Math 1 Quiz: Exponential Growth Decay Modeling

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Exponential Growth Decay ModelingQuestion 1 of 20

The population of a town decreases exponentially. In 2020, the population was 12,000. In 2023, it was 9,600. If this trend continues, what will be the population in 2028?

Approximately 6,144 people

Approximately 6,553 people

Approximately 7,372 people

Approximately 7,680 people

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Math 1 Quiz

Math 1 Quiz: Exponential Growth Decay Modeling

Practice Exponential Growth Decay Modeling in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exponential Growth Decay Modeling, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The population of a town decreases exponentially. In 2020, the population was 12,000. In 2023, it was 9,600. If this trend continues, what will be the population in 2028?

Approximately 6,144 people (correct answer)
Approximately 6,553 people
Approximately 7,372 people
Approximately 7,680 people

Explanation: Using

P(t) = P_0 e^{rt}

with

t

years after 2020. We have

9600 = 12000e^{3r}

, so

0.8 = e^{3r}

, giving

r = \frac{\ln(0.8)}{3} \approx -0.0744

. For 2028 (

t = 8

P(8) = 12000e^{8(-0.0744)} = 12000e^{-0.595} \approx 12000(0.512) \approx 6,144

. Choice B uses linear decay. Choice C uses incorrect time interval. Choice D uses wrong initial conditions.

Question 2

The value of a car depreciates exponentially. A car worth $28,000 new is worth $19,600 after 3 years. What will be its value after 8 years?

Approximately $9,680
Approximately $11,424 (correct answer)
Approximately $13,720
Approximately $15,288

Explanation: Using

V(t) = V_0 e^{rt}

19600 = 28000e^{3r}

, so

0.7 = e^{3r}

, giving

r = \frac{\ln(0.7)}{3} \approx -0.119

. For

t = 8

V(8) = 28000e^{8(-0.119)} = 28000e^{-0.952} \approx 28000(0.408) \approx 11,424

. Choice A uses linear depreciation. Choice C uses incorrect decay rate. Choice D uses wrong time calculation.

Question 3

A research study tracks the spread of an invasive plant species in a national park. The data collected shows exponential growth patterns.

The invasive plant covers 50 acres initially. After 2 years, it covers 200 acres. If growth continues at this rate, how many acres will be covered after 6 years total?

800 acres
1,200 acres
1,600 acres
3,200 acres (correct answer)

Explanation: Using

A(t) = A_0 e^{rt}

200 = 50e^{2r}

, so

4 = e^{2r}

, giving

r = \frac{\ln(4)}{2} \approx 0.693

. For

t = 6

A(6) = 50e^{6(0.693)} = 50e^{4.158} = 50(64) = 3,200

acres. Choice A assumes linear growth. Choice B uses incorrect exponential calculation. Choice C uses the wrong time period in calculations.

Question 4

A bacterial culture starts with 500 bacteria. After 3 hours, the population has grown to 4,000 bacteria. If the growth follows an exponential model $P(t) = P_0 e^{rt}$ , what will be the approximate population after 5 hours?

8,000 bacteria
10,667 bacteria
12,800 bacteria (correct answer)
16,000 bacteria

Explanation: First, find the growth rate:

4000 = 500e^{3r}

, so

8 = e^{3r}

, giving

r = \frac{\ln(8)}{3} \approx 0.693

. Then

P(5) = 500e^{5(0.693)} = 500e^{3.465} \approx 500(25.6) \approx 12,800

. Choice A assumes linear growth. Choice B uses incorrect arithmetic in the exponential calculation. Choice D incorrectly doubles the 4-hour population.

Question 5

The concentration of a medication in the bloodstream follows the model $C(t) = 20e^{-0.4t}$ mg/L, where $t$ is hours after injection. After how many hours will the concentration drop to 2.5 mg/L?

Approximately 3.2 hours
Approximately 4.6 hours
Approximately 5.2 hours (correct answer)
Approximately 6.1 hours

Explanation: Set

2.5 = 20e^{-0.4t}

. Dividing by 20:

0.125 = e^{-0.4t}

. Taking natural log:

\ln(0.125) = -0.4t

, so

t = \frac{\ln(0.125)}{-0.4} = \frac{-2.079}{-0.4} \approx 5.2

hours. Choice A uses incorrect logarithm calculation. Choice B forgets the negative sign. Choice D uses wrong decay constant.

Question 6

Two bacterial cultures start with the same initial population. Culture A doubles every 4 hours, while Culture B triples every 6 hours. After 12 hours, what is the ratio of Culture B's population to Culture A's population?

$\frac{8}{9}$
$\frac{9}{8}$ (correct answer)
$\frac{27}{8}$
$\frac{8}{27}$

Explanation: After 12 hours: Culture A undergoes

\frac{12}{4} = 3

doubling periods, so population becomes

P_0 \cdot 2^3 = 8P_0

. Culture B undergoes

\frac{12}{6} = 2

tripling periods, so population becomes

P_0 \cdot 3^2 = 9P_0

. The ratio is

\frac{9P_0}{8P_0} = \frac{9}{8}

. Choice A inverts the ratio. Choice C uses incorrect time calculations. Choice D combines both errors.

Question 7

A substance decays according to the model $N(t) = N_0 e^{-0.035t}$ , where $t$ is in days. What is the half-life of this substance?

Approximately 14.3 days
Approximately 19.8 days (correct answer)
Approximately 24.2 days
Approximately 28.6 days

Explanation: Half-life occurs when

N(t) = \frac{N_0}{2}

. Setting up:

\frac{N_0}{2} = N_0 e^{-0.035t}

. Dividing by

N_0

0.5 = e^{-0.035t}

. Taking natural log:

\ln(0.5) = -0.035t

, so

t = \frac{\ln(0.5)}{-0.035} = \frac{0.693}{0.035} \approx 19.8

days. Choice A uses incorrect logarithm value. Choice C confuses the decay constant with time. Choice D doubles the correct answer incorrectly.

Question 8

A cup of coffee at 180°F is placed in a room where the temperature is 70°F. After 5 minutes, the coffee temperature is 150°F. Using Newton's Law of Cooling, what will be the temperature after 15 minutes?

Approximately 105°F (correct answer)
Approximately 112°F
Approximately 118°F
Approximately 125°F

Explanation: Newton's Law:

T(t) = T_a + (T_0 - T_a)e^{-kt}

where

T_a = 70

T_0 = 180

. So

T(t) = 70 + 110e^{-kt}

. Using

T(5) = 150

150 = 70 + 110e^{-5k}

, so

80 = 110e^{-5k}

, giving

e^{-5k} = \frac{8}{11}

. Therefore

k = -\frac{\ln(8/11)}{5} \approx 0.0654

. Then

T(15) = 70 + 110e^{-15(0.0654)} \approx 70 + 110(0.318) \approx 105°F

. Other choices use incorrect decay constants or computational errors.

Question 9

A cup of coffee at 180°F is placed in a room at 70°F. The temperature follows Newton's law of cooling: $T(t) = 70 + 110e^{-0.08t}$ , where $t$ is in minutes. What is the coffee's temperature after 30 minutes?

79.1°F (correct answer)
81.6°F
84.2°F
87.3°F

Explanation: Substitute

t = 30

into the cooling model:

T(30) = 70 + 110e^{-0.08(30)} = 70 + 110e^{-2.4}

. Since

e^{-2.4} \approx 0.0907

, we get

T(30) = 70 + 110(0.0907) = 70 + 9.97 \approx 79.1°F

Question 10

A city's population grows from 50,000 to 65,000 over 4 years. Assuming exponential growth, what will the population be after 10 years from the initial measurement?

89,420 people (correct answer)
93,750 people
97,200 people
101,500 people

Explanation: First find the growth rate:

65,000 = 50,000 \cdot r^4

, so

r^4 = 1.3

and

r = 1.3^{1/4} \approx 1.0678

. The model is

P(t) = 50,000 \cdot (1.0678)^t

. After 10 years:

P(10) = 50,000 \cdot (1.0678)^{10} \approx 50,000 \cdot 1.7884 \approx 89,420

people.

Question 11

A virus spreads through a population according to $I(t) = 25e^{0.4t}$ , where $I(t)$ is the number of infected people after $t$ days. On which day will the number of infected people first exceed 500?

Day 6
Day 7
Day 8 (correct answer)
Day 9

Explanation: Set up the inequality:

25e^{0.4t} > 500

. Dividing by 25:

e^{0.4t} > 20

. Taking natural log:

0.4t > \ln(20) \approx 2.996

. So

t > \frac{2.996}{0.4} \approx 7.49

days. Since we need the first day when this occurs, it's day 8.

Question 12

A car's value depreciates exponentially. It loses 15% of its value each year. If the car is worth $18,000 after 3 years, what was its original value?

$28,950
$29,200
$29,650 (correct answer)
$30,100

Explanation: If the car loses 15% annually, it retains 85% of its value. The model is

V(t) = V_0(0.85)^t

. Given

V(3) = 18,000

18,000 = V_0(0.85)^3

. So

V_0 = \frac{18,000}{(0.85)^3} = \frac{18,000}{0.614125} \approx 29,650

Question 13

A bacterial culture starts with 500 bacteria and doubles every 3 hours. After 15 hours, a researcher removes 2000 bacteria from the culture. How many bacteria remain in the culture immediately after the removal?

14,000 bacteria (correct answer)
16,000 bacteria
18,000 bacteria
20,000 bacteria

Explanation: The exponential growth model is

P(t) = 500 \cdot 2^{t/3}

. After 15 hours:

P(15) = 500 \cdot 2^{15/3} = 500 \cdot 2^5 = 500 \cdot 32 = 16,000

. After removing 2000 bacteria:

16,000 - 2000 = 14,000

bacteria remain.

Question 14

A radioactive sample has the decay model $A(t) = 500e^{-0.08t}$ grams, where $t$ is in years. How much of the sample will remain after one half-life period?

125 grams
167 grams
250 grams (correct answer)
375 grams

Explanation: By definition, after one half-life, exactly half of the original amount remains. Since the initial amount is 500 grams, after one half-life there will be

\frac{500}{2} = 250

grams. This is true regardless of the specific decay constant. Choice A represents one-quarter remaining (two half-lives). Choice B represents an incorrect calculation. Choice D represents three-quarters remaining.

Question 15

The half-life of a radioactive isotope is 8 years. If a sample initially contains 240 grams of the isotope, which equation best models the amount remaining after $t$ years?

$A(t) = 240(0.5)^{t/8}$ (correct answer)
$A(t) = 240(0.5)^{8t}$
$A(t) = 240e^{-8t}$
$A(t) = 240(2)^{-t/8}$

Explanation: For half-life problems, the standard form is

A(t) = A_0(0.5)^{t/h}

where

h

is the half-life. With

A_0 = 240

and

h = 8

, we get

A(t) = 240(0.5)^{t/8}

. Choice B incorrectly places

t

in the numerator making the exponent

8t

instead of

t/8

. Choice C uses an exponential form but with the wrong decay constant. Choice D, while mathematically equivalent to A since

(2)^{-t/8} = (0.5)^{t/8}

, uses a less standard form.

Question 16

Carbon-14 has a half-life of 5,730 years. An archaeological sample contains 25% of its original Carbon-14. Approximately how old is the sample?

22,920 years old
14,325 years old
17,190 years old
11,460 years old (correct answer)

Explanation: When you encounter radioactive decay problems, you're working with exponential decay where the amount decreases by half over each half-life period. The key is determining how many half-lives have passed to reach the current amount. Start with the decay formula:

N = N_0 \times (1/2)^t

, where N is the current amount, N₀ is the original amount, and t is the number of half-lives. Here, you have 25% (or 0.25) of the original Carbon-14 remaining. Setting up the equation:

0.25 = 1 \times (1/2)^t

To solve for t, recognize that 0.25 = 1/4 = (1/2)². This means

(1/2)^t = (1/2)^2

, so t = 2 half-lives. Since each half-life is 5,730 years, the sample is 2 × 5,730 = 11,460 years old, confirming answer D. Answer A (22,920 years) represents 4 half-lives, which would leave only 6.25% of the original Carbon-14. Answer B (14,325 years) equals 2.5 half-lives, leaving about 17.7% remaining. Answer C (17,190 years) represents 3 half-lives, which would leave 12.5% of the original amount. Remember this pattern: after 1 half-life you have 50%, after 2 half-lives you have 25%, after 3 half-lives you have 12.5%, and so on. Recognizing these fractional relationships (1/2, 1/4, 1/8) will help you quickly identify how many half-lives have elapsed without complex calculations.

Question 17

A viral video has 1,200 views on day 1. The number of views grows exponentially, tripling every 4 days. Which expression represents the number of views after $d$ days?

$V(d) = 1200(3)^{d/4}$ (correct answer)
$V(d) = 1200(3)^{4d}$
$V(d) = 1200 + 3^{d/4}$
$V(d) = 1200(4)^{d/3}$

Explanation: For exponential growth that triples every 4 days, the general form is

V(d) = V_0 \cdot 3^{d/4}

where

V_0 = 1200

. This gives

V(d) = 1200(3)^{d/4}

. Choice B incorrectly places the time variable. Choice C uses addition instead of multiplication. Choice D switches the base and the time period in the exponent.

Question 18

A savings account earns 4.2% annual interest compounded continuously. If $2,500 is deposited, which inequality represents when the account balance will exceed $4,000?

$t > \frac{\ln(2500)}{0.042}$
$t > \frac{\ln(4000)}{0.042}$
$t > \frac{0.042}{\ln(1.6)}$
$t > \frac{\ln(1.6)}{0.042}$ (correct answer)

Explanation: When you encounter continuous compound interest problems, you're working with exponential growth where the formula is

A = Pe^{rt}

, where A is the final amount, P is the principal, r is the interest rate, and t is time. Here, you need to find when $2,500 grows to exceed $4,000 at 4.2% continuously compounded. Setting up the inequality: $

4000 < 2500e^{0.042t}

$ To solve for t, first divide both sides by 2500:

\frac{4000}{2500} < e^{0.042t}

, which simplifies to

1.6 < e^{0.042t}

Taking the natural logarithm of both sides:

\ln(1.6) < 0.042t

Finally, divide by 0.042:

t > \frac{\ln(1.6)}{0.042}

This matches answer choice D. Let's examine why the other options are incorrect: A)

t > \frac{\ln(2500)}{0.042}

uses the natural log of the initial deposit instead of the growth ratio. B)

t > \frac{\ln(4000)}{0.042}

uses the natural log of the target amount rather than the growth ratio. C)

t > \frac{0.042}{\ln(1.6)}

has the interest rate and natural log flipped in the fraction, which would give you the wrong units and an incorrect result. Study tip: In continuous compound interest problems, always look for the ratio between final and initial amounts (here, 4000/2500 = 1.6). The natural log of this ratio, divided by the interest rate, gives you the time needed. This pattern appears frequently on standardized tests.

Question 19

An investment of $5,000 grows continuously at an annual rate of 6.5%. Using the continuous compound interest model $ $A = Pe^{rt}$ $, after how many years will the investment first exceed $8,000?

Approximately 6.2 years
Approximately 7.3 years (correct answer)
Approximately 8.1 years
Approximately 9.4 years

Explanation: Set up

8000 = 5000e^{0.065t}

. Dividing by 5000:

1.6 = e^{0.065t}

. Taking natural log:

\ln(1.6) = 0.065t

, so

t = \frac{\ln(1.6)}{0.065} \approx \frac{0.470}{0.065} \approx 7.3

years. Choice A uses incorrect logarithm calculation. Choice C confuses annual compounding with continuous compounding. Choice D uses the wrong interest rate in calculations.

Question 20

Carbon-14 has a half-life of 5,730 years. An archaeologist finds a bone fragment with 35% of its original carbon-14 remaining. Approximately how old is the bone fragment?

7,200 years old
8,400 years old (correct answer)
9,100 years old
10,600 years old

Explanation: Use the decay model:

0.35 = (\frac{1}{2})^{t/5730}

. Taking logarithms:

\ln(0.35) = \frac{t}{5730} \ln(\frac{1}{2})

. So

t = \frac{5730 \ln(0.35)}{\ln(0.5)} = \frac{5730(-1.0498)}{-0.6931} \approx 8,680

years. The closest answer is 8,400 years.