Differential Equations Quiz: Compartment Models

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Compartment ModelsQuestion 1 of 13

The system of differential equations below models the amount of a substance in two compartments, $A$ and $B$ , in milligrams (mg), over time $t$ in minutes.

\begin{cases} \frac{dx_A}{dt} = -0.4x_A + 0.1x_B + 7 \\ \frac{dx_B}{dt} = 0.4x_A - 0.3x_B \end{cases}

What is the correct physical interpretation of the term $0.4x_A$ in this model?

The rate at which the substance flows from compartment

A

to compartment

B

The rate at which the substance flows from compartment

B

to compartment

A

The rate at which the substance in compartment

A

is eliminated from the system.

The total rate of change of the substance in compartment

A

at equilibrium.

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Differential Equations Quiz

Differential Equations Quiz: Compartment Models

Practice Compartment Models 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 Compartment Models, 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

The system of differential equations below models the amount of a substance in two compartments, $A$ and $B$ , in milligrams (mg), over time $t$ in minutes.

\begin{cases} \frac{dx_A}{dt} = -0.4x_A + 0.1x_B + 7 \\ \frac{dx_B}{dt} = 0.4x_A - 0.3x_B \end{cases}

What is the correct physical interpretation of the term $0.4x_A$ in this model?

The rate at which the substance flows from compartment $A$ to compartment $B$ . (correct answer)
The rate at which the substance flows from compartment $B$ to compartment $A$ .
The rate at which the substance in compartment $A$ is eliminated from the system.
The total rate of change of the substance in compartment $A$ at equilibrium.

Explanation: In compartment models, a term of the form

k x_i

represents a rate of transfer proportional to the amount

x_i

in compartment

i

. The term

0.4x_A

appears with a negative sign in the equation for

\frac{dx_A}{dt}

(as part of

-0.4x_A

) and with a positive sign in the equation for

\frac{dx_B}{dt}

. This indicates that this term corresponds to a flow of substance out of compartment

A

and into compartment

B

. Therefore,

0.4x_A

represents the rate at which the substance flows from compartment

A

to compartment

B

Question 2

The amounts of two interacting chemical species, $A$ and $B$ , in a reactor are given by $x(t)$ and $y(t)$ . Species $A$ converts to species $B$ at a rate $0.2x$ . Species $B$ converts back to species $A$ at a rate $0.1y$ . Additionally, species $A$ and $B$ react with each other to form an inert product, consuming both at a rate proportional to the product of their amounts, with a rate constant of $k=0.01$ . Which system of differential equations models this process?

$\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y - 0.01xy \\ \frac{dy}{dt} = 0.2x - 0.1y - 0.01xy \end{cases}$ (correct answer)
$\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y + 0.01xy \\ \frac{dy}{dt} = 0.2x - 0.1y + 0.01xy \end{cases}$
$\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y - 0.01(x+y) \\ \frac{dy}{dt} = 0.2x - 0.1y - 0.01(x+y) \end{cases}$
$\begin{cases} \frac{dx}{dt} = 0.2x - 0.1y - 0.01xy \\ \frac{dy}{dt} = -0.2x + 0.1y - 0.01xy \end{cases}$

Explanation: For species A (amount

x

): It is consumed when converting to B (term

-0.2x

), it is produced from B (term

+0.1y

), and it is consumed in the reaction with B (term

-0.01xy

). So,

\frac{dx}{dt} = -0.2x + 0.1y - 0.01xy

. For species B (amount

y

): It is produced from A (term

+0.2x

), it is consumed when converting to A (term

-0.1y

), and it is consumed in the reaction with A (term

-0.01xy

). So,

\frac{dy}{dt} = 0.2x - 0.1y - 0.01xy

. This matches choice A.

Question 3

The general solution for the amount of a substance $x(t)$ in a certain compartment is given by $x(t) = 15e^{-0.9t} - 4e^{-0.2t} + 25$ , where $t \ge 0$ . What is the initial amount of the substance in the compartment, $x(0)$ ?

$25$
$11$
$36$ (correct answer)
$0$

Explanation: To find the initial amount of the substance, we must evaluate the function

x(t)

at time

t=0

. Substituting

t=0

into the expression:

x(0) = 15e^{-0.9(0)} - 4e^{-0.2(0)} + 25

. Since

e^0 = 1

, this simplifies to:

x(0) = 15(1) - 4(1) + 25 = 15 - 4 + 25 = 11 + 25 = 36

. The value

25

represents the long-term equilibrium amount, not the initial amount. The value

11

is an intermediate calculation (

15-4

Question 4

A cascade system is modeled by:

\begin{cases} \frac{dx}{dt} = 8 - 4x \\ \frac{dy}{dt} = 4x - 2y \end{cases}

This represents a substance flowing into compartment $X$ , then to $Y$ , then out of the system. Suppose the external inflow rate of $8$ is halved to $4$ . How does the equilibrium amount of the substance in compartment $Y$ change?

It is quartered.
It is halved. (correct answer)
It does not change.
It decreases by 2 units.

Explanation: First, find the original equilibrium

(x_e, y_e)

. Set the derivatives to zero:

8 - 4x = 0 \implies x_e = 2

4x_e - 2y = 0 \implies 4(2) - 2y = 0 \implies 8 - 2y = 0 \implies y_e = 4

. Now, find the new equilibrium

(x'_e, y'_e)

with the new inflow rate of

4

4 - 4x = 0 \implies x'_e = 1

4x'_e - 2y = 0 \implies 4(1) - 2y = 0 \implies 4 - 2y = 0 \implies y'_e = 2

. The equilibrium amount in compartment

Y

changes from

y_e = 4

y'_e = 2

. Therefore, the amount is halved.

Question 5

Consider a system with two compartments, A and B, with amounts $x(t)$ and $y(t)$ and constant volumes $V_A$ and $V_B$ . A substance flows from A to B at a rate of 10 L/min. Part of the substance is then recycled from B back to A at a rate of 3 L/min. An external source adds the substance to A at a rate of 7 L/min, and the substance is removed from B at a rate of 4 L/min. Which pair of terms correctly represents the transfer of the substance between A and B in the differential equation for $\frac{dx}{dt}$ ?

$-10 x + 3 y$
$+10 \frac{x}{V_A} - 3 \frac{y}{V_B}$
$-10 \frac{x}{V_A} - 3 \frac{y}{V_B}$
$-10 \frac{x}{V_A} + 3 \frac{y}{V_B}$ (correct answer)

Explanation: When modeling compartment systems, you need to carefully track both flow rates and concentrations. The key insight is that substance transfer depends on concentration (amount per unit volume), not just the total amount in each compartment. For compartment A, you're looking at how the transfer between compartments affects

\frac{dx}{dt}

. Substance flows out of A to B at 10 L/min, so the rate of substance leaving A is

10 \times \frac{x}{V_A}

(flow rate times concentration in A). This represents a loss, so it's negative:

-10\frac{x}{V_A}

. Meanwhile, substance is recycled back from B to A at 3 L/min, giving a rate of substance entering A of

3 \times \frac{y}{V_B}

(flow rate times concentration in B). This represents a gain, so it's positive:

+3\frac{y}{V_B}

. Answer D correctly captures both terms:

-10\frac{x}{V_A} + 3\frac{y}{V_B}

. Answer A treats amounts directly without considering volumes, ignoring that flow depends on concentration, not total amount. Answer B has the wrong sign on the recycling term—it should be positive since substance flows into A from B. Answer C makes both the flow from A to B and the recycling from B to A appear as losses to compartment A, which contradicts the physical setup where recycling adds substance back to A. Remember: in compartment models, always divide amounts by volumes to get concentrations, and carefully track the direction of each flow to determine the correct signs.

Question 6

A medication is introduced into the bloodstream (compartment $x_1$ ) and subsequently absorbed by body tissues (compartment $x_2$ ). The system is modeled by:

\begin{cases} \frac{dx_1}{dt} = 10 - 0.5x_1 \\ \frac{dx_2}{dt} = 0.5x_1 - 0.1x_2 \end{cases}

where amounts are in mg and time is in hours. The term $10$ represents a constant intravenous infusion rate. Let $T(t) = x_1(t) + x_2(t)$ be the total amount of medication in the body. Which differential equation governs $T(t)$ ?

$\frac{dT}{dt} = 10 - 0.1T$
$\frac{dT}{dt} = 10 - 0.6(x_1+x_2)$
$\frac{dT}{dt} = 10 - 0.1x_2$ (correct answer)
$\frac{dT}{dt} = 10 - 0.5x_1 - 0.1x_2$

Explanation: When analyzing coupled differential equation systems, you need to understand how quantities flow between compartments and what happens to the total amount in the entire system. To find the differential equation for

T(t) = x_1(t) + x_2(t)

, differentiate both sides:

\frac{dT}{dt} = \frac{dx_1}{dt} + \frac{dx_2}{dt}

Substituting the given equations:

\frac{dT}{dt} = (10 - 0.5x_1) + (0.5x_1 - 0.1x_2)

Notice that the

0.5x_1

terms cancel out:

\frac{dT}{dt} = 10 - 0.1x_2

This makes physical sense: medication enters the system at rate 10 (intravenous infusion) and leaves only from compartment

x_2

at rate

0.1x_2

. The transfer between compartments (

0.5x_1

) doesn't affect the total amount. Answer A incorrectly assumes medication is eliminated from the total at rate

0.1T

, but elimination only occurs from compartment

x_2

, not proportionally from the total amount. Answer B uses the wrong coefficient (0.6 instead of 0.1) and incorrectly applies it to the total. Answer D simply adds both original equations without recognizing that internal transfers cancel out when considering the total. Study tip: In compartment models, when finding the rate of change of total quantity, internal transfer terms always cancel out. Only external inputs and outputs affect the total amount in the system.

Question 7

A pharmaceutical drug is administered to a patient. Let $x(t)$ be the amount of the drug in the gastrointestinal (GI) tract and $y(t)$ be the amount in the bloodstream. The drug moves from the GI tract to the blood at a rate proportional to the amount in the GI tract. It is eliminated from the blood at a rate proportional to the amount in the blood. Which system of differential equations is a plausible model for this process, assuming $k_1, k_2 > 0$ are the rate constants?

$\begin{cases} \frac{dx}{dt} = k_1 x - k_2 y \\ \frac{dy}{dt} = k_2 y \end{cases}$
$\begin{cases} \frac{dx}{dt} = k_1 x \\ \frac{dy}{dt} = -k_1 x + k_2 y \end{cases}$
$\begin{cases} \frac{dx}{dt} = -k_1 x - k_2 y \\ \frac{dy}{dt} = k_1 x \end{cases}$
$\begin{cases} \frac{dx}{dt} = -k_1 x \\ \frac{dy}{dt} = k_1 x - k_2 y \end{cases}$ (correct answer)

Explanation: When modeling drug kinetics with differential equations, you need to carefully track the rates of change based on the physical processes described. Think about what causes each compartment (GI tract and bloodstream) to gain or lose drug over time. For the GI tract,

x(t)

: The only process affecting it is drug moving out to the bloodstream at rate

k_1x

. Since drug is leaving, this creates a negative rate of change:

\frac{dx}{dt} = -k_1x

. For the bloodstream,

y(t)

: Two processes are occurring simultaneously. Drug enters from the GI tract at rate

k_1x

(positive contribution) and is eliminated at rate

k_2y

(negative contribution). This gives:

\frac{dy}{dt} = k_1x - k_2y

. Answer D correctly captures both equations:

\frac{dx}{dt} = -k_1x

and

\frac{dy}{dt} = k_1x - k_2y

. Looking at the incorrect options: Answer A has

\frac{dx}{dt} = k_1x - k_2y

, incorrectly suggesting the GI tract gains drug and is affected by blood elimination. Answer B shows

\frac{dx}{dt} = k_1x

, implying the GI tract somehow gains drug over time, which contradicts the one-way transfer described. Answer C has

\frac{dy}{dt} = k_1x

without the elimination term, ignoring that drug is removed from the bloodstream. Study tip: In compartmental models, always check that your signs make physical sense—what processes add to each compartment versus what processes remove from it. The conservation principle also helps: drug leaving one compartment should enter another at the same rate.

Question 8

A system of two compartments is described by the differential equations:

\begin{cases} \frac{dx}{dt} = 15 - 5x + y \\ \frac{dy}{dt} = 3x - 3y \end{cases}

As $t \to \infty$ , the system approaches a stable equilibrium state $(x_e, y_e)$ . What is the ratio $x_e / y_e$ at this equilibrium?

$1$ (correct answer)
$4/5$
$5/4$
$3$

Explanation: At equilibrium, the rates of change are zero:

\frac{dx}{dt} = 0

and

\frac{dy}{dt} = 0

. From the second equation,

3x - 3y = 0

, which simplifies to

3x = 3y

, or

x=y

. This directly implies that the ratio

x_e / y_e = 1

. We can also solve for the specific values to confirm. Substituting

x=y

into the first equilibrium equation gives

15 - 5x + x = 0

, which means

15 - 4x = 0

, so

x_e = 15/4

. Since

y_e=x_e

y_e = 15/4

. The ratio is

(15/4) / (15/4) = 1

Question 9

A closed system consists of two compartments, with the amount of a tracer substance in each given by $x(t)$ and $y(t)$ . In a closed system, the total amount of the substance $x(t) + y(t)$ remains constant over time. The dynamics are modeled by the linear system $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}$ . Which of the following is a necessary property of the matrix $A$ ?

The sum of the entries in each column of $A$ is zero. (correct answer)
The sum of the entries in each row of $A$ is zero.
$A$ is a symmetric matrix, i.e., $A = A^T$ .
The diagonal entries of $A$ must be zero.

Explanation: If the total amount

T = x(t) + y(t)

is constant, then its derivative must be zero:

\frac{dT}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = 0

. Let

A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

. Then

\frac{dx}{dt} = a_{11}x + a_{12}y

and

\frac{dy}{dt} = a_{21}x + a_{22}y

. Summing these gives

\frac{dT}{dt} = (a_{11}+a_{21})x + (a_{12}+a_{22})y

. For this expression to be zero for all possible values of

x

and

y

, the coefficients must be zero:

a_{11}+a_{21}=0

and

a_{12}+a_{22}=0

. These are precisely the sums of the entries in each column of

A

Question 10

A tank with a capacity of 500 liters initially contains 200 liters of pure water. A salt solution with concentration 0.1 kg/L is pumped in at a rate of 10 L/min. The well-mixed solution is pumped out at a rate of 5 L/min. Let $S(t)$ be the amount of salt in the tank at time $t$ . Which initial value problem correctly models this process?

$\frac{dS}{dt} = 1 - \frac{S}{200}, \quad S(0) = 0$
$\frac{dS}{dt} = 1 - \frac{S}{40+t}, \quad S(0) = 0$ (correct answer)
$\frac{dS}{dt} = 1 - \frac{5S}{200+5t}, \quad S(0) = 0$
$\frac{dS}{dt} = 0.1 - \frac{S}{40+t}, \quad S(0) = 0$

Explanation: When you encounter a mixing problem like this, you need to set up a differential equation based on the principle: rate of change of salt = rate of salt entering - rate of salt leaving. Let's work through this systematically. First, determine the volume of solution in the tank at time

t

. You start with 200 L, gain 10 L/min, and lose 5 L/min, so the net gain is 5 L/min. Therefore, the volume at time

t

200 + 5t

liters. For the rate equation: Salt enters at

(0.1 \text{ kg/L})(10 \text{ L/min}) = 1

kg/min. Salt leaves at a rate equal to the concentration times the outflow rate. The concentration at time

t

\frac{S(t)}{200+5t}

kg/L, so salt leaves at

\frac{S(t)}{200+5t} \times 5 = \frac{5S}{200+5t}

kg/min. This gives us

\frac{dS}{dt} = 1 - \frac{5S}{200+5t}

with

S(0) = 0

(initially pure water). Choice A incorrectly assumes constant volume (200 L) and wrong outflow calculation. Choice C has the correct volume term but is missing from our options - wait, let me recheck. Actually, choice B simplifies correctly:

\frac{dS}{dt} = 1 - \frac{5S}{200+5t} = 1 - \frac{S}{\frac{200+5t}{5}} = 1 - \frac{S}{40+t}

. Choice D uses the wrong inflow rate (should be 1, not 0.1). Study tip: Always track the changing volume in mixing problems, and remember that concentration equals amount divided by volume at any given time.

Question 11

A pharmacokinetic model describes drug absorption and elimination using two compartments: gut (G) and blood (B). Drug transfers from gut to blood at rate $k_a G$ and is eliminated from blood at rate $k_e B$ . If a 100 mg dose is given orally at $t = 0$ and $k_a = 0.8 \text{ h}^{-1}$ , $k_e = 0.2 \text{ h}^{-1}$ , what is the time when blood concentration reaches its maximum?

$t = \frac{\ln(4)}{0.6}$ hours (correct answer)
$t = \frac{\ln(4)}{0.8}$ hours
$t = \frac{\ln(0.25)}{0.6}$ hours
$t = \frac{\ln(2)}{0.6}$ hours

Explanation: The system equations are

\frac{dG}{dt} = -k_a G

and

\frac{dB}{dt} = k_a G - k_e B

. With

G(0) = 100

and

B(0) = 0

, we get

G(t) = 100e^{-k_a t} = 100e^{-0.8t}

. For blood:

\frac{dB}{dt} + k_e B = k_a G = 80e^{-0.8t}

. The solution is

B(t) = \frac{k_a}{k_a - k_e}(e^{-k_e t} - e^{-k_a t}) \cdot 100 = \frac{0.8}{0.6}(e^{-0.2t} - e^{-0.8t}) \cdot 100

. Maximum occurs when

\frac{dB}{dt} = 0

k_a G - k_e B = 0

, so

k_a e^{-k_a t} = k_e \frac{k_a}{k_a - k_e}(e^{-k_e t} - e^{-k_a t})

. Simplifying:

e^{-k_a t} = \frac{k_e}{k_a - k_e}(e^{-k_e t} - e^{-k_a t})

. After algebraic manipulation:

e^{-(k_a - k_e)t} = \frac{k_e}{k_a}

, so

t = \frac{\ln(k_a/k_e)}{k_a - k_e} = \frac{\ln(0.8/0.2)}{0.8-0.2} = \frac{\ln(4)}{0.6}

Question 12

A factory waste treatment system has four interconnected settling tanks. Contaminated water enters Tank 1 at 100 L/h with pollutant concentration 500 mg/L. Water flows from Tank 1 to Tank 2 at 80 L/h, from Tank 2 to Tank 3 at 60 L/h, from Tank 3 to Tank 4 at 40 L/h, and exits Tank 4 at 40 L/h. Each tank has overflow pipes that maintain constant volume. At steady state, if the pollutant concentration decreases by 30% in each tank due to settling, what is the pollutant removal efficiency of the entire system?

75.99% (correct answer)
82.40%
78.15%
85.20%

Explanation: If each tank reduces pollutant concentration by 30%, then 70% of the pollutant concentration remains after each tank. Starting with 500 mg/L: after Tank 1, concentration is

500 \times 0.7 = 350

mg/L; after Tank 2, concentration is

350 \times 0.7 = 245

mg/L; after Tank 3, concentration is

245 \times 0.7 = 171.5

mg/L; after Tank 4, concentration is

171.5 \times 0.7 = 120.05

mg/L. The overall removal efficiency is

\frac{500 - 120.05}{500} \times 100\% = \frac{379.95}{500} \times 100\% = 75.99\%

. Alternatively, the remaining fraction is

0.7^4 = 0.2401

, so removal efficiency is

(1 - 0.2401) \times 100\% = 75.99\%

. Choice B (82.40%) would result from

(1 - 0.7^3) \times 100\% = 65.7\%

or some other calculation error. Choice C might come from using an arithmetic average. Choice D is too high for this level of treatment.

Question 13

An epidemiological model tracks disease progression through three compartments: Susceptible (S), Infected (I), and Recovered (R). The infection rate is $\beta SI/N$ where $N$ is total population, and recovery rate is $\gamma I$ . If $\beta = 0.5$ per day, $\gamma = 0.1$ per day, and $N = 10000$ , what is the basic reproduction number $R_0$ , and what does it predict about disease spread?

$R_0 = 5$ ; disease will spread exponentially from any initial infected individual
$R_0 = 5$ ; disease spread depends on initial proportion of susceptible individuals (correct answer)
$R_0 = 0.05$ ; disease will die out regardless of initial conditions
$R_0 = 50$ ; disease will spread exponentially with very high transmission rate

Explanation: The basic reproduction number is

R_0 = \frac{\beta}{\gamma} = \frac{0.5}{0.1} = 5

. However,

R_0 > 1

doesn't guarantee exponential spread from any initial condition. The effective reproduction number is

R_{eff} = R_0 \cdot \frac{S}{N}

, so disease spread depends on the fraction of susceptible individuals. With

R_0 = 5

, if all individuals are initially susceptible, each infected person will infect 5 others on average. But if some are already immune or recovered,

R_{eff} < R_0

. Choice A is incorrect because it ignores the dependence on susceptible fraction. Choice C miscalculates

R_0

. Choice D incorrectly calculates

R_0 = \beta N / \gamma