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Pump A, pumping water at a constant rate, can fill a certain swimm....
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17 Mar 2024, 12:46
17 Mar 2024, 12:46
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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
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Pump A, pumping water at a constant rate, can fill a certain swimm....
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17 Mar 2024, 13:01
17 Mar 2024, 13:01
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bwashburn15 wrote:
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
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
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Re: Pump A, pumping water at a constant rate, can fill a certain swimm....
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17 Mar 2024, 16:49
17 Mar 2024, 16:49
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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.
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.
