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# The figure above shows the dimensions of a semicircular

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VP
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The figure above shows the dimensions of a semicircular [#permalink]

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08 Jan 2008, 06:10
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Question Stats:

48% (02:52) correct 52% (01:37) wrong based on 108 sessions

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The figure attached shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft
B. 7½ ft
C. 8 ½ ft
D. 9½ ft
E. 10 ft

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-figure-above-shows-the-dimensions-of-a-semicircular-100337.html
[Reveal] Spoiler: OA

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semi.doc [23.5 KiB]

Director
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08 Jan 2008, 07:15
Are they asking for the height of the tunnel at the EDGE of the 12 foot lane? or the height of the tunnel in the center? It's kind of vague, but I would assume they want the height of the tunnel at it's highest point (even though this isn't entirely practical, but perhaps that's where the 1/2 foot of tolerance comes in).

10 feet because the radius is 10 feet
then subtract 1/2 foot of clearance
10-1/2 = 9.5 feet
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08 Jan 2008, 07:48
eschn3am wrote:
Are they asking for the height of the tunnel at the EDGE of the 12 foot lane? or the height of the tunnel in the center? It's kind of vague, but I would assume they want the height of the tunnel at it's highest point (even though this isn't entirely practical, but perhaps that's where the 1/2 foot of tolerance comes in).

10 feet because the radius is 10 feet
then subtract 1/2 foot of clearance
10-1/2 = 9.5 feet

I think it is better to find height at the 12 foot lane, because shape of car not triangle, rather it is rectangle...
so I would have picked A, since height will be 6 and 6-1/2=5.5
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08 Jan 2008, 08:07
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That does make more sense (since automobiles aren't 2D ).

In that case:

Create a right triangle using the radius of 10 as a hypotenuse. The legs of the triangle will be 6 (1/2 of the travel lane) and X (the height at the edge of the road). 3-4-5 right triangle gives us a height of 8' at the side of the road. 8-.5 = 7.5' max
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08 Jan 2008, 09:12
eschn3am wrote:
That does make more sense (since automobiles aren't 2D ).

In that case:

Create a right triangle using the radius of 10 as a hypotenuse. The legs of the triangle will be 6 (1/2 of the travel lane) and X (the height at the edge of the road). 3-4-5 right triangle gives us a height of 8' at the side of the road. 8-.5 = 7.5' max

great, I agree with you) tellingly I never hit the right answer I am always close to it but it doesn't make me happy because real GMAT exam makes sever punishments for wrong answers whether it is close to right answer or not. what would you suggest me to do in order to decrease errors and increase precision? not to rush?
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08 Jan 2008, 09:20
kazakhb wrote:
eschn3am wrote:
That does make more sense (since automobiles aren't 2D ).

In that case:

Create a right triangle using the radius of 10 as a hypotenuse. The legs of the triangle will be 6 (1/2 of the travel lane) and X (the height at the edge of the road). 3-4-5 right triangle gives us a height of 8' at the side of the road. 8-.5 = 7.5' max

great, I agree with you) tellingly I never hit the right answer I am always close to it but it doesn't make me happy because real GMAT exam makes sever punishments for wrong answers whether it is close to right answer or not. what would you suggest me to do in order to decrease errors and increase precision? [b]not to rush[b/]?

that's it in a nutshell. Slow down, double check your work and draw out a diagram when doing geometry problems. I find I do much better on the Club Challenges when I slow down a bit. After awhile you'll know all the math you need, it's just eliminating stupid mistakes.
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08 Jan 2008, 09:37
eschn3am wrote:
That does make more sense (since automobiles aren't 2D ).

In that case:

Create a right triangle using the radius of 10 as a hypotenuse. The legs of the triangle will be 6 (1/2 of the travel lane) and X (the height at the edge of the road). 3-4-5 right triangle gives us a height of 8' at the side of the road. 8-.5 = 7.5' max

correct! ...7.5' for me too..
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08 Jan 2008, 10:44
akhi wrote:
eschn3am wrote:
That does make more sense (since automobiles aren't 2D ).

In that case:

Create a right triangle using the radius of 10 as a hypotenuse. The legs of the triangle will be 6 (1/2 of the travel lane) and X (the height at the edge of the road). 3-4-5 right triangle gives us a height of 8' at the side of the road. 8-.5 = 7.5' max

correct! ...7.5' for me too..

OA is B.

This kind of questions are easy to solve reading the text well because the are written in a tricky way! I answered D too...
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08 Jan 2008, 12:04
think of a right triangle with base 6 (12/2) and hypotnuse 9.5 (10-1/2) and find out the height=answer
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08 Jan 2008, 12:06
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Hypotenuse is 10, not 9.5.

You take the 1/2 off of the height when you find it (8-.5 =7.5), not off the hypotenuse.
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08 Jan 2008, 12:27
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Re: The figure above shows the dimensions of a semicircular [#permalink]

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30 Nov 2013, 13:50
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Re: The figure above shows the dimensions of a semicircular [#permalink]

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01 Dec 2013, 07:06
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marcodonzelli wrote:

The figure attached shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft
B. 7½ ft
C. 8 ½ ft
D. 9½ ft
E. 10 ft

See the diagram attached:

Rectangle inscribed has the length of traffic lane 12. So max height of vehicle will be 1/2 foot less than the width of this rectangle.

Now, let O be the center of the semi-circle, then OA=radius=20/2=10 and OB=12/2=6 --> $$AB=\sqrt{OA^2-OB^2}=\sqrt{10^2-6^2}=8$$.

So max height of the vehicle that are allowed to use the tunnel is 8-0.5=7.5.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-figure-above-shows-the-dimensions-of-a-semicircular-100337.html
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Re: The figure above shows the dimensions of a semicircular   [#permalink] 01 Dec 2013, 07:06
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