Pipeline A operating alone can fill a storage tank in x hours, whereas Pipeline B operatin… — GRE General Test

Pipeline A operating alone can fill a storage tank in $x$ hours, whereas Pipeline B operating alone takes $50\%$ longer than Pipeline A to fill the same tank. Drainage Pipe C operating alone can empty a full tank in $2x$ hours. If all three pipes are opened simultaneously when the tank is empty, the tank becomes completely full in $12$ hours. What is the value of $x$ ?

A
$6$
B
$12$
$14$ Answer
D
$24$
E
$26$

Answer

The value of

x

14

To find

x

, calculate the hourly rate of each pipe: Pipeline A fills

\frac{1}{x}

of the tank per hour, Pipeline B takes

1.5x = \frac{3}{2}x

hours so it fills

\frac{2}{3x}

per hour, and Pipe C empties

\frac{1}{2x}

per hour. The combined rate is

\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{7}{6x}

. Since the tank fills in

12

hours, the net rate is

\frac{1}{12}

. Setting

\frac{7}{6x} = \frac{1}{12}

gives

6x = 84

, so

x = 14

Step-by-Step Solution

Express the individual work rates per hour in terms of

x

Pipeline A rate =

\frac{1}{x}

, Pipeline B time =

1.5x = \frac{3}{2}x \implies

Pipeline B rate =

\frac{2}{3x}

, Pipe C rate =

-\frac{1}{2x}

Work rate is the reciprocal of the total time required to complete the job alone, with drainage represented as a negative rate.

Formulate the equation for the combined rate of all three pipes operating together.

Combined Rate =

\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{1}{x} \left(1 + \frac{2}{3} - \frac{1}{2}\right) = \frac{1}{x} \left(\frac{6 + 4 - 3}{6}\right) = \frac{7}{6x}

Simultaneous operation means summing the individual rates.

Equate the combined rate to the overall rate required to fill the tank in 12 hours and solve for

x

\frac{7}{6x} = \frac{1}{12} \implies 6x = 84 \implies x = 14.

The tank is filled in 12 hours, so the net rate per hour is

\frac{1}{12}

Key Concept

Combined Work Rates with Inflow and Outflow

Estimated Time:2m 0s

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