Is |xy| xy? (1) (x^2y^2)^(1/2) = xy (2) |x| + |y| = x + y

Question

Is  |xy| > xy ?

(1)  \sqrt{x^2y^2}=xy

(2)  |x|+|y|=x+y

niks18 · Accepted Answer

If both  x  and  y  are positive or negative then  |xy|=xy  and if either of the two is negative and other is positive then  |xy|>xy

Statement 1 : LHS  = \sqrt{(xy)^2}   = |xy| 
so we have  |xy|=xy . Hence we have a  NO  for our question stem.  Sufficient

Statement 2 : this implies magnitude of both  x  &  y  equals summation of both  x  &  y . Hence both  x  &  y  are positive. Hence  |xy|=xy 
Hence we have a  NO  for our question stem.  Sufficient

Option  D

Bunuel · Answer

Is  |xy| > xy ?

Notice that the question basically asks whether xy is negative. Only when xy is negative  |xy| > xy .

(1)  \sqrt{x^2y^2}=xy . This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2)  |x|+|y|=x+y . This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.

Manveetha · Answer

if either of the two is negative and other is positive then  |xy|>xy   - Can you explain this further?

niks18 · Answer

Hi  Manveetha

Take few examples, lets assume  x=-2  and  y =2 , then  xy=-2*2=-4

but  |xy|=|-4|=|4|

hence  |xy|>xy  in this case

RMD007 · Answer

Bunuel , I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. The first step of VA (Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. The question |xy| > xy is equivalent to xy < 0 . Condition 1) |xy| = xy is equivalent ot xy \ge 0 . Since the answer is no, this condition is sufficient by CMT(Common Mistake Type) 1. Condition 2) |x| + |y| = x + y => (|x| + |y|)^2 = (x + y)^2 <=> |x|^2 + 2|x||y| + |y|^2 = x^2 + 2xy + y^2 <=> x^2 + 2|x||y| + y^2 = x^2 + 2xy + y^2 <=> 2|x||y| = 2xy <=> |xy| = xy <=> xy \ge 0 Then the answer is no and "no" is also an answer. By CMT 1, this condition is also sufficient. Therefore, D is the answer.

cruiseav · Answer

Hi Bunuel . I understood Statement-2 but confused with statement-1, my logic is as below From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set. So, -(xy)>xy is possible when we have different signs of x and Y (one + and another -); then we can definitely say "Yes" to the question Is -(xy)>xy If X and Y have same signs then -(xy)xy) now in statement-1 X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient I hope you understood my thinking and plz help to correct where me going wrong with my understanding Thanks

Bunuel · Answer

For  \sqrt{x^2y^2}=xy  to be true x and y cannot have different signs because if they do, then the right hand side would become negative:  xy=negative  and it cannot equal to the square root of any number ( \sqrt{x^2y^2} ), which is always positive or 0.

lakshya14 · Answer

What if "x" and "y" are zero?

Bunuel · Answer

x = y = 0 (which certainly is one of the possible cases for x and y) gives the same NO answer as all other possible cases for each of the statements.

venkateshradiant · Answer

CAN/CANT WE CONSIDER BOTH X AND Y EQUAL ZERO??

Bunuel · Answer

I tried addressing this  here . If you meant something else, please elaborate your query.

Engineer1 · Answer

Bunuel - I understood the problem statement and what they are looking for. Stuck at option 1. Since \(\sqrt{x^2y^2}\) = xy, and not |xy| or -(xy), it proves that xy is non negative. However, I do not understand why you mentioned the sqr root of a number cannot be negative. \(\sqrt{-9}\) is not possible for a fact, but \(\sqrt{9}\) can be -3.

Appreciate any clarification.

Appreciate any clarification.

Bunuel · Answer

Mathematically,  \sqrt{...}  is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ( \sqrt{...} ) always means non-negative square root.

https://gmatclub.com/forum/download/file.php?id=39502 
 The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
 When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root.  That is:

\sqrt{9} = 3 , NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;

Notice that in contrast, the equation  x^2 = 9  has TWO solutions, +3 and -3. Because  x^2 = 9  means that  x =-\sqrt{9}=-3  or  x=\sqrt{9}=3 .

Hope it helps.

Engineer1 · Answer

Thanks for the clarification. I understand this now.

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Is |xy| > xy? (1) (x^2y^2)^(1/2) = xy (2) |x| + |y| = x + y

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