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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
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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: 2928
Given Kudos: 1257
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
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