An industrial facility has three supply pipes—Pipe A, Pipe B, and Pipe C—that can fill a s… — GRE General Test

An industrial facility has three supply pipes—Pipe A, Pipe B, and Pipe C—that can fill a storage reservoir. Pipe A operating alone can fill the empty reservoir in $a$ hours. Pipe B operating alone takes $50\%$ longer than Pipe A to fill the empty reservoir. Pipe C operating alone fills the empty reservoir at a rate equal to the combined filling rate of Pipe A and Pipe B.

Initially, the reservoir is empty. Pipe A and Pipe B are opened simultaneously. After $2$ hours, Pipe B is closed and Pipe C is opened, while Pipe A remains open. The reservoir becomes completely full exactly $4$ hours after Pipe A and Pipe B were initially opened.

Which of the following statements must be true? Select all such statements.

Pipe A operating alone would fill the empty reservoir in $8$ hours and $40$ minutes.Answer
During the first $2$ hours of the process, exactly $\frac{5}{13}$ of the total capacity of the reservoir is filled.Answer
Pipe C operating alone would fill the empty reservoir in $5$ hours and $12$ minutes.Answer
D
Pipe B operating alone would fill the empty reservoir in $11$ hours and $30$ minutes.
E
If all three pipes were opened simultaneously from the beginning, the empty reservoir would be filled in $3$ hours and $15$ minutes.

Answer

The statements asserting that Pipe A operating alone takes 8 hours and 40 minutes, that 5/13 of the capacity is filled in the first 2 hours, and that Pipe C operating alone takes 5 hours and 12 minutes are all true.

Using the rate relationships

R_A = \frac{1}{a}

R_B = \frac{2}{3a}

, and

R_C = \frac{5}{3a}

, the equation

2(R_A + R_B) + 2(R_A + R_C) = 1

simplifies to

\frac{26}{3a} = 1

, giving

a = \frac{26}{3}

hours. This verifies that Pipe A takes 8 hours 40 minutes alone, Phase 1 fills 5/13 of the reservoir capacity, and Pipe C takes 5.2 hours (5 hours 12 minutes) alone.

Step-by-Step Solution

Express the individual work rates of Pipe A, Pipe B, and Pipe C in terms of parameter

a

R_A = \frac{1}{a}

R_B = \frac{1}{1.5a} = \frac{2}{3a}

, and

R_C = R_A + R_B = \frac{1}{a} + \frac{2}{3a} = \frac{5}{3a}

Work rate is defined as the reciprocal of the time required to complete one unit of work.

Formulate an equation for total work completed over the two 2-hour phases.

2(R_A + R_B) + 2(R_A + R_C) = 1 \implies 2\left(\frac{5}{3a}\right) + 2\left(\frac{8}{3a}\right) = 1 \implies \frac{10}{3a} + \frac{16}{3a} = 1

Pipes A and B operate for the first 2 hours, followed by Pipes A and C operating for the next 2 hours to complete 1 full reservoir.

Solve the work equation for

a

and determine Pipe A's solo time.

\frac{26}{3a} = 1 \implies a = \frac{26}{3} = 8\frac{2}{3} \text{ hours} = 8 \text{ hours } 40 \text{ minutes}.

Fractional hours are converted to minutes by multiplying

\frac{2}{3}

60

Evaluate the volume filled in Phase 1 and the solo filling time for Pipe C.

\text{Phase 1 volume} = \frac{10}{3(26/3)} = \frac{5}{13}. \text{ Pipe C solo time} = \frac{1}{R_C} = \frac{3a}{5} = \frac{26}{5} = 5.2 \text{ hours} = 5 \text{ hours } 12 \text{ minutes}.

Substituting

a = \frac{26}{3}

into the respective rate expressions yields the exact time and volume parameters.

Key Concept

Formulating combined rate models and solving sequential multi-stage work problems.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic