Pump A, pumping water at a constant rate, can fill a certain swimm....

Question

Pump A, pumping water at a constant rate, can fill a certain swimming pool in 6 hours. Pump B, pumping water at a constant rate, can fill the same pool in 4 hours. If both pumps begin filling the pool simultaneously when the pool is empty and pump B breaks down 1 hour after they begin filling the pool, how many hours will it take pump A alone to finish filling the pool?

A. 1 2/5
B. 2 1/3
C. 3 1/2
D. 4
E. 4 4/5

A. 1 2/5
B. 2 1/3
C. 3 1/2
D. 4
E. 4 4/5

gmatophobia · Accepted Answer

Let's assume that the pool holds a capacity of 24 units of water.

Rate of Pump A = 4 units/hour
 
 Rate of Pump B = 6 units/hour
 
 Combined Rate = (6+4) = 10 units/hour

In the first hour, when both the pumps work together, the amount of water pumped =  10  units

Units of water remaining to be pumped =  24 - 10 = 14

Time taken by Pump A alone to finish filling the pool =  \frac{14}{4} = \frac{7}{2 }= 3.5   hours

Option C

shoshonte · Answer

Cool nice to see a new question!

My approach which took me 1 min and 41 seconds was the following:

My first question was how much proportional to the total amount of water that the pool holds did A + B working together fill?

We know from the work/rate formula that A+B working together would fill 5/12 of the pool in one hour

1.) Total proportional work done by A+B in one hour: (x_t + y_t)/ (x_t * y_t) => (4 + 6)/(4 * 6) = 5/12

So therefore A working alone will have to fill 7/12 of the pool. Since we know it fills the pool in it's entirety in 6 hours, then it fills 7/12 of the pool in 6 (7/12) = 7/2 = 3.5

Areas for improvement for me:

1.) "derived" the formula in real time from the 1/x + 1/y = total time formula then taking the inverse. Need to just memorize the (x + y)/(x * y) formula to make things faster
2.) Forget which was A and which was B at the end, so had to double check. Need to take better notes.

GmatTest2026 · Answer

Solution:   Let's take the rate of pump A = 1/6
                                   Rate of pump B = 1/4
When both pumps were working together the time taken to fill the pool = was 1 hour
So, by using the matrix approach we can solve this very easily

The extra time it took for pump A to fill the pool = t + 1

Work of pump A to fill the tank                          = 1/6t +1/6 = t+1/6
Work of pump B to fill the tank                          = 1/4 * 1     = 1/4

If we combine the work of both pumps then we have

Work of Pump A + Work of Pump B  = 1

t/6+1/6 + 1/4 = 1 ---> 3 1/2

Option C

Yes2GMAT · Answer

gmatophobia  I am really confused. How do you make this assumption? I have seen some Work-rate problems in DI where the answer is E) and this question feels similar to that. If we can assume capacity here, then isn't it possible to assume capacity in those questions?

Sorry, if my doubt is extremely basic. Thanks a lot for the solution.

KarishmaB · Answer

Rate of pump B is 1/4 (since it completes the work in 4 hours). So in 1 hour, it fills 1/4th of the tank.
Pump A must fill the entire 3/4th of the tank alone. 
Rate of pump A = 1/6 (since it completes the work in 4 hours)
Time taken by pump A = Work/Rate = (3/4)/(1/6) = 4.5 
So A will work alone for 3.5 hours from the time pump B broke down.

Answer (C)

Discussion on work rate:   https://youtu.be/88NFTttkJmA

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Pump A, pumping water at a constant rate, can fill a certain swimm....

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