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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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12 May 2017, 03:53
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Difficulty:

95% (hard)

Question Stats:

43% (01:49) correct 57% (02:04) wrong based on 289 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) > 1?  [#permalink]

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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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12 May 2017, 08:06
1
3
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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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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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
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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06 May 2018, 07:49
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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06 May 2018, 09:30
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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10 Sep 2018, 04:59
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, 04:59
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