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