If x and y are positive, is √x+y10?

Question

If x and y are positive, is  \sqrt{x+y}>10 ?

(1)  \frac{\sqrt{x}}{2} +\frac{\sqrt{y}}{2}>5

(2)  \sqrt{4x}>20

VyshakhR1995 · Answer

In Statement one 2 can be taken to the other side.After squaring LHS and RHS it can be seen that the information alone will not provide us with solution.
On squaring statement 2, we get x>100
Since x and y are positive, this statement alone is sufficient 
Hence B

ab52 · Answer

IMO B.

\sqrt{x+y}>10 
square both sides.. 
so the question becomes : Is |x+y|>100?
Since x & y are both positive x+y will be positive.

statement 1. says :  \frac{\sqrt{x}}{2} +\frac{\sqrt{y}}{2}>5  =>  \sqrt{x}+\sqrt{y}>10 
This doesn't give any information on values of x and y.

Statement 2 says:  \sqrt{4x}>20  =>  \sqrt{x}>10  squaring both sides: x>100 .

Hence this statement itself is sufficient since y is positive x+y has to be greater than 100.

Answer: B

Fdambro294 · Answer

We are given that X and Y are positive

Q: is sqrt(X + Y) < 10 ?

Rule: when both sides of an inequality are positive, if we Square both sides we maintain the inequality relationship (i.e., do not have to “reverse” the inequality sign)

Squaring both sides of the question stem we get:

Is: (X + Y) < 100 ?

(1) ( sqrt(X) + sqrt(Y) ) * (1/2) > 5

—-multiplying both sides by 2 we get ——

sqrt(X) + sqrt(Y) > 10

—then when we Square both sides, we end up the with the square of a sum quadratic template on the left side of the inequality—-

X + Y + 2 * sqrt(X) + sqrt(Y) > 100

Question: Is: (X + Y) > 100?

Since the extra value of + 2 * (sqrt (XY))
Is a positive value added to (X + Y) ——- it could be the case that the value is a lot bigger than 100, such that (X + Y) is larger than 100 also

Case 1:

X + Y + 2 * sqrt(XY) > (X + Y) > 100

Which would answer YES to the question

Case 2:

Or it could be the case that although the expression in statement 1 is larger than 100, the value of just (X + Y) is still less than 100

X + Y + 2 * sqrt(XY) > 100 > (X + Y)

Or the values could be in this order, in which case we would answer the question NO

S1 NOT Sufficient

(S2) sqrt(4X) > 20

Again, because we know that X is a positive value, when we square both sides we won’t introduce any extraneous solutions and the inequality relationship will be maintained

—-squaring both sides of statement 2 we get—-

4X > 400

X > 100

Since Y is a positive integer also, it must necessarily be true that:

(X + Y) > X > 100

In which case we can answer the question definitely YES

B S2 Sufficient ALONE

Posted from my mobile device

GMATBusters · Answer

is  \sqrt{x+y}>10 ?
or  Is x+y > 100?

St 1)  \frac{\sqrt{x}}{2} +\frac{\sqrt{y}}{2}>5 
multiplying the equation by 2 and squaring both sides, we get
 x+y+2\sqrt{xy}  >100
Now,  x+y > 100-\sqrt{xy} 
Hence, we cannot say whether  x+y > 100

NOT Suff

St 2) \sqrt{4x}>20

Squaring both sides, we get  4x> 400  or  x>100

Since, x and y are positive,  x+y > 100

Suff

Answer B

If x and y are positive, is √x+y>10?

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