text: Find all School-related info fast with the new School-Specific MBA Forum
  
   It is currently 05 May 2015, 23:51
  
  GMAT 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
  
  we will pick new questions that match your level based on your Timer History
  
  Track
  
  every week, we’ll send you an estimated GMAT score based on your performance
  
  Practice
  Pays
  
  we will pick new questions that match your level based on your Timer History
  
  Events & Promotions
  
  Events & Promotions in June
  Open Detailed Calendar
  
  If x > 1 and y > 1 , is x > y ? 1. \sqrt{x} > y
  
  Author Message
  TAGS:
  VP
  Joined: 30 Jun 2008
  Posts: 1048
  Followers: 11
  
  Kudos [?]: 348 [0], given: 1
  
  If x > 1 and y > 1 , is x > y ? 1. \sqrt{x} > y [#permalink]  30 Sep 2008, 09:02
  00:00
  
  Difficulty:
  
  (N/A)
  
  Question Stats:
  
  0% (00:00) correct 0% (00:00) wrong based on 0 sessions
  If $$x > 1$$ and $$y > 1$$ , is $$x > y$$ ?
  
  1. $$\sqrt{x} > y$$
  2. $$\sqrt{y} < x$$
  _________________
  
  "You have to find it. No one else can find it for you." - Bjorn Borg
  
  SVP
  Joined: 17 Jun 2008
  Posts: 1573
  Followers: 12
  
  Kudos [?]: 202 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 09:15
  A for me.
  
  Since, both x and y are greater than 1, hence if sqrt(x) > y then x will definitely be greater than y.
  
  stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.
  VP
  Joined: 30 Jun 2008
  Posts: 1048
  Followers: 11
  
  Kudos [?]: 348 [0], given: 1
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 09:21
  scthakur wrote:
  A for me.
  
  stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.
  
  I don't seem to understand this a bit more elaborate explanation will help
  _________________
  
  "You have to find it. No one else can find it for you." - Bjorn Borg
  
  SVP
  Joined: 17 Jun 2008
  Posts: 1573
  Followers: 12
  
  Kudos [?]: 202 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 09:27
  amitdgr wrote:
  scthakur wrote:
  A for me.
  
  stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.
  
  I don't seem to understand this a bit more elaborate explanation will help
  
  if x is between 0 and 1 then sqrt(x) will be greater than x. However, if x > 1 then sqrt(x) will always be smaller than x. We can check this with any real numbers.
  
  With this logic, if sqrt(x) > y and since x > sqrt(x) as x > 1, hence x > y.
  VP
  Joined: 30 Jun 2008
  Posts: 1048
  Followers: 11
  
  Kudos [?]: 348 [0], given: 1
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 09:46
  How do you rule out the second statement ? I understood the first statement in your first post I had asked explanation for the second one alone.
  
  Thanks
  _________________
  
  "You have to find it. No one else can find it for you." - Bjorn Borg
  
  Director
  Joined: 23 May 2008
  Posts: 840
  Followers: 3
  
  Kudos [?]: 37 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 10:11
  I get A too
  
  since x and y are greater than 1
  
  1) sqrt x>y
  x>y^2
  x has to be bigger than y since x is bigger then y*y
  
  2) sqrt y<x
  y<x^2
  x^2 > y
  in this case x is not neccessarily bigger than y
  
  for instance x=2, y=3 and x^2=4, y=3, true in this case
  but x=3, y=3 means x^2=9. x^2 is bigger than y, but x is not bigger than y so false in this case
  Director
  Joined: 27 Jun 2008
  Posts: 547
  WE 1: Investment Banking - 6yrs
  Followers: 1
  
  Kudos [?]: 48 [0], given: 92
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 10:16
  A for me - same explanation as bigtreezl.
  SVP
  Joined: 17 Jun 2008
  Posts: 1573
  Followers: 12
  
  Kudos [?]: 202 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 10:34
  amitdgr wrote:
  How do you rule out the second statement ? I understood the first statement in your first post I had asked explanation for the second one alone.
  
  Thanks
  
  Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x.
  Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x.
  Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.
  
  As a rule, if y > 1 then y > sqrt(y)
  In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.
  Manager
  Joined: 30 Sep 2008
  Posts: 111
  Followers: 1
  
  Kudos [?]: 10 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  30 Sep 2008, 23:57
  A to me
  
  1. x > 1 => x > sqrt(x) > y => x > y
  
  2. insufficent
  Manager
  Joined: 30 Sep 2008
  Posts: 111
  Followers: 1
  
  Kudos [?]: 10 [0], given: 0
  
  Re: DS: m01#5 [#permalink]  01 Oct 2008, 00:00
  scthakur wrote:
  amitdgr wrote:
  How do you rule out the second statement ? I understood the first statement in your first post I had asked explanation for the second one alone.
  
  Thanks
  
  Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x.
  Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x.
  Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.
  
  As a rule, if y > 1 then y > sqrt(y)
  In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.
  
  yes, examples in this case is the easiest & fastest way to rule out. Try 9 for y and 4 for x
  Re: DS: m01#5   [#permalink] 01 Oct 2008, 00:00
  Similar topics Replies Last post
  Similar
  Topics:
  Is x > y? 1) sqrt(x) > y 2) x^3 > y 8 29 Jun 2008, 14:09
  If x and y are positive, is x*x*x > y? (1) sqrt(x) > y 4 20 Jun 2007, 17:42
  Is x > y? (1) sqrt(x) > y (2) x^3 > y 2 13 Nov 2006, 00:25
  Is x > y? (1) sqrt(x) > y (2) x^3 > y 1 01 Sep 2006, 07:23
  Is x > y? (1) sqrt(x) > y (2) x^3 > y 1 04 Aug 2006, 11:04
  Display posts from previous: Sort by