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 -1<x<0<y<1, which of the following inequalities must be tr
[#permalink]
01 Sep 2025, 02:11
01 Sep 2025, 02:11
Expert Reply
00:00
Question Stats:
28% (02:00) correct
71% (02:08) wrong based on 7 sessions
Hide Show timer Statistics
If $\(-1<x<0<y<1\)$, which of the following inequalities must be true?
Indicate all such inequalities.
(A) $\(x+y>0\)$
(B) $\(x y<0\)$
(C) $\(x^2+y^2>x+y\)$
(D) $\(2 x+x^2<y\)$
Indicate all such inequalities.
(A) $\(x+y>0\)$
(B) $\(x y<0\)$
(C) $\(x^2+y^2>x+y\)$
(D) $\(2 x+x^2<y\)$
- Part of the project: PrepClub Daily GRE Quant Questions: Arithmetic, Algebra, Geometry & DA
- Also replying to the unanswered questions
Re: If -1<x<0<y<1, which of the following inequalities must be tr
[#permalink]
02 Oct 2025, 12:27
02 Oct 2025, 12:27
why is C not correct?
Re: If -1<x<0<y<1, which of the following inequalities must be tr
[#permalink]
02 Oct 2025, 14:53
02 Oct 2025, 14:53
Expert Reply
Since $y$ is greater than $O$ but less than 1 , it is certainly the case that $\(y>y^2\)$. On the other hand, since $x$ is negative, it's also the case that $\(x<x^2\)$. .
So we can't know whether the inequality in this answer choice is true or not: it depends on whether the difference between $\(y\)$ and $\(y^2\)$ is greater than the difference between $\(x\)$ and $\(x^2\)$.
For example: if $x\(=-\frac{1}{2}\)$ and $\(y=\frac{3}{4}\)$, then $\(x^2+y^2=\frac{13}{16}>x+y=\frac{1}{4}\)$.
True.
But if $\(x=-\frac{1}{8}\)$ and $\(y=\frac{1}{2}\)$, then
$\(x^2+y^2=\frac{17}{64}<x+y=\frac{3}{8}\)$. False.
So we can't know whether the inequality in this answer choice is true or not: it depends on whether the difference between $\(y\)$ and $\(y^2\)$ is greater than the difference between $\(x\)$ and $\(x^2\)$.
For example: if $x\(=-\frac{1}{2}\)$ and $\(y=\frac{3}{4}\)$, then $\(x^2+y^2=\frac{13}{16}>x+y=\frac{1}{4}\)$.
True.
But if $\(x=-\frac{1}{8}\)$ and $\(y=\frac{1}{2}\)$, then
$\(x^2+y^2=\frac{17}{64}<x+y=\frac{3}{8}\)$. False.
