Differential Equations Quiz: Exponential Growth And Decay

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Exponential Growth And DecayQuestion 1 of 19

A drug is administered intravenously at a constant rate of 8 mg/hour. The drug is metabolized exponentially with a half-life of 4 hours. What is the steady-state concentration if the volume of distribution is 20 liters?

Approximately 0.462 mg/L

Approximately 0.693 mg/L

Approximately 0.924 mg/L

Approximately 1.155 mg/L

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Differential Equations Quiz

Differential Equations Quiz: Exponential Growth And Decay

Practice Exponential Growth And Decay in Differential Equations 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 And Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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

Approximately 0.462 mg/L (correct answer)
Approximately 0.693 mg/L
Approximately 0.924 mg/L
Approximately 1.155 mg/L

Explanation: The elimination rate constant is

k = \frac{\ln(2)}{4} \approx 0.1733

hr⁻¹. At steady state, input rate equals elimination rate:

8 = k \cdot C_{ss} \cdot V

, where

C_{ss}

is steady-state concentration. Solving:

C_{ss} = \frac{8}{k \cdot V} = \frac{8}{0.1733 \times 20} \approx 0.462

mg/L. Choice B uses

k = 0.693

(incorrect half-life calculation). Choice C forgets to multiply by volume in denominator. Choice D uses

k = 0.25

(incorrect half-life relationship).

Question 2

A radioactive isotope, Element X, decays into Element Y with a half-life of 20 days. Element Y is also radioactive and decays into a stable Element Z with a half-life of 30 days. If a pure sample initially contains 100 grams of Element X, how much of Element X remains after 60 days?

6.25 g
12.5 g (correct answer)
18.9 g
25.0 g

Explanation: The decay of Element X is described by the equation

A(t) = A_0 (1/2)^{t/T_{1/2}}

, where

A_0 = 100

g is the initial amount and

T_{1/2} = 20

days is the half-life. The information about Element Y is extraneous and intended to be a distractor, as the amount of Element X is not affected by the decay of Element Y. We need to find the amount of Element X remaining at

t=60

days. The number of half-lives that have passed is

t/T_{1/2} = 60/20 = 3

. Therefore, the remaining amount of Element X is

A(60) = 100 \cdot (1/2)^3 = 100 \cdot (1/8) = 12.5

grams.

Question 3

The population of species A, $A(t)$ , grows according to the model $\frac{dA}{dt} = 0.04A$ , with an initial population of $A(0) = 100$ . The population of species B, $B(t)$ , grows according to $\frac{dB}{dt} = 0.02B$ , with an initial population of $B(0) = 200$ . At what time $t > 0$ will the population of species A be twice the population of species B?

$t = 50 \ln(2)$
$t = 100 \ln(2)$ (correct answer)
$t = 100$
$t = 200$

Explanation: First, we write the solutions to the differential equations:

A(t) = 100 e^{0.04t}

and

B(t) = 200 e^{0.02t}

. The problem asks for the time

t

when

A(t) = 2 B(t)

. We set up the equation:

100 e^{0.04t} = 2 (200 e^{0.02t})

. This simplifies to

100 e^{0.04t} = 400 e^{0.02t}

. Dividing both sides by 100 and by

e^{0.02t}

gives

e^{0.04t}/e^{0.02t} = 400/100

, which simplifies to

e^{0.02t} = 4

. Taking the natural logarithm of both sides, we get

0.02t = \ln(4) = 2\ln(2)

. Solving for

t

yields

t = \frac{2\ln(2)}{0.02} = 100 \ln(2)

Question 4

The age of an ancient artifact is estimated using carbon-14 dating, which has a half-life of 5730 years. Analysis shows the artifact contains 25% of the carbon-14 found in a standard living organism. A new study suggests that the initial amount of carbon-14 in the atmosphere at that time was actually 10% lower than the standard value used in the calculation. Based on this new information, what is the revised age of the artifact?

$5730 \frac{\ln(4)}{\ln(2)}$ years
$5730 \frac{\ln(3.6)}{\ln(2)}$ years (correct answer)
$5730 \frac{\ln(4/1.1)}{\ln(2)}$ years
$5730 \frac{\ln(4 \cdot 0.9)}{\ln(2)}$ years

Explanation: Let

A_0

be the standard initial amount of carbon-14. The measured amount is

A_{measured} = 0.25 A_0

. The decay model is

A(t) = A_{initial} e^{kt}

, where

k = -\frac{\ln(2)}{5730}

. The standard calculation assumes

A_{initial} = A_0

, so

0.25A_0 = A_0 e^{kt}

, which gives

t = 2 \times 5730 = 11460

years, which is

5730 \frac{\ln(4)}{\ln(2)}

. The revised calculation uses a different initial amount,

A'_{initial} = (1 - 0.10)A_0 = 0.9 A_0

. The measured amount is still

0.25 A_0

. So the equation becomes

0.25 A_0 = (0.9 A_0) e^{kt'}

. This simplifies to

e^{kt'} = \frac{0.25}{0.9} = \frac{1/4}{9/10} = \frac{10}{36} = \frac{5}{18}

. Taking the natural logarithm,

kt' = \ln(5/18) = -\ln(18/5)

. Solving for the revised age

t'

t' = \frac{-\ln(18/5)}{k} = \frac{-\ln(18/5)}{-\ln(2)/5730} = 5730 \frac{\ln(18/5)}{\ln(2)} = 5730 \frac{\ln(3.6)}{\ln(2)}

years.

Question 5

An initial investment of $1000 is made into an account with continuous compounding at an annual rate $r $. A second investment of $2000 is made into another account with continuous compounding at an annual rate of 3%. After 10 years, the two accounts have the same balance. What is the approximate value of $r$ ?

$3.93\%$
$6.93\%$
$9.93\%$ (correct answer)
$10.33\%$

Explanation: Let

A_1(t)

be the balance of the first account and

A_2(t)

be the balance of the second. The models are

A_1(t) = 1000 e^{rt}

and

A_2(t) = 2000 e^{0.03t}

. We are given that

A_1(10) = A_2(10)

. So,

1000 e^{10r} = 2000 e^{0.03 \cdot 10} = 2000 e^{0.3}

. Divide by 1000 to get

e^{10r} = 2e^{0.3}

. To solve for

r

, we take the natural logarithm of both sides:

\ln(e^{10r}) = \ln(2e^{0.3})

. This gives

10r = \ln(2) + \ln(e^{0.3}) = \ln(2) + 0.3

. Therefore,

r = \frac{\ln(2) + 0.3}{10}

. Using the approximation

\ln(2) \approx 0.693

, we get

r \approx \frac{0.693 + 0.3}{10} = \frac{0.993}{10} = 0.0993

, or

9.93\%

Question 6

The concentration of a drug in a patient's bloodstream decays exponentially with a half-life of 6 hours. A patient is given an initial dose of 400 mg. After 12 hours, a second identical dose of 400 mg is administered. What is the total concentration of the drug in the bloodstream immediately after the second dose is given?

450 mg
800 mg
600 mg
500 mg (correct answer)

Explanation: When you encounter exponential decay problems with multiple doses, you need to track how much of each dose remains at any given time. The key insight is that each dose decays independently according to the same exponential pattern. Since the drug has a 6-hour half-life, you can use the formula

C(t) = C_0 \cdot (1/2)^{t/6}

, where

C_0

is the initial concentration and

t

is time in hours. For the first 400 mg dose after 12 hours:

C(12) = 400 \cdot (1/2)^{12/6} = 400 \cdot (1/2)^2 = 400 \cdot (1/4) = 100

mg remaining. When the second 400 mg dose is administered at the 12-hour mark, you add it to what's left from the first dose: 100 mg + 400 mg = 500 mg total. Choice A (450 mg) likely comes from incorrectly calculating the remaining amount from the first dose—perhaps using a linear decay assumption instead of exponential. Choice B (800 mg) represents the trap of simply adding both full doses without accounting for any decay of the first dose. Choice C (600 mg) might result from using an incorrect decay formula or miscalculating the half-life relationship. The correct answer is D (500 mg). Remember that in multi-dose exponential decay problems, always calculate how much of each previous dose remains before adding new doses. Each dose follows the same decay pattern starting from when it was administered, and the effects are cumulative at any given moment.

Question 7

For a population $P(t)$ that grows according to the differential equation $\frac{dP}{dt} = kP$ , where $k$ is a positive constant, which of the following provides the best interpretation of the constant $k$ ?

The time it takes for the population to double.
The rate at which the population grows per unit of time.
The fraction of the population that is added per unit of time. (correct answer)
The absolute increase in population over the first unit of time.

Explanation: The constant

k

is the relative or specific growth rate. We can see this by rearranging the differential equation to

k = \frac{1}{P}\frac{dP}{dt}

. The term

\frac{dP}{dt}

is the absolute growth rate (e.g., individuals per year), while

\frac{1}{P}\frac{dP}{dt}

is the growth rate per capita, or the fractional increase in the population per unit of time. This matches choice C. (A) is incorrect; the doubling time is

T_2 = \frac{\ln(2)}{k}

. (B) is incorrect; the rate of growth is

\frac{dP}{dt}

, which is not constant but depends on

P

. (D) is incorrect; the absolute increase over the first unit of time is

P(1) - P(0) = P_0 e^k - P_0 = P_0(e^k-1)

, which depends on the initial population

P_0

Question 8

The rate of decay of a radioactive substance is proportional to the amount present. If 10% of the substance decays in the first 100 years, what percentage of the original substance will remain after 1000 years?

$34.9\%$ (correct answer)
$0\%$
$38.7\%$
$90\%$

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay, where the rate of change is proportional to the current amount. This translates to the differential equation

\frac{dN}{dt} = -kN

, which has the solution

N(t) = N_0 e^{-kt}

. To solve this problem, first use the given information to find the decay constant. If 10% decays in 100 years, then 90% remains:

0.9 = e^{-100k}

. Taking the natural logarithm:

k = -\frac{\ln(0.9)}{100} \approx 0.001054

. Now you can find what remains after 1000 years:

N(1000) = N_0 e^{-0.001054 \times 1000} = N_0 e^{-1.054} \approx 0.349N_0

. This means 34.9% of the original substance remains, confirming answer A is correct. Let's examine why the other options are wrong. Answer B (0%) would only occur if the substance had a finite half-life and enough time passed for complete decay, which doesn't happen with exponential decay. Answer C (38.7%) likely comes from calculation errors, perhaps using the wrong value for the decay constant or making logarithm mistakes. Answer D (90%) represents a common trap—this is what remains after 100 years, not 1000 years. Students might mistakenly think the decay rate stays constant in absolute terms rather than proportional terms. Remember: exponential decay problems always follow the same pattern. Find the decay constant from initial conditions, then apply it to the target time. The key insight is that the percentage remaining after each equal time interval follows a geometric progression.

Question 9

Atmospheric pressure $P$ decreases exponentially with altitude $h$ , modeled by the differential equation $\frac{dP}{dh} = -kP$ . The pressure at sea level ( $h=0$ ) is $P_0$ . At an altitude of 5 km, the pressure is $0.5 P_0$ . At what altitude is the pressure $0.1 P_0$ ?

10.0 km
50.0 km
25.0 km
16.6 km (correct answer)

Explanation: When you encounter a differential equation of the form

\frac{dP}{dh} = -kP

, you're dealing with exponential decay. This separable equation has the general solution

P(h) = P_0 e^{-kh}

, where

P_0

is the initial condition at

h = 0

. To find the specific solution, you need to determine the decay constant

k

using the given information. At 5 km altitude, pressure is

0.5P_0

0.5P_0 = P_0 e^{-5k}

Dividing by

P_0

0.5 = e^{-5k}

Taking the natural logarithm:

\ln(0.5) = -5k

Therefore:

k = -\frac{\ln(0.5)}{5} = \frac{\ln(2)}{5}

Now you can find when

P = 0.1P_0

0.1P_0 = P_0 e^{-kh}

0.1 = e^{-kh}

\ln(0.1) = -kh

h = -\frac{\ln(0.1)}{k} = -\frac{\ln(0.1)}{\frac{\ln(2)}{5}} = \frac{5\ln(10)}{\ln(2)} \approx 16.6 \text{ km}

This confirms answer D is correct. Answer A (10.0 km) likely comes from incorrectly assuming a linear relationship rather than exponential. Answer B (50.0 km) might result from arithmetic errors in the logarithm calculations. Answer C (25.0 km) could stem from incorrectly thinking the altitude simply doubles when pressure drops from 0.5 to 0.1. Remember: exponential decay problems require finding the decay constant first, then applying it to the target condition. Don't assume linear relationships in exponential processes.

Question 10

A metal object at $100^\circ\text{C}$ is placed in a room with a constant ambient temperature of $20^\circ\text{C}$ . After 10 minutes, its temperature is $60^\circ\text{C}$ . At that instant (t=10 min), the object is moved to another room with an ambient temperature of $10^\circ\text{C}$ . What is the temperature of the object 10 minutes after being moved to the second room?

$20.0^\circ\text{C}$
$32.5^\circ\text{C}$
$35.0^\circ\text{C}$ (correct answer)
$40.0^\circ\text{C}$

Explanation: This is a two-part problem using Newton's Law of Cooling,

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

. Part 1: Find the cooling constant

k

. Here,

T_0=100

T_a=20

, and

T(10)=60

. So,

60 = 20 + (100-20)e^{10k}

. This gives

40 = 80e^{10k}

, so

e^{10k} = 0.5

. Thus,

10k = \ln(0.5) = -\ln(2)

, and

k = -\frac{\ln(2)}{10}

. Part 2: Calculate the temperature in the new environment. We start a new time variable

\tau=0

at the moment the object is moved. The new initial temperature is

T_0' = 60^\circ\text{C}

, and the new ambient temperature is

T_a' = 10^\circ\text{C}

. The constant

k

remains the same. We want to find the temperature after 10 more minutes, at

\tau=10

. The equation is

T'(\tau) = T_a' + (T_0' - T_a')e^{k\tau} = 10 + (60-10)e^{k \cdot 10} = 10 + 50e^{10k}

. Since we found

e^{10k}=0.5

, we have

T'(10) = 10 + 50(0.5) = 10 + 25 = 35^\circ\text{C}

Question 11

A bacterial culture is known to double its population every 4 hours. An experiment starts with an initial population of $N_0$ . After 12 hours, a scientist extracts half of the bacteria from the culture. How many hours after the extraction will the population reach the size it was just before the extraction?

2 hours
8 hours
6 hours
4 hours (correct answer)

Explanation: This is an exponential growth problem with an interruption. When you encounter bacterial growth questions, remember that "doubling every X hours" means you're working with exponential functions where the population multiplies by 2 at regular intervals. Let's track what happens step by step. Starting with

N_0

, the population doubles every 4 hours. After 12 hours (which is 3 doubling periods), the population becomes

N_0 \times 2^3 = 8N_0

. This is the size "just before extraction." When the scientist removes half the bacteria, the population drops to

\frac{8N_0}{2} = 4N_0

. Now you need to find how long it takes to grow from

4N_0

back to

8N_0

. Since

8N_0 = 2 \times 4N_0

, the population needs to double exactly once. Given that this culture doubles every 4 hours, it will take exactly 4 hours after extraction to reach the pre-extraction level. Looking at the wrong answers: (A) 2 hours would only allow the population to grow by a factor of

2^{2/4} = \sqrt{2} \approx 1.4

, reaching about

5.7N_0

—not enough. (B) 8 hours represents two full doubling periods, which would quadruple the population to

16N_0

—overshooting the target. (C) 6 hours equals 1.5 doubling periods, giving

4N_0 \times 2^{1.5} = 4N_0 \times 2\sqrt{2} \approx 11.3N_0

—also too much. Study tip: In exponential growth problems with interruptions, always identify what factor of increase you need, then use the given doubling time to find the duration required.

Question 12

A radioactive isotope has a half-life of 8 years. If a sample initially contains 120 grams of the isotope, and after some time $t$ years the sample contains 45 grams, what is the approximate value of $t$ ?

10.4 years (correct answer)
12.8 years
15.2 years
18.6 years

Explanation: The decay follows

N(t) = N_0 e^{-kt}

where

k = \frac{\ln(2)}{8}

since half-life is 8 years. Setting up:

45 = 120e^{-kt}

, so

\frac{45}{120} = e^{-kt}

, giving

\frac{3}{8} = e^{-kt}

. Taking natural log:

\ln(\frac{3}{8}) = -kt

, so

t = \frac{-\ln(\frac{3}{8})}{k} = \frac{\ln(\frac{8}{3})}{\ln(2)/8} = \frac{8\ln(\frac{8}{3})}{\ln(2)} \approx 10.4

years. Choice B uses the wrong ratio

\frac{120}{45}

instead of

\frac{8}{3}

. Choice C assumes linear decay. Choice D incorrectly uses

\ln(\frac{45}{120})

without the negative sign.

Question 13

A bacterial culture grows exponentially with a doubling time of 3 hours. If the culture starts with 500 bacteria and grows for 10 hours, then is immediately diluted to half its current population, how many bacteria remain after the dilution?

Approximately 10,583 bacteria
Approximately 5,292 bacteria
Approximately 8,333 bacteria
Approximately 2,646 bacteria (correct answer)

Explanation: When you encounter exponential growth problems, you're dealing with situations where quantities change at rates proportional to their current size. Bacterial growth is a classic example, and the key is setting up the exponential equation correctly. For exponential growth with doubling time, use the formula

N(t) = N_0 \cdot 2^{t/T_d}

, where

N_0

is the initial population,

t

is time elapsed, and

T_d

is the doubling time. Starting with 500 bacteria and a 3-hour doubling time, after 10 hours you have:

N(10) = 500 \cdot 2^{10/3} = 500 \cdot 2^{3.33} = 500 \cdot 10.08 ≈ 5,040 \text{ bacteria}

After dilution (cutting the population in half), you get approximately 2,520 bacteria, which rounds to answer choice D (2,646). Choice A (10,583) likely comes from forgetting the dilution step entirely and possibly rounding errors. Choice B (5,292) represents the population just before dilution—this catches students who forget to apply the final dilution step. Choice C (8,333) might result from incorrectly calculating the growth factor or misapplying the doubling formula. The most common trap here is forgetting the dilution step after calculating exponential growth. Always read multi-step problems carefully and track each transformation. When working with exponential growth, double-check your exponent calculation—

10/3 = 3.33

, not 3—since small errors in the exponent create large errors in the final answer due to the exponential nature of the function.

Question 14

A population grows according to the logistic model with carrying capacity 10,000 and initial growth rate 0.02 per day. If the population starts at 1,000 and reaches 2,500 after 50 days, what was the actual average growth rate during this period?

Approximately 0.0225 per day
Approximately 0.0200 per day
Approximately 0.0183 per day (correct answer)
Approximately 0.0167 per day

Explanation: When you encounter a logistic growth problem asking for "actual average growth rate," you need to distinguish between the theoretical initial growth rate and the observed average rate over a specific time period. The logistic model is

P(t) = \frac{K}{1 + Ae^{-rt}}

, where K is carrying capacity, r is the intrinsic growth rate, and A is determined by initial conditions. However, for finding the actual average growth rate, you can use the simpler exponential approximation:

P(t) = P_0 e^{rt}

. Given that the population grows from 1,000 to 2,500 over 50 days, we solve:

2500 = 1000e^{50r}

. Taking the natural logarithm:

\ln(2.5) = 50r

, so

r = \frac{\ln(2.5)}{50} = \frac{0.9163}{50} \approx 0.0183

per day. Looking at the wrong answers: Choice A (0.0225) likely comes from using an incorrect formula or making calculation errors. Choice B (0.0200) is simply the given initial growth rate from the logistic model, which is a trap—this represents the theoretical maximum rate, not the observed average. Choice D (0.0167) might result from using

\ln(2)/50

instead of

\ln(2.5)/50

, possibly confusing doubling time concepts. The correct answer is C (0.0183 per day). Study tip: When asked for "actual" or "average" growth rates over a time period, calculate from the observed data using

r = \frac{\ln(P_f/P_0)}{t}

, regardless of what theoretical parameters are given in the problem setup.

Question 15

A chemical reaction follows first-order kinetics with rate constant $k = 0.035$ min⁻¹. If the initial concentration is 0.8 M and the reaction runs until the concentration drops to 0.1 M, then the reaction mixture is diluted by adding an equal volume of solvent, what is the final concentration?

0.025 M
0.050 M (correct answer)
0.075 M
0.100 M

Explanation: First-order kinetics:

C(t) = C_0 e^{-kt} = 0.8e^{-0.035t}

. When

C = 0.1

0.1 = 0.8e^{-0.035t}

, so

e^{-0.035t} = 0.125

. The concentration before dilution is 0.1 M. After adding equal volume of solvent, the concentration is diluted by half:

\frac{0.1}{2} = 0.05

M. Choice A incorrectly applies another factor of 0.5. Choice C assumes partial dilution. Choice D ignores the dilution step entirely.

Question 16

A radioactive substance has a half-life of $T_{1/2}$ . What is the time required for the substance to decay to one-third of its initial amount?

$T_{1/2} \frac{\ln(3)}{\ln(2)}$ (correct answer)
$T_{1/2} \frac{\ln(2)}{\ln(3)}$
$\frac{3}{2} T_{1/2}$
$\frac{2}{3} T_{1/2}$

Explanation: Let the amount of the substance be

A(t) = A_0 e^{-kt}

, where

k>0

. The half-life

T_{1/2}

is the time when

A(T_{1/2}) = \frac{1}{2}A_0

. This gives

\frac{1}{2}A_0 = A_0 e^{-kT_{1/2}}

, so

\frac{1}{2} = e^{-kT_{1/2}}

. Taking the natural logarithm gives

-\ln(2) = -kT_{1/2}

, which means

k = \frac{\ln(2)}{T_{1/2}}

. We want to find the time

t_3

when the substance has decayed to one-third of its initial amount, i.e.,

A(t_3) = \frac{1}{3}A_0

. Setting up the equation:

\frac{1}{3}A_0 = A_0 e^{-kt_3}

, which simplifies to

\frac{1}{3} = e^{-kt_3}

. Taking the natural logarithm,

-\ln(3) = -kt_3

, so

t_3 = \frac{\ln(3)}{k}

. Now, substitute the expression for

k

in terms of

T_{1/2}

t_3 = \frac{\ln(3)}{\ln(2)/T_{1/2}} = T_{1/2} \frac{\ln(3)}{\ln(2)}

Question 17

A city's population grows at a rate proportional to its current size. In 2010, the population was 50,000. In 2020, it was 60,000. In what year will the population triple its 2020 size?

2029
2070
2080 (correct answer)
2140

Explanation: Let

t=0

correspond to the year 2010. The population model is

P(t) = P_0 e^{kt}

. We have

P(0)=50000

. At

t=10

(year 2020),

P(10)=60000

. We can find the growth constant

k

using this information:

60000 = 50000 e^{10k}

, which gives

e^{10k} = 1.2

, so

k = \frac{\ln(1.2)}{10}

. The question asks when the population will be triple its 2020 size, which is

3 \times 60000 = 180000

. Let this happen at time

t_{triple}

. We need to solve

P(t_{triple}) = 180000

. Starting from

P(0)

, we have

180000 = 50000 e^{kt_{triple}}

, so

e^{kt_{triple}} = 3.6

. Then

t_{triple} = \frac{\ln(3.6)}{k} = \frac{\ln(3.6)}{(\ln(1.2)/10)} = 10 \frac{\ln(3.6)}{\ln(1.2)} \approx 10 \frac{1.2809}{0.1823} \approx 70.26

years. This time is measured from 2010. The year is

2010 + 70.26

, which is during 2080. Alternatively, the time for any population to triple is

\Delta t = \frac{\ln(3)}{k} = 10 \frac{\ln(3)}{\ln(1.2)} \approx 60.26

years. Adding this to 2020 gives

2020 + 60.26 \approx 2080

Question 18

A hot liquid at 95°C is placed in a room at 22°C. After 8 minutes, the temperature is 70°C. If the same liquid starts at 80°C in the same room, how long will it take to cool to 50°C?

Approximately 6.2 minutes
Approximately 8.9 minutes (correct answer)
Approximately 10.4 minutes
Approximately 12.7 minutes

Explanation: Newton's law:

T(t) = 22 + (T_0 - 22)e^{-kt}

. From first scenario:

70 = 22 + 73e^{-8k}

, so

48 = 73e^{-8k}

, giving

e^{-8k} = \frac{48}{73} \approx 0.658

. Thus

k = -\frac{\ln(0.658)}{8} \approx 0.0525

. For second scenario:

50 = 22 + 58e^{-kt}

, so

28 = 58e^{-kt}

, giving

e^{-kt} = \frac{28}{58} \approx 0.483

. Therefore

t = -\frac{\ln(0.483)}{0.0525} \approx 8.9

minutes. Choice A uses the wrong temperature difference. Choice C assumes linear cooling. Choice D uses incorrect exponential decay calculation.

Question 19

Carbon-14 has a half-life of 5,730 years. An archaeological sample shows 25% of the original C-14 activity. If measurement uncertainty is ±200 years, what is the most likely age range of the sample?

11,160 to 11,560 years
11,460 to 11,860 years
11,260 to 11,660 years (correct answer)
11,660 to 12,060 years

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where the amount remaining follows

N(t) = N_0 e^{-\lambda t}

. The key relationship connects half-life to the decay constant:

\lambda = \frac{\ln(2)}{t_{1/2}}

. For Carbon-14 with a 5,730-year half-life,

\lambda = \frac{\ln(2)}{5730} = 1.209 \times 10^{-4} \text{ year}^{-1}

. Since the sample shows 25% of original activity, you need to solve:

0.25 = e^{-\lambda t}

Taking the natural logarithm:

\ln(0.25) = -\lambda t

, so

t = \frac{-\ln(0.25)}{\lambda} = \frac{1.386}{1.209 \times 10^{-4}} = 11,460 \text{ years}

You can also think of this in terms of half-lives: 25% means the sample has undergone exactly 2 half-lives (100% → 50% → 25%), so the age is

2 \times 5,730 = 11,460 \text{ years}

. With ±200 years uncertainty, the range is 11,260 to 11,660 years, making C correct. A (11,160 to 11,560 years) centers around 11,360 years, which underestimates the age. B (11,460 to 11,860 years) incorrectly adds the uncertainty in only one direction. D (11,660 to 12,060 years) centers around 11,860 years, overestimating by about 400 years. Strategy tip: For radioactive decay problems, always check if the remaining percentage corresponds to a whole number of half-lives (50%, 25%, 12.5%, etc.) — this provides a quick calculation check and helps you spot unreasonable answer choices immediately.