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# A circular gear with a diameter of 24 centimeters is mounted

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Manager
Joined: 25 Jul 2006
Posts: 97
A circular gear with a diameter of 24 centimeters is mounted  [#permalink]

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11 Jul 2007, 15:35
4
A circular gear with a diameter of 24 centimeters is mounted directly on another circular gear with a diameter of 96 centimeters. Both gears turn on the same axle at their exact centers and each gear has a single notch, at the 12 o'clock position. At the same moment, the gears begin to turn at the same rate, with the larger gear moving clockwise and the smaller gear counterclockwise. How far, in centimeters, will the notch on the larger gear have traveled the second time the notches pass each other?

(A) 32.2Pi
(B) 35.6Pi
(C) 38.4Pi
(D) 39.2Pi
(E) 40.8Pi

Question from Manhattan GMAT - am not comfortable with their answer/explanation (which i'll post later). If anyone has better insight, plz explain. Thx.

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Manager
Joined: 22 May 2007
Posts: 119

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11 Jul 2007, 15:45
"mounted directly on"

Not sure I understand...
Manager
Joined: 08 Jul 2007
Posts: 167

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11 Jul 2007, 19:18
2
I get C.

The circumference of of the big circle is 96pi and the circumference of the small circle is 24pi. Their gears are traveling at the same rate, so we can pick a number, say 8pi/hr because 8pi is the LCF. Then the big circle's rate is 1/12 revolutions per hour and the small circle's rate is 1/3 revolutions per hour.

Letting d1 be the distance that the large gear is moving and d2 be the distance that the small gear is moving, we want to find out when d1+d2=192pi (ie 96pi+96pi).

So for the big wheel: (1/12)*t=d
And for the small wheel: (1/3)*t= 192pi-d

Solving for d, we get that t=12d and t=576pi-3d. This means 12d=576pi-3d. Which means 15d=576pi. Which means d=38.4pi.
Manager
Joined: 25 Jul 2006
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13 Jul 2007, 19:25
briks123 wrote:
I get C.

The circumference of of the big circle is 96pi and the circumference of the small circle is 24pi. Their gears are traveling at the same rate, so we can pick a number, say 8pi/hr because 8pi is the LCF. Then the big circle's rate is 1/12 revolutions per hour and the small circle's rate is 1/3 revolutions per hour.

Letting d1 be the distance that the large gear is moving and d2 be the distance that the small gear is moving, we want to find out when d1+d2=192pi (ie 96pi+96pi).

So for the big wheel: (1/12)*t=d
And for the small wheel: (1/3)*t= 192pi-d

Solving for d, we get that t=12d and t=576pi-3d. This means 12d=576pi-3d. Which means 15d=576pi. Which means d=38.4pi.

briks,

you are only one to respond - much appreciated!
but this was the same problem i had with their solution also. when it is given "same rate", my conclusion was "same rpm", not "same distance/hr", since we are dealing with rotational (not linear) speed. have u seen a similar problem in a reliable math text which takes the linear speed?

but for this assumption (which changes the entire solution), the rest of your solution is valid.

expert or knowledgeable person on this topic, plz comment/clarify...
Manager
Joined: 08 Jul 2007
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13 Jul 2007, 20:05
Confused by your response. So they wanted it to be same revolutions/hr? I chose a number that was same distance/hr. Do you have the explanation?

ps is this Manhattan gmat? I have been finding that they are not always clear on what they are asking for in their questions!
Manager
Joined: 25 Jul 2006
Posts: 97

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13 Jul 2007, 20:19
sorry if not clear -

actually kudos to you since they are saying the same as you do - i.e

in my mind, i think it ought to be rotational speed i.e
same rate = same rpm

which of course changes the entire solution.
that was the reason for my confusion and hence the post.
i would like some validation though - if you or anyone has seen a similar problem in a definitive math text, plz respond.

i'll copy/paste their solution sometime later.
Manager
Joined: 08 Jul 2007
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13 Jul 2007, 20:46
I think for GMAT purposes, you can assume that rate always means distance/time.

Especially since with this problem, if you assume rate is revolutions/min, then the two gears will always be at the same place on the circle so there would be nothing to solve.
Manager
Joined: 25 Jul 2006
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13 Jul 2007, 21:28
briks123 wrote:
I think for GMAT purposes, you can assume that rate always means distance/time.

Especially since with this problem, if you assume rate is revolutions/min, then the two gears will always be at the same place on the circle so there would be nothing to solve.

not if they are rotating in opposite directions as the problem states.
Manager
Joined: 25 Jul 2006
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13 Jul 2007, 21:56
Anyway, here is the official detailed answer from Manhattan GMAT:
---------------
Since the notches start in the same position and move in opposite directions towards each other, they will trace a circle together when they pass for the first time, having covered a joint total of . When the notches meet for the second time, they will have traced two full circles together for a total of .

Since the circumference of the large gear is 4 times greater than that of the small gear, the large notch will cover only 1/4 of the number of degrees that the small notch does. We can represent this as an equation (where x is the number of degrees covered by the large notch):
5x=720
x=144

So the large notch will have covered when the notches pass for the second time. Since the circumference of the large gear is , we can set up the following proportion to solve for the linear distance (call it d) covered by the large notch:
144/360 = d/96Pi

Manager
Joined: 08 Jul 2007
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14 Jul 2007, 05:31
oops wrote:
Anyway, here is the official detailed answer from Manhattan GMAT:
---------------
Since the notches start in the same position and move in opposite directions towards each other, they will trace a circle together when they pass for the first time, having covered a joint total of . When the notches meet for the second time, they will have traced two full circles together for a total of .

Since the circumference of the large gear is 4 times greater than that of the small gear, the large notch will cover only 1/4 of the number of degrees that the small notch does. We can represent this as an equation (where x is the number of degrees covered by the large notch):
5x=720
x=144

So the large notch will have covered when the notches pass for the second time. Since the circumference of the large gear is , we can set up the following proportion to solve for the linear distance (call it d) covered by the large notch:
144/360 = d/96Pi

ah! good point! oh gmats...
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Re: A circular gear with a diameter of 24 centimeters is mounted  [#permalink]

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25 Jun 2018, 10:01
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