Hi folks,

Could someone walk me through this problem? Thanks!

official answer is (E)

Abramson, whenever you are given a visual description of a problem on the Quant section, be sure to draw the problem on you scratch board. Sometimes, simply by visualizing the possibilities (Data Sufficiency) or ballparking (Problem Solving), you can arrive at the correct response without doing any math.

In this problem, the only information given is that circle C is centered at (0,0) and has a radius of 1.

Statement (1) alone is insufficient since we have no idea where line k actually crosses the x axis or what the line's slope actually is. For example, line k could cross the x axis at (2,0) and have a small (relatively horizontal) slope, thus creating an intersection with circle C. However, line k could cross the x axis at (1 million,0) and have a steep (relatively vertical) slope, thus never intersecting with circle C. Eliminate AD from your AD/BCE grid.

Statement (2) alone is insufficient as we have no idea where in the plane line k is located. Eliminate B from your BCE grid.

Together, we still suffer from the same lack of information that plagued us with Statement (1). Though we know from Statement (2) that the slope of line k is relatively horizontal, we have no idea where the line crosses the x axis - it could be at (2,0), which yields an intersection with the circle, or it could be at (1 million, 0), which yields no intersection.

Due to this lack of information, the credited response is E.

-dan

Dan-

Thanks for the explanation. When you say "Though we know from Statement (2) that the slope of line k is relatively horizontal, we have no idea where the line crosses the x axis - it could be at (2,0), which yields an intersection with the circle, or it could be at (1 million, 0), which yields no intersection" - if the circle has a radius of 1, and its centered at the origin (0,0), then how would a line that crosses at 2,0 yield an intersection with the circle? Maybe I am not understanding the question, but could you please explain?

Thanks!!

The point (2,0) is to the right of the circle with radius 1 centered at the origin. A nearly horizontal line (i.e. slope = -1/10, as given) passing thru (2,0) would intersect the circle (twice, I believe). Make sure you draw it--without a picture, it is easy to make assumptions without realizing it!

_________________
Emily Sledge
Instructor
ManhattanGMAT

Hi, I am very confused on this problem. In the geometry session with Ron, we were told that lines do not stop and that you can shift them up and down on coordinate plane. Why is B not correct then? Should a line with this negative slope eventually intersect the 2nd and 4th quadrants, therefore at some point intersect the circle? Please explain.

Thanks,
Raquel

yes.



no.

a negative slope is enough to guarantee that the line will hit somewhere in the 2nd and 4th quadrants, yeah.
however, the given circle is really, really small -- it accounts for only a tiny part of each of those quadrants. its radius is only 1; its top, bottom, leftmost, and rightmost points are at, respectively, (0, 1), (0, -1), (-1, 0), and (1, 0).

so, to prove that statement insufficient, consider the following 2 possibilities:
* line k, with slope -1/10, drawn through (0, 0) -- clearly this line will have 2 points of intersection with the circle.
* line k, with slope -1/10, drawn through, say, (0, 1000) -- this line will pass WAY over the circle, since its y-intercept is a thousand units above the x-axis.

hth.

thanks so much Ron.

sure thing

Got this problem in my GMAT prep yesterday, thought it's C too because I was thinking about the same thing as this person said:

In the geometry session with Ron, we were told that lines do not stop and that you can shift them up and down on coordinate plane.

But today, as I look back and after reading the advanced quant strategy book (believe Ron contributed a lot to it :) ), I realized the gap in my reasoning:

If you draw it based on the description, you will see that what this question is really getting is what's the y-intercept of line K. Because if the y-intercept of line K is 1 or -1, or between those two numbers, it will definitely hit circle C.

In order to find out the y intercept, under the given condition in this problem, we must be able to pin down the line, meaning we have to know both the steepness of the line (slope) and what point it goes through, so we can write a formula for it and calculate things such as y-intercept, x-intercept, what other fixed points it can pass through, what quandrant etc..

Because with the slope, the steepness of the line is fixed, but location isn't.

With the point the line passes through, it is constrained in location but not in slope.

When you combine both statements, can you really pin the line down?

No, turns out that 1 tricks us by allowing us to think that we get the location of the line but doesn't really because this is inequality, and with inequality, you can have many different cases, if you think about the extreme cases of where line k can be, its x intercept can be at 2, or at 1 million, so its location is still flexible. Thus, we cannot pin the line down even with the slope and hence we cannot write the formula for this line to find out its y-intercept.

I highly recommend the advanced quant strategy book, inside you will find a lot of good strategies for problems including geometry visualization.

:)

thanks for sharing your thoughts!

_________________
Tim Sanders
Manhattan GMAT Instructor

In a hurry I mistook slope of -1/10 as -10 and answered "C". Could you please confirm if that had been a correct answer if slope would have been -10?

no sir. still (e). in fact, the answer is still (e) even if the given slope is astronomically large (like -10,000,000).

it's a bit harder to visualize in those cases, but, if the x-intercept is close enough to the circle (like x = 1.0000000000000001) then those lines, too, will intersect the circle.

Hi there,
can you please elaborate on if the line intercepts at the x axis at 1milion, 0, with a negative slope (as stated as almost a horizontal line), how can it NOT intersect the circle? I'm having trouble visualizing this...

Thanks.

it sounds like you may have an incorrect concept of what a negative slope is. if the slope is -1, the line will definitely not intersect the circle (it will be very far from the circle in fact). however, if the slope is close enough to 0 (i.e. the line is close enough to horizontal), it can be made to pass through the circle..

_________________
Tim Sanders
Manhattan GMAT Instructor