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kevincan
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GRE 1: Q170 V170
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Working at her constant rate, Cybil takes 8 hours to rake Updated on: 13 Aug 2006, 13:53
Originally posted by kevincan on 12 Aug 2006, 00:16.
Last edited by kevincan on 13 Aug 2006, 13:53, edited 3 times in total.
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Working at her constant rate, Cybil takes 8 hours to rake Goleta Beach every Monday. One Monday, she gets Trevor to help her. They begin raking at the same time and Trevor rakes k% faster than Cybil, though he alternates 20 minute periods of raking with 20 minute periods of swimming, whereas Cybil works at her constant rate without taking breaks. For which value(s) of k would Trevor be swimming rather than raking at the moment when the raking finishes?

(I) 50
(II) 100
(III) 200



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gmatornot
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Working at her constant rate, Cybil takes 8 hours to rake 13 Aug 2006, 10:07
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This problem seems really hard...

Is is 100 ? (More of a guess than a solution)
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jeg1015
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Working at her constant rate, Cybil takes 8 hours to rake 13 Aug 2006, 12:54
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This question is evil. Take a guess and move on. :)
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agsfaltex
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Working at her constant rate, Cybil takes 8 hours to rake Updated on: 13 Aug 2006, 13:16
Originally posted by agsfaltex on 13 Aug 2006, 12:55.
Last edited by agsfaltex on 13 Aug 2006, 13:16, edited 1 time in total.
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I have no idea why I'm even attempting this one:

Say total leaves = 800
Travis = T leaves/hour
Cybil = C leaves/hour
t = time (hours) = 8

t(T+C) = 800

8(T+C) = 800

With (I) T = 1.5C, plugging in you get C = 40 leaves/hour
If C = 40, T = 60 leaves/hour

With (II) T = 2C, plugging in you get C = (100/3) leaves/hour
IF C = 100/3, T = 200/3 leaves/hour

With (II) T = 3C, plugging in you get C = 25 leaves/hour
If C = 25, T = 75 leaves/hour

Now we know that if T works for 20 min and swims for 20 min, so his actual rate of working is double his average. And if we switch everything to 20 min increments

(I) (40/3)n + (n-1)(60*2)/(3) = 800

Solving for n = 15 plus a fraction
So, on the 15 increment, T will be working because n is odd. But since n = 15 and change, T will swimming.

(II) (100/3)n + (n-1)(200*2)/3 = 800

Solving for n = 12 plus a fraction
So, on the 12 increment, T will be swimming because n is even. But since n = 12 and change, T will be working.

(III) (25/3)n + (n-1)(75*2)/3 = 800

Solving for n = 14 plus a fraction
So, on the 14 increment, T will be swimming because n is even. But since n = 14 and change, T will be working

Therefore when K = 50, when T will be swimming rather then raking when the work is completed.
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Working at her constant rate, Cybil takes 8 hours to rake 13 Aug 2006, 13:21
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now your just playing with me, man! :suspect

You're gonna drive me to :drunk



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