If x and y are integers and xy ≠ 0, is x - y > 0?

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If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2
$$\left(2\right)\ \sqrt{x^2}=\ x\ and\ \sqrt{y^2}=\ y$$

OA C

Source: Magoosh

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by fskilnik@GMATH » Mon Oct 08, 2018 10:25 am

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BTGmoderatorDC wrote:If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2
$$\left(2\right)\ \sqrt{x^2}=\ x\ and\ \sqrt{y^2}=\ y$$
Source: Magoosh
$$\left( * \right)\,\,\,\,x,y\,\,\, \ne 0\,\,\,\,{\text{ints}}\,\,\,\,\left( {xy \ne 0} \right)\,\,\,\,$$
$$x\,\,\mathop > \limits^? \,\,y$$
$$\left( 1 \right)\,\,{x \over y} < {1 \over 2}\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1, - 3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,\left\{ \matrix{
\,\left| x \right| = x\,\,\,\, \Rightarrow \,\,\,\,x \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x > 0 \hfill \cr
\,\left| y \right| = y\,\,\,\, \Rightarrow \,\,\,\,y \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,$$
$$\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
\,\,{x \over y} < {1 \over 2}\,\,\,\,\,\mathop \Rightarrow \limits_{y\, > \,\,0}^{ \cdot \,\,2y} \,\,\,\,\,2x < y \hfill \cr
\,\,\,1 < 2\,\,\,\,\,\mathop \Rightarrow \limits_{\,x\, > \,\,0}^{ \cdot \,\,x} \,\,\,\,\,\,\,x < 2x \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,x < 2x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,$$

This solution follows the notations and rationale taught in the GMATH method.

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Fabio.
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by Jay@ManhattanReview » Thu Oct 11, 2018 9:24 pm

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BTGmoderatorDC wrote:If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2
$$\left(2\right)\ \sqrt{x^2}=\ x\ and\ \sqrt{y^2}=\ y$$

OA C

Source: Magoosh
Given: x and y are integers and xy ≠ 0.

Question: Is x - y > 0?

Let's take each statement one by one.

(1) x/y < 1/2

Case 1: Say x = 2 and y = 6, we have x/y => 2/6 = 1/3 < 1/2. We see that x - y = 2 - 6 = -4 < 0. The asnwer is No.
Case 2: Say x = -2 and y = -6, we have x/y => -2/-6 = 1/3 < 1/2. We see that x - y = -2 + 6 = 4 > 0. The asnwer is Yes.

No unique answer. Insufficient.

(2) $$\left(2\right)\ \sqrt{x^2}=\ x\ and\ \sqrt{y^2}=\ y$$

Note that the square root of a number is always positive, thus, from the given information, we have x and y are positive numbers.

Thus, Case 2 discussed in Statement is not applicable, or only Case 1 is applicable. So, the answer is No, x - y is not greater than 0. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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