GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 14 Nov 2019, 14:02

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If x and y are positive, which of the following must be greater than

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 05 Oct 2008
Posts: 216
If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post Updated on: 15 Oct 2019, 04:48
7
76
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

56% (02:12) correct 44% (02:14) wrong based on 1168 sessions

HideShow timer Statistics

If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

I. \(\frac{\sqrt{x+y}}{2x}\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) I only
(C) II only
(D) I and III only
(E) II and III only

Originally posted by study on 13 Oct 2009, 12:10.
Last edited by Bunuel on 15 Oct 2019, 04:48, edited 2 times in total.
Edited the question and added the OA
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 13 Oct 2009, 14:30
39
29
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

1. \(\frac{\sqrt{x+y}}{2x}\)

2. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

3. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only

First of all \(\frac{1}{\sqrt{x+y}}\) is always positive. This by the way eliminates option III right away as \(\sqrt{x}-\sqrt{y}\) (numereator) may or may not be positive, so we should concentrate on I and II

Next:

\(\sqrt{x}+\sqrt{y}\) is always great than \(\sqrt{x+y}\) (well in fact if both \(x\) and \(y\) are 0, they are equal but it's not the case as given that \(x\) and \(y\) are positive). To check this: square them \((\sqrt{x}+\sqrt{y})^2=x+2\sqrt{xy}+y>x+y=\sqrt{x+y}^2\)

Let's proceed:

SOLUTION #1
\(\frac{1}{\sqrt{x+y}}=\frac{\sqrt{x+y}}{x+y}\)

I. \(\frac{\sqrt{x+y}}{2x}\) --> nominators are the same, obviously denominator \(2x\) may or may not be greater than \(x+y\). OUT.

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) --> denominators are the same and nominator \(\sqrt{x}+\sqrt{y}\) (as we've already discussed above) is always greater than \(\sqrt{x+y}\). OK

III. Well we can not even consider this one as our expression \(\frac{1}{\sqrt{x+y}}\) is always positive and the \(\sqrt{x}-\sqrt{y}\) (numerator) can be negative. OUT

Answer C.


SOLUTION #2
The method called cross multiplication:
Suppose we want to know which positive fraction is greater \(\frac{9}{11}\) or \(\frac{13}{15}\): crossmultiply \(9*15=135\) and \(11*13=143\) --> \(135<143\) which fraction gave us numerator for bigger value 143? \(\frac{13}{15}\)! Thus \(\frac{13}{15}>\frac{9}{11}\).

Lets do the same with our problem:
I. \(\frac{\sqrt{x+y}}{2x}\) and \(\frac{1}{\sqrt{x+y}}\) --> \(\sqrt{x+y}*\sqrt{x+y}=x+y\) and \(2x*1=2x\). \(x+y\) may or may not be greater than \(2x\). OUT

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) and \(\frac{1}{\sqrt{x+y}}\) --> \((\sqrt{x}+\sqrt{y})(\sqrt{x+y})\) and \(x+y\). Divide both sides by \(\sqrt{x+y}\) --> \(\sqrt{x}+\sqrt{y}\) and \(\sqrt{x+y}\). We know that \(\sqrt{x}+\sqrt{y}\) is always greater, which one gave the numerator for it: \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\), so \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) is always greater than \(\frac{1}{\sqrt{x+y}}\). OK

III. Well we can not even consider this one as our expression \(\frac{1}{\sqrt{x+y}}\) is always positive and the \(\sqrt{x}-\sqrt{y}\) (numerator) can be negative. OUT

Answer C.

Hope it's clear.
_________________
Most Helpful Community Reply
Senior Manager
Senior Manager
User avatar
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 327
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 10 Oct 2010, 16:33
5
TehJay wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

I. \(\frac{\sqrt{x+y}}{2x}\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) I only
(C) II only
(D) I and III
(E) II and III


Since this question is a must be true type. If we can find even one scenario wherein the condition does not hold for either I, II or III we can eliminate that choice.

Picking numbers as x=1 and y=1, we can see that only the II option satisfies the condition.

\(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) > \(\frac{1}{\sqrt{x+y}}\)
Since \(1>\frac{1}{\sqrt{2}}\). Answer is C.
_________________
:good Support GMAT Club by putting a GMAT Club badge on your blog :thanks
General Discussion
SVP
SVP
User avatar
B
Joined: 30 Apr 2008
Posts: 1593
Location: Oklahoma City
Schools: Hard Knocks
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 13 Oct 2009, 12:32
1
1
C

The easiest way to solve this problem is by picking some numbers for x and y and then solving and comparing. The problem you really have is trying to compare with radicals in the solutions which are not easy to compare.

First we're given \(\frac{1}{sqrt{x-y}}\) and asked which of the following I, II, or III MUST be larger than \(\frac{1}{sqrt{x-y}}\).

I. \(\frac{sqrt{x+y}}{2x}\)

II. \(\frac{sqrt{x} + sqrt{y}}{x+y}\)

III. \(\frac{sqrt{x} - sqrt{y}}{x+y}\)

For my numbers, I chose X = 3 and y = 1

You get \(\frac{1}{sqrt{3+1}} = \frac{1}{2}\)

Then for I, II, and III you get:

I. => \(\frac{sqrt{3+1}}{6} = \frac{2}{6} = \frac{1}{3}\)

II. => \(\frac{sqrt{2} + sqrt{1}}{3 + 1} = \frac{sqrt{2} + 1}{4}\)

III. => \(\frac{sqrt{2} - sqrt{1}}{3 + 1} = \frac{sqrt{2} + 1}{4}\)

Now we have to evaluate these numbers.

I. 1/3 is smaller than 1/2, so it does not satisfy the question of which MUST be larger.

II. Square root of 2 + 1 over 4. Even without knowing what the exact number is, we know \(\sqrt{2}\) is over 1, so add 1 to that and we get something larger than 2 over 4, which will be greater than 1/2. Can't rules out II as the answer yet. Due to the answer choices, we know the answer should be either C or E.

III. Sqrt of 2 minus 1 over 4 we know will be something under 2 over 4, so that's less than half. III cannot work either.

Something to keep in mind is that the question stem does not rule out fractions for possible values of X and Y, but due to the answers, we are able to eliminate III as a possibility and that leaves only C for the answer.

If you got to a point where both II and III was possible, you would need to also pick some fractions for values of X and Y and evaluate. It's not a really quick way to do this, but it will work and also remember that doing these problems when you're used to them is much faster than reading one of my explanations on how to do it.
_________________
------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings
SVP
SVP
User avatar
B
Joined: 30 Apr 2008
Posts: 1593
Location: Oklahoma City
Schools: Hard Knocks
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 13 Oct 2009, 14:39
4
Those are good methods and work well for someone versed in the theories and properties, but most people taking the GMAT (and those that read these forums that do not ever post) are not ones that know or even care to know the ins and outs of deep theories. The number picking on the GMAT works well, is easy, and you just have to understand what the MUST BE TRUE means, that if there is a situation where the option is greater than 2, that does not mean that it always will be. You have to consider numerous items.

All I"m saying is that for the majority of people, spending 2 min plugging in numbers will keep them better focused, working towards a solution, and on track for the rest of the GMAT. That's the most important part.
_________________
------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings
SVP
SVP
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2479
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 13 Oct 2009, 14:53
1
In exams they donot expect us to spend much time and this is a tricky question.

DONOT use number method as your first attempt.

Solution:

1/sqrt(x+y) = sqrt(x+y)/x+y // multiply den and numerator by sqrt(x+y)

For 1st option we cannot say anything.
Now for 2nd and third denominators are same that means we need to consider only numerator,and rem they have stated A n B both r positive.

now its obvious that sqrt x + sqrt y > sqrt (x+y) > sqrt x - sqrt y
( use the property.... (a+b)^2 = a^2 + b^2 + 2ab )

If any doubts pls letme know.This question shouldnt take more than 1 min if you just concentrate on basic knwledge
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1822
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 23 Nov 2010, 23:27
1
vrajesh wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}?\)

I. \(\frac{\sqrt{x+y}}{2x}\)

II.\(\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x+y}}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x+y}}\)

A. None
B. I only
C. II only
D. I and III
E. II and III

Can someone please explain what is the best way to solve this problem in under 2 minutes?

I would really like to understand how to solve these type of problems.

What makes this problem especially difficult, is the condition x and y are positive
i.e. x > 0 and y > 0

hence x and y can be any of the following 1/2, 3/4, 1, 2, 3, 4, ....., and more


Okay, so the main fraction can be simplified to the following by means of rationalization.

\(\frac{1}{\sqrt{x+y}}\) = \(\frac{\sqrt{x+y}}{x+y}\)

Now the question asks for something that MUST be true for all the values of x and y, which are positive.

So look at the options:

I: Here, your denominator is 2x. If x+y > 2x, then this option is greater than the given number. If not, it's smaller. So this can't be the answer.

II: This always has to be greater. In our simplified form, our numerator is 1, and here the numerator is \(\sqrt{x} + \sqrt{y}\) which has to be greater than 1, since the smallest possible value that x can take is 1 (Remember 0 is not a positive integer) - So this option is good.

III. \(\sqrt{x} - \sqrt{y}\). This can be greater than or lesser than one depending on the values that x and y takes, so this need not ALWAYS be greater than what's given to us. Hence incorrect.

Thus the final answer is C, just option II. Hope this helps.
Intern
Intern
avatar
Joined: 16 Jan 2012
Posts: 4
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 14 Mar 2012, 09:07
Picking numbers actually got me the wrong answer
x= 9
y=16

1/sqrt(x+y)== 1/5===0.20
1) sqrt(x+y)/2x = 5/18= 0.2777

This itself is wrong so ---- please help - GMAT TOMORROW!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 14 Mar 2012, 09:18
Ashamock wrote:
Picking numbers actually got me the wrong answer
x= 9
y=16

1/sqrt(x+y)== 1/5===0.20
1) sqrt(x+y)/2x = 5/18= 0.2777

This itself is wrong so ---- please help - GMAT TOMORROW!


The question asks which of the options MUST be greater than \(\frac{1}{\sqrt{x+y}}\), not COULD be greater than \(\frac{1}{\sqrt{x+y}}\). Hence one set of numbers showing that option (1) is greater is not enough to conclude that this option is ALWAYS greater (greater for all numbers).

Hope it's clear.
_________________
Intern
Intern
avatar
Joined: 26 Nov 2011
Posts: 13
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 02 Jul 2012, 04:09
Hi,

Please correct me where I am wrong.

I understood why I and III are false.

If {x,y} = {2,2} then 1/(x+y)^1/2 = 1/2 = 0.5

while II will come out to be 1/2*2^1/2 = 0.357

Here II is not greater than the given expression.


So, None should be an answer. isnt it?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 02 Jul 2012, 05:59
nishantmehra01 wrote:
Hi,

Please correct me where I am wrong.

I understood why I and III are false.

If {x,y} = {2,2} then 1/(x+y)^1/2 = 1/2 = 0.5

while II will come out to be 1/2*2^1/2 = 0.357

Here II is not greater than the given expression.


So, None should be an answer. isnt
it?



If \(x=y=2\), then:

\(\frac{1}{\sqrt{x+y}}=\frac{1}{2}\) and \(\frac{\sqrt{x}+\sqrt{y}}{x+y}=\frac{\sqrt{2}+\sqrt{2}}{2+2}=\frac{1}{\sqrt{2}}\) --> \(\frac{1}{2}<\frac{1}{\sqrt{2}}\).

Hope it's clear.
_________________
Intern
Intern
avatar
Joined: 08 Mar 2013
Posts: 18
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 03 May 2013, 20:22
Bunuel wrote:


SOLUTION #2
The method called cross multiplication:
Suppose we want to know which [b]positive
fraction is greater \(\frac{9}{11}\) or \(\frac{13}{15}\): crossmultiply \(9*15=135\) and \(11*13=143\) --> \(135<143\) which fraction gave us numerator for bigger value 143? \(\frac{13}{15}\)! Thus \(\frac{13}{15}>\frac{9}{11}\).

Lets do the same with our problem:
I. \(\frac{\sqrt{x+y}}{2x}\) and \(\frac{1}{\sqrt{x+y}}\) --> \(\sqrt{x+y}*\sqrt{x+y}=x+y\) and \(2x*1=2x\). \(x+y\) may or may not be greater than \(2x\). OUT

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) and \(\frac{1}{\sqrt{x+y}}\) --> \((\sqrt{x}+\sqrt{y})(\sqrt{x+y})\) and \(x+y\). Divide both sides by \(\sqrt{x+y}\) --> \(\sqrt{x}+\sqrt{y}\) and \(\sqrt{x+y}\). We know that \(\sqrt{x}+\sqrt{y}\) is always greater, which one gave the numerator for it: \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\), so \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) is always greater than \(\frac{1}{\sqrt{x+y}}\). OK

III. Well we can not even consider this one as our expression \(\frac{1}{\sqrt{x+y}}\) is always positive and the \(\sqrt{x}-\sqrt{y}\) (numerator) can be negative. OUT

Answer C.

Hope it's clear.


Major thanks for this one! I'm really bad with these types of problems, and hate plugging in numbers, but I'm very good with algebraic equations, and the cross-multiplication method just naturally makes a lot of sense to me. I think if I do a few dozen of these, there's no chance I'd go wrong on these types of questions on GMAT!
Manager
Manager
avatar
B
Joined: 29 Aug 2013
Posts: 69
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 11 Sep 2013, 02:32
1
study wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

1. \(\frac{\sqrt{x+y}}{2x}\)

2. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

3. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only


1 way to look at this problem is :-

I) Cross Multiplying LHS : x+y RHS 2x (So y can be anything greater than or less than x), hence ruled out
II) LHS : [square_root]x + [square_root]y = [square_root]x+y
Now if x were a^2, y were b^2 then LHS : a + b and RHS [square_root]a^2 + b^2 which is nothing but a right triangle with RHS as diagonal and a and b as sides of it.
And that the sum of the 2 sides of the triangle is always greater than the third side. Hence this will be true.
III) Similarly as with 2, this will become LHS : a - b and RHS : [square_root]a^2 + b^2. Any side will always be greater than the difference between the 2 sides. Hence this will be false.

Thus Answer will be C
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1553
Concentration: Finance
GMAT ToolKit User
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 05 Feb 2014, 09:49
I did the following, nothing too complicated

First statement clearly not enough since after cross multiplying we get is y>x? We don't know

For the second statement we have is (sqrt (x) + sqrt (y)(sqrt (x+y)> x+y

x + y = sqrt (x+y) ^2 so what we really need to know is if (sqrt (x) + sqrt (y) > sqrt (x+y) and yes, since sqrt (n) where n is any number is always more than 1 so >1 + >1 is always more than >1. Therefore this statement is true

For the last statement we get something a bit different but still
We have sqrt (x+y)(sqrt (x) - sqrt (y)) > x+y

Again x+y = sqrt (x+y)^2

Therefore question is ' Is sqrt (x) - sqrt (y) > sqrt (x+y)

This will never be the case as >1 - >1 will never be >1

Therefore our only correct answer choice is the second statement

Hope its clear
Cheers
J
Intern
Intern
User avatar
Status: It's time...
Joined: 29 Jun 2015
Posts: 7
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q48 V25
GPA: 3.59
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 20 Jun 2016, 05:25
Hi Bunuel,
Can you please give me the link to a set of these type of problems to practice?

Thanks in Advance.
_________________
Perseverance, the answer to all our woes. Amen.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 20 Jun 2016, 05:30
1
Manager
Manager
avatar
B
Joined: 14 Jun 2016
Posts: 70
Location: India
GMAT 1: 610 Q49 V21
WE: Engineering (Manufacturing)
Reviews Badge
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 29 Aug 2017, 20:24
Hi Bunuel,

For option 2, do we need to consider the negative values of of sqrt(x) and sqrt(y) as well?
If x is positive sqrt(x) can be positive or negative-but if we take the negative values then option 2 is not working.
For ex, I took x=16 & y = 9 & calculated putting the negative values but it didn't work.

Please guide me. Thanks.
_________________
Please help me with Kudos dear friends-need few more to reach the next level. Thanks. :)
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 29 Aug 2017, 22:02
buan15 wrote:
Hi Bunuel,

For option 2, do we need to consider the negative values of of sqrt(x) and sqrt(y) as well?
If x is positive sqrt(x) can be positive or negative-but if we take the negative values then option 2 is not working.
For ex, I took x=16 & y = 9 & calculated putting the negative values but it didn't work.

Please guide me. Thanks.


When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{16}=4\), NOT +4 or -4. Even roots have only a positive value on the GMAT.

In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
_________________
Manager
Manager
avatar
B
Joined: 06 Aug 2017
Posts: 78
GMAT 1: 570 Q50 V18
GMAT 2: 610 Q49 V24
GMAT 3: 640 Q48 V29
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 29 Aug 2017, 22:43
study wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

1. \(\frac{\sqrt{x+y}}{2x}\)

2. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

3. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only


Pick easy substitutions x=2 and y=2 and you will get the answer straight to C
_________________
-------------------------------------------------------------------------------
Kudos are the only way to tell whether my post is useful.

GMAT PREP 1: Q50 V34 - 700

Veritas Test 1: Q43 V34 - 630
Veritas Test 2: Q46 V30 - 620
Veritas Test 3: Q45 V29 - 610
Veritas Test 4: Q49 V30 - 650

GMAT PREP 2: Q50 V34 - 700

Veritas Test 5: Q47 V33 - 650
Veritas Test 5: Q46 V33 - 650
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59039
Re: If x and y are positive, which of the following must be greater than  [#permalink]

Show Tags

New post 29 Aug 2017, 23:23
Jabjagotabhisavera wrote:
study wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

1. \(\frac{\sqrt{x+y}}{2x}\)

2. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

3. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only


Pick easy substitutions x=2 and y=2 and you will get the answer straight to C


The question asks which of the options MUST be greater than \(\frac{1}{\sqrt{x+y}}\), not COULD be greater than \(\frac{1}{\sqrt{x+y}}\). Hence one set of numbers showing that option (2) is greater is not enough to conclude that this option is ALWAYS greater (greater for all numbers).

Hope it's clear.
_________________
GMAT Club Bot
Re: If x and y are positive, which of the following must be greater than   [#permalink] 29 Aug 2017, 23:23

Go to page    1   2    Next  [ 21 posts ] 

Display posts from previous: Sort by

If x and y are positive, which of the following must be greater than

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne