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

Question

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

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

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

Bunuel · Accepted Answer

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.

gshashank18 · Answer

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.

HKD1710 · Answer

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.

Heseraj · Answer

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

sleepynut · Answer

(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

chetan2u · Answer

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

dabaobao · Answer

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."

niks18 · Answer

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

KarishmaB · Answer

Recall number line concepts. When 0 < x < 1, we know that \sqrt{x} > x Statement (1) \sqrt{x+y}>1 Since we have the principal square root here on LHS and it is greater than 1, we can simply square both sides (because both sides are positive) to get x + y > 1 If either of x and y is greater than 1, its square root will be smaller but still greater than 1. Square root of a number greater than 1 is always greater than 1. Hence \sqrt{x}+\sqrt{y}>1 holds If both are less than 1, \sqrt{x} > x and \sqrt{y} > y Since x + y > 1, \sqrt{x}+\sqrt{y}>1 holds Sufficient alone. Statement (2) x>y>\frac{1}{4} Square root of 1/4 is 1/2. If both x and y are greater than 1/4, their square roots will be greater than 1/2 and hence sum of square roots will be greater than 1. Sufficient alone. Answer (D) Check a similar but trickier question here: https://anaprep.com/algebra-importance- ... -plugging/

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

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