SVTTCGMAT wrote:
Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24
Solving this question quickly requires to go through each statement thoroughly and interpret an equation from all of those. I am following the same strategy to solve this question,
Quote:
Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours.
This means that pump X can pump out the entire water in a tank in 8 hours.
Quote:
Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours.
The rest of the water here means 3/4th of the water in the tank. Thus, the complete water will be pumped out by (x+y+z) in 4 hours.
Quote:
If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water.
This means that 3/4th of the water in the tank is pumped out in 18 hours. Thus, the entire water will be pumped out by Y in 24 hours.
Q: how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?1/X + 1/Y + 1/Z = 1/4
1/8 + 1/24 + 1/Z = 1/4
1/Z = 1/4 - 1/8 - 1/24 = (6 - 3 -1)/24 = 2/24 = 1/12
Thus, 12 hours is the answer! Bingo.