| Author |
Message |
|
Guest
|
Post subject: In the figure above, if x and y are each less than 90 and PS Posted: Mon Jun 23, 2008 10:47 am |
|
|
|
|
In the figure above, if x and y are each less than 90 degrees and PS ll QR, is the length of segment PQ less than the length of segment SR?
____p_________s____
/ /
____q_x__________r_y_
Data suff-gmat prep exam 1
in the above figure y is outside the slanted line and x is within the slanted line/
1 x>y slanted line runs from p to q and s to r
2. x+y>90
ANSWER IS A.
How can i go about solving this problem?[/url]
|
|
 |
|
 |
|
RonPurewal
|
Post subject: Re: In the figure above, if x and y are each less than 90 an Posted: Thu Jun 26, 2008 4:04 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
|
hi -
that was a valiant attempt, but these forums auto-format your text in such a way that the diagrams won't look the same way they do when you're typing them. furthermore, i know of no way to disable this auto-formatting.
therefore, we'd appreciate it if you could post an image of this problem. if you don't know how to do that, post back and i'll tell you.
|
|
 |
|
 |
|
Guest
|
Post subject: Posted: Thu Jun 26, 2008 11:19 am |
|
|
|
|
How can I go about posting this image?
Greatly appreciated...
|
|
 |
|
 |
|
RonPurewal
|
Post subject: Posted: Sun Jun 29, 2008 3:39 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
Anonymous wrote: How can I go about posting this image? Greatly appreciated...
here's the easiest way i know of:
* take a screen shot of the problem. (i'm a mac user, and you're definitely not (because gmatprep doesn't run on macs) so i don't know exactly what application you'd use to do this)
* make sure that the screen shot is saved as a JPG, GIF, or PNG graphic file. (if it's stored as something else, such as .bmp or .tiff, then you might have to open it up in a graphics program and re-save it in the different format)
* after you've saved the file, go to an image hosting website. there are tons of these, but two that i've found particularly reliable are
http://xs.to
and
http://supload.com
* go to 'upload images', pick the file off your desktop/hard drive, click the box that says you accept their terms and conditions, and you're ready to go.
* after you upload the image, copy and paste the url's that appear.
|
|
 |
|
 |
|
Guest
|
Post subject: Posted: Sun Jun 29, 2008 9:02 pm |
|
|
|
Hi Ron,
I have uploaded the diagram you requested.
Thanks

|
|
 |
|
 |
|
RonPurewal
|
Post subject: Posted: Fri Jul 04, 2008 6:25 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
|
here's a way of thinking about it if you don't know trigonometry.
statement (1)
since angle X is bigger than angle Y, it follows that segment PQ is steeper (i.e., has a greater slope) than segment RS.
imagine drawing perpendiculars (which in this diagram would be vertical lines) down from P and S, and considering the right triangles thereby formed.
the vertical legs of those right triangles would have the same length, because they're drawn between the same parallel lines.
the horizontal leg of the triangle with hypotenuse PQ would be shorter, though, because the slope (= rise/run) is greater. since "rise" is identical, as just mentioned, the fact that (rise/run) is greater means that "run" must be smaller.
because the vertical legs have the same length and the horizontal leg of the left-hand triangle is shorter, it follows that the left-hand hypotenuse (i.e., PQ) is shorter.
sufficient.
(2)
this statement is symmetric in x and y, meaning that you can switch x and y without consequence.
consider two cases in which this happens: say, x = 40 and y = 60, and then x = 60 and y = 40.
in the latter case, the reasoning is the same as for statement (1); in the former case, it's the opposite, and PQ is now longer.
insufficient.
answer = a
--
alternatively, you could just try SKETCHING A BUNCH OF DIAGRAMS satisfying each of the statements. if you draw sketches that are halfway decent, it will soon be apparent that it's impossible to draw statement (1) without pq being shorter.
|
|
 |
|
 |
|
Guest
|
Post subject: Posted: Sat Jul 05, 2008 11:23 am |
|
|
|
|
Hi Ron,
Your feedback on this problem is greatly appreciated. I have a question with respects to a portion of your explanation, that is, the horizontal leg of the triangle with hypotenuse PQ would be shorter, though, because the slope (= rise/run) is greater. since "rise" is identical, as just mentioned, the fact that (rise/run) is greater means that "run" must be smaller. If the rise is identical, then how does one know that the run must be smaller? Is it because there is a direct relationship between the angle and the slant? As you mentioned, the greater the angle the steeper the line(the greater the slope), making the line larger. Also, is the reverse of what you have stated with respects to angle x also true for angle y, that is, the smaller the angle the less steep the line(the smaller the slope), making line SR larger?
Your feed back is greatly appreciated... Once again thanks for your help....
|
|
 |
|
 |
|
RonPurewal
|
Post subject: Posted: Sat Jul 12, 2008 4:46 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
Anonymous wrote: If the rise is identical, then how does one know that the run must be smaller? pure algebra; this has nothing to do with geometry. if a, b, and c are positive numbers and the ratio a/b is bigger than the ratio a/c, then b must be smaller than c. or: smaller denominators make bigger fractions. Anonymous wrote: As you mentioned, the greater the angle the steeper the line(the greater the slope), making the line larger. no, that would make the hypotenuse shorter, not longer. try drawing a couple of lines out yourself: one that's almost perpendicular to the 2 horizontal lines (meaning the angle is close to 90º), and one that's sharply oblique to them (angle 45º or less). you'll notice that the one making an angle close to 90º is clearly shorter than the other one. or: imagine that you have to run from one end of a football field to the other end. if you want the shortest path, what do you do? that's right: you run perpendicular to the boundary lines. the more diagonal your path is (= smaller angle), the longer the path. Anonymous wrote: Also, is the reverse of what you have stated with respects to angle x also true for angle y, that is, the smaller the angle the less steep the line(the smaller the slope), making line SR larger?
correct this time: the smaller angles create lines that are more oblique, and therefore longer.
|
|
 |
|
 |
|
shantascherla
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Mon May 16, 2011 4:01 am |
|
 |
| Course Students |
|
|
Posts: 21
|
|
Ron/Jamie
Is there another approach to this question -
|
|
 |
|
 |
|
RonPurewal
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Sat May 21, 2011 4:34 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
shantascherla wrote: Ron/Jamie
Is there another approach to this question - please be more specific than this, as there is already a wealth of information in this thread. "another approach" besides what? i.e., what techniques are off limits for you? what do you understand and/or not understand from the preceding explanations? thanks.
|
|
 |
|
 |
|
fegoga
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Fri Jun 17, 2011 2:36 pm |
|
 |
| Students |
|
|
Posts: 1
|
|
shantascherla,
here is another approach, as in another, as in different; perhaps you mean concise, aphoristic, and/or succint, or maybe allegorical, so here it goes:
Since bottmom and top line are parallel, they could be thought of as floor and roof of a typical adobe house somewhere near the equator. If you put two poles that are at 90 degrees respective to the floor inside the house ( or it could be in the porch too), then they must be the same length to reach the roof.
So, if the sum of the lenght of the angles of both your poles respective to the floor is greater than 90 degrees, it just means that they both (angles) may or may not be each 90 degrees. So (2) is not sufficient.
Next scene: if you incline your "y" pole to the side, then the angle it makes with the floor becomes smaller, thus making "x" and "y" not equal, and the pole must become larger to still touch the roof, this makes (1) sufficient.
Hope that helps
|
|
 |
|
 |
|
tim
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Thu Jul 21, 2011 6:51 pm |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 1779 Location: Southwest Airlines, seat 21C
|
|
Interesting. i'll leave it to shantascherla to let us know if that helped.. :)
_________________ Tim Sanders Manhattan GMAT Instructor
|
|
 |
|
 |
|
rahuljawale
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Sun Nov 13, 2011 10:04 am |
|
 |
| Students |
|
|
Posts: 1
|
|
Sorry to dig an old thread.
I believe there is certain bit of ambiguity here. My understand of the problem was that that the angle x could increase beyond 90 degrees, in which case, at the complementing angle of y , PQ would have the same length as that of SR and thus, the statement 1 may not help us to provide a definite answer.
Please correct me if I am wrong.
|
|
 |
|
 |
|
RonPurewal
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Wed Nov 23, 2011 6:33 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 6765
|
rahuljawale wrote: Sorry to dig an old thread.
I believe there is certain bit of ambiguity here. My understand of the problem was that that the angle x could increase beyond 90 degrees no. the very first words of the problem statement (and the title of this forum thread) are, "in the figure above, if x and y are each less than 90 degrees..."
|
|
 |
|
 |
|
rachelhong2012
|
Post subject: Re: In the figure above, if x and y are each less than 90 and PS Posted: Fri Dec 02, 2011 5:38 pm |
|
 |
| Course Students |
|
|
Posts: 39
|
|
Hi,
I wonder if my approach to this problem is right. It took a long time to figure out though:
I tried to rephrase the question:
The figure PQRS seems like a parallelogram to me and the fact that question is asking whether length of segment PQ is less than length of segment SR makes me think of the rule that in parallelogram, opposite sides are equal. But since the question only tells us that PS is parallel to QR, and in order to prove this is a parallelogram, I need to know whether side PQ is parallel to side SR. If I know that angle X is equal to angle Y, then PQ must be parallel to SR, thus making this a parallologram and side PQ equal to side SR and giving us a definite answer for the question.
Statement 1 tells me the two angles aren't equal, thus a no to all the conditions stated above, so it's sufficient.
Statement 2 doesn't help answering the question of if angle X is equal to angle Y, insufficent.
Is this right?
Thanks!
|
|
 |
|
 |
|