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As shown in the figure above, a thin conveyor belt 15 feet long is [#permalink]

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05 May 2008, 18:13

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As shown in the figure above, a thin conveyor belt 15 feet long is drawn tightly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels?

Re: As shown in the figure above, a thin conveyor belt 15 feet long is [#permalink]

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05 May 2008, 19:51

GMAT_700 wrote:

A thin conveyor belt 15feet long is drawn tightly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels?

The correct answer is (15-pi)/ 2

the circumference covered by the belt of each circle = 1/2 (2 pi r/2) = 1/2 (2 pi 1/2) = pi/2 the total circumference of both circles covered by the belt = pi the lenth of the distance belt that doesnot cover the circumferences of 2 circles = 15 - pi so the distance between the centers of the circles = (15 - pi)/2

drawing a picture would be an easy way to understand why half of (15 - pi) is the distance between the centers of two circles. _________________

As shown in the figure above, a thin conveyor belt 15 feet long is dra [#permalink]

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29 Jul 2009, 19:35

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As shown in the figure above, a thin conveyor belt 15 feet long is drawn tightly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels?

A) \(\frac{15-\pi}{2}\) B) \(\frac{5\pi}{4}\) C) \(15-2\pi\) D) \(15-\pi\) E) \(2\pi\)

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Conveyer Belt.jpg [ 5.75 KiB | Viewed 3711 times ]

Last edited by Engr2012 on 16 Jul 2015, 06:56, edited 1 time in total.

Re: As shown in the figure above, a thin conveyor belt 15 feet long is dra [#permalink]

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29 Jul 2009, 20:07

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The length of the belt = (semi perimeter of circle1) + 2* distance between the centres of the circles + (semi perimeter of circle2)

let's say the distance between the centres of the circles = d. since both the circles have diameter = 1 ft ,therefore their radius = .5ft therefore semiperimeter of each of the circles = (1/2) * 2* pi * r = pi* .5 so, 15 = (pi* .5) + 2d + (pi*.5) 15 = pi + 2d therefore d = (15-pi)/2

Re: As shown in the figure above, a thin conveyor belt 15 feet long is dra [#permalink]

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29 Jul 2009, 20:16

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It is always better to include the picture rather than just reference the picture. You're more likely to get responses by putting in the actual question.

Attachment:

ConveyorBelt.png [ 7.3 KiB | Viewed 5957 times ]

As shown in the figure above, a thin coveyor belt 15 feet long is drawn tightly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels.

a) [m]\frac{15-∏}{2}

b) [m]\frac{5∏}{4}

c) 15 - 2∏

d) 15 - ∏

e) 2∏

A is correct because you have 15 total feet around the belt. There are 2 wheels, but instead of subtracting the circumference of both wheels, we only have to do the circumference of 1 wheel because the belt is only touching 1/2 of each wheel. The diameter is 1 foot, so in terms of circumference, that is ∏ * diameter or ∏. See the picture below.

Attachment:

ConveyorBeltRed.png [ 7.31 KiB | Viewed 5957 times ]

The red represents what has not been determined by the circumference of the wheel. The length of each red segment is the same length as the distance between the centers of the wheel. Because these 2 red segments are part of the belt, we know it's length if 15 feet long, minus the portion we just figured for around each half of the wheel, or 15 - ∏, but this is the length for both of them, and the distance between the center of the wheels is the same as the length of only one (1) of them, so we need to divide by 2. That give us a) for the answer. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: As shown in the figure above, a thin conveyor belt 15 feet long is dra [#permalink]

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30 Jul 2009, 19:59

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Total length=15 ft both cricles are consuming= pi length=(half circumfrence of c1 + half circumfrence of c2) Circumference= 2*pi*r r=d/2=1/2 Half circumference=pi/2 Both circles consume=2*pi/2=pi Now to find distance between two centers=(15-pi) is twice distance from centers of two circles. therefore actual distance=(15-pi)/2

Re: As shown in the figure above, a thin conveyor belt 15 feet long is [#permalink]

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16 Jul 2015, 05:58

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Re: As shown in the figure above, a thin conveyor belt 15 feet long is dra [#permalink]

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16 Jul 2015, 06:07

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Re: As shown in the figure above, a thin conveyor belt 15 feet long is [#permalink]

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16 Jul 2015, 08:53

Expert's post

GMAT_700 wrote:

As shown in the figure above, a thin conveyor belt 15 feet long is drawn tightly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels?

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