What a popular post this turned out to be: 0 timer attempts in 48 hours. I thought the question was decent at least, but of course, I might be biased. Anyway, this question builds off a simple concept, as many GMAT™ questions do, but then masks that concept to make you think a little harder. So, what is the core concept being tested? The slope-intercept form of a line: y = mx + b. If the question asks about either intercept, all we really need is a point on the line (x, y) and the slope,
m.
(1) We know triangle BCE is a right triangle because sides BC and CE are parts of different squares. Hence, we can use the Pythagorean theorem to solve for the third side:
\(6^2 + b^2 = \sqrt{52}^2\)
\(36 + b^2 = 52\)
\(b^2 = 16\)
\(b = 4\)
Since we know segment CE is 4 units and segment CD must be 6 units, we know that each side of the larger square is 10 units, and point E must therefore lie at (-10, 10). We can further deduce that point B lies at (-16, 6). With two points on the line, we could definitely calculate the slope and the x-intercept, so
statement (1) must be SUFFICIENT. If you need to appease your inner fact-checker, however, you could actually work out the numbers:
\(m = \frac{(10 - 6)}{(-10 - -16)}\)
\(m = \frac{4}{6}\)
\(m = \frac{2}{3}\)
Thus, our linear equation resembles the form
\(y = \frac{2}{3}x + b\)
We can plug in either point that we know to work out the value of
b. How about (-10, 10)?
\((10) = \frac{2}{3}(-10) + b\)
\(10 = \frac{-20}{3} + b\)
\(10 + \frac{20}{3} = b\)
\(\frac{50}{3} = b\)
Thus, the full equation of the line is
\(y = \frac{2}{3}x + \frac{50}{3}\)
To answer the question, substitute 0 for
y:
\((0) = \frac{2}{3}x + \frac{50}{3}\)
\(-\frac{50}{3} = \frac{2}{3}x\)
\(-50 = 2x\)
\(-25 = x\)
The x-intercept is at -25.
(2) Knowing that the area of triangle BCE is 12 square units does not help us find the x-intercept because we have no way of pinning down either points B or E. For all we know, the figure could not be drawn to scale, and segment BC could be 3 units and segment EC 8 units. This would then make square DEFG an 11 by 11, and point E would then lie at (-11, 11), while point B would lie at (-14, 3). The slope would not be the same:
\(m = \frac{(3 - 11)}{(-14 - -11)}\)
\(m = \frac{-8}{-3}\)
\(m = \frac{8}{3}\)
If the slope were not the same and the points C and E were also not the same, the x-intercept would be different from the one we calculated before. But since everything could also be the same,
statement (2) is NOT sufficient.
We can comfortably say that
the answer must be (A). I had fun with this question, even if I wonder whether I should have labeled a bit more and added a note for good measure, something to the effect of, "Figure may not be drawn to scale."
Anyway, good luck with your studies, everyone.
- Andrew
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