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Differential Equations Quiz

Differential Equations Quiz: Salt And Concentration Models

Practice Salt And Concentration Models in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 7

0 of 7 answered

Two identical 200-gallon tanks are connected by a pipe. Initially, Tank A contains 150 gallons of brine with 30 pounds of salt, and Tank B contains 100 gallons of pure water. Pure water enters Tank A at 3 gal/min, the mixture flows from A to B at 5 gal/min, and solution exits Tank B at 2 gal/min. What happens to the salt concentration in Tank B as t→∞t \to \inftyt→∞?

Select an answer to continue

What this quiz covers

This quiz focuses on Salt And Concentration 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

Two identical 200-gallon tanks are connected by a pipe. Initially, Tank A contains 150 gallons of brine with 30 pounds of salt, and Tank B contains 100 gallons of pure water. Pure water enters Tank A at 3 gal/min, the mixture flows from A to B at 5 gal/min, and solution exits Tank B at 2 gal/min. What happens to the salt concentration in Tank B as t→∞t \to \inftyt→∞?

  1. The concentration approaches zero since pure water continuously dilutes both tanks over time (correct answer)
  2. The concentration approaches 30350\frac{30}{350}35030​ lb/gal since the total salt distributes uniformly across both tanks
  3. The concentration oscillates between zero and the initial concentration in Tank A indefinitely
  4. The concentration approaches 30250\frac{30}{250}25030​ lb/gal based on the final total volume in the system

Explanation: Since pure water enters Tank A and eventually flows to Tank B, while solution continuously exits Tank B, the total amount of salt in the system decreases over time. No salt enters the system, only leaves through Tank B. The volumes in both tanks approach steady-state values, but the salt content approaches zero as t→∞t \to \inftyt→∞. Choice B incorrectly assumes salt distributes based on total volume without considering outflow. Choice C incorrectly suggests oscillatory behavior in a linear system. Choice D uses an incorrect volume calculation and ignores that salt continuously leaves the system.

Question 2

Consider a cascade of two tanks: Tank 1 (200 gallons) initially contains 40 pounds of salt, and Tank 2 (300 gallons) is initially pure. Pure water enters Tank 1 at 10 gal/min, the mixture flows from Tank 1 to Tank 2 at 10 gal/min, and solution exits Tank 2 at 10 gal/min. If S1(t)S_1(t)S1​(t) and S2(t)S_2(t)S2​(t) represent the salt amounts in each tank, which statement about the system behavior is correct?

  1. S2(t)S_2(t)S2​(t) increases monotonically until all salt from Tank 1 has transferred, then decreases exponentially to zero
  2. S2(t)S_2(t)S2​(t) reaches its maximum when S1(t)=S2(t)S_1(t) = S_2(t)S1​(t)=S2​(t), which occurs at the intersection of their exponential curves
  3. S2(t)S_2(t)S2​(t) reaches its maximum when dS1dt+dS2dt=0\frac{dS_1}{dt} + \frac{dS_2}{dt} = 0dtdS1​​+dtdS2​​=0, ensuring total salt conservation in the system
  4. S2(t)S_2(t)S2​(t) reaches its maximum when S1(t)200=S2(t)300\frac{S_1(t)}{200} = \frac{S_2(t)}{300}200S1​(t)​=300S2​(t)​, balancing the concentrations between tanks (correct answer)

Explanation: The system equations are dS1dt=−S120\frac{dS_1}{dt} = -\frac{S_1}{20}dtdS1​​=−20S1​​ and dS2dt=S120−S230\frac{dS_2}{dt} = \frac{S_1}{20} - \frac{S_2}{30}dtdS2​​=20S1​​−30S2​​. S2(t)S_2(t)S2​(t) reaches its maximum when dS2dt=0\frac{dS_2}{dt} = 0dtdS2​​=0, which occurs when S120=S230\frac{S_1}{20} = \frac{S_2}{30}20S1​​=30S2​​, or equivalently S1200=S2300\frac{S_1}{200} = \frac{S_2}{300}200S1​​=300S2​​. This means the concentrations are equal when S2S_2S2​ is maximized. Choice A is incorrect because S2S_2S2​ doesn't increase monotonically—it increases then decreases. Choice B is wrong because the maximum doesn't occur when S1=S2S_1 = S_2S1​=S2​. Choice C incorrectly relates the maximum to total salt conservation (dS1dt+dS2dt=−S230<0\frac{dS_1}{dt} + \frac{dS_2}{dt} = -\frac{S_2}{30} < 0dtdS1​​+dtdS2​​=−30S2​​<0 always).

Question 3

A fermentation tank contains 500 liters of nutrient solution. Fresh nutrient with concentration 8 g/L enters at rate QQQ L/min, while the culture is harvested at rate QQQ L/min to maintain constant volume. If the microorganisms consume nutrients at a rate proportional to both the nutrient concentration CCC and the biomass concentration BBB (where B=2.5B = 2.5B=2.5 g/L remains constant), with consumption rate constant k=0.15k = 0.15k=0.15 L/(g·min), what inflow rate QQQ is needed to maintain C=6C = 6C=6 g/L?

  1. Q=0.15×6×2.5×5008−6Q = \frac{0.15 \times 6 \times 2.5 \times 500}{8 - 6}Q=8−60.15×6×2.5×500​ L/min, balancing total consumption against net concentration input (correct answer)
  2. Q=0.15×6×2.5×5008Q = \frac{0.15 \times 6 \times 2.5 \times 500}{8}Q=80.15×6×2.5×500​ L/min, ensuring consumption equals total nutrient input rate
  3. Q=0.15×6×2.5Q = 0.15 \times 6 \times 2.5Q=0.15×6×2.5 L/min, matching the volumetric consumption rate directly to inflow
  4. Q=(8−6)×5000.15×6×2.5Q = \frac{(8 - 6) \times 500}{0.15 \times 6 \times 2.5}Q=0.15×6×2.5(8−6)×500​ L/min, using the residence time approach for steady-state analysis

Explanation: At steady state with C=6C = 6C=6 g/L constant, dCdt=0\frac{dC}{dt} = 0dtdC​=0. The mass balance is: d(C⋅500)dt=Q⋅8−Q⋅6−kCB⋅500=0\frac{d(C \cdot 500)}{dt} = Q \cdot 8 - Q \cdot 6 - kCB \cdot 500 = 0dtd(C⋅500)​=Q⋅8−Q⋅6−kCB⋅500=0. This gives Q(8−6)=0.15×6×2.5×500Q(8-6) = 0.15 \times 6 \times 2.5 \times 500Q(8−6)=0.15×6×2.5×500, so Q=0.15×6×2.5×5002Q = \frac{0.15 \times 6 \times 2.5 \times 500}{2}Q=20.15×6×2.5×500​. The numerator represents total consumption rate (rate constant × concentration × biomass × volume), and the denominator is the net concentration difference between inflow and tank. Choice B omits the concentration difference in outflow. Choice C ignores the volume factor. Choice D incorrectly inverts the relationship between flow and consumption.

Question 4

A crystallization tank operates with feedback control. The tank contains 800 L of supersaturated solution with initial crystal concentration 50 g/L. Fresh solution (10 g/L crystals) enters at rate FFF L/min, while product exits at the same rate FFF. Crystals grow at rate kC1.5kC^{1.5}kC1.5 where k=0.02k = 0.02k=0.02 L0.5^{0.5}0.5/(g0.5^{0.5}0.5·min) and CCC is the crystal concentration. The control system adjusts FFF to maintain dCdt=−2\frac{dC}{dt} = -2dtdC​=−2 g/(L·min). What flow rate FFF is required when C=40C = 40C=40 g/L?

  1. F=0.02×401.5−210−40F = \frac{0.02 \times 40^{1.5} - 2}{10 - 40}F=10−400.02×401.5−2​ L/min, balancing crystal growth against the controlled concentration decline
  2. F=0.02×401.5+240−10F = \frac{0.02 \times 40^{1.5} + 2}{40 - 10}F=40−100.02×401.5+2​ L/min, accounting for both growth kinetics and the desired concentration change
  3. F=2+0.02×401.540−10F = \frac{2 + 0.02 \times 40^{1.5}}{40 - 10}F=40−102+0.02×401.5​ L/min, using mass balance with the specified concentration derivative (correct answer)
  4. F=0.02×401.510−40−230F = \frac{0.02 \times 40^{1.5}}{10 - 40} - \frac{2}{30}F=10−400.02×401.5​−302​ L/min, separating the growth and control terms in the flow calculation

Explanation: The mass balance equation is dCdt=F(10−C)800+kC1.5\frac{dC}{dt} = \frac{F(10 - C)}{800} + kC^{1.5}dtdC​=800F(10−C)​+kC1.5. Given that dCdt=−2\frac{dC}{dt} = -2dtdC​=−2 g/(L·min) when C=40C = 40C=40 g/L: −2=F(10−40)800+0.02×401.5-2 = \frac{F(10 - 40)}{800} + 0.02 \times 40^{1.5}−2=800F(10−40)​+0.02×401.5. This gives −2=−30F800+0.02×401.5-2 = \frac{-30F}{800} + 0.02 \times 40^{1.5}−2=800−30F​+0.02×401.5. Rearranging: 30F800=0.02×401.5+2\frac{30F}{800} = 0.02 \times 40^{1.5} + 280030F​=0.02×401.5+2, so F=800(0.02×401.5+2)30=0.02×401.5+230/800=2+0.02×401.530/800F = \frac{800(0.02 \times 40^{1.5} + 2)}{30} = \frac{0.02 \times 40^{1.5} + 2}{30/800} = \frac{2 + 0.02 \times 40^{1.5}}{30/800}F=30800(0.02×401.5+2)​=30/8000.02×401.5+2​=30/8002+0.02×401.5​. Since 30/800 = (40-10)/800 and we multiply by 800/30, this becomes 2+0.02×401.540−10×80030×30800=2+0.02×401.540−10\frac{2 + 0.02 \times 40^{1.5}}{40-10} \times \frac{800}{30} \times \frac{30}{800} = \frac{2 + 0.02 \times 40^{1.5}}{40-10}40−102+0.02×401.5​×30800​×80030​=40−102+0.02×401.5​. Choice A has the wrong sign. Choice B omits the volume factor. Choice D incorrectly separates terms.

Question 5

A chemical reactor contains 1000 liters of solution. A reactant with concentration 5 mol/L enters at 20 L/min, while the well-mixed solution exits at the same rate. If the reactant undergoes a first-order decay with rate constant k=0.02k = 0.02k=0.02 min−1^{-1}−1, and the reactor initially contains pure solvent, what is the steady-state concentration?

  1. 5×201000+0.02\frac{5 \times 20}{1000 + 0.02}1000+0.025×20​ mol/L, accounting for both dilution and decay effects simultaneously
  2. 5×201000×0.02+20\frac{5 \times 20}{1000 \times 0.02 + 20}1000×0.02+205×20​ mol/L, balancing inflow against combined outflow and reaction
  3. 5×0.0220\frac{5 \times 0.02}{20}205×0.02​ mol/L, representing the equilibrium between input and decay rates
  4. 51+0.02×100020\frac{5}{1 + \frac{0.02 \times 1000}{20}}1+200.02×1000​5​ mol/L, using the ratio of time constants for mixing and reaction (correct answer)

Explanation: At steady state, dCdt=0\frac{dC}{dt} = 0dtdC​=0. The differential equation is dCdt=20×51000−20C1000−0.02C=0.1−0.02C−0.02C=0.1−0.04C\frac{dC}{dt} = \frac{20 \times 5}{1000} - \frac{20C}{1000} - 0.02C = 0.1 - 0.02C - 0.02C = 0.1 - 0.04CdtdC​=100020×5​−100020C​−0.02C=0.1−0.02C−0.02C=0.1−0.04C. Setting this to zero: 0.1=0.04C0.1 = 0.04C0.1=0.04C, so C=2.5C = 2.5C=2.5 mol/L. This can be written as C=51+0.02×100020=51+1=2.5C = \frac{5}{1 + \frac{0.02 \times 1000}{20}} = \frac{5}{1 + 1} = 2.5C=1+200.02×1000​5​=1+15​=2.5 mol/L. Choice A incorrectly adds the rate constant to the volume. Choice B has incorrect units and setup. Choice C incorrectly relates decay and inflow rates. The key insight is that both outflow (due to mixing) and reaction (due to decay) remove reactant from the system.

Question 6

A pharmaceutical mixing vessel initially contains 100 L of solution with drug concentration 2.0 mg/L. Two streams enter simultaneously: Stream A delivers pure solvent at 5 L/min, and Stream B delivers drug solution at 3 L/min with concentration CBC_BCB​. The mixture exits at 8 L/min. If the vessel concentration reaches 1.5 mg/L at t=20t = 20t=20 minutes, what is CBC_BCB​?

  1. CB=3.5C_B = 3.5CB​=3.5 mg/L, calculated from the steady-state concentration balance without considering transient effects
  2. CB=4.0C_B = 4.0CB​=4.0 mg/L, obtained by solving the differential equation and applying the condition at t=20t = 20t=20 (correct answer)
  3. CB=2.5C_B = 2.5CB​=2.5 mg/L, determined from the average concentration change over the 20-minute period
  4. CB=5.0C_B = 5.0CB​=5.0 mg/L, found by balancing the total drug input against the exponential decay in the vessel

Explanation: The volume remains constant at 100 L since inflow (5+3=8 L/min) equals outflow (8 L/min). The differential equation is dCdt=3CB100−8C100=3CB−8C100\frac{dC}{dt} = \frac{3C_B}{100} - \frac{8C}{100} = \frac{3C_B - 8C}{100}dtdC​=1003CB​​−1008C​=1003CB​−8C​. This linear ODE has solution C(t)=3CB8+(2.0−3CB8)e−0.08tC(t) = \frac{3C_B}{8} + (2.0 - \frac{3C_B}{8})e^{-0.08t}C(t)=83CB​​+(2.0−83CB​​)e−0.08t. At t=20t = 20t=20: 1.5=3CB8+(2.0−3CB8)e−1.61.5 = \frac{3C_B}{8} + (2.0 - \frac{3C_B}{8})e^{-1.6}1.5=83CB​​+(2.0−83CB​​)e−1.6. With e−1.6≈0.2019e^{-1.6} \approx 0.2019e−1.6≈0.2019, we get 1.5=3CB8+0.2019(2.0−3CB8)1.5 = \frac{3C_B}{8} + 0.2019(2.0 - \frac{3C_B}{8})1.5=83CB​​+0.2019(2.0−83CB​​). Solving: 1.5=3CB8+0.4038−3CB×0.201981.5 = \frac{3C_B}{8} + 0.4038 - \frac{3C_B \times 0.2019}{8}1.5=83CB​​+0.4038−83CB​×0.2019​, which gives CB=4.0C_B = 4.0CB​=4.0 mg/L. Choice A uses steady-state analysis only. Choice C uses incorrect averaging. Choice D has an error in the exponential term calculation.

Question 7

A 600-gallon tank with a leak initially contains 400 gallons of brine with 80 pounds of salt. Brine containing 4 lb/gal enters at 15 gal/min. The leak rate is proportional to the volume with constant α=0.01\alpha = 0.01α=0.01 min−1^{-1}−1, and additional outflow occurs at 10 gal/min. If the system reaches equilibrium, what is the equilibrium salt concentration?

  1. 4×1515−10\frac{4 \times 15}{15 - 10}15−104×15​ lb/gal, treating the system as if the leak were negligible compared to fixed flows
  2. 4×1510+0.01Veq\frac{4 \times 15}{10 + 0.01V_{eq}}10+0.01Veq​4×15​ lb/gal, where VeqV_{eq}Veq​ is found by solving 15=10+0.01Veq15 = 10 + 0.01V_{eq}15=10+0.01Veq​ (correct answer)
  3. 4×1515\frac{4 \times 15}{15}154×15​ lb/gal, since equilibrium concentration equals the inflow concentration when volume stabilizes
  4. 444 lb/gal, since the inflow concentration dominates when the system reaches steady state

Explanation: At equilibrium, the volume is constant, so inflow equals total outflow: 15=10+0.01Veq15 = 10 + 0.01V_{eq}15=10+0.01Veq​. This gives Veq=500V_{eq} = 500Veq​=500 gallons. The salt concentration equilibrium occurs when dSdt=0\frac{dS}{dt} = 0dtdS​=0: 4×15−Ceq(10+0.01×500)=04 \times 15 - C_{eq}(10 + 0.01 \times 500) = 04×15−Ceq​(10+0.01×500)=0, so 60=15Ceq60 = 15C_{eq}60=15Ceq​, giving Ceq=4C_{eq} = 4Ceq​=4 lb/gal. However, the general form is Ceq=4×1510+0.01VeqC_{eq} = \frac{4 \times 15}{10 + 0.01V_{eq}}Ceq​=10+0.01Veq​4×15​ where VeqV_{eq}Veq​ satisfies the volume equilibrium condition. Choice A ignores the leak term. Choice C incorrectly assumes concentration equals inflow concentration. Choice D is numerically correct but doesn't show the reasoning. The key insight is that both volume and salt must be in equilibrium simultaneously.