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.
If 0<p<q<1
[#permalink]
05 Aug 2026, 03:24
05 Aug 2026, 03:24
Expert Reply
00:00
Question Stats:
88% (01:30) correct
11% (00:31) wrong based on 9 sessions
Hide Show timer Statistics
If \(0<p<q<1 \)
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 |
\(\frac{1-p}{1+q}\) |
\(\frac{1-q}{1+p}\) |
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: 2942
Given Kudos: 1287
GPA: 3.39
Re: If 0<p<q<1
[#permalink]
07 Aug 2026, 12:06
07 Aug 2026, 12:06
2
Expert Reply
Solution
Compare
\(\frac{1-p}{1+q}\) and \(\frac{1-q}{1+p}\).
Since all denominators are positive, cross-multiply:
\((1-p)(1+p)\) vs.\((1-q)(1+q)\).
Using \((1-a)(1+a)=1-a^2\),
\(1-p^2\) vs. \(1-q^2\).
Given \(0<p<q,\)
\(p^2<q^2 ⇒1-p^2>1-q^2\).
Therefore,
\(\frac{1-p}{1+q}>\frac{1-q}{1+p}\).
Answer: A
Compare
\(\frac{1-p}{1+q}\) and \(\frac{1-q}{1+p}\).
Since all denominators are positive, cross-multiply:
\((1-p)(1+p)\) vs.\((1-q)(1+q)\).
Using \((1-a)(1+a)=1-a^2\),
\(1-p^2\) vs. \(1-q^2\).
Given \(0<p<q,\)
\(p^2<q^2 ⇒1-p^2>1-q^2\).
Therefore,
\(\frac{1-p}{1+q}>\frac{1-q}{1+p}\).
Answer: A
gmatclubot
Re: If 0<p<q<1 [#permalink]
07 Aug 2026, 12:06
Moderators:
