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# Is |xy| > xy?

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Math Expert
Joined: 02 Sep 2009
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13 Sep 2017, 20:48
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Is $$|xy| > xy$$?

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

(2) $$|x|+|y|=x+y$$
[Reveal] Spoiler: OA

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13 Sep 2017, 21:17
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Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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 $$D$$
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Re: Is |xy| > xy? [#permalink]

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26 Sep 2017, 21:14
niks18 wrote:
Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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 $$D$$

if either of the two is negative and other is positive then $$|xy|>xy$$ - Can you explain this further?
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Re: Is |xy| > xy? [#permalink]

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26 Sep 2017, 21:18
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Manveetha wrote:
niks18 wrote:
Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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 $$D$$

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

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
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Re: Is |xy| > xy? [#permalink]

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27 Sep 2017, 20:54
Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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?
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27 Sep 2017, 21:26
We have to just figure out whether the signs of x and y are same or different.
If they have the same signs, then the two sides would be equal. If one of the two values or both are zero, then the two sides are equal and not less than or greater than.

Both the statements answer this question: The signs are same.
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Re: Is |xy| > xy? [#permalink]

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27 Sep 2017, 22:09
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RMD007 wrote:
Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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?

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.

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Re: Is |xy| > xy? [#permalink]

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01 Oct 2017, 12:59
Bunuel wrote:
Is $$|xy| > xy$$?

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

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

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)
$$|x| + |y| = x + y$$
=> $$(|x| + |y|)^2 = (x + y)^2$$
<=> $$|x|^2 + 2|x||y| + |y|^2 = x^2 + 2xy + y^2$$
<=> $$x^2 + 2|x||y| + y^2 = x^2 + 2xy + y^2$$
<=> $$2|x||y| = 2xy$$
<=> $$|xy| = xy$$
<=> $$xy \ge 0$$
By CMT 1, this condition is also sufficient.

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Re: Is |xy| > xy?   [#permalink] 01 Oct 2017, 12:59
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