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Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
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11 Mar 2014, 23:39
5
1
The pentagon can be split in to 3 triangles: PQR, PRS and PST Consider Triangle PQR: Since PQ= 3, QR = 2; we can say that 1<PR<5 - Based on the property of triangles: The length of any side of a triangle must be smaller than the sum of the other 2 sides and greater than the difference of the other 2 sides.
Consider Triangle PRS: We can say that 1<PS<9;
Consider Triangle PST: We can say that PT is definitely less than 15. So, eliminate B, D and E. 4<PT<14;
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
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14 Aug 2014, 02:23
Bunuel wrote:
SOLUTION
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
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).
Bunuel, thanks for your explanation! I understand the solution and I agree, but I have one concern.
On the figure drawn we can see that 1) the direction of lines PQ, QR and RS is to the right from the point P. The sum of of these lines is only 9. 2) the direction of line ST is opposite (or to the left / back to point P).
So, if taking into account this fact, it appears that the line TP is <=9.
Is this reasoning incorrect only because it is written that the figure is drawn to scale?
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
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09 Oct 2017, 14:52
Divide pentagon into three triangles: Triangle RST, Triangle QRT, Triangle PQT In RST, RS = 4, ST =5; RT has to be less than their sum, and greater than their difference; therefore, RT could be 8-2; just start with 8, and if it does not work, go back and try 7-2 In QRT, QR = 2, RT = 8, and QT is unknown. QT < 10; QT >6; therefore, QT = 9-7 In PQT, PQ = 3, QT = 9, PT unknown. PT<12; PT>6; therefore, 15 cannot be PT, thus eliminating answer choices B, D, and E Suppose QT is 8; PT<11; PT>3, so it could still be 5, or 10 Suppose QT is 7; PT<10; PT>4. 10 is eliminated here, but since if QT equals 8, 10 is still possible, as with 5. Supposing RT is 2, then QT<4; if QT is 3, then PT<6; therefore, 5 has to be the answer since we used the smallest possible outcomes. Ten could be eliminated, however, under other conditions, it is permissible.
The only answer choice that was eliminated across every possible condition is 15. The conditions showed us that it can be 5 or 10. Therefore, the answer is (C) 5 and 10 only
A helpful general rule: questions that involve ranges, work with extremes such as biggest or smallest.
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
[#permalink]
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13 Dec 2017, 09:24
Bunuel wrote:
SOLUTION
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
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).
I followed the same logic and eliminated 15. My concern was about 5 and 10, the official solution contains quite a long and detailed explanation but should I bother? Is there any need in a question like that and dig further - proving that 10 or 5 can be the answer?
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
[#permalink]
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13 Dec 2017, 23:29
Erjan_S wrote:
Bunuel wrote:
SOLUTION
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
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).
I followed the same logic and eliminated 15. My concern was about 5 and 10, the official solution contains quite a long and detailed explanation but should I bother? Is there any need in a question like that and dig further - proving that 10 or 5 can be the answer?
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
[#permalink]
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09 Jan 2018, 10:47
2
Erjan_S wrote:
Bunuel wrote:
SOLUTION
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
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).
I followed the same logic and eliminated 15. My concern was about 5 and 10, the official solution contains quite a long and detailed explanation but should I bother? Is there any need in a question like that and dig further - proving that 10 or 5 can be the answer?
Hi Erjan,
Please find a picture attached for easy understanding.
Divide the pentagon PQRST into three triangles- triangle PQR, triangle PRS and triangle PST.
The range for the lengths of PR, PS and PT comes from one of the properties of triangles that the length of the third side will be less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.
I hope this helps.
Aiena.
Attachments
Pentagon PQRST.jpg [ 681.92 KiB | Viewed 9083 times ]
Re: In pentagon PQRST, PQ= 3, QR = 2, RS = 4, and ST = 5. Which
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14 Jan 2018, 20:36
Fcma wrote:
Would you not say this is a 700 question? I find it quite tedious
You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty: 600-700 Level.
_________________
Can we derive a general rule: length of any side of a polygon (including regular polygons) must be < sum (n-1) sides. Also, can we derive a rule for the minimum? Say > difference of two smallest lengths?
close
66% (01:59) correct 
