Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 05 Oct 2008
Posts: 216

If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
Updated on: 15 Oct 2019, 04:48
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
Official Answer and Stats are available only to registered users. Register/ Login.
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




Math Expert
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
13 Oct 2009, 14:30
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.
_________________




Senior Manager
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
10 Oct 2010, 16:33
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.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog




SVP
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
13 Oct 2009, 12:32
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 \(\frac{1}{sqrt{xy}}\) and asked which of the following I, II, or III MUST be larger than \(\frac{1}{sqrt{xy}}\). 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
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
13 Oct 2009, 14:39
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
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2479
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
13 Oct 2009, 14:53
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 fightMoney Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership  big benefits and savingsGmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1822
Concentration: General Management, Nonprofit

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
23 Nov 2010, 23:27
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
Joined: 16 Jan 2012
Posts: 4

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Joined: 26 Nov 2011
Posts: 13

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Joined: 08 Mar 2013
Posts: 18

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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 crossmultiplication 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
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
11 Sep 2013, 02:32
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
Joined: 06 Sep 2013
Posts: 1553
Concentration: Finance

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Status: It's time...
Joined: 29 Jun 2015
Posts: 7
Location: India
Concentration: Operations, General Management
GPA: 3.59

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
20 Jun 2016, 05:30
oloz wrote: Hi Bunuel, Can you please give me the link to a set of these type of problems to practice?
Thanks in Advance. Check must or could be questions here: search.php?search_id=tag&tag_id=193
_________________



Manager
Joined: 14 Jun 2016
Posts: 70
Location: India
WE: Engineering (Manufacturing)

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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 negativebut 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 friendsneed few more to reach the next level. Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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 negativebut 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
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
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
Joined: 02 Sep 2009
Posts: 59039

Re: If x and y are positive, which of the following must be greater than
[#permalink]
Show Tags
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.
_________________




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 ]



