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# In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the

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Manager
Joined: 26 Mar 2017
Posts: 101
Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the  [#permalink]

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13 May 2017, 06:55
victory47 wrote:
I do not see this question from og books.

pls tell me the number of this question.

OG question is official guide.

No. 149 Official Guide Math Review 2017
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I hate long and complicated explanations!
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Joined: 26 Mar 2017
Posts: 101
Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the  [#permalink]

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03 Jun 2017, 00:15
abhi47 wrote:
Attachment:
Pentagon PQRST.PNG
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 10 only
C 5 and 10 only
D 10 and 15 only
E 5, 10 and 15

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/in-pentagon-p ... 68634.html

hey,

is it also true that PT must be > 4 ?
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I hate long and complicated explanations!
Intern
Joined: 07 Jun 2017
Posts: 10
GMAT 1: 560 Q42 V25
Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the  [#permalink]

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29 Aug 2017, 09:06
Bunuel wrote:
pbull78 wrote:
bunuel any other wayout for this problem. i have drawn 3 triangles and done that way, is it ok. thanks

Yes, you can solve it with a triangle property which says that the length of any side of a triangle must be smaller than the sum of the other two sides.

Though you can naturally apply this property to any polygon and you won't need intermediary steps for solving.

This is easier to understand than in explanation in OG.
Thanks a lot!

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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the   [#permalink] 29 Aug 2017, 09:06

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