High School Maths · Year 2 · Unit 14 · Lesson 2 of 5
Depreciation and Working Backwards: How Fast Does a New Car Lose Value?
A new car costs $24,000. It loses 15% of its value every year.
After 7 years, 7 × 15% = 105%. Is the car worth nothing by then?
No. Each year the car keeps 85% of the value it had at the start of that year, so each year’s loss is smaller than the last. After 7 years it is worth 24,000 × 0.85⁷ = $7,693.85. Its value keeps falling, but it never reaches $0.
Master the essentials
Depreciation
Losing value over time. A fall of r% multiplies the value by (1 − r ÷ 100).
12% fall → × 0.88
The formula
Value after n years: V = P(1 − r/100)ⁿ, where P is the starting value.
9,000 × 0.9³ = $6,561
Loss in value
Loss = starting value − value now. It is not r% × n.
9,000 − 6,561 = $2,439
Finding the original
Work backwards: divide by the multiplier once for each year.
P × 1.05² = 4,410 → P = 4,410 ÷ 1.05² = 4,000
Finding the rate
Divide to get the multiplier for all the years. For 2 years, take the square root.
$1,600 → $1,444 in 2 years: √(1,444 ÷ 1,600) = 0.95, a 5% fall
Finding the years
Use trial: multiply year by year until you pass the target.
500 × 1.2³ = 864 and 500 × 1.2⁴ = 1,036.80, so 4 years to pass $1,000
First below, first above
The answer is the first whole year that passes the target. Check the year before as well.
$10,000 falling 20%: year 3: $5,120, year 4: $4,096. First below $5,000: year 4
Growth and decay
Prices, populations and bacteria grow or decay in the same way. Growth: × (1 + r/100). Decay: × (1 − r/100).
8,000 people, +2% a year: × 1.02
See it done
1. A machine costs $45,000. It depreciates by 12% per year. Find its value after 4 years, to the nearest cent.
- A 12% fall multiplies by 1 − 0.12 = 0.88.
- V = 45,000 × 0.88⁴.
- = 26,986.2912.
- To the nearest cent: $26,986.29.
$26,986.29
2. A flat’s value rose by 6% each year for 3 years. It is now worth $297,754. What was it worth 3 years ago?
- Now = P × 1.06³.
- Work backwards: P = 297,754 ÷ 1.06³.
- = 297,754 ÷ 1.191016.
- = $250,000. (Not 297,754 × 0.94³: taking 6% off does not undo adding 6%.)
$250,000
3. A colony of 6,000 insects falls by 18% each week. After how many whole weeks is it first below 2,000?
- The multiplier is 1 − 0.18 = 0.82.
- Trial: after 5 weeks, 6,000 × 0.82⁵ = 2,224.4…, still above 2,000.
- After 6 weeks, 6,000 × 0.82⁶ = 1,824.0…, now below 2,000.
- So it is first below 2,000 after 6 weeks.
| Weeks | Insects (6,000 × 0.82ⁿ) |
|---|---|
| 4 | 2,712.73056 |
| 5 | 2,224.4… |
| 6 | 1,824.0… |
Trial: stop at the first week below 2,000.
6 weeks
Common mistake: Adding the percentage back to undo a fall. If a price fell by 20% to $640, the original is NOT 640 × 1.2 = $768. The $640 is 80% of the original, so divide: 640 ÷ 0.8 = $800.
10 questions with instant feedback
Choose the correct answer. A calculator is allowed. Find the multiplier first. To work backwards, divide by it; to find the years, try one year at a time. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.
Warm-up (The basic skill: one or two steps.)
1.A laptop costs $1,250. It loses 20% of its value each year. What is it worth after 2 years?
Correct.
Solution
- Multiplier: 1 − 0.2 = 0.8.
- After 1 year: 1,250 × 0.8 = $1,000.
- After 2 years: 1,000 × 0.8 = $800.
Not quite. This takes off 2 × 20% = 40% at once. The second 20% is of a smaller value.
Solution
- Multiplier: 1 − 0.2 = 0.8.
- After 1 year: 1,250 × 0.8 = $1,000.
- After 2 years: 1,000 × 0.8 = $800.
Answer: A. $800.
Not quite. This is the value after 1 year only.
Solution
- Multiplier: 1 − 0.2 = 0.8.
- After 1 year: 1,250 × 0.8 = $1,000.
- After 2 years: 1,000 × 0.8 = $800.
Answer: A. $800.
Not quite. This is the loss in value. The question asks what the laptop is worth.
Solution
- Multiplier: 1 − 0.2 = 0.8.
- After 1 year: 1,250 × 0.8 = $1,000.
- After 2 years: 1,000 × 0.8 = $800.
Answer: A. $800.
2.A motorbike worth $8,400 depreciates by 9% per year. Which calculation gives its value after 5 years?
Correct.
Solution
- A 9% fall multiplies by 1 − 0.09 = 0.91.
- This happens 5 times.
- Value = 8,400 × 0.91⁵.
Not quite. This is growth. A fall of 9% multiplies by 0.91.
Solution
- A 9% fall multiplies by 1 − 0.09 = 0.91.
- This happens 5 times.
- Value = 8,400 × 0.91⁵.
Answer: A. 8,400 × 0.91⁵.
Not quite. This takes off 5 × 9% = 45% at once.
Solution
- A 9% fall multiplies by 1 − 0.09 = 0.91.
- This happens 5 times.
- Value = 8,400 × 0.91⁵.
Answer: A. 8,400 × 0.91⁵.
Not quite. 0.09 is the part lost. The bike keeps 0.91 of its value each year.
Solution
- A 9% fall multiplies by 1 − 0.09 = 0.91.
- This happens 5 times.
- Value = 8,400 × 0.91⁵.
Answer: A. 8,400 × 0.91⁵.
3.A boat is worth $36,000. It depreciates by 11% per year. Find its value after 3 years, to the nearest cent.
Not quite. This takes off 3 × 11% = 33% at once.
Solution
- Multiplier: 0.89.
- V = 36,000 × 0.89³ = 25,378.884.
- To the nearest cent: $25,378.88.
Answer: B. $25,378.88.
Correct.
Solution
- Multiplier: 0.89.
- V = 36,000 × 0.89³ = 25,378.884.
- To the nearest cent: $25,378.88.
Not quite. This adds 11% each year. Depreciation is a fall: × 0.89.
Solution
- Multiplier: 0.89.
- V = 36,000 × 0.89³ = 25,378.884.
- To the nearest cent: $25,378.88.
Answer: B. $25,378.88.
Not quite. This is the loss in value, not the value.
Solution
- Multiplier: 0.89.
- V = 36,000 × 0.89³ = 25,378.884.
- To the nearest cent: $25,378.88.
Answer: B. $25,378.88.
Exam standard (Several steps, like real exam questions.)
4.A phone loses 25% of its value each year. After 2 years it is worth $405. What did it cost when new?
Not quite. This adds 25% twice. The $405 is 75% of 75% of the price, so divide by 0.75².
Solution
- Value now = P × 0.75².
- P = 405 ÷ 0.75² = 405 ÷ 0.5625.
- P = $720. Check: 720 × 0.75 = 540 and 540 × 0.75 = 405 ✓.
Answer: B. $720.
Correct.
Solution
- Value now = P × 0.75².
- P = 405 ÷ 0.75² = 405 ÷ 0.5625.
- P = $720. Check: 720 × 0.75 = 540 and 540 × 0.75 = 405 ✓.
Not quite. This takes 2 × 25% = 50% as one fall. Use 0.75² = 0.5625.
Solution
- Value now = P × 0.75².
- P = 405 ÷ 0.75² = 405 ÷ 0.5625.
- P = $720. Check: 720 × 0.75 = 540 and 540 × 0.75 = 405 ✓.
Answer: B. $720.
Not quite. This adds 50%. Working backwards means dividing by the multiplier.
Solution
- Value now = P × 0.75².
- P = 405 ÷ 0.75² = 405 ÷ 0.5625.
- P = $720. Check: 720 × 0.75 = 540 and 540 × 0.75 = 405 ✓.
Answer: B. $720.
5.A painting was bought for $12,500. Two years later it was worth $15,125. Its value rose by the same percentage each year. Find that yearly percentage.
Not quite. 1.21 is the multiplier for both years together. Take the square root to get one year.
Solution
- Multiplier for 2 years: 15,125 ÷ 12,500 = 1.21.
- One year: √1.21 = 1.1.
- A multiplier of 1.1 means a 10% rise each year.
Answer: C. 10%.
Not quite. This halves the 21% rise. The rises multiply, so use a square root.
Solution
- Multiplier for 2 years: 15,125 ÷ 12,500 = 1.21.
- One year: √1.21 = 1.1.
- A multiplier of 1.1 means a 10% rise each year.
Answer: C. 10%.
Correct.
Solution
- Multiplier for 2 years: 15,125 ÷ 12,500 = 1.21.
- One year: √1.21 = 1.1.
- A multiplier of 1.1 means a 10% rise each year.
Not quite. 1.1 is the multiplier. The rise is 0.1, which is 10%.
Solution
- Multiplier for 2 years: 15,125 ÷ 12,500 = 1.21.
- One year: √1.21 = 1.1.
- A multiplier of 1.1 means a 10% rise each year.
Answer: C. 10%.
6.The table shows the value of a van at the end of each year. It falls by the same percentage every year. What will it be worth at the end of year 5, to the nearest cent?
| End of year | Value of the van |
|---|---|
| 0 | $30,000 |
| 1 | $25,500 |
| 2 | $21,675 |
Correct.
Solution
- 25,500 ÷ 30,000 = 0.85, so the van loses 15% each year.
- Year 5: 30,000 × 0.85⁵.
- = 13,311.159375, which is $13,311.16.
Not quite. This takes off $4,500 every year. The fall is 15% of a smaller value each year.
Solution
- 25,500 ÷ 30,000 = 0.85, so the van loses 15% each year.
- Year 5: 30,000 × 0.85⁵.
- = 13,311.159375, which is $13,311.16.
Answer: A. $13,311.16.
Not quite. This is the value at the end of year 4.
Solution
- 25,500 ÷ 30,000 = 0.85, so the van loses 15% each year.
- Year 5: 30,000 × 0.85⁵.
- = 13,311.159375, which is $13,311.16.
Answer: A. $13,311.16.
Not quite. This starts from the year 2 value and goes 5 more years. Start from $30,000.
Solution
- 25,500 ÷ 30,000 = 0.85, so the van loses 15% each year.
- Year 5: 30,000 × 0.85⁵.
- = 13,311.159375, which is $13,311.16.
Answer: A. $13,311.16.
7.A town has 42,000 people. Its population grows by 3.5% each year. After how many whole years does it first pass 50,000?
Not quite. After 5 years there are about 49,883 people: not yet past 50,000.
Solution
- Multiplier: 1.035.
- Trial: 42,000 × 1.035⁵ ≈ 49,883, which is below 50,000.
- 42,000 × 1.035⁶ ≈ 51,629, which is above 50,000. So 6 years.
Answer: B. 6 years.
Correct.
Solution
- Multiplier: 1.035.
- Trial: 42,000 × 1.035⁵ ≈ 49,883, which is below 50,000.
- 42,000 × 1.035⁶ ≈ 51,629, which is above 50,000. So 6 years.
Not quite. After 6 years there are already about 51,629 people.
Solution
- Multiplier: 1.035.
- Trial: 42,000 × 1.035⁵ ≈ 49,883, which is below 50,000.
- 42,000 × 1.035⁶ ≈ 51,629, which is above 50,000. So 6 years.
Answer: B. 6 years.
Not quite. 3.5% = 0.035, so the multiplier is 1.035, not 1.35.
Solution
- Multiplier: 1.035.
- Trial: 42,000 × 1.035⁵ ≈ 49,883, which is below 50,000.
- 42,000 × 1.035⁶ ≈ 51,629, which is above 50,000. So 6 years.
Answer: B. 6 years.
Challenge (Harder problems that mix ideas or need reasoning.)
8.A car costs $18,000 and loses 14% of its value each year. Which working shows that after 5 years it is worth less than half of what it cost?
Not quite. This takes off 5 × 14% = 70% at once. Each year’s 14% is of a smaller value.
Solution
- Half of $18,000 is $9,000.
- After 5 years: 18,000 × 0.86⁵ = 8,467.686….
- $8,467.69 < $9,000 ✓.
Answer: C. 18,000 × 0.86⁵ = 8,467.686… < 9,000.
Not quite. 0.14 is the part lost. The car keeps 0.86 of its value each year.
Solution
- Half of $18,000 is $9,000.
- After 5 years: 18,000 × 0.86⁵ = 8,467.686….
- $8,467.69 < $9,000 ✓.
Answer: C. 18,000 × 0.86⁵ = 8,467.686… < 9,000.
Correct.
Solution
- Half of $18,000 is $9,000.
- After 5 years: 18,000 × 0.86⁵ = 8,467.686….
- $8,467.69 < $9,000 ✓.
Not quite. This multiplies by 5. The 5 years make a power: 0.86⁵.
Solution
- Half of $18,000 is $9,000.
- After 5 years: 18,000 × 0.86⁵ = 8,467.686….
- $8,467.69 < $9,000 ✓.
Answer: C. 18,000 × 0.86⁵ = 8,467.686… < 9,000.
9.A computer’s value fell from $2,500 to $1,280 in 3 years. It fell by the same percentage each year. Find the yearly percentage decrease.
Not quite. This shares the total fall of 48.8% equally between the 3 years. The falls multiply, so use a cube root.
Solution
- Multiplier for 3 years: 1,280 ÷ 2,500 = 0.512.
- One year: ∛0.512 = 0.8.
- A multiplier of 0.8 means a 20% fall each year.
Answer: D. 20%.
Not quite. This is the total fall over 3 years, not the fall each year.
Solution
- Multiplier for 3 years: 1,280 ÷ 2,500 = 0.512.
- One year: ∛0.512 = 0.8.
- A multiplier of 0.8 means a 20% fall each year.
Answer: D. 20%.
Not quite. 0.8 is the multiplier: the computer keeps 80% each year. The fall is 20%.
Solution
- Multiplier for 3 years: 1,280 ÷ 2,500 = 0.512.
- One year: ∛0.512 = 0.8.
- A multiplier of 0.8 means a 20% fall each year.
Answer: D. 20%.
Correct.
Solution
- Multiplier for 3 years: 1,280 ÷ 2,500 = 0.512.
- One year: ∛0.512 = 0.8.
- A multiplier of 0.8 means a 20% fall each year.
10.Car A costs $20,000 and loses 18% of its value each year. Car B costs $16,000 and loses 10% of its value each year. After how many whole years is car A first worth less than car B?
Not quite. After 2 years car A ($13,448) is still worth more than car B ($12,960).
Solution
- Multipliers: 0.82 for car A and 0.9 for car B.
- After 2 years: A = $13,448, B = $12,960. A is still more.
- After 3 years: A = $11,027.36, B = $11,664. A is less, so the answer is 3 years.
Answer: D. 3 years.
Not quite. Car A is already worth less after 3 years: $11,027.36 against $11,664.
Solution
- Multipliers: 0.82 for car A and 0.9 for car B.
- After 2 years: A = $13,448, B = $12,960. A is still more.
- After 3 years: A = $11,027.36, B = $11,664. A is less, so the answer is 3 years.
Answer: D. 3 years.
Not quite. Car A loses value faster, so it must drop below car B at some point.
Solution
- Multipliers: 0.82 for car A and 0.9 for car B.
- After 2 years: A = $13,448, B = $12,960. A is still more.
- After 3 years: A = $11,027.36, B = $11,664. A is less, so the answer is 3 years.
Answer: D. 3 years.
Correct.
Solution
- Multipliers: 0.82 for car A and 0.9 for car B.
- After 2 years: A = $13,448, B = $12,960. A is still more.
- After 3 years: A = $11,027.36, B = $11,664. A is less, so the answer is 3 years.
Buying a used van
A van costs $28,500 new. It loses 16% of its value each year. Give money answers to the nearest cent.
1. Find the value of the van after 3 years.
2. After how many whole years is it first worth less than $10,000? Show your trial values.
3. A dealer says, "In 3 years the van loses 3 × 16% = 48% of its value." Is he right? Explain.
- V = 28,500 × 0.84³ = 16,892.064, so $16,892.06.
- Year 6: 28,500 × 0.84⁶ = $10,011.99 (just above $10,000). Year 7: 28,500 × 0.84⁷ = $8,410.07 (below). So 7 years.
- No. After 3 years the van keeps 0.84³ = 0.592704 of its value, so it loses 1 − 0.592704 = 0.407296, about 40.7%, not 48%. Each year’s 16% is taken from a smaller value.
Next: the Lesson 2 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.