to explain why the same annual interest can come from different P/r pairs. Keep fees, product conditions and graph intersections outside the core.
Language / EAL/D	Keep a word bank with initial value, gradient, linear and per year beside the annotated graph. Rehearse explanations orally before writing and allow home-language notes (NESA, 2022b).
Access and assessment	Provide large printed axes and keyboard-friendly cells. For a documented access need, supply a plotted graph for annotation or record an oral interpretation; record the adjustment with the checkpoint evidence. Do not prompt a mathematical solution during the independent checkpoint.

Vocabulary

Term	Definition	Image
Initial value	The amount when n = 0. For an amount graph, it is the principal P.	Annotated graph below
Gradient / rate of change	Change in amount divided by change in time. With n in years, the unit is dollars per year.	Annotated graph below
Linear	The amount changes by equal increments for equal time intervals, so the graph is a straight line.	Annotated graph below
Coefficient of n	The number multiplying n. In A = 2000 + 120n, 120 is the yearly dollar increase Pr.	Annotated graph below
Representation	A way to show a relationship, such as a rule, table or graph.	Annotated graph below
Annual percentage rate	The yearly interest as a percentage of the principal. It differs from the dollar increase Pr.	Annotated graph below

Visual: prepared for these lessons. It connects the words to quantities, units or a mathematical operation.

Sources acknowledged in this lesson

The simple-interest representation connects to Palmer et al. (2024, Section 1F, p. 41). All investigation values, matching cards, graphs and checkpoint questions are prepared for these lessons. The workbook uses the same mathematical relationships as the paper task.

Ainsworth, S. (2006). DeFT: A conceptual framework for considering learning with multiple representations. Learning and Instruction, 16(3), 183–198. https://doi.org/10.1016/j.learninstruc.2006.03.001
NSW Education Standards Authority. (2022a). Mathematics K–10 syllabus. https://curriculum.nsw.edu.au/learning-areas/mathematics/mathematics-k-10-2022/content/stage-5/facb3aa952
NSW Education Standards Authority. (2022b). Vocabulary and language in Mathematics 7–10 [Support material]. https://curriculum.nsw.edu.au/file/0e513738-cc88-4ff1-9984-ef0b842faf18/mathematics-7-10-supporting-language-and-vocabulary.docx
Palmer, S., McDaid, K., Greenwood, D., Woolley, S., Goodman, J., & Vaughan, J. (2024). CambridgeMATHS NSW Stage 5 – Core and Standard Paths Year 10 (3rd ed.). Cambridge University Press & Assessment.