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GMAT Practice Test 1: Question 12

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by birdalways44 » Sat Mar 22, 2008 8:44 am
As shown in the figure, a thin conveyor belt 15 felt long is drawn tighly around two circular wheels each 1 foot in diameter. What is the distance, in feet, between the centers of the two wheels?

Here are the answer choices:

(15-Pi)/2

(5Pi)/4

15-2Pi

15-Pi

2Pi

Im not sure how to approach this problem. Any suggestions would be helpful. Thanks
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Hoping for a correct answer

by Leo2008 » Sat Mar 22, 2008 1:56 pm
Hi

The answer to this problem from my calculation comes out to (A) i.e. the first choice.

Explanation is as follows:

Its tight around the two circular wheels therefore the length of the belt on the wheels is 2.pi.R (R=1/2 foot) The remaining is 15-.pi divided by 2

Therefore the answer (15-pi)/2......Let me know if its correct.
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by II » Sun Mar 23, 2008 1:54 pm
Just a side note ... does everyone get the same questions when they take the GMATPrep CAT? This was also question 12 for me on my first GMATPrep !

My approach to answer this:
Diameter = 1ft ... so Radius = 1/2ft.

Lets work out the circumference (c) of 1 wheel:
c=2*pi*r => 2*pi*(1/2) = pi

So circumference of 2 wheels is 2*pi.

But we are not interested in the full circumference ... since the conveyor belt doesnt go all the way around the wheel. It only goes around half of the wheel. So it covers half the circumference ... which 1/2 * pi.
For the 2 wheels the conveyor belt covers pi.
So this enables us to come to 15-pi, which is the length of the conveyor belt not tightly against the wheel.

15-pi gives us the length of the conveyor belt which goes along the top AND the bottom (not the part which is around the wheels (i.e. pi).

The question is asking for the distance between the centers of the 2 wheels ... so hence you have to divide 15-pi by 2 to get the length between the centers of the 2 wheels.

Final answer: (15-pi) / 2

Hope this makes sense.
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by gmatwizard » Sat Oct 03, 2009 4:03 am
Good explanation GD
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by bhumika.k.shah » Tue Feb 02, 2010 7:57 am
This was ! # 12 in my gmat prep too .. FYI!
II wrote:Just a side note ... does everyone get the same questions when they take the GMATPrep CAT? This was also question 12 for me on my first GMATPrep !

My approach to answer this:
Diameter = 1ft ... so Radius = 1/2ft.

Lets work out the circumference (c) of 1 wheel:
c=2*pi*r => 2*pi*(1/2) = pi

So circumference of 2 wheels is 2*pi.

But we are not interested in the full circumference ... since the conveyor belt doesnt go all the way around the wheel. It only goes around half of the wheel. So it covers half the circumference ... which 1/2 * pi.
For the 2 wheels the conveyor belt covers pi.
So this enables us to come to 15-pi, which is the length of the conveyor belt not tightly against the wheel.

15-pi gives us the length of the conveyor belt which goes along the top AND the bottom (not the part which is around the wheels (i.e. pi).

The question is asking for the distance between the centers of the 2 wheels ... so hence you have to divide 15-pi by 2 to get the length between the centers of the 2 wheels.

Final answer: (15-pi) / 2

Hope this makes sense.
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by sumanr84 » Tue Feb 02, 2010 8:55 am
I haven't seen this Q in 2 of my 1st GMATPrep tests.
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by gmatmachoman » Thu Jul 22, 2010 10:43 pm
sumanr84 wrote:I haven't seen this Q in 2 of my 1st GMATPrep tests.
Here is my geometrical approach using Length of arc approach!!

Plz see the drawing :

Length of th wire = Arc AC + AB + Arc BD +DC

Arc AC = (180/360) * 2 * pi * (1/2)
= Pi/2

Arc BD = Pi/2

Let AB = DC= X ( also the distance between the centres of 2 circle)

Length of th wire = Arc AC + AB + Arc BD +DC

15 = Pi/2 +X+Pi/2+X

15- Pi = 2X

X= (15- Pi)/2

Since X is equal to the distance between the centres of 2 circle , gives the answer (15- Pi)/2
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