A municipal water treatment facility uses two intake pipes, Pipe X and Pipe Y, to fill a m… — GRE General Test

A municipal water treatment facility uses two intake pipes, Pipe $X$ and Pipe $Y$ , to fill a main reservoir. Operating alone at its constant rate, Pipe $X$ can fill the empty reservoir in $10$ hours. Operating alone at its constant rate, Pipe $Y$ can fill the empty reservoir in $15$ hours. Pipe $X$ is turned on first and operates alone for $3$ hours. Then, Pipe $Y$ is also turned on, and both pipes operate together until the reservoir is completely full. What is the total number of hours Pipe $X$ operates from the moment it is turned on until the reservoir is completely filled?

A
$4.2$
B
$4.8$
C
$6.0$
$7.2$ Cevap
E
$7.5$

Cevap

The total number of hours Pipe X operates is

7.2

hours.

The correct answer is

7.2

hours. Pipe X completes

3/10

of the reservoir in 3 hours, leaving

7/10

of the reservoir to be filled. The combined rate of Pipe X and Pipe Y is

1/10 + 1/15 = 1/6

per hour. The joint time required to fill the remaining

7/10

(7/10) / (1/6) = 4.2

hours. Adding the initial 3 hours Pipe X worked alone yields a total of

3 + 4.2 = 7.2

hours.

Adım Adım Çözüm

Determine the individual hourly work rates of Pipe X and Pipe Y.

Rate of Pipe X =

\frac{1}{10}

reservoir per hour; Rate of Pipe Y =

\frac{1}{15}

reservoir per hour.

The rate is the reciprocal of the total time required to complete the job individually.

Calculate the fraction of the reservoir filled by Pipe X during its initial 3-hour solo operation.

Work done in first 3 hours =

3 \times \frac{1}{10} = \frac{3}{10}

of the reservoir.

Multiplying Pipe X's rate by its solo operating time gives the completed portion of the work.

Find the remaining fraction of the reservoir that needs to be filled.

Remaining work =

1 - \frac{3}{10} = \frac{7}{10}

of the reservoir.

Subtracting the completed portion from the whole (1) leaves the uncompleted portion.

Determine the combined hourly rate when both pipes operate simultaneously.

Combined rate =

\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}

of the reservoir per hour.

Simultaneous operation rates are additive.

Calculate the time

t

during which both pipes operate together to finish the remaining work.

t = \frac{\frac{7}{10}}{\frac{1}{6}} = \frac{7}{10} \times 6 = \frac{42}{10} = 4.2

hours.

Dividing the remaining work by the combined rate yields the joint operation time.

Add Pipe X's solo time to the joint operation time to get Pipe X's total operating time.

Total time =

3 + 4.2 = 7.2

hours.

Pipe X was active during both the initial solo period and the joint operating period.

Anahtar Kavram

Combined Rate and Staggered Work Modeling

Tahmini Süre:2m 0s

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Cevap

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