boomgoesthegmat wrote:Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5
We are given that one inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the one inlet pipe is (1/2)/3 = 1/6.
We are also given that a second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate is (2/3)/6 = 1/9.
We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity.
If we let t = the time in hours the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t and the work of the second inlet pipe = (1/9)t.
Since the tank is filled, we can set total work to 1 and create the following equation:
(1/6)t + (1/9)t = 1
Multiplying the entire equation by 18, we obtain:
3t + 2t = 18
5t = 18
t = 18/5 = 3.6
Answer:
B
