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.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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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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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Find out what's new at GMAT Club - latest features and updates

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.

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.

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.

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.

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!

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?

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!