The shape of the lesson
By the end of the lesson, learners can
- Simple interest: In the lab, choose Simple and compound and set the rate and number of years.
- Compound interest: With compound interest the interest is added to the account each year, and next year's interest is worked out on the new total.
- Comparing accounts: To compare offers, work out each total for the same principal and time, then compare.
- Simple: \(I = \dfrac{PRT}{100}\), the same every year.
- Compound: amount \(= P \times \left(1 + \dfrac{r}{100}\right)^{n}\).
- Interest or amount? Subtract \(P\) if only the interest is asked for.
- Number of years: try values of \(n\) until the target is passed.
- Round at the end to the nearest cent.
One picture every idea returns to
Simple interest: \(I = \dfrac{PRT}{100}\), the same every year. Compound interest: amount \(= P\left(1 + \dfrac{r}{100}\right)^{n}\); subtract \(P\) if the question asks for the interest only.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Interest on interest.. Simple interest adds the same amount every year. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Simple interest | In the lab, choose Simple and compound and set the rate and number of years. Project tab 4 of the lab (Ratio, rates and percentages lab, interest) and let a student drive it while the class predicts. Model the worked example Simple interest on the board, one line at a time. Your turn: How many years does it take for \(\$2500\) to earn \(\$500\) simple interest at 4% per year? | Whole class, then pairs |
| 13 to 21 min | Teach and check: Compound interest | With compound interest the interest is added to the account each year, and next year's interest is worked out on the new total. Model the worked example Compound interest on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Comparing accounts | To compare offers, work out each total for the same principal and time, then compare. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Comparing accounts, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Check whether the question asks for the INTEREST or the final AMOUNT. | Giving \(\$336\) when the total \(\$2736\) was asked for loses the mark. |
| Compound interest is NOT the simple interest for one year multiplied by the number of years. \(2400 \times 1.035 \times 4\) is meaningless; the power goes on the multiplier: \(2400 \times 1.035^{4}\). | Return to Compound interest: ask a pair to state the correct version and show it on the lab before moving on. |
| Do not round the multiplier or round year by year. | Keep the full calculator value and round only the final money answer to the nearest cent. |
Simple and compound interest: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Calculate the simple interest on \(\$1200\) invested for 5 years at 3% per year.
\(\$1200\) is invested for 5 years at 3% per year compound interest. Calculate the interest earned.
\(\$2500\) is invested at 4% per year simple interest. How many years will it take to earn \(\$500\) in interest?
Bank A pays 4.5% per year simple interest. Bank B pays 4.2% per year compound interest. Amir invests \(\$6000\) for 6 years.
Which bank pays more interest, and by how much?
(Extended) \(\$2000\) is invested at \(r\%\) per year compound interest. After 2 years it is worth \(\$2205\). Find \(r\).
Nadia invests \(\$5000\) at 3.5% per year compound interest. Find the number of complete years until her investment is first worth more than \(\$6000\).
Answer key and marking notes
\(I = \dfrac{1200 \times 3 \times 5}{100} = \$180\)
- Amount \(= 1200 \times 1.03^{5} = 1391.128...\)
- Interest \(= 1391.13 - 1200 = \$191.13\)
- Interest per year \(= 0.04 \times 2500 = \$100\).
- \(500 \div 100 = 5\) years.
- Bank A: \(6000 \times 0.045 \times 6 = \$1620\).
- Bank B: \(6000 \times 1.042^{6} - 6000 = \$1679.94\).
- Bank B pays \(\$59.94\) more.
- \(2000 \times m^{2} = 2205\), so \(m^{2} = 1.1025\).
- \(m = \sqrt{1.1025} = 1.05\).
- \(r = 5\)
- After 5 years: \(5000 \times 1.035^{5} = \$5938.43\), not enough.
- After 6 years: \(5000 \times 1.035^{6} = \$6146.28\), more than \(\$6000\).
- 6 years.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
\(\$3200\) is invested at 2.75% per year simple interest for 5 years. How much interest is earned?
- A\(\$440\)
- B\(\$3640\)
- C\(\$88\)
- D\(\$464.87\)
\(\$1500\) is invested at 4% per year compound interest. Find the value of the investment after 3 years.
- A\(\$1687.30\)
- B\(\$1680.00\)
- C\(\$187.30\)
- D\(\$4680.00\)
\(\$2000\) is invested at 5% per year compound interest. After how many complete years will it first be worth more than \(\$2500\)?
- A5 years
- B4 years
- C6 years
- D3 years
Leo has \(\$5000\) to invest for 3 years. Account A pays 4% per year simple interest. Account B pays 3.8% per year compound interest. Which account gives more money, and by how much?
- AAccount A, by \(\$8.07\)
- BAccount B, by \(\$8.07\)
- CAccount A, by \(\$30.00\)
- DAccount A, by \(\$32.39\)
(Extended) \(\$6400\) grows to \(\$7225\) in 2 years at compound interest. Find the annual rate of interest.
- A6.25%
- B6.45%
- C12.9%
- D1.06%
(Extended) An amount was invested at 3% per year compound interest. After 4 years it is worth \(\$5627.54\). How much was invested?
- A\(\$5000.00\)
- B\(\$4982.02\)
- C\(\$5024.59\)
- D\(\$5463.63\)
Quiz key
\(I = \dfrac{3200 \times 2.75 \times 5}{100} = \$440\). The tempting \(\$3640\) is the total amount, not the interest asked for. \(\$88\) is one year's interest only, and \(\$464.87\) is the compound interest, which this account does not pay.
Amount \(= 1500 \times 1.04^{3} = 1687.296\ldots = \$1687.30\). The tempting \(\$1680.00\) is the simple interest answer (12% of 1500 added once). \(\$187.30\) is the interest only, and \(\$4680.00\) comes from \(1500 \times 1.04 \times 3\), multiplying by 3 instead of using a power.
\(2000 \times 1.05^{4} = \$2431.01\) (not yet) and \(2000 \times 1.05^{5} = \$2552.56\) (more than 2500), so 5 years. The tempting 6 years comes from simple interest: \(\$100\) a year reaches exactly \(\$2500\) after 5 years, so it would need 6 to be MORE than \(\$2500\). Compound interest grows faster than that. 4 years still gives less than \(\$2500\).
A: \(5000 + \dfrac{5000 \times 4 \times 3}{100} = \$5600\). B: \(5000 \times 1.038^{3} = \$5591.93\). A gives \(\$8.07\) more. The tempting answer B assumes compound interest always wins, but over only 3 years the higher simple rate is still ahead. \(\$30.00\) treats B as simple interest too, and \(\$32.39\) treats A as compound interest.
\(m^{2} = \dfrac{7225}{6400} = 1.12890625\), so \(m = \sqrt{1.12890625} = 1.0625\), a rate of 6.25%. The tempting 6.45% halves the total growth of 12.9%, which is the simple interest method. 12.9% is the growth over both years, and 1.06% misreads the multiplier as the rate.
Original \(= 5627.54 \div 1.03^{4} = \$5000.00\). The tempting \(\$4982.02\) multiplies by \(0.97^{4}\), taking 3% off the larger amounts, which is not the reverse of adding 3%. \(\$5024.59\) divides by 1.12 (simple interest thinking), and \(\$5463.63\) undoes only one year.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.