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# Geometry problem from QR 2nd PS150

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In pentagon PQRST, PQ=3 , QR=2 , RS=4 , and ST=5 . Which of [#permalink]  16 Aug 2010, 21:37
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In pentagon PQRST, PQ=3, QR=2, RS=4, and ST=5. Which of the lengths 5, 10, and 15 could be the value of PT?

(A) 5 only
(B) 15 only
(C) 5 and 10 only
(D) 10 and 15 only
(E) 5, 10, and 15
[Reveal] Spoiler: OA

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problem.jpg [ 36.98 KiB | Viewed 2017 times ]

Last edited by jpr200012 on 17 Aug 2010, 07:01, edited 1 time in total.
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Re: Geometry problem from QR 2nd DS150 [#permalink]  16 Aug 2010, 21:49
I guess if you added all the given lengths you get 15, so if the figure has to loop back to P then 15 is not possible, 5 and 10 are
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Re: Geometry problem from QR 2nd DS150 [#permalink]  17 Aug 2010, 00:01
mainhoon wrote:
I guess if you added all the given lengths you get 15, so if the figure has to loop back to P then 15 is not possible, 5 and 10 are

It only adds to 14. You must use triangles.
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Re: Geometry problem from QR 2nd DS150 [#permalink]  17 Aug 2010, 00:27
I guess you are referring to the fact that sum of two sides > third side? What is QR 2nd DS150? I have the green book for quant review, there is no 150 in DS there...
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Re: Geometry problem from QR 2nd DS150 [#permalink]  17 Aug 2010, 07:02
mainhoon wrote:
I guess you are referring to the fact that sum of two sides > third side? What is QR 2nd DS150? I have the green book for quant review, there is no 150 in DS there...

Sorry, it is PS 150.

Bunnel, can you move this topic?
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Re: Geometry problem from QR 2nd PS150 [#permalink]  17 Aug 2010, 07:13
I suppose the Logic given by @mainhoo is perfect. Even if it adds to 14, then even 15 is not possible.
Lets take it this way..If all points are assumed to be in straight line, then only PT can attain its maximum length. So its clear that PT=15 is not possible. Rest all values can be obtained at different configuration.
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Re: Geometry problem from QR 2nd PS150 [#permalink]  17 Aug 2010, 08:31
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Here is my version ...

Step1:
Given: PQ = 3, QR = 2
So, Using Traingles Sides theorm, PR can be: > (3-2) & < (3+2). i.e, PR can be >1 && <5.
So, possible values for PR are: 2, 3, 4

Step2:
Given: RS = 4, ST = 5,
So, RT can be >1 && < 9
So, possible values for RT are: 2, 3, 4, 5, 6, 7, 8

Step3: PRT forms a traingle
From Step1: PR can be: 2, 3, 4
From Step2: RT can be: 2, 3, 4, 5, 6, 7, 8
So, PT can take values till less than 12.

Answer: 5 & 10 are possible but 15 is not possible. The maximum side has < 12 (i.e, 8 + 4)

Cheers!
Ravi

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Re: Geometry problem from QR 2nd PS150 [#permalink]  17 Aug 2010, 10:03
Kudos to all. Nice post in Geometry
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Re: Geometry problem from QR 2nd PS150 [#permalink]  17 Aug 2010, 10:21
Good question. thanks nravi4 for the explanation.
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Re: Geometry problem from QR 2nd PS150 [#permalink]  19 Aug 2010, 12:24
You rule Ravi!

I have a question, how did you know that you had to use triangles?
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Re: Geometry problem from QR 2nd PS150 [#permalink]  19 Aug 2010, 23:48
Practice ...Practice ...Practice...!!!!
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Re: Geometry problem from QR 2nd PS150 [#permalink]  20 Aug 2010, 09:42
nravi4 wrote:
Practice ...Practice ...Practice...!!!!

You're right. It's the only way!
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Re: Geometry problem from QR 2nd PS150 [#permalink]  20 Aug 2010, 19:51
Is there an alternative way to solve this question?
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Re: Geometry problem from QR 2nd PS150 [#permalink]  21 Aug 2010, 01:50
put the 4 segment for which you know the length in a straight line...that's the limit configuration and it gives you the maximum length of the 5th segment is <14.

Something like this:

P---Q--R----S-----T

Then you will see that by bending the segments of the line all values <14 are possible for the 5th side.
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Re: Geometry problem from QR 2nd PS150 [#permalink]  21 Aug 2010, 12:44
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Answer is C i.e. 5 and 10 because the sum of all the other sides is 14 and one side can never be greater than sum of other sides
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Re: Geometry problem from QR 2nd PS150 [#permalink]  27 Aug 2010, 07:28
The length of any side of a closed polygon must be shorter than the sum of all the other sides:

Re: Geometry problem from QR 2nd PS150   [#permalink] 27 Aug 2010, 07:28
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