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Can anyone help me out with this problem

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by VS2013 » Thu Jul 22, 2010 4:08 pm
Thanks for the insight GMAT_Mumbai.

Here's my re-take on the answer:

Figure below shows the arrangement of the circles whereby the width of the strip is minimum. My apologies for the quality of the drawing.

Image


The width of the paper= sum of green lines + purple line
We know that the triangle is equilateral since it connects the center of the 3 circles and each side = 4 since the radius of each circle is 2 (given diameter=4).
The height of the triangle (purple line) = 2root3 using the 30 60 90 triangle rule (side ratio of 1: root3:2)
Each green line = radius of circle = 2
Therefore the width of paper = 2+(2root3)+2= 4+2root3

Answer= B
Last edited by VS2013 on Thu Jul 22, 2010 9:09 pm, edited 2 times in total.
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by 4GMAT_Mumbai » Thu Jul 22, 2010 8:09 pm
Hi,

Interesting problem. Thanks.

Perhaps, it could be done better.

The 5 circles could be like the olympic rings - only that the rings are disjoint in our case. In that case, our height will be less than 8 units. (Pls refer the attachment)

Here, AB = 4 units

Angle A of the triangle = 30 degrees

Angle B of the triangle = 60 degrees

One can figure the height of the triangle using the relationship 1:root 3: 2

Guess it would be 2 root 3.

The width of the paper would be the height of the triangle + (2 times radius).

Hope this helps. Thanks.
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by outreach » Thu Jul 22, 2010 8:29 pm
@4GMAT_Mumbai

AB should be 4,bcz diameter of circle is 4 ???
4GMAT_Mumbai wrote:Hi,

Interesting problem. Thanks.

Perhaps, it could be done better.

The 5 circles could be like the olympic rings - only that the rings are disjoint in our case. In that case, our height will be less than 8 units. (Pls refer the attachment)

Here, AB = 8 units

Angle A of the triangle = 30 degrees

Angle B of the triangle = 60 degrees

One can figure the height of the triangle using the relationship 1:root 3: 2

Guess it would be 4 root 3.

The width of the paper would be the height of the triangle + (2 times radius).

Hope this helps. Thanks.
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by 4GMAT_Mumbai » Thu Jul 22, 2010 8:33 pm
Hi Outreach,

Yes ... AB should be 4 ...

My poor excuse: Concentration on process rather than the numbers !! Just kidding ... Thanks for the correction.
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by VS2013 » Thu Jul 22, 2010 8:36 pm
4GMAT_Mumbai wrote:Hi,

Interesting problem. Thanks.

Perhaps, it could be done better.

The 5 circles could be like the olympic rings - only that the rings are disjoint in our case. In that case, our height will be less than 8 units. (Pls refer the attachment)

Here, AB = 8 units

Angle A of the triangle = 30 degrees

Angle B of the triangle = 60 degrees

One can figure the height of the triangle using the relationship 1:root 3: 2

Guess it would be 4 root 3.

The width of the paper would be the height of the triangle + (2 times radius).

Hope this helps. Thanks.
Hey there, you are right. I see your point. I edited my reply above.
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by GMATGuruNY » Fri Jul 23, 2010 6:18 am
Pinku wrote:suppose five circles, each 4 inches in diameter are cut from a rectangular strip of paper 12 inches long. if the least of paper is to be wasted, what is the width of the paper?
a) 5
b)4+[2(root3)]
c)8
d)4[1+(root3)]
Arranging the 5 circles so that they resemble the Olympics symbol, we can see that the width of the paper has to be a little less than 4 + 4 = 8 (the sum of 2 diameters).

Only answer choice B works: 4 + 2root3 = 4 + 2(1.7) = 7.4 approximately.

The correct answer is B.

Always look at the answer choices! It can save you a lot of time.
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