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

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

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New post 12 May 2017, 03:53
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A
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D
E

Difficulty:

  95% (hard)

Question Stats:

37% (01:15) correct 63% (01:31) wrong based on 182 sessions

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

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New post 12 May 2017, 04:22
The answer to this question is D.

Both the statements are sufficient to answer this question.
Let's start with statement 2; it is given that the value of x and y is greater than .25. By considering the value of y as .25 + any least possible value. If we take the square root of this value, the answer comes out to be at least .5 + some value (.5 is the square root of .25). with this, it is clear that the value of x^(1/2) is definitely more than .5 because x>y. and the value of x^(1/2) + y^(1/2) is definitely greater than 1.

With statement 1, it is clear that the sum of x and y is more than 1 because square root of any value <1 can not be greater than 1. by considering the value of x and y as .5 and .51 or .99 and .02 or 0 and 1.01 or any possible combination, it is clear that the statement given in question stem is always greater than 1.
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Re: If x and y are positive, is x^(1/2) + y^(1/2) > 1? [#permalink]

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New post 12 May 2017, 08:06
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stmt-1:

\sqrt{x + y} > 1

x = 0.10
y = 0.99

x+y = 1.09

sqrt of any number greater than 1 will be greater than the number itself.

sufficient

stmt-2:
x > y > 1/4

sqrt of 1/4 is 1/2

if both x and y are 1/4 then 1/2 + 1/2 will be EQUAL TO 1. observe that 1/4 < 1/2 but sqrt of 1/4 = 1/2 > 1/4

if y and x are both greater than 1/4 then their roots will be greater than 1/2. sum of two values greater than 1/2 will be > 1.

sufficient.

Trick/Pattern/Takeaway:
If a number is between 0 & 1, then its square root will be a bigger number itself but would remain < 1.
sqrt of any number greater than 1 will be greater than the number itself.
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Re: If x and y are positive, is x^(1/2) + y^(1/2) > 1? [#permalink]

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New post 13 May 2017, 04:39
D

2ned Statement:
The least possible values for X and Y should be larger than 1/4. If we consider both of them equal to 1/4 and consider the original question stem it will be equal to one. But our values are definitely larger than 1/4. Therefore, the statement two is sufficient.

1st Statement:
We can square both sides and get (x+y)>1. We can test different numbers. We can check 0.02 and 0.99. Considering these numbers in the original question stem, we should know that squaring a number between 0<x<1, will increase the number while taking the square root will decrease the number. Therefore, statement is also sufficient.

Statement one and two are sufficient: D
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Re: If x and y are positive, is x^(1/2) + y^(1/2) > 1? [#permalink]

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New post 19 May 2017, 07:24
Bunuel wrote:
If x and y are positive, is \(\sqrt{x}+\sqrt{y}>1\)?


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

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


(1)
since x & y are positive,we have x > 0 and y > 0
Thus \(\sqrt{x}\) + \(\sqrt{y}\) > \(\sqrt{x+y}\) SUFFICIENT


(2)
As x > y > \(\frac{1}{4}\),we will have \(\sqrt{x}\) > \(\frac{1}{2}\) and \(\sqrt{y}\) > \(\frac{1}{2}\).
Thus,\(\sqrt{x}\)+\(\sqrt{y}\) > \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 SUFFICIENT
[Reveal] Spoiler:
Ans : D
Re: If x and y are positive, is x^(1/2) + y^(1/2) > 1?   [#permalink] 19 May 2017, 07:24
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