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

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Joined: 07 Aug 2011
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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, 07:49
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.

Thanks Bunuel walker . Thanks for pointing out use of extreme cases for ruling out options.
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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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Re: If x and y are positive, which of the following must be [#permalink]

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27 Jan 2016, 12:16
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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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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
Expert's post
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

[Reveal] Spoiler:
E

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Re: If x and y are positive, which of the following must be   [#permalink] 25 Oct 2017, 06:13

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