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Considering a semicircular cross section of a one-way tunnel

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Considering a semicircular cross section of a one-way tunnel  [#permalink]

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New post 19 Apr 2008, 01:12
Considering a semicircular cross section of a one-way tunnel with a diameter of 20 feet. The single traffic lane is 12 feet wide and is equidistant from the side 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?

5 ½ ft
7 ½ ft
8 ½ ft
9 ½ ft
10 ft

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Re: A tunnel  [#permalink]

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New post 19 Apr 2008, 02:05
Capthan wrote:
Considering a semicircular cross section of a one-way tunnel with a diameter of 20 feet. The single traffic lane is 12 feet wide and is equidistant from the side 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?

5 ½ ft
7 ½ ft
8 ½ ft
9 ½ ft
10 ft


D

You should make a draw and you will see that D
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Re: A tunnel  [#permalink]

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New post 19 Apr 2008, 02:48
(B) for me: we should consider the height on the sides of traffic lane.
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Re: A tunnel  [#permalink]

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New post 19 Apr 2008, 03:21
I think D too -> height should be 1/2 less than the radius(10) thus 9.5, isn't it ?
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Re: A tunnel  [#permalink]

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New post 19 Apr 2008, 04:39
I too feel it's B....coz height at corner of road will b 8'........
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Re: A tunnel  [#permalink]

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New post 19 Apr 2008, 05:11
prasannar wrote:
I think D too -> height should be 1/2 less than the radius(10) thus 9.5, isn't it ?

yes, right. But the height will decrease from the center of the traffic line to its side. If we take 6 ft from the center (1/2 of traffic line), the height there will be only 8, which gives us maximum height of vehicles of 7 1/2ft
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Re: A tunnel  [#permalink]

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New post 20 Apr 2008, 10:24
if H = maximum height of the car, what you end up with here is a right triangle with sides of 10,6 and H+0.5

solving using pythagoras, H+0.5 = 8 therefore H = 7.5 ---> B
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Re: A tunnel  [#permalink]

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New post 22 Apr 2008, 04:10
1
For me, the height should be within another radius of 9.5 (10-.5)

Also, we know that on one side of the road, the side extends to 6 ft from the center.

Using pythagoras theorem, we have, hypotenuse =9.5, 1 side(base) =6

Therefore, the other side (height) = sqrt(9.5 ^ 2 - 6 ^2)= sqrt(54.25)= < 7.5 ft

Therefore I chose A.
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Re: A tunnel  [#permalink]

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New post 22 Apr 2008, 14:28
for me this is 7.5 B
-------------
- -
- -
- -
-----4---6--10-----4----

dont mind my drawing but..

the base of the trinagle is 6..hyp is 10..then the prep is 8..we are told that the max clearing should be 1/2 feet below this..so 7.5 feet is the answer
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Re: A tunnel  [#permalink]

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New post 22 Apr 2008, 14:57
Capthan wrote:
Considering a semicircular cross section of a one-way tunnel with a diameter of 20 feet. The single traffic lane is 12 feet wide and is equidistant from the side 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?

5 ½ ft
7 ½ ft
8 ½ ft
9 ½ ft
10 ft


B.

Here is why. (See attachment)

Let me know whether you cant see it. Hrmm maybe I cannot attach it... Owell. Here is my reasoning.

The line from the center of the semicircle is the same distance as the slanted line from the center of the semicircle.

This line is now Excluding the ½ foot. We now have 9 1/2feet as our hypotenous of the square. We get the square from the greater rectangle. Split the rectangle in two and we get a perfect square. (Assuming the entire semicircle is perfectly symmetrical).

Now we need the height of the rectangle or the height of our square. This is 9 ½ ft/sqrt2

~ 7 ½ ft
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Re: A tunnel  [#permalink]

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New post 22 Apr 2008, 22:11
GMATBLACKBELT wrote:
Capthan wrote:
Considering a semicircular cross section of a one-way tunnel with a diameter of 20 feet. The single traffic lane is 12 feet wide and is equidistant from the side 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?

5 ½ ft
7 ½ ft
8 ½ ft
9 ½ ft
10 ft


B.

Here is why. (See attachment)

Let me know whether you cant see it. Hrmm maybe I cannot attach it... Owell. Here is my reasoning.

The line from the center of the semicircle is the same distance as the slanted line from the center of the semicircle.

This line is now Excluding the ½ foot. We now have 9 1/2feet as our hypotenous of the square. We get the square from the greater rectangle. Split the rectangle in two and we get a perfect square. (Assuming the entire semicircle is perfectly symmetrical).

Now we need the height of the rectangle or the height of our square. This is 9 ½ ft/sqrt2

~ 7 ½ ft


Can you make the attachment? It will helpful to the discusstion! Thanks
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Re: A tunnel  [#permalink]

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New post 23 Apr 2008, 04:57
sondenso wrote:
GMATBLACKBELT wrote:
Capthan wrote:
Considering a semicircular cross section of a one-way tunnel with a diameter of 20 feet. The single traffic lane is 12 feet wide and is equidistant from the side 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?

5 ½ ft
7 ½ ft
8 ½ ft
9 ½ ft
10 ft


B.

Here is why. (See attachment)

Let me know whether you cant see it. Hrmm maybe I cannot attach it... Owell. Here is my reasoning.

The line from the center of the semicircle is the same distance as the slanted line from the center of the semicircle.

This line is now Excluding the ½ foot. We now have 9 1/2feet as our hypotenous of the square. We get the square from the greater rectangle. Split the rectangle in two and we get a perfect square. (Assuming the entire semicircle is perfectly symmetrical).

Now we need the height of the rectangle or the height of our square. This is 9 ½ ft/sqrt2

~ 7 ½ ft


Can you make the attachment? It will helpful to the discusstion! Thanks


I made it but it wont attach. =(... NM finally it attached. Use the above explanation. I cannot enter in text for the attached or it goes nuts on me... very strange.

--== Message from the GMAT Club Team ==--

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This discussion does not meet community quality standards. It has been retired.


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Re: Considering a semicircular cross section of a one-way tunnel  [#permalink]

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New post 27 Sep 2018, 23:32
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Re: Considering a semicircular cross section of a one-way tunnel &nbs [#permalink] 27 Sep 2018, 23:32
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