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Nasahtahir
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If x and y are positive, which of the following must be 29 Nov 2016, 07:23
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Hi.

Would you mind explaining why I got the wrong answer? I assumed x as 3 and y as 6 as we know they are positive integers. Is it wrong for me to assume this?
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If x and y are positive, which of the following must be 29 Nov 2016, 07:43
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Nasahtahir
Hi.

Would you mind explaining why I got the wrong answer? I assumed x as 3 and y as 6 as we know they are positive integers. Is it wrong for me to assume this?
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The question asks which of the following MUST be greater than ... So, which is ALWAYS greater than ... If it's greater for some particular set of numbers it does not mean that it will be greater for other sets.
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If x and y are positive, which of the following must be 29 Nov 2016, 09:13
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noboru
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. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

A. I only
B. II only
C. III only
D. I and II only
E. None
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Plug in an try -

\(x = 9\) & \(y = 16\)

\(\frac{1}{\sqrt{x+y}}\) = \(\frac{1}{25}\) \(= 0.04\)


I. \(\frac{\sqrt{x+y}}{2}\) \(= \frac{5}{2} =2.5\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{2}\) \(= \frac{3 + 4}{2}\) \(= 3.5\)

Option III is a bit different , there can be 2 cases -

\(x = 9\) & \(y = 16\) & \(x = 16\) & \(y = 9\)

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

Thus there can be multiple possible solutions for option (III)

Hence, we are confident about I & II, but option III , may or may not be > \(\frac{1}{\sqrt{x+y}}\) , so answer will be (E) None of the above.
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If x and y are positive, which of the following must be 22 Feb 2020, 00:40
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VeritasKarishma
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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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Hi,

It states that X and Y are positive numbers. So, How can we consider x + y = 0 which leads to infinity.
Moreover, If I consider x=y=1 then I am getting answer B as 2/2= 1 which is greater.
Could you please tell me what I am doing wrong ?
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If x and y are positive, which of the following must be 05 Apr 2021, 03:09
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aarushisingla
VeritasKarishma
walker
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.

Hi,

It states that X and Y are positive numbers. So, How can we consider x + y = 0 which leads to infinity.
Moreover, If I consider x=y=1 then I am getting answer B as 2/2= 1 which is greater.
Could you please tell me what I am doing wrong ?
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aarushisingla, of course x+y can never be exactly 0. But we can take it pretty close to 0. x+y could be equal to 0.00000000000000000000001.
Now, this would give you an unusually large value for the question stem. Veritaskarishma and walker are talking about values approaching infinity and not exactly infinity.
Also, the question states "must be greater". You're not trying to find a particular value for which it would be greater. But they're asking if the options would be greater than the stem for ALL the values.

I realise that this is more than a year later LOL.
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If x and y are positive, which of the following must be Updated on: 07 Apr 2021, 22:34
Originally posted by Fdambro294 on 06 Apr 2021, 18:42.
Last edited by Fdambro294 on 07 Apr 2021, 22:34, edited 1 time in total.
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To test the values, the first thing we want to find is easy numbers we can work quickly with:

The given expression takes the square root of 2 Numbers

And in the Roman numeral II, we have individual square roots of X and Y being taken, then added


So ideally we would want 2 numbers that are themselves perfect squares AND ALSO Sum to a Perfect Square

Immediately the Pythagorean Triplet of 3-4-5 should comes to mind.

Sqrt(9 + 16) = 5

Sqrt(9) + sqrt(16) = 3 + 4 = 7


Also, we are given that X and Y are positive. We should test 2 Cases:

Case 1: when X and Y are positive proper fractions ——> (1/9) and (1/16)

Case 2: when X and Y are greater than > 1 ———-> 9 and 16


III is the easiest to start with

The given expression: 1 / square root of X and Y ———> must always be positive, because X and Y themselves are positive and you can not take the Square Root of a negative value on the GMAT

However, in III

Sqrt(x) - Sqrt(y)
_____________
2

If sqrt(Y) > Sqrt(X) ———-> the value can be (-)Negative

So III can be less than


I and II

Let X = 1/9 ———> Sqrt(1/9) = 1/3
Let Y = 1/16———-> Sqrt(1/16) = 1/4
X and Y Sum ———> Sqrt(1/25) = 1/5


Our given expression will be:
1 / (1/5) = 5


I
(1/5) / 2 = 1/10

Less than


II
(1/3 + 1/4)
_________
2

= 7/24

Less than


So we have found a case for all 3 Roman Numerals in which the value can be less than the Given Expression


(E) NONE

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If x and y are positive, which of the following must be 07 Apr 2021, 10:40
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BrentGMATPrepNow
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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. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

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

Let's test some values.

x = 1 and y = 1
1/√(x + y) = 1/√(1 + 1) = 1/√2

I. √(x + y)/2 = √(1 + 1)/2 = √2/2
Notice that, if we take 1/√2 and multiply top and bottom by √2, we get: √2/2, which is the same as quantity I
Since quantity I is not greater than 1/√2, statement I is not true

II. (√x + √y)/2 = (√1 + √1)/2 = (1 + 1)/2 = 2/2 = 1
Since 1 IS greater than 1/√2, we cannot say for certain whether quantity II will always be greater than √(x + y)/2

III. (√x - √y)/(x + y) = (√1 - √1)/(1 + 1) = (1 - 1)/2 = 0/2 = 0
Since 0 is not greater than 1/√2, statement III is not true

So, statements I and III are definitely not true, and we aren't yet 100% certain about statement II
Let's try another pair of values for x and y


x = 0.25 and y = 0.25
1/√(x + y) = 1/√(0.25 + 0.25) = 1/√0.5
Let's further simplify 1/√0.5
Since 1 = √1, we can say: √1/√0.5
Then we'll use a rule that says (√k)/(√j) = √(k/j)
So, √1/√0.5 = √(1/0.5) = √2
We see that, when x = 0.25 and y = 0.25, 1/√(x + y) = √2

II. (√x + √y)/2 = (√0.25 + √0.25)/2 = (0.5 + 0.5)/2 = 1/2
Since 1/2 is NOT greater than √2, statement II is not true

Answer:
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E
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Hi, I took x=12 and y=4 and I saw that option A was greater than the fraction in question. Can you tell me where am I going wrong?
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If x and y are positive, which of the following must be 13 Jul 2025, 11:04
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