BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Prep 13
Source: Beat The GMAT — Data Sufficiency |
-
jackcrystal
- Senior | Next Rank: 100 Posts
- Posts: 31
- Thanked: 1 times
X-Y > -2 => x>-2 +Y
X-2y<-6 => x<-6+2Y => -x> 6-2Y
Add them 0 >4-Y => y>4
x>-2 +Y = > X must be positive since Y >4
so C
X-2y<-6 => x<-6+2Y => -x> 6-2Y
Add them 0 >4-Y => y>4
x>-2 +Y = > X must be positive since Y >4
so C
Last edited by Tryingmybest on Thu Nov 20, 2008 8:33 am, edited 1 time in total.
C
explanation:
for xy>0 you need both x,y with same sign (i.e. both positive or both negative).
(1) x-y > -2 ==> x+2>y
this doesn't lead to x,y with same sign, so A and D are out.
(2) x-2y<-6 ==> x+6<2y
again, doesn't lead to x,y with same sign, so B is out.
need to eliminate C, so let's check both statements together.
start with (2):
x-2y<-6
x+6<2y
x+2+4<2y
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<2y
2<y
so you have:
y>2
and
x+2>y
x+2>y>2
x+2>2
x>0
so both x, y are positive, thus xy>0 .
explanation:
for xy>0 you need both x,y with same sign (i.e. both positive or both negative).
(1) x-y > -2 ==> x+2>y
this doesn't lead to x,y with same sign, so A and D are out.
(2) x-2y<-6 ==> x+6<2y
again, doesn't lead to x,y with same sign, so B is out.
need to eliminate C, so let's check both statements together.
start with (2):
x-2y<-6
x+6<2y
x+2+4<2y
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<2y
2<y
so you have:
y>2
and
x+2>y
x+2>y>2
x+2>2
x>0
so both x, y are positive, thus xy>0 .
Dilbert, check your calculations. Be careful with inserting one ineq into another. According your solutiondilbert wrote:C
so you have:
y>2
x>0
so both x, y are positive, thus xy>0 .
y>2 and x>0 BUT y=3 and x=1 WILL NOT satisfy both of equations.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
logitech -
thanks for pointing this out!
indeed, I had a mistake in my calculations. here is the corrected one:
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<y <== this is corrected line, earlier I had mistake here
so you have:
y>4
and
x+2>y
x+2>y>4
x+2>4
x>2
so the correct solution for the two ineq is
x>2
y>4
and answer is C .
we see from the graph logitech attached that the intersection is (2,4), so x>2;y>4 is the area that solves both ineq.
again - thank you logitech for paying attention to my fault.
this time I was lucky to get C with a mistake, but I won't be that lucky in the real test...
thanks for pointing this out!
indeed, I had a mistake in my calculations. here is the corrected one:
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<y <== this is corrected line, earlier I had mistake here
so you have:
y>4
and
x+2>y
x+2>y>4
x+2>4
x>2
so the correct solution for the two ineq is
x>2
y>4
and answer is C .
we see from the graph logitech attached that the intersection is (2,4), so x>2;y>4 is the area that solves both ineq.
again - thank you logitech for paying attention to my fault.
this time I was lucky to get C with a mistake, but I won't be that lucky in the real test...
