SOLUTION: A water tank can be emptied by using one pump for 5 hours. A second, smaller pump can empty the tank in 7 hours. If the larger pump is started at 1:00 P.M., how much time should pa
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Question 899874: A water tank can be emptied by using one pump for 5 hours. A second, smaller pump can empty the tank in 7 hours. If the larger pump is started at 1:00 P.M., how much time should pass until the smaller pump is started so that the tank will be emptied at 5:00 P.M.? Found 2 solutions by stanbon, richwmiller:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A water tank can be emptied by using one pump for 5 hours. A second, smaller pump can empty the tank in 7 hours. If the larger pump is started at 1:00 P.M., how much time should pass until the smaller pump is started so that the tank will be emptied at 5:00 P.M.?
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Larger pump rate:: 1/5 tank/hr
Smaller pump rate:: 1/7 tank/hr
Together rate:: (1/5)+(1/7) = 12/35 tank/hr
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Equation:
Larger for x hrs + both for (4-x) hrs = 1 job
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(1/5)x + (12/35)(4-x) = 1 tank
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Multiply thru by 35 to get:
7x + 12(4-x) = 35
7x + 48 - 12x = 35
-5x = -13
x = 13/5 hrs = 2 hrs 36 minutes
4-x = 1 hr 24 minutes
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Cheers,
Stan H.
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You can put this solution on YOUR website!
Since the larger pump can do the job in five hours and the whole job together will take on 4 hours the larger pump will do 4/5 of the job which leaves 1/5 for the smaller pump
4/5+x/7=1
x/7=1/5
x=7/5
x=1.4 hours
1 2/5 hours = 1 hour 24 minutes
5:00-1:24
4:60-1:24=3:36 pm
Wait 2 hours and 36 minutes and turn on the smaller pump at 3:36 pm
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