Differential Equations Quiz: Mixing Problems
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Mixing ProblemsQuestion 1 of 18

A tank initially contains 200 gallons of pure water. Brine containing 0.5 pounds of salt per gallon flows in at 4 gallons per minute, while the well-mixed solution flows out at 6 gallons per minute. If S(t)S(t) represents the amount of salt in the tank at time tt minutes, which differential equation correctly models this situation?

dSdt=2−6S200−2t\frac{dS}{dt} = 2 - \frac{6S}{200-2t}
dSdt=2−3S100−t\frac{dS}{dt} = 2 - \frac{3S}{100-t}
dSdt=2−4S200−2t\frac{dS}{dt} = 2 - \frac{4S}{200-2t}
dSdt=0.5−6S200−2t\frac{dS}{dt} = 0.5 - \frac{6S}{200-2t}
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Differential Equations Quiz

Differential Equations Quiz: Mixing Problems

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

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This quiz focuses on Mixing Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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Question 1

A tank initially contains 200 gallons of pure water. Brine containing 0.5 pounds of salt per gallon flows in at 4 gallons per minute, while the well-mixed solution flows out at 6 gallons per minute. If S(t)S(t) represents the amount of salt in the tank at time tt minutes, which differential equation correctly models this situation?

  1. dSdt=2−6S200−2t\frac{dS}{dt} = 2 - \frac{6S}{200-2t} (correct answer)
  2. dSdt=2−3S100−t\frac{dS}{dt} = 2 - \frac{3S}{100-t}
  3. dSdt=2−4S200−2t\frac{dS}{dt} = 2 - \frac{4S}{200-2t}
  4. dSdt=0.5−6S200−2t\frac{dS}{dt} = 0.5 - \frac{6S}{200-2t}
Explanation: The rate of salt entering is (0.5 lb/gal)(4 gal/min) = 2 lb/min. The volume at time t is 200 + 4t - 6t = 200 - 2t gallons. The concentration is S/(200-2t) lb/gal, so salt leaves at rate (6 gal/min)(S/(200-2t) lb/gal) = 6S/(200-2t) lb/min. Thus dS/dt = 2 - 6S/(200-2t). Choice B incorrectly simplifies the outflow term. Choice C uses inflow rate instead of outflow rate. Choice D uses concentration instead of rate for inflow.

Question 2

A 500-liter tank initially contains 200 liters of brine with 10 kg of dissolved salt. Brine containing 0.05 kg/L of salt flows into the tank at a rate of 10 L/min. The well-stirred mixture flows out at a rate of 5 L/min. What is the amount of salt, in kg, in the tank at the moment it begins to overflow?

  1. 20−10e−1.520 - 10e^{-1.5}
  2. 25 (correct answer)
  3. 34
  4. 50
Explanation: Let A(t)A(t) be the amount of salt in kg and V(t)V(t) be the volume of brine in L at time tt in minutes. The initial conditions are A(0)=10A(0) = 10 and V(0)=200V(0) = 200. The volume changes according to V(t)=200+(10−5)t=200+5tV(t) = 200 + (10 - 5)t = 200 + 5t. The tank overflows when V(t)=500V(t) = 500, which gives 200+5t=500200 + 5t = 500, so t=60t = 60 minutes. The differential equation for the amount of salt is dAdt=(rate in)−(rate out)=(10)(0.05)−5⋅A(t)V(t)=0.5−5A200+5t\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}) = (10)(0.05) - 5 \cdot \frac{A(t)}{V(t)} = 0.5 - \frac{5A}{200+5t}. This is a first-order linear ODE: dAdt+140+tA=0.5\frac{dA}{dt} + \frac{1}{40+t}A = 0.5. The integrating factor is I(t)=e∫140+tdt=40+tI(t) = e^{\int \frac{1}{40+t}dt} = 40+t. Multiplying by I(t)I(t) gives ((40+t)A)′=0.5(40+t)((40+t)A)' = 0.5(40+t). Integrating yields (40+t)A=20t+0.25t2+C(40+t)A = 20t + 0.25t^2 + C. Using A(0)=10A(0)=10, we find 40(10)=C40(10) = C, so C=400C=400. The solution is A(t)=20t+0.25t2+40040+tA(t) = \frac{20t + 0.25t^2 + 400}{40+t}. At t=60t=60, A(60)=20(60)+0.25(60)2+40040+60=1200+900+400100=2500100=25A(60) = \frac{20(60) + 0.25(60)^2 + 400}{40+60} = \frac{1200 + 900 + 400}{100} = \frac{2500}{100} = 25 kg.

Question 3

A 100-gallon tank is initially half-full of pure water. A solution with a time-varying salt concentration of c(t)=1t+1c(t) = \frac{1}{t+1} lbs/gal is pumped in at a rate of 5 gal/min. The mixture is kept uniform by stirring and is drained at a rate of 3 gal/min. Which initial value problem correctly models the amount of salt A(t)A(t) in lbs in the tank for t≥0t \ge 0, while the tank is filling?

  1. dAdt=5t+1−3A50+2t,A(0)=0\frac{dA}{dt} = \frac{5}{t+1} - \frac{3A}{50+2t}, \quad A(0)=0 (correct answer)
  2. dAdt=5t+1−3A50,A(0)=0\frac{dA}{dt} = \frac{5}{t+1} - \frac{3A}{50}, \quad A(0)=0
  3. dAdt=5t+1−3A100,A(0)=0\frac{dA}{dt} = \frac{5}{t+1} - \frac{3A}{100}, \quad A(0)=0
  4. dAdt=5−3A50+2t,A(0)=0\frac{dA}{dt} = 5 - \frac{3A}{50+2t}, \quad A(0)=0
Explanation: The rate of change of salt is dAdt=(rate in)−(rate out)\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}). The initial volume is V0=100/2=50V_0 = 100/2 = 50 gal. The volume at time tt is V(t)=V0+(rin−rout)t=50+(5−3)t=50+2tV(t) = V_0 + (r_{in}-r_{out})t = 50 + (5-3)t = 50+2t. The rate of salt flowing in is rin⋅cin(t)=5⋅1t+1=5t+1r_{in} \cdot c_{in}(t) = 5 \cdot \frac{1}{t+1} = \frac{5}{t+1}. The concentration of salt in the tank is A(t)V(t)=A50+2t\frac{A(t)}{V(t)} = \frac{A}{50+2t}. The rate of salt flowing out is rout⋅A(t)V(t)=3⋅A50+2tr_{out} \cdot \frac{A(t)}{V(t)} = 3 \cdot \frac{A}{50+2t}. Combining these gives the differential equation dAdt=5t+1−3A50+2t\frac{dA}{dt} = \frac{5}{t+1} - \frac{3A}{50+2t}. Since the tank initially contains pure water, the initial amount of salt is A(0)=0A(0)=0.

Question 4

A 200 L tank initially contains 100 L of water with 5 kg of salt. For the first 10 minutes, brine with a concentration of 0.2 kg/L flows in at 4 L/min, and the mixture flows out at 2 L/min. After 10 minutes, the inflow is switched to pure water at 4 L/min, and the outflow rate is changed to 5 L/min. Let A10A_{10} be the amount of salt at t=10t=10 min. Which expression represents the amount of salt A(t)A(t) for t>10t > 10?

  1. A(t)=A10(110−t100)5A(t) = A_{10} \left( \frac{110-t}{100} \right)^5
  2. A(t)=A10e−5(t−10)/120A(t) = A_{10} e^{-5(t-10)/120}
  3. A(t)=5(130−t120)5A(t) = 5 \left( \frac{130-t}{120} \right)^5
  4. A(t)=A10(130−t120)5A(t) = A_{10} \left( \frac{130-t}{120} \right)^5 (correct answer)
Explanation: For t>10t>10, let τ=t−10\tau = t-10. At t=10t=10, the volume is V(10)=100+(4−2)(10)=120V(10) = 100 + (4-2)(10) = 120 L. The amount of salt is A(10)=A10A(10) = A_{10}. For t>10t>10 (or τ>0\tau>0), the new parameters are cin=0c_{in}=0, rin=4r_{in}=4, rout=5r_{out}=5. The volume is V2(τ)=120+(4−5)τ=120−τV_2(\tau) = 120 + (4-5)\tau = 120 - \tau. The differential equation is dAdτ=0−5A120−τ\frac{dA}{d\tau} = 0 - 5 \frac{A}{120-\tau}. This is a separable equation: dAA=−5120−τdτ\frac{dA}{A} = -\frac{5}{120-\tau}d\tau. Integrating gives ln⁡(A)=5ln⁡(120−τ)+C\ln(A) = 5\ln(120-\tau) + C, so A(τ)=K(120−τ)5A(\tau) = K(120-\tau)^5. The initial condition is A(0)=A10A(0) = A_{10} (at τ=0\tau=0), so A10=K(120)5A_{10} = K(120)^5, which means K=A10/1205K = A_{10}/120^5. Thus, A(τ)=A10(120−τ)51205=A10(120−τ120)5A(\tau) = A_{10} \frac{(120-\tau)^5}{120^5} = A_{10} (\frac{120-\tau}{120})^5. Substituting τ=t−10\tau = t-10 gives A(t)=A10(120−(t−10)120)5=A10(130−t120)5A(t) = A_{10} (\frac{120-(t-10)}{120})^5 = A_{10} (\frac{130-t}{120})^5.

Question 5

The amount of salt A(t)A(t) in kilograms in a tank is described by the function A(t)=50−30e−0.05tA(t) = 50 - 30e^{-0.05t}, where tt is in minutes. The volume of the tank is constant at 200 L. Which physical setup does this function describe?

  1. A 200 L tank, initially with 50 kg of salt, where brine enters at 10 L/min with concentration 0.25 kg/L and the mixture leaves at the same rate.
  2. A 200 L tank, initially with 20 kg of salt, where brine enters at 5 L/min with concentration 0.5 kg/L and the mixture leaves at the same rate.
  3. A 200 L tank, initially with 20 kg of salt, where pure water enters at 10 L/min and the mixture leaves at the same rate.
  4. A 200 L tank, initially with 20 kg of salt, where brine enters at 10 L/min with concentration 0.25 kg/L and the mixture leaves at the same rate. (correct answer)
Explanation: From the solution A(t)=50−30e−0.05tA(t) = 50 - 30e^{-0.05t}, we can deduce the parameters. The initial amount is A(0)=50−30e0=20A(0) = 50 - 30e^0 = 20 kg. The steady-state (limiting) amount is Ass=lim⁡t→∞A(t)=50A_{ss} = \lim_{t \to \infty} A(t) = 50 kg. The general solution for a constant volume mixing problem is A(t)=Ass+(A0−Ass)e−(r/V)tA(t) = A_{ss} + (A_0 - A_{ss})e^{-(r/V)t}. Comparing this with the given function, we have A0=20A_0=20, Ass=50A_{ss}=50, and the exponent is r/V=0.05r/V = 0.05. Since V=200V=200 L, the flow rate is r=0.05×200=10r = 0.05 \times 200 = 10 L/min. The steady-state amount is given by Ass=cinVA_{ss} = c_{in}V, so 50=cin×20050 = c_{in} \times 200, which gives the inflow concentration cin=50/200=0.25c_{in} = 50/200 = 0.25 kg/L. Therefore, the setup is a 200 L tank, initially with 20 kg of salt, with inflow/outflow at 10 L/min and an inflow concentration of 0.25 kg/L.

Question 6

A tank initially holds 100 L of brine containing 20 kg of salt. Brine with a salt concentration of 0.1 kg/L enters the tank at 2 L/min. The well-stirred mixture leaves the tank at 4 L/min. What is the concentration of salt in the tank at the moment it becomes empty?

  1. 0 kg/L
  2. 0.1 kg/L (correct answer)
  3. 0.15 kg/L
  4. 0.2 kg/L
Explanation: Let A(t)A(t) be the amount of salt and V(t)V(t) be the volume. V(t)=100+(2−4)t=100−2tV(t) = 100 + (2-4)t = 100-2t. The tank empties when V(t)=0V(t)=0, so at t=50t=50 min. The DE for the amount of salt is dAdt=(2)(0.1)−4A100−2t=0.2−2A50−t\frac{dA}{dt} = (2)(0.1) - 4 \frac{A}{100-2t} = 0.2 - \frac{2A}{50-t}. This is a linear ODE. Its solution is A(t)=0.2(50−t)+C(50−t)2A(t) = 0.2(50-t) + C(50-t)^2. Using A(0)=20A(0)=20, we find 20=0.2(50)+C(50)220 = 0.2(50) + C(50)^2, which gives 10=2500C10 = 2500C, so C=1/250C=1/250. Thus, A(t)=0.2(50−t)+1250(50−t)2A(t) = 0.2(50-t) + \frac{1}{250}(50-t)^2. The concentration is C(t)=A(t)V(t)=0.2(50−t)+1250(50−t)22(50−t)=0.1+1500(50−t)C(t) = \frac{A(t)}{V(t)} = \frac{0.2(50-t) + \frac{1}{250}(50-t)^2}{2(50-t)} = 0.1 + \frac{1}{500}(50-t). We want the concentration as t→50t \to 50. lim⁡t→50C(t)=0.1+1500(50−50)=0.1\lim_{t \to 50} C(t) = 0.1 + \frac{1}{500}(50-50) = 0.1 kg/L.

Question 7

A vat contains 100 gallons of cider, initially with no sugar. A solution of 2 pounds of sugar per gallon is pumped into the vat at a rate of 3 quarts per minute. The mixture is pumped out at the same rate. How much sugar, in pounds, is in the vat after one hour? (Note: 1 gallon = 4 quarts)

  1. 200(1−e−1.8)200(1-e^{-1.8})
  2. 200(1−e−0.03)200(1-e^{-0.03})
  3. 200(1−e−0.45)200(1-e^{-0.45}) (correct answer)
  4. 200(1−e−0.0075)200(1-e^{-0.0075})
Explanation: First, ensure all units are consistent. Let's use gallons and minutes. The volume is V=100V=100 gal. The inflow concentration is cin=2c_{in} = 2 lbs/gal. The flow rate is r=3r = 3 quarts/min =34= \frac{3}{4} gal/min =0.75= 0.75 gal/min. The initial amount of sugar is A(0)=0A(0)=0. The differential equation is dAdt=r⋅cin−rAV=(0.75)(2)−0.75A100=1.5−0.0075A\frac{dA}{dt} = r \cdot c_{in} - r \frac{A}{V} = (0.75)(2) - 0.75 \frac{A}{100} = 1.5 - 0.0075A. The solution has the form A(t)=Ass+(A0−Ass)e−ktA(t) = A_{ss} + (A_0 - A_{ss})e^{-kt}, where k=0.0075k=0.0075 and Ass=1.5/0.0075=200A_{ss} = 1.5/0.0075 = 200 lbs. Since A0=0A_0=0, the solution is A(t)=200(1−e−0.0075t)A(t) = 200(1 - e^{-0.0075t}). We need to find the amount after one hour, which is t=60t=60 minutes. A(60)=200(1−e−0.0075×60)=200(1−e−0.45)A(60) = 200(1 - e^{-0.0075 \times 60}) = 200(1 - e^{-0.45}).

Question 8

A 400-liter tank is full of a 25% salt solution by mass. The solution has a density of 1.2 kg/L. Pure water is run into the tank at a rate of 10 L/min, and the mixture is pumped out at the same rate. How long does it take for the amount of salt in the tank to be halved?

  1. 40ln⁡(2)40 \ln(2) min (correct answer)
  2. 20ln⁡(2)20 \ln(2) min
  3. 40ln⁡(2)\frac{40}{\ln(2)} min
  4. 4040 min
Explanation: First, calculate the initial amount of salt, A(0)A(0). The total mass of the solution is Volume ×\times density = 400400 L ×1.2\times 1.2 kg/L = 480480 kg. The salt is 25% of this mass, so A(0)=0.25×480=120A(0) = 0.25 \times 480 = 120 kg. The differential equation for the amount of salt is dAdt=(rate in)−(rate out)\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}). Since pure water flows in, cin=0c_{in}=0. So, dAdt=(10)(0)−10A400=−A40\frac{dA}{dt} = (10)(0) - 10 \frac{A}{400} = -\frac{A}{40}. This is an exponential decay model with solution A(t)=A(0)e−t/40A(t) = A(0)e^{-t/40}. We want to find the time tt when the amount of salt is halved, i.e., A(t)=A(0)/2A(t) = A(0)/2. So, A(0)2=A(0)e−t/40\frac{A(0)}{2} = A(0)e^{-t/40}, which simplifies to 12=e−t/40\frac{1}{2} = e^{-t/40}. Taking the natural logarithm of both sides gives ln⁡(1/2)=−t/40\ln(1/2) = -t/40, or −ln⁡(2)=−t/40-\ln(2) = -t/40. Therefore, t=40ln⁡(2)t = 40 \ln(2) minutes.

Question 9

A tank's volume is changing over time. Brine flows in with concentration cinc_{in} at rate rinr_{in}, and the mixture flows out at rate routr_{out}. The amount of salt in the tank, A(t)A(t), is observed to reach a local maximum value at some time tmax>0t_{max} > 0. What is the concentration of salt in the tank at time tmaxt_{max}?

  1. cinc_{in}
  2. cinrinroutc_{in} \frac{r_{in}}{r_{out}} (correct answer)
  3. cinroutrinc_{in} \frac{r_{out}}{r_{in}}
  4. A(0)V(0)\frac{A(0)}{V(0)}
Explanation: Let A(t)A(t) be the amount of salt and V(t)V(t) be the volume. The differential equation governing the amount of salt is dAdt=rincin−routA(t)V(t)\frac{dA}{dt} = r_{in}c_{in} - r_{out} \frac{A(t)}{V(t)}. A local maximum of A(t)A(t) occurs when dAdt=0\frac{dA}{dt} = 0. Setting the equation to zero gives rincin−routA(tmax)V(tmax)=0r_{in}c_{in} - r_{out} \frac{A(t_{max})}{V(t_{max})} = 0. The concentration at time tmaxt_{max} is C(tmax)=A(tmax)V(tmax)C(t_{max}) = \frac{A(t_{max})}{V(t_{max})}. Substituting this into the equation gives rincin−routC(tmax)=0r_{in}c_{in} - r_{out}C(t_{max}) = 0. Solving for C(tmax)C(t_{max}) yields C(tmax)=rincinroutC(t_{max}) = \frac{r_{in}c_{in}}{r_{out}}.

Question 10

A tank initially holds 400 L of brine. Brine containing 0.1 kg/L of salt flows in at 10 L/min, and the mixture flows out at the same rate. After 20 minutes, the concentration of salt in the tank is 0.075 kg/L. What was the initial amount of salt A0A_0 in the tank?

  1. 40−10e0.540 - 10e^{0.5} kg (correct answer)
  2. 40−10e−0.540 - 10e^{-0.5} kg
  3. 40+10e0.540 + 10e^{0.5} kg
  4. 30e0.530e^{0.5} kg
Explanation: The DE is dAdt=(10)(0.1)−10A400=1−A40\frac{dA}{dt} = (10)(0.1) - 10\frac{A}{400} = 1 - \frac{A}{40}. The general solution is A(t)=40+Ce−t/40A(t) = 40 + Ce^{-t/40}. The initial condition A(0)=A0A(0)=A_0 gives A0=40+CA_0 = 40 + C, so C=A0−40C = A_0 - 40. The solution is A(t)=40+(A0−40)e−t/40A(t) = 40 + (A_0-40)e^{-t/40}. We are given that the concentration at t=20t=20 is 0.0750.075 kg/L. The amount of salt at this time is A(20)=0.075×400=30A(20) = 0.075 \times 400 = 30 kg. Plugging this into the solution: 30=40+(A0−40)e−20/40=40+(A0−40)e−0.530 = 40 + (A_0-40)e^{-20/40} = 40 + (A_0-40)e^{-0.5}. Solving for A0A_0: −10=(A0−40)e−0.5-10 = (A_0-40)e^{-0.5}, so A0−40=−10/e−0.5=−10e0.5A_0-40 = -10/e^{-0.5} = -10e^{0.5}. Therefore, A0=40−10e0.5A_0 = 40 - 10e^{0.5} kg.

Question 11

A large tank contains 1000 L of pure water. A salt solution with a concentration of 0.2 kg/L is pumped in at a rate of rr L/min, and the well-mixed solution is pumped out at the same rate. The amount of salt A(t)A(t) in the tank approaches a limiting value as t→∞t \to \infty. If the inflow rate rr were doubled, how would this limiting amount of salt change?

  1. It would be halved.
  2. It would remain the same. (correct answer)
  3. It would double.
  4. It would be four times as large.
Explanation: The differential equation for the amount of salt is dAdt=r⋅cin−r⋅AV\frac{dA}{dt} = r \cdot c_{in} - r \cdot \frac{A}{V}. In this case, dAdt=r(0.2)−rA1000\frac{dA}{dt} = r(0.2) - r\frac{A}{1000}. The limiting amount of salt, AlimA_{lim}, is the equilibrium solution where dAdt=0\frac{dA}{dt} = 0. So, 0=0.2r−rAlim10000 = 0.2r - \frac{r A_{lim}}{1000}. Since r>0r>0, we can divide by rr to get 0=0.2−Alim10000 = 0.2 - \frac{A_{lim}}{1000}, which gives Alim=0.2⋅1000=200A_{lim} = 0.2 \cdot 1000 = 200 kg. Notice that the rate rr cancels out, meaning the limiting amount is independent of the flow rate. Doubling the rate rr would cause the system to approach this limit faster, but the limit itself would remain the same.

Question 12

Consider a system of two interconnected 50-liter tanks. Tank 1 initially contains pure water, while Tank 2 initially contains 5 kg of salt dissolved in 50 L of water. Pure water flows into Tank 1 at 10 L/min. The mixture from Tank 1 flows into Tank 2 at 10 L/min. The mixture from Tank 2 flows out of the system at 10 L/min. Let A1(t)A_1(t) and A2(t)A_2(t) be the amount of salt (in kg) in Tank 1 and Tank 2, respectively. Which is the correct system of differential equations for this scenario?

  1. dA1dt=−A15;dA2dt=A1−A25\frac{dA_1}{dt} = -\frac{A_1}{5}; \quad \frac{dA_2}{dt} = \frac{A_1 - A_2}{5} (correct answer)
  2. dA1dt=A2−A15;dA2dt=−A25\frac{dA_1}{dt} = \frac{A_2 - A_1}{5}; \quad \frac{dA_2}{dt} = -\frac{A_2}{5}
  3. dA1dt=10−A15;dA2dt=A1−A25\frac{dA_1}{dt} = 10 - \frac{A_1}{5}; \quad \frac{dA_2}{dt} = \frac{A_1 - A_2}{5}
  4. dA1dt=−10A1;dA2dt=10A1−10A2\frac{dA_1}{dt} = -10A_1; \quad \frac{dA_2}{dt} = 10A_1 - 10A_2
Explanation: For Tank 1, the volume is constant at V1=50V_1=50 L. The inflow is pure water, so the rate of salt in is 0. The mixture flows out at 10 L/min with concentration A1/V1=A1/50A_1/V_1 = A_1/50. So, rate of salt out is 10⋅(A1/50)=A1/510 \cdot (A_1/50) = A_1/5. Thus, dA1dt=0−A15=−A15\frac{dA_1}{dt} = 0 - \frac{A_1}{5} = -\frac{A_1}{5}. For Tank 2, the volume is constant at V2=50V_2=50 L. The inflow is from Tank 1, so the rate of salt in is the rate of salt out from Tank 1, which is A1/5A_1/5. The mixture flows out of Tank 2 at 10 L/min with concentration A2/V2=A2/50A_2/V_2 = A_2/50. So, rate of salt out is 10⋅(A2/50)=A2/510 \cdot (A_2/50) = A_2/5. Thus, dA2dt=A15−A25=A1−A25\frac{dA_2}{dt} = \frac{A_1}{5} - \frac{A_2}{5} = \frac{A_1 - A_2}{5}.

Question 13

A tank contains a brine solution. New brine flows in, and the mixture flows out, with the volume of solution in the tank increasing over time. The initial concentration of salt in the tank, C0C_0, is greater than the concentration of the incoming brine, cinc_{in}. Under what condition will the amount of salt in the tank, A(t)A(t), initially increase?

  1. A(t)A(t) must always decrease since C0>cinC_0 > c_{in}.
  2. rin>routr_{in} > r_{out}
  3. rincin>routC0r_{in}c_{in} > r_{out}C_0 (correct answer)
  4. rinC0>routcinr_{in}C_0 > r_{out}c_{in}
Explanation: The rate of change of the amount of salt is given by the differential equation dAdt=rincin−routC(t)\frac{dA}{dt} = r_{in}c_{in} - r_{out}C(t), where C(t)C(t) is the concentration in the tank at time tt. We want to find the condition under which the amount of salt is initially increasing, which means dAdt\frac{dA}{dt} at t=0t=0 must be positive. At t=0t=0, the concentration in the tank is C(0)=C0C(0) = C_0. So we need dAdt∣t=0=rincin−routC0>0\left.\frac{dA}{dt}\right|_{t=0} = r_{in}c_{in} - r_{out}C_0 > 0. This inequality simplifies to rincin>routC0r_{in}c_{in} > r_{out}C_0. Even if the incoming concentration is lower (cin<C0c_{in} < C_0), the amount of salt can increase if the inflow rate is sufficiently larger than the outflow rate.

Question 14

A tank contains 500 gallons of brine with 50 pounds of salt. Fresh water flows in at rate rr gal/min and mixture flows out at the same rate rr gal/min. If the salt concentration decreases to 10% of its initial value in exactly 30 minutes, what is the inflow rate rr?

  1. r=500ln⁡(10)3r = \frac{500\ln(10)}{3} gal/min
  2. r=50ln⁡(10)3r = \frac{50\ln(10)}{3} gal/min
  3. r=500ln⁡(10)30r = \frac{500\ln(10)}{30} gal/min (correct answer)
  4. r=50ln⁡(10)r = 50\ln(10) gal/min
Explanation: This is a classic mixing problem that requires setting up a differential equation for the rate of change of salt in the tank. When you see problems involving substances flowing in and out of containers, always think: rate of change = rate in - rate out. Let S(t)S(t) be the amount of salt at time tt. Since fresh water flows in (no salt), the rate of salt entering is 0. The rate of salt leaving equals the concentration times the outflow rate: S(t)500⋅r\frac{S(t)}{500} \cdot r. This gives us the differential equation dSdt=−r500S\frac{dS}{dt} = -\frac{r}{500}S. Solving this separable equation: S(t)=S0e−rt/500S(t) = S_0 e^{-rt/500} where S0=50S_0 = 50 pounds initially. The initial concentration is 50500=0.1\frac{50}{500} = 0.1 lb/gal. After 30 minutes, the concentration should be 10% of this: 0.010.01 lb/gal, meaning S(30)=5S(30) = 5 pounds. Substituting: 5=50e−30r/5005 = 50e^{-30r/500}, which simplifies to 0.1=e−30r/5000.1 = e^{-30r/500}. Taking the natural logarithm: ln⁡(0.1)=−30r500\ln(0.1) = -\frac{30r}{500}. Since ln⁡(0.1)=−ln⁡(10)\ln(0.1) = -\ln(10), we get r=500ln⁡(10)30r = \frac{500\ln(10)}{30}. Choice A incorrectly uses 3 instead of 30 in the denominator. Choice B uses the wrong numerator (50 instead of 500) and wrong denominator. Choice D is missing the denominator entirely, representing a failure to properly solve the exponential equation. Remember: in mixing problems, always track whether the volume stays constant, and set up your rate equation carefully by identifying what flows in versus what flows out.

Question 15

A mixing tank problem involves brine with salt concentration c(t)c(t) flowing into a tank at rate RinR_{in} and mixture flowing out at rate RoutR_{out}. If the tank volume is V(t)=V0+(Rin−Rout)tV(t) = V_0 + (R_{in} - R_{out})t and salt amount is S(t)S(t), which expression correctly represents the salt concentration in the outflow at time tt?

  1. c(t)⋅RinRoutc(t) \cdot \frac{R_{in}}{R_{out}}
  2. dS/dtRout\frac{dS/dt}{R_{out}}
  3. S(t)V0+(Rin−Rout)t\frac{S(t)}{V_0 + (R_{in} - R_{out})t} (correct answer)
  4. S(t)⋅RoutV0+(Rin−Rout)t\frac{S(t) \cdot R_{out}}{V_0 + (R_{in} - R_{out})t}
Explanation: When you encounter mixing tank problems, you're dealing with the fundamental relationship between amount, volume, and concentration. The key insight is that concentration equals the total amount of substance divided by the total volume containing it. The correct answer is C because salt concentration in the outflow equals the current concentration throughout the tank, which is the total salt amount S(t)S(t) divided by the current tank volume V(t)=V0+(Rin−Rout)tV(t) = V_0 + (R_{in} - R_{out})t. Since the mixture is well-stirred, the concentration leaving the tank must equal the concentration inside the tank at that moment. Let's examine why the other options fail: A incorrectly assumes the outflow concentration depends on the inflow concentration multiplied by a flow rate ratio. This ignores the actual mixing that occurs in the tank and doesn't account for the accumulated salt. B represents the rate of change of salt amount divided by outflow rate. While dSdt\frac{dS}{dt} tells us how fast salt is changing, dividing by RoutR_{out} doesn't give us concentration—it gives us something with wrong units. D multiplies the correct concentration by RoutR_{out}, which would give you the rate at which salt leaves the tank (amount per unit time), not the concentration of that outflow. Study tip: In mixing problems, always remember that concentration = amount ÷ volume. The outflow concentration equals the current tank concentration when the mixture is well-stirred. Don't confuse rates of change with the actual concentrations or amounts.

Question 16

A 250-gallon tank initially contains brine with 25 pounds of salt. Pure water enters at 8 gal/min and mixture exits at 10 gal/min. The differential equation dSdt=−10S250−2t\frac{dS}{dt} = -\frac{10S}{250-2t} models the salt amount S(t)S(t). At what time does the tank become empty, and what happens to the salt concentration just before this occurs?

  1. Tank empties at t=125t = 125 min; concentration approaches 25250=0.1\frac{25}{250} = 0.1 lb/gal as t→125−t \to 125^-
  2. Tank empties at t=125t = 125 min; concentration approaches 0 as t→125−t \to 125^- (correct answer)
  3. Tank empties at t=125t = 125 min; concentration approaches ∞\infty as t→125−t \to 125^-
  4. Tank empties at t=100t = 100 min; concentration approaches 0 as t→100−t \to 100^-
Explanation: Tank mixing problems involve tracking both the volume and concentration of a substance over time. When inflow and outflow rates differ, you need to determine when the tank empties and analyze the limiting behavior of concentration. First, let's find when the tank empties. The volume changes at rate: inflow (8 gal/min) minus outflow (10 gal/min) = -2 gal/min. So V(t)=250−2tV(t) = 250 - 2t. The tank empties when V(t)=0V(t) = 0, giving us 250−2t=0250 - 2t = 0, so t=125t = 125 minutes. Now let's solve the differential equation dSdt=−10S250−2t\frac{dS}{dt} = -\frac{10S}{250-2t}. Separating variables: dSS=−10250−2tdt\frac{dS}{S} = -\frac{10}{250-2t}dt. Integrating both sides: ln⁡S=−10ln⁡∣250−2t∣/(−2)+C=5ln⁡∣250−2t∣+C\ln S = -10 \ln|250-2t|/(-2) + C = 5\ln|250-2t| + C. This gives S(t)=K(250−2t)5S(t) = K(250-2t)^5 for some constant KK. Using initial condition S(0)=25S(0) = 25: 25=K(250)525 = K(250)^5, so K=252505K = \frac{25}{250^5}. The concentration is c(t)=S(t)V(t)=K(250−2t)5250−2t=K(250−2t)4c(t) = \frac{S(t)}{V(t)} = \frac{K(250-2t)^5}{250-2t} = K(250-2t)^4. As t→125−t \to 125^-, we have (250−2t)→0+(250-2t) \to 0^+, so c(t)→0c(t) \to 0. Choice A incorrectly uses the initial concentration without considering dilution over time. Choice C suggests concentration approaches infinity, but the salt amount decreases faster than volume. Choice D uses the wrong emptying time (would need a -5 gal/min net rate). Remember: in mixing problems, always track both the substance amount and volume separately, then find their ratio for concentration.

Question 17

In a cascade of two tanks, pure water enters Tank 1 (100 gal) at 5 gal/min, overflow goes to Tank 2 (150 gal) at 5 gal/min, and mixture exits Tank 2 at 5 gal/min. Initially, Tank 1 has 20 lb salt and Tank 2 has 30 lb salt. After a long time, what will be the total amount of salt in both tanks combined?

  1. 25 pounds
  2. 0 pounds (correct answer)
  3. 50 pounds
  4. The salt amount oscillates indefinitely
Explanation: When you encounter a cascade tank problem, you're dealing with a system where the long-term behavior depends on what's flowing in versus what's flowing out of the entire system. The key insight is to think about the overall input and output, not just individual tanks. Since pure water enters Tank 1 at 5 gal/min and mixture exits Tank 2 at 5 gal/min, the volumes remain constant. More importantly, no salt enters the system - only pure water. Meanwhile, salt continuously leaves through Tank 2's outflow. This means the total salt in the system must decrease over time and approach zero as t→∞t \to \infty. You can verify this mathematically by setting up the differential equations. Let x1(t)x_1(t) and x2(t)x_2(t) be the salt amounts in tanks 1 and 2. Then: dx1dt=−x120\frac{dx_1}{dt} = -\frac{x_1}{20} and dx2dt=x120−x230\frac{dx_2}{dt} = \frac{x_1}{20} - \frac{x_2}{30} The total salt S(t)=x1(t)+x2(t)S(t) = x_1(t) + x_2(t) satisfies dSdt=−x230<0\frac{dS}{dt} = -\frac{x_2}{30} < 0, confirming continuous decrease toward zero. Answer A (25 pounds) incorrectly assumes some equilibrium between tanks. Answer C (50 pounds) mistakenly thinks salt is conserved, ignoring the outflow. Answer D (oscillation) confuses this with systems having periodic inputs or feedback loops. Study tip: In cascade problems, always track what enters and leaves the entire system first. If no salt enters but salt continuously exits, the long-term amount must be zero, regardless of initial distributions between tanks.

Question 18

Two tanks are connected: Tank A (100 gal) initially has pure water, Tank B (150 gal) initially has 30 lb of salt. Liquid flows from A to B at 2 gal/min, from B to A at 3 gal/min, and pure water enters A at 1 gal/min while mixture leaves B at 1 gal/min. If SA(t)S_A(t) and SB(t)S_B(t) are the amounts of salt in tanks A and B respectively, which system correctly models this situation?

  1. dSAdt=3SB149+t−2SA101−t\frac{dS_A}{dt} = \frac{3S_B}{149+t} - \frac{2S_A}{101-t}, dSBdt=2SA101−t−4SB149+t\frac{dS_B}{dt} = \frac{2S_A}{101-t} - \frac{4S_B}{149+t}
  2. dSAdt=3SB150−2SA100\frac{dS_A}{dt} = \frac{3S_B}{150} - \frac{2S_A}{100}, dSBdt=2SA100−3SB150−SB150\frac{dS_B}{dt} = \frac{2S_A}{100} - \frac{3S_B}{150} - \frac{S_B}{150}
  3. dSAdt=3SB150−3SA100\frac{dS_A}{dt} = \frac{3S_B}{150} - \frac{3S_A}{100}, dSBdt=3SA100−4SB150\frac{dS_B}{dt} = \frac{3S_A}{100} - \frac{4S_B}{150}
  4. dSAdt=3SB150−2SA100\frac{dS_A}{dt} = \frac{3S_B}{150} - \frac{2S_A}{100}, dSBdt=2SA100−4SB150\frac{dS_B}{dt} = \frac{2S_A}{100} - \frac{4S_B}{150} (correct answer)
Explanation: When modeling connected tank systems, you need to track salt flow rates carefully. The rate of change of salt in each tank equals salt flowing in minus salt flowing out, where concentration equals amount of salt divided by current volume. First, determine the volumes at time t. Tank A starts with 100 gallons, gains 1 gal/min (pure water in) and loses 2 gal/min (to B), so it loses 1 gal/min net, giving volume 100-t. Wait - that's decreasing! Actually, A gains 3 gal/min from B and 1 gal/min pure water (4 total in) while losing 2 gal/min to B, so net gain is 2 gal/min. Tank A volume: 100+2t. Similarly, Tank B starts with 150 gallons, gains 2 gal/min from A, loses 3 gal/min to A and 1 gal/min out, so net loss is 2 gal/min. Tank B volume: 150-2t. Actually, let me recalculate more carefully. The problem states volumes remain constant at 100 and 150 gallons respectively, which means we use these fixed volumes. For Tank A: Salt enters at rate 3SB150\frac{3S_B}{150} (from B) and exits at rate 2SA100\frac{2S_A}{100} (to B). For Tank B: Salt enters at rate 2SA100\frac{2S_A}{100} (from A) and exits at rates 3SB150\frac{3S_B}{150} (to A) and SB150\frac{S_B}{150} (outflow), totaling 4SB150\frac{4S_B}{150}. Choice A incorrectly uses time-varying volumes. Choice B has the wrong outflow rate from A (should be 2, not 3). Choice C incorrectly shows 3 gal/min leaving A instead of 2. Choice D correctly captures all flow rates with constant tank volumes. Study tip: Always verify that inflow and outflow rates match the problem description exactly, and check whether tank volumes are constant or changing with time.