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21 Jan 2010, 17:24
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Working independently at their respective constant rates, pumps X and Y took 48 minutes to fill an empty tank with water. What fraction of the water in the full tank came from pump X ?
(1) Working alone at its constant rate, pump X would have taken 80 minutes to fill the tank with water.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank with water.
(1) Working alone at its constant rate, pump X would have taken 80 minutes to fill the tank with water.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank with water.
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23 Jul 2010, 14:44
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1. Working independently at their respective rate, pump X and Y took 48 minutes to fill an empty tank with water. what fraction of the water in the full tank came from pump X?
Let the rate of pump X be \(x\) liters per minute, the rate of pump Y be \(y\) liters per minute and the capacity of the tank be \(c\) liters.
Given: \((x+y)*48=c\). Question: \(\frac{48x}{c}=?\) (As from the pump X in 48 minutes would come \(48x\) liters)
(1) Working along at its constant rate, pump X would have taken 80 minutes to fill the tank with water --> \(80x=c\) --> \(\frac{48x}{c}=\frac{48x}{80x}=\frac{3}{5}\). Sufficient.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank --> \(120y=c\) --> as \((x+y)*48=c\), then \((x+y)*48=120y\) and \(48x=72y\) --> \(\frac{48x}{c}=\frac{72y}{120y}=\frac{3}{5}\). Sufficient.
Answer: D.
Hope it's clear.
Let the rate of pump X be \(x\) liters per minute, the rate of pump Y be \(y\) liters per minute and the capacity of the tank be \(c\) liters.
Given: \((x+y)*48=c\). Question: \(\frac{48x}{c}=?\) (As from the pump X in 48 minutes would come \(48x\) liters)
(1) Working along at its constant rate, pump X would have taken 80 minutes to fill the tank with water --> \(80x=c\) --> \(\frac{48x}{c}=\frac{48x}{80x}=\frac{3}{5}\). Sufficient.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank --> \(120y=c\) --> as \((x+y)*48=c\), then \((x+y)*48=120y\) and \(48x=72y\) --> \(\frac{48x}{c}=\frac{72y}{120y}=\frac{3}{5}\). Sufficient.
Answer: D.
Hope it's clear.
25 Jan 2011, 20:18
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luminousmocha
You can look at it in another way
Stmnt 1: Time taken by X alone : Time taken by X and Y = 80:48 = 5:3
Rate of work of X : Rate of work of X and Y = 3:5 (Work done in both cases is the same. If X alone takes more time, its rate of work is smaller)
So X does 3/5 of total work. Sufficient.
Stmnt 2: Same calculations for y alone with give you that Y does 2/5 of the total work. Sufficient.
Answer D
