|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 24 Jun 2003
Posts: 54
Followers: 0
Kudos [?]:
1
[0], given: 0
|
In the attached figure, what is the length of PQ times the [#permalink]
10 Aug 2003, 14:36
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
In the attached figure, what is the length of PQ times the length of RS?
(1) The length of PQ is 5
(2) The length of QR times the length of PR is equal to 12
What is sufficient? (1), (2), (both together), (both by themselves), (neither is sufficient)
I think there is an error in PR's solution to the problem.
This problem is from the 2004 edition (problem set)
Let me know what you think, and then i'll post the PR solution, as well as my solution.
Thanks,
Htown
|
|
|
|
|
|
|
Manager
Joined: 24 Jun 2003
Posts: 54
Followers: 0
Kudos [?]:
1
[0], given: 0
|
That is correct.
However, I chose D because I felt that A was sufficient as well.
Since RS is perpendicular to QP, can't you assume that QS is = to 2.5 (half of 5)? If that is the case, you can assume that SR is also 2.5 because Triangle QRS is a 90, 45, 45.
Now you have PQ as stated in (1) and SR which will give you what you need.
Obviously PR has a problem with me defining QS as 2.5
What do you think?
Htown
|
|
|
|
|
|
GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
Kudos [?]:
9
[0], given: 0
|
htown wrote: That is correct.
However, I chose D because I felt that A was sufficient as well. Since RS is perpendicular to QP, can't you assume that QS is = to 2.5 (half of 5)? If that is the case, you can assume that SR is also 2.5 because Triangle QRS is a 90, 45, 45.
Now you have PQ as stated in (1) and SR which will give you what you need.
Obviously PR has a problem with me defining QS as 2.5
What do you think?
Htown
There is no justification for you to assume that QS is 2.5, even if it "looks" like it in the picture. All you know is that the hypotenuese of a right triangle is 5. The legs could be anything, hence the QS could be anything.
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
|
|
|
|
|
|
Manager
Joined: 24 Jun 2003
Posts: 54
Followers: 0
Kudos [?]:
1
[0], given: 0
|
I agree that the legs could be "anything," but if you have a 45, 45, 90 triangle and you know one length, you have them all. How can RS form a perpendicular with PQ and not intersect at the midpoint of PQ?
Thanks
Htown
|
|
|
|
|
|
GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
Kudos [?]:
9
[0], given: 0
|
htown wrote: I agree that the legs could be "anything," but if you have a 45, 45, 90 triangle and you know one length, you have them all. How can RS form a perpendicular with PQ and not intersect at the midpoint of PQ?
Thanks
Htown
Draw ANY right triangle with its hypotenuese on the bottom. Now simply drop a line from the tip of the triangle to the hypotenuese, which is simply the attitude. Unless your triangle was a 45-45-90, the line you drop will certainly not bisect the hypotenuese but still be perpendicular to it.
Note: don't just ask questions, try some of these out for yourself.
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|