A water reservoir is filled by Pipe A and Pipe B operating simultaneously at their respect… — GRE General Test

A water reservoir is filled by Pipe A and Pipe B operating simultaneously at their respective constant rates. Operating together at their original rates, the two pipes can fill the empty reservoir completely in $12$ hours. On a certain day, both pipes begin filling the empty reservoir together at their original rates. After $4$ hours, Pipe A's rate decreases by $25\%$ , while Pipe B's rate increases by $50\%$ . Operating at these new constant rates, the two pipes require an additional $7$ hours to fill the remainder of the reservoir. How many hours would it take Pipe A, operating alone at its original rate, to fill the entire reservoir?

Answer: 25.2 hours

Answer

It would take Pipe A 25.2 hours operating alone at its original rate to fill the entire reservoir.

By defining the original work rates

a

and

b

in reservoirs per hour, the initial condition yields

a + b = \frac{1}{12}

. In the first 4 hours,

\frac{1}{3}

of the job is completed, leaving

\frac{2}{3}

. Setting up the equation for the remaining job with modified rates

0.75a

and

1.5b

over 7 hours produces

7(0.75a + 1.5b) = \frac{2}{3}

. Solving this system of two linear equations yields

a = \frac{5}{126}

reservoirs per hour. Taking the reciprocal gives the time required for Pipe A alone to fill the reservoir, which is

25.2

hours.

Step-by-Step Solution

Set up equations for the original rates of Pipe A (

a

) and Pipe B (

b

The combined original rate is

a + b = \frac{1}{12}

reservoir per hour.

Together they complete

1

reservoir in

12

hours.

Determine the fraction of the reservoir filled in the first 4 hours and the remaining fraction.

Work completed =

4 \times \frac{1}{12} = \frac{1}{3}

; Remaining work =

\frac{2}{3}

The pipes worked at their original combined rate for 4 hours.

Set up an equation for the work done during the remaining 7 hours at the adjusted rates.

7 \left(0.75a + 1.5b\right) = \frac{2}{3} \implies 5.25a + 10.5b = \frac{2}{3} \implies 63a + 126b = 8

Pipe A's rate decreases by

25\%

0.75a

, and Pipe B's rate increases by

50\%

1.5b

Solve the system of linear equations for

a

a = \frac{5}{126}

reservoir per hour.

Multiplying

a + b = \frac{1}{12}

126

yields

126a + 126b = 10.5

. Subtracting

63a + 126b = 8

gives

63a = 2.5

, so

a = \frac{2.5}{63} = \frac{5}{126}

Calculate the time for Pipe A alone to fill the entire reservoir.

Time

= \frac{1}{a} = \frac{126}{5} = 25.2

hours.

Time equals total work divided by individual rate.

Key Concept

Algebraic modeling of combined work and rates with mid-process rate modifications

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