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 ?
Work = Rate * TimeRate of ( X + Y + Z ) = Rate of X + Rate of Y + Rate of ZLet's use these 2 basic concepts to solve this question.
Most of the above explanations have used the fractional method. So, let's try the
LCM approach here.
X pumped out ¼ of the water in a tank in 2 hours
i.e X pumped out full water in 2* 4 hours = 8 hours
Let's assume that total volume = 8 L
Why should I select 8 as the total volume?
The reason is very simple. A number that is divisible by 8. You can also try 16 or 24 .. . The final answer will not change. Remember, in fraction method we are assuming total volume as 1 . Taking LCM as the total volume can help you to reduce the calculation time.
Rate of pump X = 8 L/ 8 hrs = 1L/hr
X pumped out 2 L in 2hrs, so the remaining water in the tank = 8-2 = 6 L
X, Y, and Z working together pumped out the rest of the water (6L) in 3 hours.
Rate of ( X + Y + Z ) = 6/3 = 2 L/hr
If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water.
i.e Pump Y alone can pump out the rest of the water (6L) in 18 hours
Rate of Y = 6/18 = 1/3 L/hr
Rate of ( X + Y + Z ) = Rate of X + Rate of Y + Rate of Z
2 = 1 + 1/3 + Rate of Z
Rate of Z = 2/3 L/hr
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?
Time taken by pump Z = Work/Rate of Z =\( 8/\frac{2}{3}=\) 4 *3 =12 hours
Option B is the answer.Thanks,
Clifin J Francis,
GMAT QUANT SME