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

Redeem
TTP OnDemand GMAT Video Course
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
Start Free 5-Day Trial Read 559 Reviews →

gmat prep question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member
Location: Silicon valley, California
Thanked: 30 times

Re: gmat prep question

by jayhawk2001 » Sun Apr 15, 2007 8:42 pm
yvonne12 wrote:If Xand Y are positive, which of the following would be greater than 1/ sqrt x+y?

1. Sqrt x+y/2x

2. sqrt x + sqrt y/ x+y

3. Sqrt X - sqrt y /x+y
I think this is best solved by subst values for x and y.

Take x = 1, y = 8

1/sqrt x+y = 1/3 which is < 1

1 - sqrt (9/2) > 1

2 - sqrt (1) + sqrt (8/9) > 1

3 - sqrt (1) - sqrt (8/9) = 1 - 2*1.4/3 = .2/3
This is less than 1/3


I think 1 and 2 alone satisfy the conditions.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member
Thanked: 1 times

by f2001290 » Mon May 28, 2007 9:07 am
Jay

What is the difference between

"Which of them would be greater" and "Which of them must be greater"
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member
Location: Silicon valley, California
Thanked: 30 times

by jayhawk2001 » Mon May 28, 2007 5:40 pm
f2001290 wrote:Jay

What is the difference between

"Which of them would be greater" and "Which of them must be greater"
In the context of the question, I guess "would be" and "must be"
have the same meaning.

In general "must be" is a condition that should ALWAYS be true.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member
Thanked: 1 times

by f2001290 » Tue May 29, 2007 8:24 am
Jay

OA for this question is 2. Refer the attachment. I got it from some other forum.
Attachments
1.JPG
1.JPG
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member
Location: Silicon valley, California
Thanked: 30 times

by jayhawk2001 » Tue May 29, 2007 12:13 pm
f2001290 wrote:Jay

OA for this question is 2. Refer the attachment. I got it from some other forum.
Since we are asked prove "must", we have to ensure that the condition
holds true for all values.

Looking at I, II and III, you can infer that calculations will involve a lot
of root.

To simplify, Take a pythogorean triplet, x=64, y=36, so that x+y = 100

I and III immediately fall apart. You can eliminate B, D and E.

To evaluate II,

sqrt-x + sqrt-y / (x + y)
= (sqrt-x + sqrt-y) / sqrt (x+y)*sqrt(x+y) ..... eqn-a

To see if II is greater than 1/sqrt(x+y), we have to prove

sqrt-x + sqrt-y / sqrt(x+y) > 1 i.e. numerator of eqn-a should be > 1

i.e. prove sqrt-x + sqrt-y > sqrt(x+y)

Since we are dealing with positive values for x and y, we know
the above will hold true. Hence sufficient.

So, II it is.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member
Thanked: 1 times

by f2001290 » Tue Jun 05, 2007 11:10 am
Is there any methodical approach to this problem.

At times, I feel that substituting values is a tedious job and there is ample scope for getting such inequality problems wrong.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1