Given that xy 0, which of the following must also be greater than ze

Question

Given that xy > 0, which of the following must also be greater than zero?

(A) \(x + y\)
(B) \(x^2 + y\)
(C) \((x + y)^2\)
(D) \(x^2 - y^2 + 100\)
(E) \((x - y)^2\)

For a discussion of plugging in numbers, as well as a full solution to this question, see:
https://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike :-)

SIDDHARTH1901 · Accepted Answer

XY>0 , it means both are positive or both are negative.
Here > sign is there means they are having non-zero values.

(x-y)^2 can be a zero , when x=y
But 
(x+y)^2 will always have a positive value.

Posted from my mobile device

Bunuel · Answer

Given that xy > 0, which of the following must also be greater than zero?

(A)  x + y 
(B)  x^2 + y 
(C)  (x + y)^2 
(D)  x^2 - y^2 + 100 
(E)  (x - y)^2

Notice two things:
1. The question asks which if the following MUST be greater than zero, not COULD be greater than zero.
2. xy > 0 means that x and y have the same sign, they are either both positive or both negative (and neither of them could be zero).

Right away we can notice that  (x + y)^2  (option C) is either  (positive + positive)^2=positive^2=positive  or  (negative + negative )^2=negative ^2=positive . So,  (x + y)^2  is always greater than zero.

As for other options:

(A)  x + y . If both x and y are negative, then x + y = negative.

(B)  x^2 + y . If say x = -1 and y = -100,000, then x^2 + y = negative.

(D)  x^2 - y^2 + 100 . If say x = 1 and y = 100,000, then x^2 - y^2 + 100 = negative.

(E)  (x - y)^2 . If x = y, then (x - y)^2 = (x - x)^2 = 0^2 = 0. Not positive.

So, only C MUST be greater than 0.

Answer: C.

Hope it helps.

Vips0000 · Answer

Actually if u just go through the options.. you'll notice that option C, is a perfect square.. which has to be greater than or equal to zero. Now looking back at the question, we know xy>0, which implies x and y are not equal to 0 and are of same sign. therefore x+y will be a non zero number and its square would be greater than 0.

Ans C it is.

rohit8865 · Answer

Given that xy > 0, which of the following must also be greater than zero?

(A) \(x + y\)
(B) \(x^2 + y\)
(C) \((x + y)^2\)
(D) \(x^2 - y^2 + 100\)
(E) \((x - y)^2\)

We are not told that X & Y were integers.
Consider X=Y=1/2

only option C fits in.....

mikemcgarry · Answer

Dear rohit8865 , Quite true, my friend: the values of x & y are not necessarily integers and not necessarily positive. The two perfect squares, (C) & (E), must be greater than or equal to zero. As you figured out, if the two variables are equal non-zero numbers (whether positive or negative), then (E) will equal zero and not work. Only (C) works. Good job! Mike :-)

mysteriouswoman · Answer

Bunuel  please explain how is C only the right option

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Given that xy > 0, which of the following must also be greater than ze

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