Asked by Meredith
Pump A, working alone, can fill a tank in 3 hours, and pump B can fill the same tank in 2 hrs. If the tank is empty to start and pump A is switched on for one hour, after which pump B is also switched on and the two work together, how many minutes will pump B have been working by the time the pool is filled?
a)48
b)50
c)54
d)60
e)64
a)48
b)50
c)54
d)60
e)64
Answers
Answered by
Damon
pump A, 1/3 tank /hr
pump B, 1/2 tank/hr
in first hour pump A fills 1/3 tank
so together they fill 2/3 tank
x hours (1/3 tank/hr + 1/2 tank/hr) = 2/3 tank
x (5/6) = 2/3 = 4/6
5 x = 4
x = 4/5 hr = (4/5)60 = 48 minutes
pump B, 1/2 tank/hr
in first hour pump A fills 1/3 tank
so together they fill 2/3 tank
x hours (1/3 tank/hr + 1/2 tank/hr) = 2/3 tank
x (5/6) = 2/3 = 4/6
5 x = 4
x = 4/5 hr = (4/5)60 = 48 minutes
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