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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, 04:53
2
16
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

41% (01:51) correct 59% (02:03) wrong based on 267 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, 05:22
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, 09:06
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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, 05:39
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, 08:24
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
Ans : 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 06 May 2018, 08:49
HKD1710 wrote:
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.



Could someone please explain how x + y > 1 proves that sqrt(x) + sqrt(y) > 1?

I'm feel that the explanation taken from the quote above "sqrt of any number greater than 1 will be greater than the number itself." might be wrong? Please correct me if that's not the case.

I fee it should be lesser than and not greater than.
For example:
Let x = 4
sqrt (4) = 2 => sqrt of any number greater than 1
Therefore, x > sqrt (x), when x > 1 => "sqrt of any number greater than 1 will be lesser than the number itself."
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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 06 May 2018, 10:30
1
Bunuel wrote:
If x and y are positive, is \(\sqrt{x}+\sqrt{y}>1\)?


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

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


So here's the Algebraic approach -

Is \(\sqrt{x}+\sqrt{y}>1\), square both sides to get the rephrased question stem as

Is \(x+y+2\sqrt{xy}>1\)

Statement 1: \(\sqrt{x+y}>1\), square both sides to get

\(x+y>1\), Now add \(2\sqrt{xy}\) to both sides to get

\(x+y+2\sqrt{xy}>1+2\sqrt{xy}\), Now as \(x\) & \(y\) are positives so \(2\sqrt{xy}\) will be positive, hence when a positive number is added to \(1\), it will definitely be greater than \(1\)

So we have \(x+y+2\sqrt{xy}>1\). Sufficient

Statement 2: \(x>\frac{1}{4}\) take square root of both sides to get

\(\sqrt{x}>\frac{1}{2}\)----------------------(1)

\(y>\frac{1}{4}\), take square root of both sides to get

\(\sqrt{y}>\frac{1}{2}\)----------------------(2), Now add equations (1) & (2) to get

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

Option 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 10 Sep 2018, 05:59
dabaobao wrote:
HKD1710 wrote:
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.



Could someone please explain how x + y > 1 proves that sqrt(x) + sqrt(y) > 1?

I'm feel that the explanation taken from the quote above "sqrt of any number greater than 1 will be greater than the number itself." might be wrong? Please correct me if that's not the case.

I fee it should be lesser than and not greater than.
For example:
Let x = 4
sqrt (4) = 2 => sqrt of any number greater than 1
Therefore, x > sqrt (x), when x > 1 => "sqrt of any number greater than 1 will be lesser than the number itself."


From S(1) we get x+y>1 therefore 0<x<1 and 0<y<1, since booth are positive as mentioned in the question stem.

You are correct that the above mentioned rule is wrong! It should be: "The square root of any number smaller than 1, but bigger than 0 (0<n<1) will be greater than the number itself."

So with this rule it can be seen that \(\sqrt{x}\)+\(\sqrt{y}\)>1.

Is it clear?
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Re: If x and y are positive, is x^(1/2) + y^(1/2) > 1? &nbs [#permalink] 10 Sep 2018, 05:59
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