Exponential Growth and Decay (Copy)
Practice Questions — 1.17 Exponential Growth and Decay
Question 1
A car is bought for Rs. 2 400 000. It depreciates at 12% per year.
(a) Find its value after 1 year.
(b) Find its value after 3 years.
Question 2
The population of a village is 8000 and increases by 5% each year.
(a) Find the population after 1 year.
(b) Find the population after 4 years.
Question 3
A machine worth Rs. 150 000 depreciates by 8% per year.
Find its value after 5 years.
Question 4
The number of bacteria in a culture doubles every 6 hours. If there are 300 bacteria at the start,
(a) How many will there be after 6 hours?
(b) How many will there be after 24 hours?
Question 5
A city’s population of 2.5 million grows at a rate of 3% per year.
Find the population after 10 years, correct to the nearest thousand.
Written and Compiled By Sir Hunain Zia, World Record Holder With 154 Total A Grades, 7 Distinctions and 11 World Records For Educate A Change O Level And IGCSE Mathematics Full Scale Course
Question 6
The value of a phone is Rs. 160 000. It loses 18% of its value each year.
Find its value after 2 years.
Question 7
The half-life of a radioactive substance is 10 hours. A sample initially has 640 g.
(a) Find the mass after 10 hours.
(b) Find the mass after 30 hours.
Question 8
A house is worth Rs. 5 200 000. It appreciates at 4% per year.
Find its value after 6 years.
Question 9
The population of a country decreases by 2% per year. If the population is currently 50 million,
(a) Find the population after 1 year.
(b) Find the population after 5 years.
Question 10
A car is bought for Rs. 1 800 000. It loses 15% of its value each year.
After how many years will it first fall below Rs. 1 000 000?
Written and Compiled By Sir Hunain Zia, World Record Holder With 154 Total A Grades, 7 Distinctions and 11 World Records For Educate A Change O Level And IGCSE Mathematics Full Scale Course
Answer key and explanations — 1.17 Exponential Growth and Decay
1.
Value after 1 year = 2 400 000 × 0.88 = Rs. 2 112 000
Value after 3 years = 2 400 000 × 0.88³ = 2 400 000 × 0.681472 = Rs. 1 635 533
Explanation: Depreciation factor = (1 − 0.12) = 0.88, apply repeatedly for each year.
2.
Population after 1 year = 8000 × 1.05 = 8400
Population after 4 years = 8000 × 1.05⁴ = 8000 × 1.21550625 = 9724 (nearest whole number)
Explanation: Growth factor = 1 + 0.05 = 1.05, raise to number of years.
3.
Value after 5 years = 150 000 × 0.92⁵ = 150 000 × 0.659081 ≈ Rs. 98 862
Explanation: Depreciation factor = 1 − 0.08 = 0.92.
4.
(a) After 6 hours (1 doubling) = 300 × 2 = 600
(b) After 24 hours = 24 ÷ 6 = 4 doublings → 300 × 2⁴ = 300 × 16 = 4800
Explanation: For doubling, use 2ⁿ where n is the number of periods.
5.
Population after 10 years = 2 500 000 × 1.03¹⁰
= 2 500 000 × 1.343916 ≈ 3 359 790 ≈ 3 360 000 (nearest thousand)
Explanation: Growth factor applied for 10 years.
Written and Compiled By Sir Hunain Zia, World Record Holder With 154 Total A Grades, 7 Distinctions and 11 World Records For Educate A Change O Level And IGCSE Mathematics Full Scale Course
6.
Value after 2 years = 160 000 × 0.82²
= 160 000 × 0.6724 = Rs. 107 584
Explanation: Depreciation factor = 1 − 0.18 = 0.82.
7.
(a) After 10 h (1 half-life): 640 ÷ 2 = 320 g
(b) After 30 h (3 half-lives): 640 ÷ 2³ = 640 ÷ 8 = 80 g
Explanation: Each half-life divides the quantity by 2.
8.
Value after 6 years = 5 200 000 × 1.04⁶
= 5 200 000 × 1.265319 ≈ Rs. 6 579 659
Explanation: Appreciation factor = 1 + 0.04 = 1.04.
9.
(a) After 1 year = 50 000 000 × 0.98 = 49 000 000
(b) After 5 years = 50 000 000 × 0.98⁵
= 50 000 000 × 0.9039208 = 45 196 040
Explanation: Decay factor = 1 − 0.02 = 0.98.
10.
Car value = 1 800 000 × 0.85ⁿ < 1 000 000
0.85ⁿ < 1 000 000 ÷ 1 800 000 = 0.5556
By trial:
0.85⁴ = 0.5220 (< 0.5556), 0.85³ = 0.6141 (> 0.5556)
So after 4 years the value first falls below Rs. 1 000 000.
Written and Compiled By Sir Hunain Zia, World Record Holder With 154 Total A Grades, 7 Distinctions and 11 World Records For Educate A Change O Level And IGCSE Mathematics Full Scale Course