An empty swimming pool can be filled to capacity through an inlet pipe

Question

An empty swimming pool can be filled to capacity through an inlet pipe in 3 hours, and it can be completely drained by a drainpipe in 6 hours. If both pipes are fully open at the same time, in how many hours will the empty pool be filled to capacity?

(A) 4

(B) 4.5

(C) 5

(D) 5.5

(E) 6

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Matthyrou · Answer

I'm not sure on this one but for me, after 3hours, the pipe would fill a full capacity when att the same time, the drainpipe will drain only half of the capacity (because, it drains fully in 6 hours), so half of the pool will be empty in 3 hours, so because it's the same principe for all of the duration, it will be fully filled in 6hours

Purnank · Answer

Yes correct!

another way to look at it by assuming numbers,
Here we have 2 numbers 3Hrs and 6Hrs
Lets take their LCM = 6 
Assume volume of Pool as 6 units
Efficiency of filling pipe = 6/3 = 2 units/hr
Efficiency of draining pipe = 6/6 = 1 units/hr
Now when both are working simultaneously draining pipe acting as negative work for filling pipe.
So Combined efficiency = 2 - 1 = 1 units/hr
Now total time required for filling when combined will be = 6 / 1 = 6 hr.

Answer is E.

tgsankar10 · Answer

Flow rate of inlet pipe,  f=\frac{1}{3}

Flow rate of drain pipe,  e=\frac{1}{6} . Since drain pipe does negative work,  e=-\frac{1}{6}

Both pipes working together,

f+e=\frac{1}{t}

\frac{1}{3}-\frac{1}{6}=\frac{1}{t}

\frac{2-1}{6}=\frac{1}{6}=\frac{1}{t}

t=6

Answer: E

Bismuth83 · Answer

1. We are asked to find how long the pool will take to fill if both pipes are open. Let's mark this as t and the maximum capacity of the pool as L.

2. Now model the pool and the two pipes. Let the speed of filling be equal to v\(_f\) and the speed of draining be equal to v\(_d\).

3. The pool takes 3 hours to fill and 6 hours to drain. In other words, v\(_f\) \(* 3 = \) L and L - v\(_d\) \(* 6 = \) 0. This can be turned into v\(_f\) = \(\frac{L}{3}\) and v\(_d\) = \(\frac{L}{6}\), respectively.

4. Our question can be turned into: (v\(_f\) - v\(_d\))\( * \)t = L \(\rightarrow\) t = \(\frac{L}{v_f - v_d}\) = \(\frac{L}{\frac{L}{3} - \frac{L}{6}}\) = \(\frac{1}{\frac{1}{3} - \frac{1}{6}}\) = \(\frac{1}{\frac{1}{6}}\) = 6.

5. So, our answer is 6 hours.

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Abdelaali91 · Answer

(work done filling the swimming pool)-(work done draining the pool)=1 (full swimming pool)

ie (t/3)-(t/6)=1, so (2t/6)-(t/6)=1

finally t/6=1 ie t=6 hours

An empty swimming pool can be filled to capacity through an inlet pipe

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An empty swimming pool can be filled to capacity through an inlet pipe

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