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.
Apr 24Get an Extra 30% Off Target Test Prep Plans
08:00 PM EDT
-11:59 PM EDT
Grab 30% off any Target Test Prep GRE plan during our Flash Sale and join Brian and many other students who have used Target Test Prep to score high on the GRE. Just enter the coupon code FLASH30 at checkout to save big.
Apr 26Get a Top GRE Score with Target Test Prep
12:00 PM EDT
-01:00 PM EDT
The Target Test Prep course represents a quantum leap forward in GRE preparation, a radical reinterpretation of the way that students should study. In fact, more and more Target Test Prep students are scoring 329 and above on the GRE.
Real numbers x, y, and z satisfy the inequalities
[#permalink]
28 Jan 2022, 07:20
28 Jan 2022, 07:20
Expert Reply
00:00
Question Stats:
85% (01:55) correct
14% (04:05) wrong based on 7 sessions
Hide Show timer Statistics
Real numbers x, y, and z satisfy the inequalities \(0<x<1\), \(-1<y<0\), and \(1<z<2\). Which of the following numbers is necessarily positive?
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Real numbers x, y, and z satisfy the inequalities
[#permalink]
28 Jan 2022, 08:10
28 Jan 2022, 08:10
Carcass wrote:
Real numbers x, y, and z satisfy the inequalities \(0<x<1\), \(-1<y<0\), and \(1<z<2\). Which of the following numbers is necessarily positive?
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
Given:
\(-1<y<0\)
\(1<z<2\)
Since the inequality symbols are facing the same direction, we can add the two inequalities to get: \(0<y+z<2\)
So, it must be the case that \(y+z\) is positive.
Answer: E
Re: Real numbers x, y, and z satisfy the inequalities
[#permalink]
17 Feb 2023, 09:11
17 Feb 2023, 09:11
Hi Brent GreenlightTestPrep
How about the x in the inequalities 0<x<1 ?
How do we know only to use the other two inequalities? Could you help clarify? Thanks Brent
How about the x in the inequalities 0<x<1 ?
How do we know only to use the other two inequalities? Could you help clarify? Thanks Brent
