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

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In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  01 Feb 2012, 18:23
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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 10 only
C 5 and 10 only
D 10 and 15 only
E 5, 10 and 15
Attachment:

Pentagon PQRST.PNG [ 14.5 KiB | Viewed 7478 times ]

Could someone please provide a simpler solution to this problem ? The actual solution is bit confusing and also it is time consuming.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 Feb 2012, 06:46, edited 2 times in total.
Edited the question and moved to PS subforum
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  01 Feb 2012, 18:37
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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 10 only
C 5 and 10only
D 10 and 15 only
E 5, 10 and 15

The length of any side of a triangle must smaller than the sum of the other two sides.

The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

PQ+QR+RS+ST=3+2+4+5=14, so the length of the fifths side can not be more than 14.

Answer: C (5 and 10 only).

P.S. Please do not reword the questions while posting.

Also please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  29 Feb 2012, 06:37
bunuel any other wayout for this problem. i have drawn 3 triangles and done that way, is it ok. thanks
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  29 Feb 2012, 06:50
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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.
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  01 May 2012, 02:01
Thanks for the response.
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  08 Sep 2012, 02:29
As mentioned by Bunuel, I will give another way to solve this problem using Triangle property.

Join P & R
In triangle PQR, the known two sides are 3 & 2 units so third side (Let x) must be between 1<x<5 i.e. it can take any value 2,3 or 4
In other words the maximum & minimum values possible are 4 & 2 respectively

Join T & R
In triangle PQR, the known two sides are 4 & 5 units so third side (Let y) must be between 1<x<9 i.e. it can take any value 2,3,4,5,6,7 or 8
In other words the maximum & minimum values possible are 8 & 2 respectively

We want to know the values possible for PT
In triangle PRT, the range of two sides is 1<x<5 & 1<x<9 units so third side (PT) must be between (2-2)< PT <(8+4) or 0< PT<12
i.e. it can take any value 1,2,3,4,5,6,7....10,11 (assuming sides can take only integer values)

Thus both 5 & 10 are possible

Hope it helps.
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  04 Jan 2013, 23:37
Bunuel wrote:
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 10only
D 10 and 15 only
E 5, 10 and 15

The length of any side of a triangle must smaller than the sum of the other two sides.

The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

PQ+QR+RS+ST=3+2+4+5=14, so the length of the fifths side can not be more than 14.

Answer: C (5 and 10 only).

P.S. Please do not reword the questions while posting.

Also please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

The length of any side of a triangle must smaller than the sum of the other two sides.

We also have
The length of any side of a triangle must be greater than the subtraction of the other two sides.

Don;t we have a similar property for other polygons?
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  05 Jan 2013, 05:50
Bunuel wrote:
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 10only
D 10 and 15 only
E 5, 10 and 15

The length of any side of a triangle must smaller than the sum of the other two sides.

The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

PQ+QR+RS+ST=3+2+4+5=14, so the length of the fifths side can not be more than 14.

Answer: C (5 and 10 only).

P.S. Please do not reword the questions while posting.

Also please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

Hi Bunuel,

The length of any side of a triangle must smaller than the sum of the other two sides.
The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

is this true only for triangle and pentagon or any other polygon?
How can we determine it?
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  31 Jan 2013, 01:26
I do not see this question from og books.

pls tell me the number of this question.

OG question is official guide.
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  18 Apr 2013, 06:50
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Bunuel wrote:
The length of any side of a triangle must smaller than the sum of the other two sides.
The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ?
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  19 Apr 2013, 02:55
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jainpiyushjain wrote:
Bunuel wrote:
The length of any side of a triangle must smaller than the sum of the other two sides.
The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ?
--
Thank you

Yes, the length of any side of a polygon must be less than the sum of the lengths of the other sides.
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Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]  17 May 2013, 06:32
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SrinathVangala wrote:
In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5.Which of the lengths 5,10 and 15 could be the value of PT?

Attachment:
The attachment Untitled.jpg is no longer available

A. 5 only
B. 15 only
C. 5 and 10 only
D.10 and 15 only
E. 5 , 10 and 15

Good one! The answer seems to be [C].

The easy way to approach this would be to consider the pentagon be comprised of 3 triangles as in the figure.

Let x and y be the diagonals under the quad which make the 3 triangles possible.
Now by triangle property, we know x < QR + PQ = 8 ......(1)
and y < RS + ST = 7 ...........(2)

Hence now in the interior triangle with sides x,y and PT, we further apply the triangle property:
PT < x+y
From above we already know by adding (1) and (2) x+y < 15, Hence we can say, that PT < 15. That rules out 15 off the options and 5, 10 can be possible values of the side!

Hope my answer and process are accurate!

Regards,
Arpan
Attachments

image-quad.jpg [ 7.2 KiB | Viewed 4949 times ]

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Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]  17 May 2013, 06:36
Since there is no more information is given about the pentagon, only condition which has to be satisfied is that sum of any 4 sides must be larger than the fifth side.
Now, 4 lengths are given as: 2,3,5,5.
Fifth side must be smaller than sum of these 4, i.e must be smaller than 15, so 15 can't be the answer.
About 5 and 10, there can not be any restriction, as the above rule is not broken in any of these cases.
So, answer is c) 5 and 10 only.
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Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]  17 May 2013, 06:40
Expert's post
SrinathVangala wrote:
In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5.Which of the lengths 5,10 and 15 could be the value of PT?

Attachment:
Untitled.jpg

A. 5 only
B. 15 only
C. 5 and 10 only
D.10 and 15 only
E. 5 , 10 and 15

Assuming OR is actually QR above, then this problem hinges on the fact that the fifth side of the pentagon must be shorter than the sum of the other four sides. If the four sides add up to 15 (3+5+2+5) then the fifth side cannot be 15, as this would form a line. Similarly, it cannot be greater than 15. It could, however, easily be 5 or 10, depending on the angles of the other arcs. Answer choice C.

Hope this helps!
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Re: In Pentagon PQRST, PQ = 3,OR =5,RS =2 and ST =5 [#permalink]  17 May 2013, 06:54
RULE:
In every polygon each side has to be less than the sum of the others

The sum is $$3+5+2+5=15$$

So only 5 and 10.

C
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Re: In the pentagon PQRST which is attached as a jpg file , PQ=3 [#permalink]  16 Jan 2014, 02:33
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Bunuel wrote:
jainpiyushjain wrote:
Bunuel wrote:
The length of any side of a triangle must smaller than the sum of the other two sides.
The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides.

Dear Bunuel, does this property apply to all quadrilaterals ?
--
Thank you

Yes, the length of any side of a polygon must be less than the sum of the lengths of the other sides.

Hi Bunnuel,

Is there a rule that pertains to the minimum length of polygons? I'm assuming that there is only one for triangles...
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  04 Apr 2014, 12:00
Hi Experts,

Yes please could you tell us if there is a rule for the minimum possible length in a polygon (like the one we have for triangle)?

Thanks!
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  05 Apr 2014, 01:55
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MensaNumber wrote:
Hi Experts,

Yes please could you tell us if there is a rule for the minimum possible length in a polygon (like the one we have for triangle)?

Thanks!

Please check here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1051688 an here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1214245

Hope this helps.
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Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the [#permalink]  25 May 2014, 13:08
Bunuel wrote:
MensaNumber wrote:
Hi Experts,

Yes please could you tell us if there is a rule for the minimum possible length in a polygon (like the one we have for triangle)?

Thanks!

Please check here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1051688 an here: in-the-pentagon-pqrst-which-is-attached-as-a-jpg-file-pq-126943.html#p1214245

Hope this helps.

Hi Bunuel,

To piggy back on this question -- the links that you have provided state that the length of the 5th side has to be less than the sum of the other 4 sides. Is there anything governing the lower end? For example, in a triangle, the third side must be bigger than the difference of the two sides. Does a similar rule apply to pentagons and if they do, how do you compare which 2 sides?

My question just lies as to why we didn't prove/justify that 5 wasn't too small? In this case, we don't have an answer choice that just states that "10" is the value but if it did, are we to assume that 5 will automatically work?
Re: In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the   [#permalink] 25 May 2014, 13:08
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# In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the

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