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

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If x and y are positive, which of the following must be [#permalink]

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13 Oct 2009, 11:10
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Feb 2012, 00:43, edited 1 time in total.
Edited the question and added the OA
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13 Oct 2009, 13:30
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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

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

Hope it's clear.
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13 Oct 2009, 13:39
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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.
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13 Oct 2009, 13:53
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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
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Re: If x and y are positive, which of the following must be [#permalink]

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23 Jun 2016, 14:08
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Bunuel wrote:
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.

Isn't it the other way around?

1/2 < 1/sqr(2) ?
0.5 < 0.71

Thanks!
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04 Aug 2010, 12:04
Hey Bunuel,

For this problem, if we were to test by picking numbers then ideally we should check with integers as well as with fractions...right?? Because the problem does not say that we have to stick with integers only.
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Re: PS: If x and y are positive, [#permalink]

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23 Nov 2010, 22:27
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.
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Re: PS: If x and y are positive, [#permalink]

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23 Nov 2010, 23:00
How do you handle for fractions, since fractions are still positive??
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Re: PS: If x and y are positive, [#permalink]

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23 Nov 2010, 23:05
What do you mean for fractions?

The question asks about the options that MUST be greater. So even if you can find one positive x or y for which the fraction in the answer is lesser than the fraction given, your answer choice is ruled out.
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Re: PS: If x and y are positive, [#permalink]

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23 Nov 2010, 23:09
Since x is positive x can be 1/2 and same goes for y. Y can be 2/3 or some other fraction since the questionb does say x and y are integers
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Re: PS: If x and y are positive, [#permalink]

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23 Nov 2010, 23:17
Ah, good point. Are you sure that the question doesn't say it's an integer? I didn't consider that possibility, for option II. There must be some way to prove that it works even for fractions.
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30 Apr 2011, 19:53
Let x = 2, y = 2

1/sqrt(2+2) = 1/sqrt(4) = 1/2

(sqrt(x+y))/2x = sqrt(4)/4 = 1/2

(sqrtx + sqrty)/(x+y) = 2sqrt(2)/4 = sqrt(2)/2 = 1/sqrt(2) > 1/2

sqrt(2) - sqrt(2)/(2+2) = 0

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Re: If x and y are positive, which of the following must be [#permalink]

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14 Mar 2012, 08:07
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

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Re: If x and y are positive, which of the following must be [#permalink]

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14 Mar 2012, 08:18
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

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

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02 Jul 2012, 03:09
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?
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If x and y are positive, which of the following must be [#permalink]

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02 Jul 2012, 04:59
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.
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03 May 2013, 19:22
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

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!
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Re: If x and y are positive, which of the following must be [#permalink]

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04 Jul 2013, 18:43
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

plugin values

put x = 3 and y = 1, u may also try out x = 2 and y = 2 but we get zero in one choice, which may again hold one's horses with doubt,

once u substitute we get only c as our answer and lastly remember we are looking for choices which are greater than but not equal to
Re: If x and y are positive, which of the following must be   [#permalink] 04 Jul 2013, 18:43

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