A chemical processing tank receives two liquid solutions, Solution X and Solution Y, from … — GRE General Test

A chemical processing tank receives two liquid solutions, Solution $X$ and Solution $Y$ , from separate inlet pipes.

- Solution $X$ contains chemical $A$ and water in a volume ratio of $3:2$ and enters the tank at a constant rate of $150$ liters per hour.
- Solution $Y$ contains chemical $A$ and water in a volume ratio of $1:4$ and enters the tank at a constant rate of $250$ liters per hour.

Both inlet pipes run simultaneously into an initially empty tank for $4$ hours. After $4$ hours, the inlet pipes are shut off. To adjust the mixture, pure chemical $A$ is added to the tank at a constant rate of $50$ liters per hour, while water is continuously drained from the tank at a constant rate of $30$ liters per hour.

How many hours must this adjustment process run until the volume of chemical $A$ in the tank is equal to the volume of water in the tank?

Answer: 6 hours

Answer

To find the time

t

when the volumes of chemical A and water in the tank are equal, first determine the initial quantities contributed by both solutions during the 4-hour filling period. Solution X provides

150 \times 4 = 600

liters total, containing

\frac{3}{5} \times 600 = 360

liters of chemical A and

\frac{2}{5} \times 600 = 240

liters of water. Solution Y provides

250 \times 4 = 1000

liters total, containing

\frac{1}{5} \times 1000 = 200

liters of chemical A and

\frac{4}{5} \times 1000 = 800

liters of water. Adding these amounts yields

360 + 200 = 560

liters of chemical A and

240 + 800 = 1040

liters of water. In the adjustment phase of

t

hours, chemical A increases at

50

L/hr to

560 + 50t

, while water decreases at

30

L/hr to

1040 - 30t

. Equating the two expressions gives

560 + 50t = 1040 - 30t

, which simplifies to

80t = 480

, resulting in

t = 6

hours.

Step-by-Step Solution

Determine the volumes of chemical A and water supplied by Solution X during the first 4 hours.

Solution X delivers 600 liters in total, consisting of 360 liters of chemical A and 240 liters of water.

Solution X flows at 150 L/hr for 4 hours (150 * 4 = 600 L) with a 3:2 chemical A to water ratio, meaning chemical A represents 3/5 of the total volume and water represents 2/5.

Determine the volumes of chemical A and water supplied by Solution Y during the first 4 hours.

Solution Y delivers 1000 liters in total, consisting of 200 liters of chemical A and 800 liters of water.

Solution Y flows at 250 L/hr for 4 hours (250 * 4 = 1000 L) with a 1:4 chemical A to water ratio, meaning chemical A represents 1/5 of the total volume and water represents 4/5.

Calculate the total initial quantities of chemical A and water present in the tank prior to the adjustment phase.

Total chemical A = 560 liters; Total water = 1040 liters.

Sum the quantities from both solutions: Chemical A = 360 + 200 = 560 L; Water = 240 + 800 = 1040 L.

Formulate linear expressions representing the total volume of chemical A and water after t hours of adjustment.

Chemical A volume = 560 + 50t; Water volume = 1040 - 30t.

Pure chemical A is added at 50 L/hr, increasing its total volume, while water is drained at 30 L/hr, reducing its total volume.

Set the two component volume expressions equal to each other and solve for t.

t = 6 hours.

Solving 560 + 50t = 1040 - 30t leads to 80t = 480, which yields t = 6.

Key Concept

Multi-stream mixture rate integration and ratio equality modeling

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