Data Sufficiency

Hi All,

We're told that in the xy-plane, Line K has NEGATIVE slope. We're asked if the y-intercept of Line K is positive. This is a YES/NO question. It can be solved in a couple of different ways, including by TESTing VALUES. To start, with a negative slope, BOTH the x-intercept and y-intercept of Line K will have the same 'sign' (meaning they are both positive, both negative or both 0).

1) The x-intercept of Line K is less than the y-intercept of Line K.

Since Line K has a NEGATIVE slope, we know that the line goes "down and to the right", but we know nothing about where it crosses the X-axis or Y-axis. With the information in Fact 1, the x-intercept and y-intercept could be:
x-intercept = 2 y-intercept = 3, the slope = -3/2 and the answer to the question is YES
x-intercept = -2 y-intercept = -1, the slope = -1/2 and the answer to the question is NO
Fact 1 is INSUFFICIENT

2) The slope of line k is less than -2.

Fact 2 tells us NOTHING about where the line appears on the graph, so it's possible that the y-intercept could be positive (a YES answer) or 0 (a NO answer) or negative (also a NO answer).
Fact 2 is INSUFFICIENT

Combined, we know...
-The x-intercept of Line K is less than the y-intercept of Line K.
-The slope of line k is less than -2.

For the slope to be LESS than -2 AND the x-intercept to be LESS than the y-intercept, the only possibility exists when both intercepts are POSITIVE. If the two intercepts are negative, then the x-intercept would actually be GREATER than the y-intercept (and that option does not 'fit' what we're told). Thus, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

In the image attached, we have proved that each statement alone is insufficient through GEOMETRIC BIFURCATIONS.

\[\left( {1 + 2} \right)\,\]
\[{x_{\,k}}\,\, = \,\,{\left( {x - {\text{intercept}}} \right)_{\,k}}\]
\[{y_{\,k}}\,\, = \,\,{\left( {y - {\text{intercept}}} \right)_{\,k}}\,\,\]
\[{\text{slop}}{{\text{e}}_{\,k}}\,\,\, < \,\,\,\,0\]
\[{y_{\,k}}\,\,\mathop > \limits^? \,\,\,0\]
\[\underline {{x_k} = 0} \,\,\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\left\{ \begin{gathered}
{y_k} > 0 \hfill \\
{\text{slop}}{{\text{e}}_{\,k}}\,\,\, < \,\,\,\,0 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\underline {{x_k} > 0} \,\,\,\,\,\,{\text{impossible}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_k} \ne 0\]
\[\left( {{x_k} \ne 0} \right)\,\,\,\,\, - \frac{{{y_{\,k}}}}{{{x_{\,k}}}}\,\, = \,\,\,\frac{{{y_{\,k}} - 0}}{{0 - {x_{\,k}}\,}}\,\,\,{\text{ = }}\,\,\,\,{\text{slop}}{{\text{e}}_{\,k}}\,\,\,\left\{ \begin{gathered}
< \,\,\,\,0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{y_k}\,\,,\,\,{x_k}\,\,{\text{same}}\,\,{\text{signs}}\,\,\,\,\,\left( {{\text{both}}\,\, \ne 0\,\,} \right)\,\,\,\,\,\,\,\,\left( * \right) \hfill \\
\mathop < \limits^{\left( 2 \right)} - 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\, - \frac{{{y_{\,k}}}}{{{x_{\,k}}}} < - 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{{y_{\,k}}}}{{{x_{\,k}}}} > 2\,\,\,\,\,\,\,\,\,\left( {**} \right)\, \hfill \\
\end{gathered} \right.\]
\[\underline {{y_k} < 0} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{x_k} < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,{y_{\,k}} < \,\,\,2\,\,{x_{\,k}}\,\,\,\mathop < \limits^{{x_k}\,\, < \,\,0} \,\,\,{x_k}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\underline {\,\left( 1 \right)\,\,\,{\text{contradicted}}\,} \]
\[\left. \begin{gathered}
{y_k} < 0\,\,\,{\text{false}}\,\,\, \hfill \\
{y_k}\,\, \ne 0\,\,\,\,\left( * \right) \hfill \\
\end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,{y_k} > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\]

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

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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