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

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Intern
Joined: 01 Oct 2006
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If x and y are positive, which of the following must be [#permalink]

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11 Oct 2006, 19:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x and y are positive, which of the following must be greater than 1 / sqt root of (x + y)

I) sqt root of (x+y) / 2x
II) sqt root of (x) + sqt root of (y) / x + y
III) sqt root of (x) - sqt root of (y) / x + y

A) none
B) I
C) II only
D) I and III
E) II amd III

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Director
Joined: 23 Jun 2005
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GMAT 1: 740 Q48 V42

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11 Oct 2006, 19:17
c?

Squaring the question, we get what is greater than 1/(x+y)

Squaring the options
I) sqt root of (x+y) / 2x = (x+y)/4*x^2. Without knowing values of x and y, not possible to decide if this is greater than 1/(x+y) .

II) sqt root of (x) + sqt root of (y) / x + y = x+y+2*sqrt(xy)/x+y.
Will always be greater than 1/(x+y)

III) sqt root of (x) - sqt root of (y) / x + y = x+y - 2*sqrt(xy)/x+y. not possible to determine

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Intern
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11 Oct 2006, 19:29
for II and III, don't you need to square the denominator as well which gives you x^2 + 2xy + y2?

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Director
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11 Oct 2006, 19:47
eagerbeaver wrote:
for II and III, don't you need to square the denominator as well which gives you x^2 + 2xy + y2?

Yes, I shouldn't be doing any math today evening.

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Senior Manager
Joined: 08 Jun 2006
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11 Oct 2006, 21:09
I am also getting C

As both x and y are positive, 1 / sqt root of (x + y) has to be positive.

If y > x, (sqt root of (x) - sqt root of (y)) can be negative, so III can be negative. III is not the answer. Eliminate answer choices D and E.

Remaining answers are A) none B) I and C) II Only

1 / sqt root of (x + y) can be written as sqt root of (x + y) / (x + y)

Now sqt root of (x) + sqt root of (y) / x + y and sqt root of (x + y) / (x + y) have same denominator. The question is â€“ which numerator is greater?

Is {sqt root of (x) + sqt root of (y)} > sqt root of (x + y)
Is {sqt root of (x) + sqt root of (y)} ^ 2 > { sqt root of (x + y)} ^ 2
Is x + y + 2 sqt root of (xy) > x + y
Is 2 sqt root of (xy) > 0, this is true because both x and y are positive.

II must be one of the answers. Now from the answer choices C is the only possible answer. We donâ€™t need to check I.

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Director
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11 Oct 2006, 23:42
Lets say A = 1/sqrt(x +y)

I) (x +y)/(2*x) * 1/sqrt(x + y) = (x + y)/(2*x) * A
what if y > 2? Say NO to I.

II) ( SR(x) + SR(y) )/SR(x + y) * 1/SR(x + y)
SR(x) + SR(y)/SR(x+y) * A

Now SR(a) + SR(b) > SR(a + b)
So, II is GOOD.

III) NOT GOOD.

Only II. C.

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VP
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12 Oct 2006, 00:12
plugging in, got C.

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Director
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12 Oct 2006, 07:27
anandsebastin wrote:
Yes, I shouldn't be doing any math today evening.

After that, it turns out that my answer is still right. I guess that is what you would call a "stroke of luck".

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12 Oct 2006, 12:00
If x and y are positive, which of the following must be greater than 1 / sqt root of (x + y)

I) sqt root of (x+y) / 2x
II) sqt root of (x) + sqt root of (y) / x + y
III) sqt root of (x) - sqt root of (y) / x + y

A) none B) I C) II only D) I and III E) II amd III

MULTIPLY STEM BY SQRT(X+Y) / SQRT(X+Y) IT WILL LOOK BETTER

IE: SQRT(X+Y) /( X+Y)

FROM ONE COMPARING (X+Y) TO 2X DEPENDS.......OUT (B,D OUT)

NOW WE COMPARE SQRT X + SQRT Y TO SQRT( X+Y) ....

ASSUME X = 9 ,Y = 4
5 > SQRT 13

ASSUME X = 16 Y = 9

7>5

ASSUME X= 1/4 , Y = 1/9

1/2+ 1/3 = 5/6 TO SQRT13/36 = 3.X/6 APPRPX 0.5

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Senior Manager
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12 Oct 2006, 12:42
Take x=y=1 and clearly I and III are out.

So it must be II only
_________________

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12 Oct 2006, 12:42
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# If x and y are positive, which of the following must be

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