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

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Manager
Joined: 29 Aug 2013
Posts: 74
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
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WE: Programming (Computer Software)
Re: If x and y are positive, which of the following must be  [#permalink]

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11 Sep 2013, 01:32
1
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.

Senior Manager
Joined: 21 Jan 2010
Posts: 253
Re: If x and y are positive, which of the following must be  [#permalink]

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11 Sep 2013, 23:11
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

I am not sure why everyone is deriving complex formulas for this problem. I answered this question in 30 seconds. Put x=y=1.
Only 2 survives. You can test by a couple of more values. It will always hold good. Hence C.
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Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance
Re: If x and y are positive, which of the following must be  [#permalink]

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05 Feb 2014, 08:49
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
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Posts: 61
Concentration: Strategy, Marketing
WE: Research (Pharmaceuticals and Biotech)
Re: If x and y are positive, which of the following must be  [#permalink]

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30 Jan 2015, 20:27
Hi all,

If I assume x = 3 and y = 6, then:

Original expression = 1/3

1 evaluates to 1/2
2 evaluates to (-1/9)*something
3 evaluates to (-1/9)*something

Thus, I get only 1 as an answer.

TO
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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31 Jan 2015, 04:25
thorinoakenshield wrote:
Hi all,

If I assume x = 3 and y = 6, then:

Original expression = 1/3

1 evaluates to 1/2
2 evaluates to (-1/9)*something
3 evaluates to (-1/9)*something

Thus, I get only 1 as an answer.

TO

If x=3 and y=6, then option 2 gives approximately 1/2, which is greater than 1/3.

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Joined: 07 Jul 2010
Posts: 7
Re: If x and y are positive, which of the following must be  [#permalink]

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15 Jun 2015, 03:17
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

Interesting .....
It's not a good question
What if i take x = y = 2 ....Then c option would work out to be 0 /4 .
Nothing in the prompt prevents me from considering x and y to be equal and doing that Option C is not possible as well
I think the question should have made it very clear that x is not equal to y
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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15 Jun 2015, 03:19
amlan009 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

Interesting .....
It's not a good question
What if i take x = y = 2 ....Then c option would work out to be 0 /4 .
Nothing in the prompt prevents me from considering x and y to be equal and doing that Option C is not possible as well
I think the question should have made it very clear that x is not equal to y

The question asks "which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$", not "which of the following could be greater than $$\frac{1}{\sqrt{x+y}}$$".
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Joined: 07 Jul 2010
Posts: 7
Re: If x and y are positive, which of the following must be  [#permalink]

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17 Jun 2015, 23:28
Bunuel wrote:
amlan009 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

Interesting .....
It's not a good question
What if i take x = y = 2 ....Then c option would work out to be 0 /4 .
Nothing in the prompt prevents me from considering x and y to be equal and doing that Option C is not possible as well
I think the question should have made it very clear that x is not equal to y

The question asks "which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$", not "which of the following could be greater than $$\frac{1}{\sqrt{x+y}}$$".

Firstly , i am a huge fan and love the way you help people ...it's amazing
Thanks bunuel , and now for the topic
I think i have not explained my stance properly ..i am not very happy with the question because must implies true in all cases whereas in the specific case when x and Y are equal positive integers , i don't see the third option to be valid for the question stem anymore

Let me explain , Since x and Y are positive integers , i take x = y= 2 ( which i am surely allowed to )
Now , question translates to what is greater than 1 / \sqrt{4} = 1 /2 = 0.5

When the expression c with this numbers translates to 0/(2+2) =0/4 i.e 0
Surely , 0.5 is not greater than 0 .

Now, therefore when x and y are same , it is clearly not true .

Am i assuming wrong thing's somewhere ?
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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18 Jun 2015, 00:39
amlan009 wrote:
Firstly , i am a huge fan and love the way you help people ...it's amazing
Thanks bunuel , and now for the topic
I think i have not explained my stance properly ..i am not very happy with the question because must implies true in all cases whereas in the specific case when x and Y are equal positive integers , i don't see the third option to be valid for the question stem anymore

Let me explain , Since x and Y are positive integers , i take x = y= 2 ( which i am surely allowed to )
Now , question translates to what is greater than 1 / \sqrt{4} = 1 /2 = 0.5

When the expression c with this numbers translates to 0/(2+2) =0/4 i.e 0
Surely , 0.5 is not greater than 0 .

Now, therefore when x and y are same , it is clearly not true .

Am i assuming wrong thing's somewhere ?

The correct answer to the question is C, which says that only II option must be greater than $$\frac{1}{\sqrt{x}+\sqrt{y}}$$. Why are talking about the third option?
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Status: It's time...
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Posts: 7
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q48 V25
GPA: 3.59
Re: If x and y are positive, which of the following must be  [#permalink]

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20 Jun 2016, 04:25
Hi Bunuel,
Can you please give me the link to a set of these type of problems to practice?

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Perseverance, the answer to all our woes. Amen.

Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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20 Jun 2016, 04:30
1
oloz wrote:
Hi Bunuel,
Can you please give me the link to a set of these type of problems to practice?

Check must or could be questions here: search.php?search_id=tag&tag_id=193
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Location: India
GMAT 1: 610 Q49 V21
WE: Engineering (Manufacturing)
Re: If x and y are positive, which of the following must be  [#permalink]

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29 Aug 2017, 19:24
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 negative-but 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.

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Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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29 Aug 2017, 21:02
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 negative-but 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.

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 $$\sqrt[3]{-64} =-4$$.
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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Aug 2017, 21:43
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
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Posts: 52294
Re: If x and y are positive, which of the following must be  [#permalink]

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29 Aug 2017, 22:23
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.
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Re: If x and y are positive, which of the following must be  [#permalink]

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22 Oct 2017, 07:38
I used a simple approach of number picking. pick a small number say 2. it will take about 1 minute to eliminate the wrong answers.
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If x and y are positive, which of the following must be  [#permalink]

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25 Oct 2017, 10:33
Most Simple Solution:-

Keep x = y amd solve the equation
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Re: If x and y are positive, which of the following must be  [#permalink]

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27 Oct 2018, 00:34
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