csingh158 wrote:
Bunuel wrote:
Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?
A. 24
B. 36
C. 48
D. 50
E. 64
Two concerns with the question
1. If both the pumps fill the container then the time taken should be less, not more when both pumps are open.
Therefore this statement doesn't make sense "The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container." 2. Both pumps will take 36 minutes and Pump x will take 144 mins, which is not in the options.
The above highlighted statement makes sense as Pump X is filling the container and Pump Y is emptying it. This is the reason that when both are open they will take more time to fill up compared to when Pump X is only open.
Solution:
Let rate of pump X = x, rate of pump Y = y, time taken by Pump Y alone to empty the container = 48 minutes, time taken by Pump X alone to fill the container = t =?
Work done = filling the tank or emptying the tank = 1 [Work done is taken as 1 when unknown]y * 48 = 1 [Emptying the container by pump Y]
=> y = \(\frac{1}{48}\)
x * t = 1 [Filling the container by pump X]
=> x = \(\frac{1}{t}\)
Also,
(x - y) * 4t = 1 [Filling the container when both pumps are open]
=> \((\frac{1}{t} - \frac{1}{48}) * 4t = 1\)
=> 4 - \(\frac{t}{12}\) = 1
=> 3 = \(\frac{t}{12}\)
=> t = 12 * 3 = 36 minutes.
So, correct answer is option B.