ACT Math Exponential Growth and Decay: 14 Practice Problems with Step-by-Step Explanations
Exponential growth and decay problems on the ACT model quantities that change by a fixed percentage over repeated equal time intervals — population growth, investment returns, depreciating value, doubling times, and radioactive-style decay all follow the same underlying structure: A(t) = A0·bt, where A0 is the starting amount, b is the growth or decay factor, and t counts how many intervals have passed. The entire chapter comes down to correctly identifying these three pieces from a word problem.
For growth, the factor b is 1 plus the percent increase (written as a decimal) — a 3% annual increase gives b=1.03. For decay, b is 1 minus the percent decrease — a 15% annual decrease gives b=0.85. When a problem describes doubling or tripling instead of a percentage, b is simply 2 or 3, and t becomes the number of doubling (or tripling) periods that have elapsed, found by dividing the total elapsed time by the length of one period — this division doesn't always come out to a whole number, which is completely fine and expected.
Common traps include mixing up units (a growth rate given "per minute" needs the time variable converted to minutes, not left in hours), forgetting that "decreases by one third" means the quantity retains two-thirds each period (b=2/3), not that it decreases to one-third, and on problems describing growth in a table rather than a stated rate, forgetting to first calculate the growth factor from the given data points before building the exponential model. Work through the 14 problems below, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Modeling percent growth
- Problem 2 — Doubling time
- Problem 3 — Modeling percent decay
- Problem 4 — Decay over a fixed number of periods
- Problem 5 — Evaluating a growth model
- Problem 6 — Fractional decay rate
- Problem 7 — Evaluating a pricing model
- Problem 8 — Finding the growth factor
- Problem 9 — Doubling time as an expression
- Problem 10 — Applying a decay factor
- Problem 11 — Non-whole-number doubling periods
- Problem 12 — Building a model from a table
- Problem 13 — Solving for the exponent variable
- Problem 14 — Matching units in the exponent
Practice Problems
Trayvon is buying 5,436 fish to start his fish farm. The number of fish in his pond are estimated to increase at a rate of 3% per year. Which equation models the total number of fish, P, in the pond t years from now?
- A) P(t) = 5,436(1.03)t
- B) P(t) = 5,436(0.03)t
- C) P(t) = 0.03(5,436)t
- D) P(t) = 1.03(5,436)t
For a 3% annual increase, the growth factor is 1 + 0.03 = 1.03. The starting amount, 5,436, is the base multiplier, and the growth factor is raised to the power t.
P(t) = 5,436(1.03)t.
The number of ants in a colony grows exponentially, doubling every 5 days. What is the population of the colony, in thousands, exactly 30 days after the population first reaches a population of 3,000 ants?
- A) 48
- B) 75
- C) 96
- D) 192
30 days is 30/5 = 6 doubling periods. Population = 3,000 × 26 = 3,000 × 64 = 192,000.
In thousands, that's 192.
A surfboard depreciates at an annual rate of 15%. If the initial value of the surfboard is $995, which of the following functions P correctly models the price of the surfboard t years from purchase?
- A) P(t) = 0.15(995)t
- B) P(t) = 0.85(995)t
- C) P(t) = 995(1.15)t
- D) P(t) = 995(0.85)t
For a 15% annual decrease, the decay factor is 1 − 0.15 = 0.85. The starting value, 995, is the base multiplier.
P(t) = 995(0.85)t.
Priscilla selects a random number on Monday. Every day, the number decreases by 40%. If Priscilla selects 340 on Monday, which of the following expressions is equal to the number the following Monday?
- A) 340(0.6)7
- B) 340(0.4)7
- C) 340(1.4)7
- D) 340 − 7(0.4)
A 40% decrease each day means the number retains 60% (1−0.40=0.60) of its previous value each day. The following Monday is 7 days later.
340(0.6)7.
The population of otters in a lake, l, at the beginning of each year can be modeled by the equation l(x) = 5(2x), where x represents the number of years after the beginning of the year 2010. For example, x=0 represents the beginning of the year 2010, x=1 represents the beginning of the year in 2011, and so on. According to the model, how many otters were in the lake at the beginning of the year 2015?
- A) 160
- B) 320
- C) 640
- D) 1,280
2015 is 5 years after 2010, so x=5.
l(5) = 5(25) = 5(32) = 160.
An invasive species of fish that preys on snails was introduced to Lake Michigan in 2008. Since the introduction of the invasive species, the population of snails has decreased by one third each year. If the population of snails was estimated to be 81,000 in January 2018, what was the population estimated to be in January 2022?
- A) 36,000
- B) 27,000
- C) 24,000
- D) 16,000
Decreasing by one third each year means the population retains two-thirds (2/3) of its previous value each year. From January 2018 to January 2022 is 4 years.
81,000 × (2/3)4 = 81,000 × (16/81) = 16,000.
Cocoa Daily sells dark chocolate almonds in bulk online. The price per pound, P(x), for a store to purchase x pounds of dark chocolate almonds is given by the function below.
P(x) = 6.5 + 0.95x
Which of the following dollar values is closest to the total price to purchase 90 pounds of dark chocolate almonds from Cocoa Daily?
- A) $499.50
- B) $586.00
- C) $594.50
- D) $670.50
First find the price per pound at x=90: P(90) = 6.5 + 0.9590. Since 0.9590 is very small (about 0.0099), P(90) ≈ 6.51 dollars per pound.
Total price for 90 pounds: 90 × 6.51 ≈ $585.89, which is closest to $586.00.
The function f(x) = 75,000bx models a stock's annual value, in dollars, x years after being placed on the stock market, where b is a constant. If the stock's value increases 6% per year, what is the value of b?
- A) 0.06
- B) 0.6
- C) 1.06
- D) 1.6
For a 6% annual increase, the growth factor is 1 + 0.06 = 1.06.
Starting in 1950, the number of people living in Alaska doubled every 20 years. The population of Alaska was 150,000 in 1950. Which of the following expressions gives the population of Alaska in 2010?
- A) 150,000(2)3
- B) 150,000(2)20
- C) 150,000(2)60
- D) 150,000(2)(60)
From 1950 to 2010 is 60 years, which is 60/20 = 3 doubling periods.
150,000(2)3.
Each year after a car is purchased, the price is estimated to be 15% less than the value the previous year. If the initial purchase price of a car was $29,500, which of the following is closest to the price of the car 3 years after it was purchased?
- A) $13,000
- B) $16,000
- C) $16,225
- D) $18,100
The decay factor is 1 − 0.15 = 0.85. After 3 years: 29,500(0.85)3 ≈ 29,500(0.614) ≈ $18,117.
This is closest to $18,100.
The population of humpback whales in the North Pacific have been doubling every 11 years since 1980. If the population of humpback whales was estimated to be 5,500 in 1980, the population of humpback whales in 2010 was estimated to be:
- A) 5,500 to 11,000
- B) 11,000 to 22,000
- C) 22,000 to 33,000
- D) 33,000 to 44,000
From 1980 to 2010 is 30 years, which is 30/11 ≈ 2.73 doubling periods (this doesn't need to be a whole number).
5,500 × 230/11 ≈ 5,500 × 6.62 ≈ 36,400, which falls in the range 33,000 to 44,000.
From the 2nd month to the 6th month of its life, a golden retriever experiences exponential growth. The table below shows the weight of a golden retriever every 2 months.
| Age (months) | Weight (pounds) |
|---|---|
| 2 | 8 |
| 4 | 18 |
| 6 | 40.5 |
Which of the following equations most closely models the weight of the golden retriever, W(m), m months after 2 months, for ages of 2 months to 6 months?
- A) W(m) = 8(2.25)m/2
- B) W(m) = 8(2.25)2m
- C) W(m) = 8(1.25)m/2
- D) W(m) = 8(1.25)2m
Find the growth factor between table values: 18/8=2.25, and 40.5/18=2.25 — the weight multiplies by 2.25 every 2-month interval, confirming exponential growth.
Since m counts months since age 2 (with starting weight 8), and the growth factor of 2.25 applies once per 2-month interval, the exponent should be m/2: W(m) = 8(2.25)m/2.
The amount of nitrogen, in milligrams per kilogram of soil, in a soil sample is given by the formula N = N0(32m), where N is the total nitrogen in a sample m miles from a site where a water treatment plant just opened and N0 is the initial amount of nitrogen in the soil before the treatment plant opened. Which of the following expressions gives the number of miles where an initial nitrogen concentration of 4 milligram per kilogram is now measured to be 108 milligrams per kilogram?
- A) 0.5
- B) 1.5
- C) 2
- D) 3
Substitute the given values: 108 = 4(32m). Divide both sides by 4: 27 = 32m.
Since 27=33, this means 2m=3, so m=1.5.
The equation below models the number of bacteria, in thousands, on a petri dish h hours after the dish has been inoculated. According to the model, the number of bacteria is predicted to increase by 2% every n minutes. What is the value of n?
F(h) = 62(1.02)h/5
- A) 5
- B) 12
- C) 120
- D) 300
h hours equals 60h minutes. If the population increases 2% every n minutes, the number of n-minute intervals in 60h minutes is 60h/n, which should match the given exponent h/5.
Set 60h/n = h/5: 60/n = 1/5, so n = 60 × 5 = 300.
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