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Difficulty: HardAlgebraic Word Problems and Modeling

A pharmaceutical laboratory produces a custom solution in a mixing tank using two automated pumps, Pump A and Pump B. Initially, the tank contains 600600 liters of a solution that is 20%20\% active reagent by volume. Pump A adds pure active reagent to the tank at a constant rate of 1515 liters per minute, while Pump B simultaneously adds distilled water (0%0\% active reagent) to the tank at a constant rate of 2525 liters per minute. At the exact instant when the mixture in the tank reaches 35%35\% active reagent by volume, Pump B is turned off while Pump A continues to add pure active reagent at 1515 liters per minute. How many total minutes, from the moment both pumps were initially started, does it take for the solution in the tank to reach 50%50\% active reagent by volume?

Answer: 174 minutes

Answer

The total elapsed time is 174 minutes.
The problem requires setting up a two-stage rational rate model. In the first stage, initial reagent volume is 120120 liters (20%20\% of 600600). With Pump A adding 1515 L/min of reagent and Pump B adding 2525 L/min of water, the mixture concentration after t1t_1 minutes is 120+15t1600+40t1=0.35\frac{120 + 15t_1}{600 + 40t_1} = 0.35, yielding t1=90t_1 = 90 minutes. At t1=90t_1 = 90, the total volume is 4,2004,200 liters and the reagent volume is 1,4701,470 liters. In the second stage, only Pump A is active, so concentration after t2t_2 additional minutes is 1470+15t24200+15t2=0.50\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50, yielding t2=84t_2 = 84 minutes. Summing both intervals gives 90+84=17490 + 84 = 174 total minutes.

Step-by-Step Solution

1
Determine initial quantities of active reagent and total solution.
Initial total volume V0=600V_0 = 600 liters. Initial volume of active reagent A0=0.20×600=120A_0 = 0.20 \times 600 = 120 liters.
Establishing the baseline volume and solute concentration before any pumps are turned on.
2
Model Phase 1 where both pumps operate concurrently for t1t_1 minutes to reach 35%35\% concentration.
Total volume equation: V(t1)=600+(15+25)t1=600+40t1V(t_1) = 600 + (15 + 25)t_1 = 600 + 40t_1. Reagent volume equation: A(t1)=120+15t1A(t_1) = 120 + 15t_1. Set concentration to 0.350.35: 120+15t1600+40t1=0.35  ⟹  t1=90\frac{120 + 15t_1}{600 + 40t_1} = 0.35 \implies t_1 = 90 minutes.
Both pumps add volume, but only Pump A adds active reagent. Solving 120+15t1600+40t1=720\frac{120 + 15t_1}{600 + 40t_1} = \frac{7}{20} gives 20(120+15t1)=7(600+40t1)  ⟹  2400+300t1=4200+280t1  ⟹  20t1=1800  ⟹  t1=9020(120 + 15t_1) = 7(600 + 40t_1) \implies 2400 + 300t_1 = 4200 + 280t_1 \implies 20t_1 = 1800 \implies t_1 = 90 minutes.
3
Calculate the accumulated total volume and reagent volume at the end of Phase 1.
Total volume V1=600+40(90)=4,200V_1 = 600 + 40(90) = 4,200 liters. Active reagent volume A1=120+15(90)=1,470A_1 = 120 + 15(90) = 1,470 liters.
These amounts serve as the starting parameters for Phase 2.
4
Model Phase 2 where only Pump A operates for t2t_2 additional minutes to reach 50%50\% concentration.
In Phase 2, volume increases at 1515 L/min and reagent increases at 1515 L/min. Set concentration to 0.500.50: 1470+15t24200+15t2=0.50  ⟹  t2=84\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50 \implies t_2 = 84 minutes.
Solving 2(1470+15t2)=4200+15t2  ⟹  2940+30t2=4200+15t2  ⟹  15t2=1260  ⟹  t2=842(1470 + 15t_2) = 4200 + 15t_2 \implies 2940 + 30t_2 = 4200 + 15t_2 \implies 15t_2 = 1260 \implies t_2 = 84 minutes.
5
Sum the durations of Phase 1 and Phase 2 to find the total time.
Total time T=t1+t2=90+84=174T = t_1 + t_2 = 90 + 84 = 174 minutes.
The question asks for the total minutes from the initial start of both pumps.

Key Concept

Multi-Phase Algebraic Rate and Mixture Modeling
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Topics

Algebraic Word Problems and Modeling

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