BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Students Height

Expert replies
by BlindVision » Tue Jun 16, 2009 4:29 pm
For the students in class A, the range of their heights is r centimeters and the greatest height is g centimeters. For the students in class B, the range of their heights is s centimeters and the greatest height is h centimeters. Is the least height of the students in class A greater than the least height of the students in class B?


1) r < s

2) g > h


OA = C
Life is a Test
Join the discussion
Source: — Data Sufficiency |

by mehravikas » Tue Jun 16, 2009 7:08 pm
You can pick number for these problems:

Start from statement 2: g > h, no other information is given about r, s and least height.

Answer can be: A, C, E

Statement 1: r < s, let r = 8, s = 10
r = 14 - 6
s = 15 - 5 or 17 - 7

Not sufficient.

combine both the statements and pick numbers satisfying both the answer choices.
r = 8, s = 10, g = 14, h = 12
r = 14 - x1
s = 12 - x2

we know that r < s therefore, x1 is greater than x2
Join the discussion

by BlindVision » Tue Jun 16, 2009 7:53 pm
Thanks for the help but I got lost on the combining part. Can you please help me better understand that last part?

P.S. When I combine, I get: r < s < h < g . Is that right?

P.S.S. Maybe I'm not understanding the question stem: Is it asking 'Which of the two classes has the most shortest students'?
Life is a Test
Join the discussion

by tohellandback » Tue Jun 16, 2009 8:18 pm
BlindVision wrote:Thanks for the help but I got lost on the combining part. Can you please help me better understand that last part?

P.S. When I combine, I get: r < s < h < g . Is that right?

P.S.S. Maybe I'm not understanding the question stem: Is it asking 'Which of the two classes has the most shortest students'?
you already have got the individual parts. so lets see what happens when we combine both
plug in numbers
let r=15
g=20
so least height in class A is 5
now plug in numbers for B
r<s so lets say s is 16
g>h
so lets say h=19
lest height in B is 3 i.e less than 5
The powers of two are bloody impolite!!
Join the discussion

by aj5105 » Tue Jun 16, 2009 11:58 pm
Combining both, least element in ClassB has to be less than ClassA.

(C)
Join the discussion

by aj5105 » Wed Jun 17, 2009 12:03 am
Ian's post:

dragonxiam- when you chose numbers here, you made r larger than s, when it should be smaller than s, which is why you've arrived at the wrong answer.

Neither statement is sufficient on its own- clearly we need to know about r, s, g and h here. Together we know:

s > r
g > h

These inequalities are in the same direction, so we can add them:

s+g > r+h
g-r > h-s

So we've shown that the smallest height in class A, which is g-r, is larger than the smallest height in class B, which is h-s. C.
Join the discussion