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

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11 Sep 2013, 01:32

1

This post was BOOKMARKED

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.

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.

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

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

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 \(\frac{1}{\sqrt{x+y}}\)".
_________________

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 .

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?
_________________

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.

Please guide me. Thanks.
_________________

If you appreciate my post then please click +1Kudos

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.

Please guide me. Thanks.

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 \(\sqrt[3]{-64} =-4\).
_________________

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).