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

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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, 07:43
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, 09:13
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, 07:13
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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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14 Nov 2019, 23:53
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Re: If x and y are positive, which of the following must be   [#permalink] 14 Nov 2019, 23:53

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