1. Increase or Decrease at a Constant Rate

2. Increase at a Constant Rate
A fish is now 8 cm long.
The length of the fish increases at a rate of 10% per month.
How long will it be after n months?
Increases by 10%
After the 1st month
Increases by 10%
After the 2nd month …
8 cm
This question involves successive percentage changes at a constant rate.

3. Increase at a Constant Rate
Increases by 10%
After the 1st month
Increases by 10%
After the 2nd month …
8 cm
Length after the 1st month 8 cm
10%) (1 + × =
Length after the 2nd month = 8
10%) (1 + × cm
10%) (1 + × 8
10%)2 cm (1 + × =
Length after the 3rd month = + 8
10%)2 (1 ×
10%) cm (1 + × 8
10%)3 cm (1 + × = … ∴
Length after the n th month 8
10%)n cm (1 + × =

4. If a value P increases at a constant rate r % in each period, the new value A after n periods can be calculated by the following formula:
A = P  (1 + r %)n
(1 + r %) is called the growth factor, which must be greater than 1.

5. The number of cells after 3 hours = 1000  (1 + 20%)3
For example,
there are 1000 cells at the beginning of an experiment and the number of cells grows at a rate of 20% every hour.
The number of cells after 3 hours =
1000  (1 + 20%)3
Substitute P = 1000, r = 20 and n = 3
Into the formula A = P  (1 + r %)n.
= 1000  1.23
= 1728

6. Follow-up question
1. Amy is now 100 cm tall. If her height increases at a rate of 5% per year, find Amy’s height after 2 years.
Solution
Amy’s height after 2 years
= 100  (1 + 5%)2 cm
= 100  cm
= cm

7. Follow-up question (cont’d)
2. The volume of a certain gas in a room is 40 mL. If the volume increases at a rate of 6% per second, find the volume of the gas in the room after 5 seconds. (Give your answer correct to 2 decimal places.)
Solution
The volume of the gas in the room after 5 seconds
= 40  (1 + 6%)5 mL
= 40  mL
= mL
(cor. to 2 d.p.)

8. Decrease at a Constant Rate
Similarly, if a value P decreases at a constant rate r % in each period, the new value A after n periods can be calculated by the following formula:
A = P  (1 – r %)n
(1 – r %) is called the decay factor, which must be less than 1.

9. the length of a pencil was 10 cm. The length of
For example,
the length of a pencil was 10 cm.
The length of
the pencil was decreased by 3% after each use.
The length of the pencil after being used 2 times =
10  (1 – 3%)2 cm
Substitute P = 10, r = 3 and n = 2
Into the formula A = P  (1 – r %)n.
= 10  cm
= cm

10. and the percentage decrease is called the depreciation rate.
In our daily life, the values of machines and electrical appliances decrease after they have been used. Such a drop in value is called depreciation,
and the percentage decrease is called the depreciation rate.
For example,
the value of a washing machine is $6000 now and its depreciation rate is 30% per year.
$6000
The value of the washing machine 3 years later
30%) (1
$6000 -  = 3
Depreciation means decrease.
0.7
$6000  = 3
$2058 =

11. Follow-up question
1. The population of a city is If the population is decreased at a rate of 15% per year, find the population of the city after 3 years.
Solution
The population of the city after 3 years
=  (1 – 15%)3
=  0.853 =

12. Follow-up question (cont’d)
2. The value of a car is $ now and its depreciation rate is 8% per year. Find its value after 5 years. (Give your answer correct to the nearest dollar.)
Solution
The value of the car after 5 years
= $  (1 – 8%)5
= $  0.925
= $
(cor. to the nearest dollar)