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Cutting Board

by smclean23 » Mon Aug 04, 2008 5:50 pm
The figure above shows the dimensions of a rectangular board that is to be cut into four identical pieces by making cuts at points A, B, and C, as indicated. If x = 45, what is the length AB ?
(1 foot = 12 inches)

(A) 5 ft 6 in
(B) 5 ft 3 sq rt. 2 in
(C) 5 ft 3 in
(D) 5 ft
(E) 4 ft 9 in
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by pepeprepa » Tue Aug 05, 2008 1:31 am
Well, I am not at all in the 2 minutes...

First 20 feet=240inches
The small triangle I drew with vertex A is 45 45 90 so we know the ratio is x : x : xsqrt(2)
Therefore we know one side which is 6 so I marked on paint a red 6 for the small part of the segment AB which is as well 6.
Area K + Area J is equal to (240*6)/4 (question tells us that all parts are equal so let's use it), which can be written as follow:
(6*6)/2 + y*6 = 360
y=57 inches

So AB is 57+6=63 inches
Let's see in feet
63/12=5*12 + 3

The length of AB is 5ft 3in
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by smclean23 » Tue Aug 05, 2008 5:58 pm
Thanks for your time, but how do we know that the first triangle is 45 45 90 and not 30 60 90?

Could you give a clearer explanation or a "dummied down" version for me?
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by pepeprepa » Wed Aug 06, 2008 12:49 am
Well, the question tells you x=45 and the triangle we build is a right triangle so the last angle is 45.
Tell me if you have problems.
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by amitdgr » Wed Aug 06, 2008 2:11 am
great explanation :) it surely comes under difficult question category for me ... BTW wat is the difficulty level of this q ?
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by pepeprepa » Wed Aug 06, 2008 2:19 am
The concepts used are not hard... but personnally it took time for me to solve it, I was far from the 2' limit!
I let experts judge if it is an easy or hard question but I noticed it is pretty hard to categorize gmat questions in terms of easy/middle/hard.
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by augusto » Wed Aug 06, 2008 3:44 am
I think it's an easy to medium question, and you don't need to calculate any area like pepeprepa did

you know that you have one big rectangle and you have to cut it in 4 identical pieces. So point B is going to be in the middle of the 20 feet.

So now the problem is to get 2 identical pieces from a rectangle of 10 feet by 6 inches. (or 120 in per 6 in)

If x is 45 degrees, then you can think that you'll have a small square in the middle of the rectangle, and each piece will be the sum of the rectangles in the sides plus a piece of the triangle. This is
So each side of the square in the middle has 6 inches, so each side-board will be 57in long.
So
57+7 = 63 in = 5f 3in

I hope the drawing helps to explain how I solved it
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by reply2spg » Thu Apr 22, 2010 12:04 pm
I solved this problem 2-3 times, but today I got it properly. thanks buddy it really helped....I think it's level might be 710 +
pepeprepa wrote:Well, I am not at all in the 2 minutes...

First 20 feet=240inches
The small triangle I drew with vertex A is 45 45 90 so we know the ratio is x : x : xsqrt(2)
Therefore we know one side which is 6 so I marked on paint a red 6 for the small part of the segment AB which is as well 6.
Area K + Area J is equal to (240*6)/4 (question tells us that all parts are equal so let's use it), which can be written as follow:
(6*6)/2 + y*6 = 360
y=57 inches

So AB is 57+6=63 inches
Let's see in feet
63/12=5*12 + 3

The length of AB is 5ft 3in
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by eaakbari » Thu Apr 22, 2010 11:40 pm
augusto wrote:I think it's an easy to medium question, and you don't need to calculate any area like pepeprepa did

you know that you have one big rectangle and you have to cut it in 4 identical pieces. So point B is going to be in the middle of the 20 feet.

So now the problem is to get 2 identical pieces from a rectangle of 10 feet by 6 inches. (or 120 in per 6 in)

If x is 45 degrees, then you can think that you'll have a small square in the middle of the rectangle, and each piece will be the sum of the rectangles in the sides plus a piece of the triangle. This is
So each side of the square in the middle has 6 inches, so each side-board will be 57in long.
So
57+7 = 63 in = 5f 3in

I hope the drawing helps to explain how I solved it
You meant 6 there instead of 7. Nice solution btw
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by Nilakshi1001 » Fri Apr 23, 2010 2:38 am
I have attached the explanation document with my answer.

E.
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by govind_raj_76 » Sat Apr 24, 2010 4:41 am
I have joined this forum recently. I have become an active reader for the PS problems. In this one, I cannot see the diagram from the attached document to understand the problem. Let me know how do I see the diagram.

Thanks,
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by govind_raj_76 » Sat Apr 24, 2010 5:56 am
I was able to see the diagram and found a way to download. For some reason, it is downloading as .php extension, I changed the files to be a .doc extension. I was able to decipher the answers. Thanks everone.
Govind
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