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# Working independently at their respective rate, pump X and Y

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Joined: 16 Apr 2010
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Working independently at their respective rate, pump X and Y [#permalink]

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23 Jul 2010, 14:19
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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?

A. Working along at its constant rate, pump X would have taken 80 minutes to fill the tank with water
B. Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank

Ans-D

2. If K #0, 1 or -1 is 1/k >0

A. 1/k-1 >0
B. 1/k+1>0

Ans-A
[Reveal] Spoiler: OA
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23 Jul 2010, 14:44
Expert's post
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.

2. If k#0, 1 or -1 is $$\frac{1}{k}> 0$$?

(1) $$\frac{1}{k-1}> 0$$ --> $$k-1>0$$--> $$k>1$$, hence $$\frac{1}{k}>0$$. Sufficient.

(2) $$\frac{1}{k+1}> 0$$--> $$k+1>0$$ --> $$k>-1$$, hence $$k$$ can be negative as well as positive: $$\frac{1}{k}$$ may or may not be $$>0$$. Not sufficient.

Hope it's clear.

PLEASE POST ONE QUESTION PER TOPIC.
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Re: Pls explain   [#permalink] 23 Jul 2010, 14:44
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