Three water pumps, A, B, and C, are used to drain a large industrial reservoir. Operating … — GRE General Test

Three water pumps, A, B, and C, are used to drain a large industrial reservoir. Operating alone at its constant rate, Pump A can drain the reservoir in $x$ hours, where $x > 0$ . Pump B operating alone takes $x + 4$ hours to drain the reservoir. When operating together for 2 hours, Pump A and Pump B complete the exact same fraction of the total job that Pump C completes operating alone in 3 hours. If all three pumps working simultaneously at their respective constant rates can drain the entire reservoir in $\frac{9}{4}$ hours, what is the value of $x$ ?

A
4
6Answer
C
8
D
10
E
12

Answer

The value of

x

is 6.

The correct answer 6 is derived by properly formulating the work rates

R_A = \frac{1}{x}

R_B = \frac{1}{x+4}

, and

R_C = \frac{4(x+2)}{3x(x+4)}

. Setting their sum

\frac{10(x+2)}{3x(x+4)}

equal to the combined rate

\frac{4}{9}

forms the quadratic

2x^2 - 7x - 30 = 0

, which yields the unique positive solution

x = 6

Step-by-Step Solution

Express the individual work rates of Pump A and Pump B in terms of

x

R_A = \frac{1}{x}

and

R_B = \frac{1}{x+4}

. Combined rate

R_{A+B} = \frac{1}{x} + \frac{1}{x+4} = \frac{2x+4}{x(x+4)}

Work rate is defined as the fraction of the job completed per hour.

Determine the work rate of Pump C using the given relationship.

Work done by A and B in 2 hours is

2 \cdot \frac{2x+4}{x(x+4)} = \frac{4x+8}{x(x+4)}

. Since Pump C does this in 3 hours,

R_C = \frac{1}{3} \cdot \frac{4x+8}{x(x+4)} = \frac{4(x+2)}{3x(x+4)}

Pump C's hourly rate is one-third of the total work completed by A and B in 2 hours.

Sum all three rates to find the total combined rate and set it equal to the given combined rate.

R_{total} = \frac{2x+4}{x(x+4)} + \frac{4x+8}{3x(x+4)} = \frac{3(2x+4) + 4x+8}{3x(x+4)} = \frac{10(x+2)}{3x(x+4)}

. Given total time is

\frac{9}{4}

hours, total rate is

\frac{4}{9}

. Thus,

\frac{10(x+2)}{3x(x+4)} = \frac{4}{9}

The sum of individual rates equals the inverse of total time required when working simultaneously.

Solve the algebraic equation for

x

Multiply both sides by 9 to get

\frac{90(x+2)}{3x(x+4)} = 4 \implies \frac{30(x+2)}{x(x+4)} = 4 \implies 15(x+2) = 2x(x+4) \implies 15x + 30 = 2x^2 + 8x \implies 2x^2 - 7x - 30 = 0

. Factoring gives

(2x + 5)(x - 6) = 0

. Since

x > 0

x = 6

Solving the quadratic yields the valid positive real root for time

x

Key Concept

Work Rate Modeling and Rational Equation Systems

Question

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