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# I understand this problem, but the solution is based on the

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Intern
Joined: 15 Jun 2008
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I understand this problem, but the solution is based on the [#permalink]

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13 Jul 2008, 14:15
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I understand this problem, but the solution is based on the distance of the drive wheel=distance of the mixing wheel. No problem there. Thats 2 wheels, what happened to the 3rd wheel?
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Joined: 15 Dec 2008
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Schools: HBS(08) - Ding. HBS, Stanford, Kellogg, Tuck, Stern, all dings. Yale - Withdrew App. Emory Executive -- Accepted, Matriculated, Withdrewed (yes, I spelled it wrong on purpose). ROSS -- GO BLUE 2011.
Re: m05 question 23 - what happened to the 3rd wheel [#permalink]

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17 Mar 2009, 08:14
dariusbanasik wrote:
I understand this problem, but the solution is based on the distance of the drive wheel=distance of the mixing wheel. No problem there. Thats 2 wheels, what happened to the 3rd wheel?

The third wheel "remains in the middle."

The distance the motor and mixing wheels travel is equal -- you stated you understand that.

The distance the middle wheel travels is also the same.

If the middle wheel had a radius of 6,000,000 miles, it would turn the same "distance" as a middle wheen with radius of 1 nanometer.

The number of rotations would be vastly different, but in the end it would still turn the mixing wheel the same "distance" that the motor wheel turn it (the middle wheel).

Therefore since they all turn the same distance, you can essentially ignore the middle (third) wheel.

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Re: m05 question 23 - what happened to the 3rd wheel   [#permalink] 17 Mar 2009, 08:14
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