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If x and y are positive, is x^(1/2) + y^(1/2) ?

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If x and y are positive, is x^(1/2) + y^(1/2) ?  [#permalink]

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New post 06 Nov 2017, 06:00
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Question Stats:

41% (01:07) correct 59% (01:46) wrong based on 88 sessions

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If x and y are positive, is \(\sqrt{x}+ \sqrt{y} > 1\) ?


(1) \(\sqrt{x+y}>1\)

(2) \(x > y > \frac{1}{4}\)
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Re: If x and y are positive, is x^(1/2) + y^(1/2) ?  [#permalink]

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New post 06 Nov 2017, 06:37
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If x and y are positive, is \(\sqrt{x}+ \sqrt{y} > 1\) ?

Is \(\sqrt{x}+ \sqrt{y} > 1\) ?
Square: is \(x+ 2\sqrt{xy}+y > 1\) ?

(1) \(\sqrt{x+y}>1\). Square: x + y > 1. Since x and y are positive, then \(2\sqrt{xy}>0\), and thus \(x+ 2\sqrt{xy}+y=(x+y)+2\sqrt{xy}=(number \ more \ than \ 1)+(positive \ value)>1\). Sufficient.


(2) \(x > y > \frac{1}{4}\). Since both x and y are greater than 1/4, then both \(\sqrt{x}\) and \(\sqrt{y}\) are greater than 1/2. Hence, \(\sqrt{x}+ \sqrt{y} > 1\). Sufficient.

Answer: D.
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Re: If x and y are positive, is x^(1/2) + y^(1/2) ?  [#permalink]

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New post 06 Nov 2017, 06:38
leshwarnag wrote:
If x and y are positive, is \(\sqrt{x}+ \sqrt{y} > 1\) ?


(1) \(\sqrt{x+y}>1\)

(2) \(x > y > \frac{1}{4}\)



hi...

\(\sqrt{x}+ \sqrt{y} > 1\)
square both sides....
\((\sqrt{x}+ \sqrt{y})^2 > 1^2.............x+y+2\sqrt{xy}>1\)

lets see statements

(1) \(\sqrt{x+y}>1\)
square both sides..
\(x+y>1\)...
if \(x+y >1\), when you add another positive value \(2\sqrt{xy}\), it will surely be >1
so \(x+y+2\sqrt{xy}>1\)
suff

(2) \(x > y > \frac{1}{4}\)
lets take both as 1/4
so \(\sqrt{x}+ \sqrt{y} = \sqrt{1/4}+\sqrt{1/4}=1/2+1/2=1\)
so the equation will be >1
suff

D
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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If x and y are positive, is x^(1/2) + y^(1/2) ?  [#permalink]

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New post 06 Nov 2017, 06:42
Square root of a positive fraction is greater than itself

1. \(\sqrt{x+y} > 1\)
For the above statement to be true: x+y > 1
You can check with any combination of x and 1-x
0.5, 0.5 : \(\sqrt{0.5} + \sqrt{0.5} => 1.4\)
0.1, 0.9 : \(\sqrt{0.1} + \sqrt{0.9} => 0.31 + 0.94 => 1.25\)
0.01, 0.99 : \(\sqrt{0.01} + \sqrt{0.99} => 0.1 + 0.995 => 1.095\)
It would always be greater than 1
Yes

2. \(x > y > \frac{1}{4}\)

\(\sqrt{1/4} = 0.5\)
So \(\sqrt{x} > \sqrt{y} > 0.5\)
\(\sqrt{x} + \sqrt{y} > 1\)
YES

D.
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If x and y are positive, is x^(1/2) + y^(1/2) ? &nbs [#permalink] 06 Nov 2017, 06:42
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