Losing value
Many things are worth less as they get older: a phone, a bike, a car. Losing value over time is called depreciation.
When a value falls by 20% in a year, it keeps 100% − 20% = 80% of itself. Keeping 80% is one multiplication by 0.8, the same multiplier as 20% off a price.
0%: × 1, the amount is unchanged, 100 × 1 = 100
Set −20% and read the one multiplier: 100 × ?
The dial opens at 0%, where 100 stays 100. Turn it down to −20%: 20 of the 100 fall away, 80 are left, and the multiplier is × 0.8.
The first year
A new phone costs $500, and it loses 20% of its value in its first year. 20% of 500 is 100, so after one year it is worth 500 − 100 = $400. In one step: 500 × 0.8 = 400.
Each part is $20. The new phone is worth 25 parts; after one year it is worth 20 parts, $400.
The second year starts from less
In its second year the phone loses 20% again, but 20% of what it is worth at the start of that year, which is $400. 20% of 400 is 80, so it falls to 400 − 80 = $320. In one step: 400 × 0.8 = 320.
The two years together are 500 × 0.8 × 0.8 = 320. Each year is one more multiplication by 0.8.
The first year takes 5 parts away, $100. The second year takes only 4 parts, $80, because 20% of a smaller value is a smaller amount.
Not the same amount every year
The usual mistake is to take $100 off every year: $500, then $400, then $300. That takes 20% of the first price each year. But each year's 20% is a share of the value at the start of that year, so after two years the phone is worth $320, not $300.
The amount lost shrinks every year. The phone loses $100, then $80, then 20% of 320, which is $64. After three years it is worth 500 × 0.8 × 0.8 × 0.8 = $256.
Each year the phone keeps 80% of its value, so each hop down is shorter than the one before.
Worked example: A Bakery Oven Losing a Fixed Percentage of Its Value Each Year
Question A bakery bought an oven for $25000. Each year, the oven loses 20% of the value it had at the start of that year. (a) What is the oven worth at the end of the third year? (b) By what percentage of the price the bakery paid has the oven's value fallen over the three years?
1.Year 1: $25000 is 5 units of $5000. One unit is lost, so 4 units remain: 4 × 5000 = $20000.
Year 1: $25000 is 5 units of $5000. One is lost, so $20000 remains. 2.Year 2: $20000 is 5 units of $4000. One unit is lost, so 4 × 4000 = $16000 remains.
Year 2: $20000 is 5 new units of $4000, so $16000 remains. 3.Year 3: $16000 is 5 units of $3200. One unit is lost, so 4 × 3200 = $12800 remains.
Year 3: $16000 is 5 units of $3200, so $12800 remains. 4.(a) At the end of the third year, the oven is worth $12800.
(a) At the end of the third year the oven is worth $12800. 5.The value lost over the three years is 5000 + 4000 + 3200 = $12200, which agrees with 25000 − 12800 = 12200.
Lost over three years: 5000 + 4000 + 3200 = $12200. 6.(b) As a percentage of the price paid: 1220025000 × 100% = 48.8%. This is less than 3 × 20% = 60%, because each year's 20% is taken from a smaller value.
(b) 1220025000 × 100% = 48.8% of the price paid.
Answer: (a) $12800; (b) 48.8%
Common mistakes
- Taking $5000, which is 20% of the price paid, off every year to reach $10000. Each year's 20% is of the value at the start of that year, and that value gets smaller every year.
- Adding the three percentages to say that the value fell by 60%. The three 20% drops are of three different values, so they cannot be added as percentages of the price paid.