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# Jane makes toy bears. When she works with an assistant she

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Manager
Joined: 14 May 2006
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Jane makes toy bears. When she works with an assistant she [#permalink]  22 Aug 2007, 07:35
Jane makes toy bears. When she works with an assistant she makes 80% more bears per week, and works 10 percent fewer hours each week. Having an assistant increases Jane's output of toy bears per hour by what percent?

A. 20%
B. 80%
C. 100%
D. 180%
E. 200%
Intern
Joined: 19 Aug 2007
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Re: Toy Bears [#permalink]  22 Aug 2007, 07:52
BCC145 wrote:
Jane makes toy bears. When she works with an assistant she makes 80% more bears per week, and works 10 percent fewer hours each week. Having an assistant increases Jane's output of toy bears per hour by what percent?

A. 20%
B. 80%
C. 100%
D. 180%
E. 200%

B, because she makes 80% more per week. we have no idea about how many hours she worked in a week to begin with.
Manager
Joined: 15 Aug 2007
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Re: Toy Bears [#permalink]  22 Aug 2007, 08:03
BCC145 wrote:
Jane makes toy bears. When she works with an assistant she makes 80% more bears per week, and works 10 percent fewer hours each week. Having an assistant increases Jane's output of toy bears per hour by what percent?

A. 20%
B. 80%
C. 100%
D. 180%
E. 200%

Lets x=total toyes per week
t= hours in week
y=jane's output per hour
Befor assistant - he makes toyes x=t*y => y=t/x
After assistand he makes toyes 1.8x=0.9t*y =>y =2t/x

So 100 % ... C is the answer
Manager
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Correct is c, right? [#permalink]  22 Aug 2007, 08:10
C is the correct answer to this, right?
Manager
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I come up with 2w/t after assistant hence 200%
Manager
Joined: 03 Sep 2006
Posts: 233
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I can't get an equation for this, but I used POI:
if Jane + assistant = 180% and Jane works less with assistant (-10%), than it's obvious, that assistant adds more then 80% to Janes work, so A and B is out.
180% - Jane do together with assistant, so - it is too much.
Hence, assistant = 100%
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