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An empty water tank can be filled to capacity through an inlet Pipe $A
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10 Jun 2025, 23:42
10 Jun 2025, 23:42
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An empty water tank can be filled to capacity through an inlet Pipe $A$ in 4 hours, and the same filled tank can be completely drained by an outlet Pipe $B$ in 12 hours. If both the Pipes $A$ and $B$ are fully open at the same time, in how many hours will the empty tank be filled to capacity?
(A) 4
(B) 5
(C) 6
(D) 8
(D) 12
(A) 4
(B) 5
(C) 6
(D) 8
(D) 12
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Re: An empty water tank can be filled to capacity through an inlet Pipe $A
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13 Jun 2025, 04:00
13 Jun 2025, 04:00
Expert Reply
This is a classic "pipes and cisterns" problem, which is a type of work/rate problem.
- Pipe A (Inlet): Fills the tank in 4 hours.
- In 1 hour, Pipe A fills $\frac{1}{4}$ of the tank. This is its filling rate.
- Pipe B (Outlet): Drains the tank in 12 hours.
- In 1 hour, Pipe B drains $\(\frac{1}{12}\)$ of the tank. This is its draining rate.
When both pipes are open simultaneously, Pipe A is filling and Pipe B is draining. So, their effective rate is the filling rate minus the draining rate.
- Combined Rate: Rate of Pipe A - Rate of Pipe B
- Combined Rate $\(=\frac{1}{4}-\frac{1}{12}\)$
To subtract these fractions, find a common denominator, which is 12: Combined Rate $\(=\frac{3}{12}-\frac{1}{12}=\frac{3-1}{12}=\frac{2}{12}=\frac{1}{6}\)$
This means that when both pipes are open, $\(\frac{1}{6}\)$ of the tank is filled every hour.
To find the total time it takes to fill the empty tank to capacity, we take the reciprocal of the combined rate.
$$
\begin{aligned}
& \text { Time }=\frac{1}{\text { Combined Rate }} \\
& \text { Time }=\frac{1}{\frac{1}{6}}=6 \text { hours. }
\end{aligned}
$$
So, it will take 6 hours to fill the empty tank to capacity.
- Pipe A (Inlet): Fills the tank in 4 hours.
- In 1 hour, Pipe A fills $\frac{1}{4}$ of the tank. This is its filling rate.
- Pipe B (Outlet): Drains the tank in 12 hours.
- In 1 hour, Pipe B drains $\(\frac{1}{12}\)$ of the tank. This is its draining rate.
When both pipes are open simultaneously, Pipe A is filling and Pipe B is draining. So, their effective rate is the filling rate minus the draining rate.
- Combined Rate: Rate of Pipe A - Rate of Pipe B
- Combined Rate $\(=\frac{1}{4}-\frac{1}{12}\)$
To subtract these fractions, find a common denominator, which is 12: Combined Rate $\(=\frac{3}{12}-\frac{1}{12}=\frac{3-1}{12}=\frac{2}{12}=\frac{1}{6}\)$
This means that when both pipes are open, $\(\frac{1}{6}\)$ of the tank is filled every hour.
To find the total time it takes to fill the empty tank to capacity, we take the reciprocal of the combined rate.
$$
\begin{aligned}
& \text { Time }=\frac{1}{\text { Combined Rate }} \\
& \text { Time }=\frac{1}{\frac{1}{6}}=6 \text { hours. }
\end{aligned}
$$
So, it will take 6 hours to fill the empty tank to capacity.
gmatclubot
Re: An empty water tank can be filled to capacity through an inlet Pipe $A [#permalink]
13 Jun 2025, 04:00
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