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.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
Which of the following inequalities, if true, is sufficient
[#permalink]
20 Mar 2020, 23:23
20 Mar 2020, 23:23
Expert Reply
00:00
Question Stats:
0% (00:00) correct
100% (00:42) wrong based on 2 sessions
Hide Show timer Statistics
Which of the following inequalities, if true, is sufficient alone to show that \(\sqrt[3]{x} < \sqrt[5]{x}\) ?
(A) \(-1 < x < 0\)
(B) \(x > 1\)
(C) \(|x| < - 1\)
(D) \(|x| > 1\)
(E) \(x < -1\)
(A) \(-1 < x < 0\)
(B) \(x > 1\)
(C) \(|x| < - 1\)
(D) \(|x| > 1\)
(E) \(x < -1\)
Re: Which of the following inequalities, if true, is sufficient
[#permalink]
08 May 2020, 13:22
08 May 2020, 13:22
1
Expert Reply
Raising to the odd power won't change the sign of inequality.
\([\sqrt[3]{x}]^{15} < [\sqrt[5]{x}]^{15}\)
\(x^5 < x^3\)
\(x^3(1-x^2) <0\)
case 1
if x<0, then \(1 - x^2>0\)
x<-1
case 2
if x >0, then \(1-x^2<0\)
0<x<1
E is the answer
\([\sqrt[3]{x}]^{15} < [\sqrt[5]{x}]^{15}\)
\(x^5 < x^3\)
\(x^3(1-x^2) <0\)
case 1
if x<0, then \(1 - x^2>0\)
x<-1
case 2
if x >0, then \(1-x^2<0\)
0<x<1
E is the answer
gmatclubot
Re: Which of the following inequalities, if true, is sufficient [#permalink]
08 May 2020, 13:22
Moderators:
