Rates SIM Motion

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Sun Jun 14, 2009 3:15 pm

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) 32.2 pi
(B)35.6 pi
(C) 38.4 pi
(D) 39.2 pi
(E) 40.8 pi

OA is C

Abdulla

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Source: Beat The GMAT — Data Sufficiency | Sun Jun 14, 2009 3:15 pm

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: Rates SIM Motion

by Vemuri » Mon Jun 15, 2009 4:52 am

Let C(l) - represent the larger gear & C(s) - represent the smaller gear.

Given, both the gears turn at the same rate. Lets assume the rate is 96cms/min. From this we know:

C(l) - covers 96pi cms in a min (which is 1 revolution clockwise)
C(s) - covers 96pi cms in a min (which is 4 revolutions anti-clockwise)

From this we know that in a min the notches meet 4 times in one revolution of the larger gear.

This is how far I could get

On the D day I might guess an answer for this type of a question.

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jakesing: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Fri Jun 12, 2009 2:10 pm; Thanked: 1 times

Re: Rates SIM Motion

by jakesing » Mon Jun 15, 2009 7:37 am

Tricky one, but I figured it out. Here is how I did it:

First, the circumferences of the circles are 24pi and 96pi, for the small and big circles respectively. When the notch meets for the first time it will create the same angle with respect to the starting point on each of the circles. Call that angle x. Since, as the previous post mentions, the smaller circle rotates 4 times in each of the bigger circle's rotations, this meeting will take place in the upper right quadrant of the circle (the small one will not yet have finished one full rotation, so the bigger one will not yet have finished one quarter of a rotation), which helps for visualization.

At the meeting point, the bigger circle will have traveled (x/360)*96pi centimeters. The smaller circle, whose angle of travel is (360-x), has traveled ((360-x)/360)*24pi - don't forget it traveled counterclockwise. Since the rate and the time of travel is the same (they started at the same time and they met, obviously, at the same time), these distances are equal as well. So you set up the equation:

(x/360)*96pi=((360-x)/360)*24pi ---> x=72.

If x=72, that means that the bigger circle has traveled (72/360)*96pi centimeters, or (96/5)pi, or 19.2 pi. This is after the first meeting of the two notches. Simply double this to find the distance traveled after 2 meetings. The answer, then, is 38.4 pi. Does this look alright?

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pg850: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Sat Jun 13, 2009 12:33 pm; Location: Mumbai, India

by pg850 » Mon Jun 15, 2009 8:31 am

It is indeed tricky..but well solved. I had a slightly different approach:

To find x (angle at which the two meet),

At angle x, the angle for the smaller gear would be (360 - 4*x)
Since these 2 have to be equal when the two gears meet,
x=360-4x
=> x = 72 degrees

Therefore, distance travelled by larger gear when they first meet would be (72/360)*2*pi*48
Multiple this no. by 2 to get the answer: 38.4pi

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