The shape of the lesson
By the end of the lesson, learners can
- One multiplier, many years: Open the lab on Population growth, then Depreciation.
- How many years?: To find when a value first passes a target, try whole numbers of years in the formula.
- Working backwards: If you know the value after the change, divide by the multiplier to the power \(n\) to get the start value.
- Multiplier: growth \(1 + \dfrac{r}{100}\); decay \(1 - \dfrac{r}{100}\).
- After n steps: start \(\times\) multiplier\(^{n}\).
- Number of steps: try values of \(n\) and show the values either side of the target.
- Backwards: divide by multiplier\(^{n}\).
- Rate: multiplier \(= \sqrt[n]{\text{end} \div \text{start}}\).
One picture every idea returns to
Value after \(n\) years \(= \text{start} \times \left(1 \pm \dfrac{r}{100}\right)^{n}\). Growth uses a multiplier bigger than 1, decay a multiplier between 0 and 1. The power is the number of times the change happens.
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, The same multiplier, again.. Extended. 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: One multiplier, many years | Open the lab on Population growth, then Depreciation. Project tab 4 of the lab (Ratio, rates and percentages lab, growth and decay) and let a student drive it while the class predicts. Model the worked example Depreciation on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: How many years? | To find when a value first passes a target, try whole numbers of years in the formula. Model the worked example First year below a target on the board, one line at a time. Your turn: Bacteria double every 20 minutes. There are 300 bacteria at the start. How many are there after 2 hours? | Pairs |
| 21 to 30 min | Teach and check: Working backwards | If you know the value after the change, divide by the multiplier to the power \(n\) to get the start value. Model the worked example The original value on the board, one line at a time. | Numbered Heads Together |
| 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, 15 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 Working backwards, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Exponential decay is NOT a fixed amount off each year. \(18\,000 - 4 \times 0.15 \times 18\,000 = \$7200\) is wrong, because each year's 15% is taken from a smaller value. | Return to One multiplier, many years: ask a pair to state the correct version and show it on the lab before moving on. |
| The power is the number of times the change happens, not the number of hours. | In 24 hours a 6-hour halving happens 4 times, so the power is 4, not 24. |
| To undo 3 years of 10% decay, do not multiply by \(1.1^{3}\). | Losing 10% and gaining 10% are not opposites; divide by \(0.9^{3}\) instead. |
Exponential growth and decay: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Write down the multiplier for (a) growth of 7% per year (b) a decrease of 12% per year.
A town has a population of 24 000. It grows by 3.5% each year. Calculate the population after 10 years, to the nearest hundred.
A machine costs \(\$45\,000\). Its value decreases by 22% each year. Calculate its value after 5 years.
A colony of 300 bacteria doubles every 20 minutes. How many bacteria are there after 2 hours?
A house is worth \(\$250\,000\). Its value increases by \(r\%\) each year. After 3 years it is worth \(\$281\,216\). Find \(r\).
There are 12 000 fish in a lake. The number decreases by 6% each year.
(a) How many fish are there after 5 years?
(b) After how many complete years is the number of fish first less than half of 12 000?
Answer key and marking notes
- (a) 1.07
- (b) 0.88
- \(24\,000 \times 1.035^{10} = 33\,854.37...\)
- 33 900 to the nearest hundred
- Multiplier 0.78.
- \(45\,000 \times 0.78^{5} = \$12\,992.28\)
- 2 hours \(= 6\) lots of 20 minutes.
- \(300 \times 2^{6} = 19\,200\)
- \(250\,000 \times m^{3} = 281\,216\), so \(m^{3} = 1.124\,864\).
- \(m = \sqrt[3]{1.124\,864} = 1.04\).
- \(r = 4\)
- (a) \(12\,000 \times 0.94^{5} = 8806.8...\), about 8810 fish (3 s.f.).
- (b) Half is 6000. \(12\,000 \times 0.94^{11} = 6075.6\), still above 6000.
- \(12\,000 \times 0.94^{12} = 5711.0\), below 6000.
- 12 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.
A van costs \(\$24\,000\). Its value falls by 12% each year. Find its value after 3 years.
- A\(\$16\,355.33\)
- B\(\$15\,360.00\)
- C\(\$33\,718.27\)
- D\(\$41.47\)
A town has a population of 15 000. The population grows by 2.8% each year. Estimate the population after 6 years, to the nearest whole number.
- A17 703
- B17 520
- C65 971
- D92 520
A lake has 12 000 fish. The number of fish falls by 6% each year. After how many complete years is the number of fish first below 8000?
- A7 years
- B6 years
- C8 years
- D5 years
The amount of a drug in a patient's blood halves every 5 hours. A patient is given 640 mg. How much remains after 20 hours?
- A40 mg
- B80 mg
- C32 mg
- D0.000610 mg
An antique increases in value by 5% each year. After 4 years it is worth \(\$12\,155.06\). What was it worth 4 years ago?
- A\(\$10\,000.00\)
- B\(\$9900.37\)
- C\(\$10\,129.22\)
- D\(\$11\,576.25\)
The value of a machine falls from \(\$50\,000\) to \(\$40\,500\) in 2 years, by the same percentage each year. What is the percentage decrease each year?
- A10%
- B9.5%
- C19%
- D90%
Quiz key
Multiplier \(= 1 - 0.12 = 0.88\). Value \(= 24\,000 \times 0.88^{3} = \$16\,355.33\) (to the nearest cent). The tempting \(\$15\,360.00\) takes off a fixed 12% of \(\$24\,000\) each year (36% in total), but each year's 12% is taken from a smaller value. \(\$33\,718.27\) uses growth (1.12), and \(\$41.47\) uses 0.12 as the multiplier.
Population \(= 15\,000 \times 1.028^{6} = 17\,703.1\ldots \approx 17\,703\). The tempting 17 520 adds the same 2.8% of 15 000 each year (simple growth). 65 971 uses 1.28 as the multiplier (a 28% rise), and 92 520 multiplies by 1.028 and then by 6 instead of using a power.
\(12\,000 \times 0.94^{6} = 8278\) (still above 8000) and \(12\,000 \times 0.94^{7} = 7782\) (below), so 7 years. The tempting 6 years comes from taking a fixed 720 fish off each year, which gives a fall of 4000 after only 5.6 years; but each year's 6% is of a smaller number, so the decay is slower than that.
In 20 hours there are \(20 \div 5 = 4\) halvings, so the amount is \(640 \times 0.5^{4} = 40\) mg. The tempting 0.000610 mg uses the number of hours (20) as the power instead of the number of halvings. 80 mg halves only 3 times, and 32 mg divides 640 by 20.
Original \(= 12\,155.06 \div 1.05^{4} = \$10\,000.00\). The tempting \(\$9900.37\) multiplies by \(0.95^{4}\), but losing 5% does not undo gaining 5%. \(\$10\,129.22\) divides by 1.2 (a fixed 5% a year for 4 years), and \(\$11\,576.25\) undoes only one year.
\(m^{2} = \dfrac{40\,500}{50\,000} = 0.81\), so \(m = \sqrt{0.81} = 0.9\). A multiplier of 0.9 means a 10% decrease each year. The tempting 9.5% halves the total fall of 19%, which ignores that the second year's fall is on a smaller value. 19% is the total fall over two years, and 90% is the multiplier written as a percentage.
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.