GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
P and Q are two different points on line l
[#permalink]
31 Aug 2026, 08:16
31 Aug 2026, 08:16
Expert Reply
00:00
Question Stats:
75% (03:45) correct
25% (02:07) wrong based on 4 sessions
Hide Show timer Statistics
P and Q are two different points on line l , and R is a point not on line l such that triangle P Q R is isosceles with P R=Q R . Point T is the midpoint of segment P Q , and the distance from T to R is 6 .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The perimeter of triangle P Q R |
28 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2932
Given Kudos: 1283
GPA: 3.39
Re: P and Q are two different points on line l
[#permalink]
04 Sep 2026, 10:54
04 Sep 2026, 10:54
Expert Reply
Solution
Since \(PR=QR\), triangle \(PQR\) is isosceles, and the midpoint \(T\) of the base \(PQ\) makes \(RT\) perpendicular to \(PQ\).
So we have a right triangle with:
\(RT=6\)
Let \(PT=TQ=x\). Then:
\(PR=QR=\sqrt{x^2+6^2}=\sqrt{x^2+36}\)
Thus the perimeter is
\(2x+2\sqrt{x^2+36}\)
We need to compare this with 28. Set the perimeter equal to 28:
\(2x+2\sqrt{x^2+36}=28 x+\sqrt{x^2+36}=14\sqrt{x^2+36}=14-x\)
Squaring:
\(x^2+36=196-28x+x^2 28x=160 x=\frac{40}{7}\)
This is possible, so the perimeter can equal 28. But \(P\) and \(Q\) are only specified as different points, so \(x\) is not fixed. The perimeter can be less than, equal to, or greater than 28.
For example:
If \(x=1,\) perimeter \(\approx 14.6<28.\)
If \(x=10\), perimeter \(\approx 48.3>28.\)
Answer: D
Since \(PR=QR\), triangle \(PQR\) is isosceles, and the midpoint \(T\) of the base \(PQ\) makes \(RT\) perpendicular to \(PQ\).
So we have a right triangle with:
\(RT=6\)
Let \(PT=TQ=x\). Then:
\(PR=QR=\sqrt{x^2+6^2}=\sqrt{x^2+36}\)
Thus the perimeter is
\(2x+2\sqrt{x^2+36}\)
We need to compare this with 28. Set the perimeter equal to 28:
\(2x+2\sqrt{x^2+36}=28 x+\sqrt{x^2+36}=14\sqrt{x^2+36}=14-x\)
Squaring:
\(x^2+36=196-28x+x^2 28x=160 x=\frac{40}{7}\)
This is possible, so the perimeter can equal 28. But \(P\) and \(Q\) are only specified as different points, so \(x\) is not fixed. The perimeter can be less than, equal to, or greater than 28.
For example:
If \(x=1,\) perimeter \(\approx 14.6<28.\)
If \(x=10\), perimeter \(\approx 48.3>28.\)
Answer: D
gmatclubot
Re: P and Q are two different points on line l [#permalink]
04 Sep 2026, 10:54
Moderators:
