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Re: If x and y are positive, which of the following must be [#permalink]
08 Aug 2009, 20:38

1

This post received KUDOS

we have to keep in mind that as 'x' is +ve; it can be integer or fracrtion. my approach is to plug values in both situations (integer or fracrtion) and check the greater than condition....try to plug integers for sqrt sign & plug fraction for rest.

case I) [sqrt(x+y)]/2 say [sqrt(x+y)] is fractional value = 0.9 ( its not imporatant to assume x & y values as both eqns have [sqrt(x+y)] as common)

hence [sqrt(x+y)]/2 < 1/sqrt(x+y)..hence not greater

case II) [sqrt(x)+sqrt(y)]/2 [sqrt(64)+sqrt(36)]/2 = 50 1/sqrt(36+64) = 0.1

hence [sqrt(x+y)]/2 > 1/sqrt(x+y)..hence greater

case III) [sqrt(x)-sqrt(y)]/(x+y)[/b]

[sqrt(100)-sqrt(64)]/(164)= 36/164 ~ 0.2yz ( dont calculate) 1/sqrt(100+64) ~ we know that 13^2= 169 thus 1/13 = 0.07 hence value will be surely less than ...hence hence greater

Ans II & III _________________

Bhushan S. If you like my post....Consider it for Kudos

Re: If x and y are positive, which of the following must be [#permalink]
08 Aug 2009, 20:46

9

This post received KUDOS

Expert's post

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity. Now, let's see what do our options at x,y ---> 0.

I) \frac{\sqrt{x+y}}{2} goes to 0 at x,y ---> 0.

II) \frac{\sqrt{x}+\sqrt{y}}{2} goes to 0 at x,y ---> 0.

III) \frac{\sqrt{x}-sqrt{y}}{x+y} hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

Re: If x and y are positive, which of the following must be [#permalink]
10 Dec 2010, 19:26

walker wrote:

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity. Now, let's see what do our options at x,y ---> 0.

I) \frac{\sqrt{x+y}}{2} goes to 0 at x,y ---> 0.

II) \frac{\sqrt{x}+\sqrt{y}}{2} goes to 0 at x,y ---> 0.

III) \frac{\sqrt{x}-sqrt{y}}{x+y} hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

Can you explain how you used this in more detail, and perhaps give me some idea of how to apply this same principle going forward. Thanks a lot, i appreciate the help.

Re: If x and y are positive, which of the following must be [#permalink]
10 Dec 2010, 21:57

3

This post received KUDOS

Expert's post

I don't see what could I add to my explanation but I can say a couple of words about the approach.

Using extreme examples often helps clearly figure out what is wrong and it has simple explanation.

Let's say we have something like that: 0.01*x^4 + 34.2x^2 + 12x + 123 How can we assess the expression at x -> infinity? if x is very big, x^4 will essentially larger than all other members and ~ 0.01x^4 is a good approximation for the expression at large x. So, instead of a complex expression, we use simple one.

I remember that I saw a couple of problems in OG12 that could be solved using this approach in 10-20sec instead of 2 min.... For example, OG12 PS149. Assuming that x=0 (v should be 60) and x=100 (v should be 40), it takes up to 20 sec to figure out what option is correct without solving the problem.

Sometimes extreme examples make obvious contradictions in statements, for instance, in geometry, assuming angle is ~180 or 0 for a triangle
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Re: If x and y are positive, which of the following must be [#permalink]
10 Dec 2010, 23:39

3

This post received KUDOS

Expert's post

mmcooley33 wrote:

walker wrote:

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity. Now, let's see what do our options at x,y ---> 0.

I) \frac{\sqrt{x+y}}{2} goes to 0 at x,y ---> 0.

II) \frac{\sqrt{x}+\sqrt{y}}{2} goes to 0 at x,y ---> 0.

III) \frac{\sqrt{x}-sqrt{y}}{x+y} hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

Can you explain how you used this in more detail, and perhaps give me some idea of how to apply this same principle going forward. Thanks a lot, i appreciate the help.

Just to add couple of words to Walker's great solution:

Note that we are asked "which of the following MUST be greater than \frac{1}{\sqrt{x+y}}?" not COULD be greater.

"MUST BE TRUE" questions: These questions ask which of the following MUST be true (must be greater in our case), or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions: The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, if we find even one set of x and y for which \frac{1}{\sqrt{x+y}} is greater than option I for example then it'll mean that option I is not ALWAYS greater then \frac{1}{\sqrt{x+y}}.

How can we increase the value of \frac{1}{\sqrt{x+y}}? Testing extreme examples: if x and y are very small (when their values approach zero) then denominator approaches zero and thus the value of the fraction goes to +infinity, becomes very large. In this case:

I. \frac{\sqrt{x+y}}{2}: nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction; II. \frac{\sqrt{x}+\sqrt{y}}{2}: the same here - nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;

As for: III. \frac{\sqrt{x}-\sqrt{y}}{x+y}, if x=y then this fraction equals to zero and \frac{1}{\sqrt{x+y}} has some value more than zero, so this option also is not always more than given fraction;

Answer: none of the options must be greater than the given fraction.

Re: If x and y are positive, which of the following must be [#permalink]
13 Dec 2010, 20:35

Expert's post

walker wrote:

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity.

Nice approach... I think your first two lines given above are sufficient to answer the question. If the given expression \frac{1}{\sqrt{x+y}} approaches infinity for some values of x and y, there is no way any other expression can be always greater than this expression. Some other expression can equal this expression by approaching infinity for the same values of x and y (e.g. the third expression), but cannot be greater than this.

So the moment one realizes that \frac{1}{\sqrt{x+y}} tends to infinity, the answer is clear.
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Re: If x and y are positive, which of the following must be [#permalink]
14 Dec 2010, 06:01

Expert's post

walker wrote:

I think we still have to consider all options. For example, \frac{2}{\sqrt{x+y}} is always greater than \frac{1}{\sqrt{x+y}}

Only if it is given that (x+y) is not equal to 0. Since that is not mentioned in the question, when (x+y) is 0, both the expressions will be the same.
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Re: If x and y are positive, which of the following must be [#permalink]
23 Mar 2011, 23:39

1

This post received KUDOS

noboru 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}}{2} II. \frac{\sqrt{x}+\sqrt{y}}{2} III. \frac{\sqrt{x}-\sqrt{y}}{x+y}

walker wrote:

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity. Now, let's see what do our options at x,y ---> 0.

I) \frac{\sqrt{x+y}}{2} goes to 0 at x,y ---> 0.

II) \frac{\sqrt{x}+\sqrt{y}}{2} goes to 0 at x,y ---> 0.

III) \frac{\sqrt{x}-sqrt{y}}{x+y} hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

if x and y are very small values, then the inverse of their sum or sqrt{sum} will be very big

II. (sqrt{x}+sqrt{y})/2 = (0.0003+0.0004)/2 = 0.0007/2 = 0.00035 < 2000

III. sqrt{x}-sqrt{y}/(x+y)= -ve < 2000

the idea is; if x and y are very small values, then the inverse of their sum or sqrt{sum} will be very big

the inverse will be an even bigger number if we consider x=0.00000000000000000000000000000000000001 and y=0.0000000000000000000000001
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Re: If x and y are positive, which of the following must be [#permalink]
28 Mar 2011, 11:49

Bunuel wrote:

mmcooley33 wrote:

walker wrote:

Let's consider original statement: \frac{1}{\sqrt{x+y}}

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \frac{1}{\sqrt{x+y}} goes to infinity. Now, let's see what do our options at x,y ---> 0.

I) \frac{\sqrt{x+y}}{2} goes to 0 at x,y ---> 0.

II) \frac{\sqrt{x}+\sqrt{y}}{2} goes to 0 at x,y ---> 0.

III) \frac{\sqrt{x}-sqrt{y}}{x+y} hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

Can you explain how you used this in more detail, and perhaps give me some idea of how to apply this same principle going forward. Thanks a lot, i appreciate the help.

Just to add couple of words to Walker's great solution:

Note that we are asked "which of the following MUST be greater than \frac{1}{\sqrt{x+y}}?" not COULD be greater.

"MUST BE TRUE" questions: These questions ask which of the following MUST be true (must be greater in our case), or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions: The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, if we find even one set of x and y for which \frac{1}{\sqrt{x+y}} is greater than option I for example then it'll mean that option I is not ALWAYS greater then \frac{1}{\sqrt{x+y}}.

How can we increase the value of \frac{1}{\sqrt{x+y}}? Testing extreme examples: if x and y are very small (when their values approach zero) then denominator approaches zero and thus the value of the fraction goes to +infinity, becomes very large. In this case:

I. \frac{\sqrt{x+y}}{2}: nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction; II. \frac{\sqrt{x}+\sqrt{y}}{2}: the same here - nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;

As for: III. \frac{\sqrt{x}-\sqrt{y}}{x+y}, if x=y then this fraction equals to zero and \frac{1}{\sqrt{x+y}} has some value more than zero, so this option also is not always more than given fraction;

Answer: none of the options must be greater than the given fraction.

Hope it's clear.

taking value x=y=1 I am getting C. What is wrong please help.
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Re: If x and y are positive, which of the following must be [#permalink]
09 Apr 2012, 21:26

Expert's post

ad20 wrote:

I got none too, I used x=4 and y =0 but GMAT prep says answer is II only...can anyone help?

The answer is none to the question given in this post. Is the GMAT prep question a little different? Does it say 'x and y are positive integers'?
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Bunuel what can't the cross multiplication trick that you suggested in the linked post be used in this particular problem? Or can it be?

You can use that approach for options I and II, since both those fractions are non-negative. Though I'd solve this question the way presented above.
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