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

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Intern
Joined: 26 Mar 2013
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Location: India
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Re: If x and y are positive, which of the following must be  [#permalink]

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17 Jan 2015, 11:26
Ans E

stat 1 & 2 were straight....but 3 though was easily simplified....took smtym to confirm

Good tricky qtn
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If x and y are positive, which of the following must be  [#permalink]

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28 Jan 2016, 09:41
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. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$

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

In "MUST BE TRUE" questions, you need to use POE and be ready to prove that none of the options/statements are possible. Even 1 not possible scenario will make that option not allowed.

Had this question been "could be true" instead, II only would have been correct with x=y=1. But as this is a MUST BE TRUE question, you need to make sure to find 1 set of (x,y) that will negate the given conditions.

i) and iii) can be easily POE-d by assuming x=y=1. In both these cases the resulting values will NOT BE GREATER than $$1/(x+y)^{0.5}$$.

For ii), you can clearly see x=y=1 gives you a value greater than $$1/(x+y)^{0.5}$$ but what about x=y=4 again you get a value greater. So lets take x=y=0.25. In this case you will end up getting ii) < $$1/(x+y)^{0.5}$$. Hence this expression as well is NOT always true and is hence eliminated.

As you eliminated all the 3 possible options, OA must be E (none).

Hope this helps.
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Joined: 30 Sep 2016
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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Nov 2016, 06:23
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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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Nov 2016, 06:43
Nasahtahir wrote:
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?

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

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29 Nov 2016, 08:13
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. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$

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

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

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25 Oct 2017, 06:13
Top Contributor
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. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$

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

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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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Re: If x and y are positive, which of the following must be   [#permalink] 27 Oct 2018, 00:34

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