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Question 1207767: A 5-horsepower (hp) pump can empty a pool in 5 hours.A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool.After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool?
Found 5 solutions by josgarithmetic, mananth, math_tutor2020, ikleyn, mccravyedwin:
Answer by josgarithmetic(39630)   (Show Source):
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website!
A 5-horsepower (hp) pump can empty a pool in 5 hours.A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool.After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool?
Consider emptying the pool is 1 job
A 5-horsepower (hp) pump can empty a pool in 5 hours. rate of Emptying = 1/5 of the job
A smaller, 2-hp pump empties the same pool in 8 hours.1/8 of the job
The pumps are used together for two hours
2(1/5 +1/8) of the job is done in two hours =26/40 = 13/20
Balance job = 1- 13/20 = 7/20
1/5 of job the 5Hp does in 1 hour
7/20 of the job it will do in (7/20)/ (1/5)= 35/20 = 7/4 hours
It will take the 5-hp pump an additional 1.75 hours to empty the pool after the 2-hp pump breaks down.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Answer: 1.75 hours
Explanation
The horsepower values are a distraction. Ignore them entirely.
Consider a 4000 gallon pool.
The larger pump, which I'll call pump A, can empty the entire pool in 5 hours when working alone.
A's unit rate is 4000/5 = 800 gallons per hour.
Formula used: rate = (amountDone)/time
The smaller pump B can empty the entire pool in 8 hours when working alone.
B's unit rate is 4000/8 = 500 gallons per hour.
When the two pumps work together, without either pump slowing down the other, their combined rate is 800+500 = 1300 gallons per hour.
The two pumps work together for 2 hours. That drains 1300*2 = 2600 gallons and leaves 4000-2600 = 1400 gallons remaining.
Pump B breaks down at the 2 hour marker.
Pump A now works alone to empty the remaining 1400 gallons of water.
rate*time = amountDone
time = amountDone/rate
time = 1400/800
time = 14/8
time = (2*7)/(2*4)
time = 7/4
time = 1.75 hours is the amount of time needed for the larger pump to finish the job of emptying the pool.
The 4000 value mentioned isn't special. Feel free to change it to any other positive number you want. The final answer will still be the same regardless of the pool volume.
I picked this value based on the LCM of 5 and 8, which is 40. Then I tacked on a few extra zeros to lead to a slightly more realistic pool volume. Even then 4000 is probably on the smaller side of things.
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Another approach
one job = emptying the pool
Pump A does one job in 5 hours when working alone
A's unit rate is 1/5 of a job per hour
Pump B does one job in 8 hours when working alone
B's unit rate is 1/8 of a job per hour
Their combined unit rate is 1/8 + 1/5 = 5/40 + 8/40 = 13/40 of a job per hour.
This assumes neither pump slows the other down.
In 2 hours the pumps work together to handle 2*13/40 = 13/20 of the job.
1 - (13/20) = 20/20 - 13/20 = 7/20 of the job remains.
x = amount of extra time, in hours, pump A needs to work alone to finish the job
This is after pump B stops working
rate*time = amountDone
(1/5 of a job per hour)*(x hours) = 7/20 of a job remains
(1/5)x = 7/20
x = 5*7/20
x = 5*7/(5*4)
x = 7/4
x = 1.75 hours
Extra info:
1.75 hours = 60*1.75 = 105 minutes
1.75 hours = 1 hour & 45 minutes since 0.75 hr = 0.75*60 = 45 min
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
This problem was solved at this forum many years ago (perhaps, 20 - 25 years ago).
For the solution, see the link
https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.5399.html
Answer by mccravyedwin(409)  (Show Source):
You can put this solution on YOUR website! A 5-horsepower (hp) pump can empty a pool in 5 hours.
A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used
together to begin emptying this pool.
The LCM of 5 and 8 hours is 40 hours,
So the 5-horsepower (hp) pump can empty 8 pools in 40 hours.
The smaller, 2-hp pump can empty 5 pools in 40 hours.
So in 40 hours both pumps cane empty 13 pools,
Since 2 hours is 1/20 of 40 hours, in 2 hours together they can empty 1/20 of 13
pools. That's what they've pumped out together in those first 2 hours.
So the question becomes: how long does it take the larger pump to
pomp out the remaining 7/20 of 1 pool?
It can empty 1 pool in 5 hours.
So it can empty 7/20 of a pool in 7/20ths of 5 hours, or 35/20ths 7/4ths of an
hour, 1 3/4 hours or 1 hour and 45 minutes.
Edwin
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