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29 Sep 2009, 01:21
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (01:54) correct
45%
(01:49)
wrong
based on 569
sessions
History
Date
Time
Result
Not Attempted Yet
If lines k and j are in the xy–coordinate plane, is the slope of line k greater than the slope of line j ?
(1) The x-intercept of line k is greater than the x-intercept of line j
(2) Lines k and j intersect at (7, 2)
(1) The x-intercept of line k is greater than the x-intercept of line j
(2) Lines k and j intersect at (7, 2)
03 Oct 2009, 13:27
Kudos
Bookmarks
Sorry to disappoint you guys, but it seems that the answer should be E.
First of all, to solve this question, think that the best way is the graphic approach:
We all agree that (1) and (2) alone are not sufficient.
Let's combine --> draw only the X-axis. The k intercept is greater than the j intercept --> the k intercept (k1) point is right to the j intercept point (j1) --> The interception of lines is above the X-axis (point 7;2) --> it will give us only three possible ways to draw the lines:
So (1)+(2) is not sufficient. Answer E.
BUT the way you were solving also should get E:
f(k) = mx + n
f(j) = tx + p
(1) The x-intercept of line k is greater than the x-intercept of line j --> 0 = mx + n; 0 = tx + p --> -n/m > -p/t or n/m < p/t --> not sufficient.
(2) Lines k and j intersect at (7, 2)
2 = 7m + n; 2 = 7t + p --> 7m + n = 7t + p --> not sufficient.
Combining the way you did:
n = 2 - 7m --> n/m = 2/m - 7
p = 2 - 7t --> p/t = 2/t - 7
n/m < p/t --> 2/m - 7 < 2/t - 7 --> 1/m < 1/t --> (t - m)/(mt) < 0
And here is the catch, from the above statement, you cannot determine whether m > t or not. (t = 1) < (m = 3) statement is true and (t = 1) > (m = -3) statement is also true.
Answer E.
First of all, to solve this question, think that the best way is the graphic approach:
We all agree that (1) and (2) alone are not sufficient.
Let's combine --> draw only the X-axis. The k intercept is greater than the j intercept --> the k intercept (k1) point is right to the j intercept point (j1) --> The interception of lines is above the X-axis (point 7;2) --> it will give us only three possible ways to draw the lines:
- A. j has a negative slope, k has a negative slope --> the slope of k > slope of j, (because j is closer to the vertical line);
B. j has a positive slope, k has a negative slope --> the slope of k < slope of j;
C. j has a positive slope, k has a positive slope --> the slope of k > slope of j, (because k is closer to the vertical line).
So (1)+(2) is not sufficient. Answer E.
BUT the way you were solving also should get E:
f(k) = mx + n
f(j) = tx + p
(1) The x-intercept of line k is greater than the x-intercept of line j --> 0 = mx + n; 0 = tx + p --> -n/m > -p/t or n/m < p/t --> not sufficient.
(2) Lines k and j intersect at (7, 2)
2 = 7m + n; 2 = 7t + p --> 7m + n = 7t + p --> not sufficient.
Combining the way you did:
n = 2 - 7m --> n/m = 2/m - 7
p = 2 - 7t --> p/t = 2/t - 7
n/m < p/t --> 2/m - 7 < 2/t - 7 --> 1/m < 1/t --> (t - m)/(mt) < 0
And here is the catch, from the above statement, you cannot determine whether m > t or not. (t = 1) < (m = 3) statement is true and (t = 1) > (m = -3) statement is also true.
Answer E.
General Discussion
29 Sep 2009, 05:26
Kudos
Bookmarks
I get B but I am not sure. OA pls?
Following is the reasoning:
stmt 1: we just have two points on the x-axis. we can draw many lines via these two points and having different slopes. Insuff.
stmt2: we have the intersection point in the first quadrant. Now, draw a line perpendicular from (7,2) to the x-axis. This will be a vertical line. Now select two points on the axis just to right and left of (7,0) on x-axis. Let these points be the x-intercepts of the lines. Now the slope of line with a greater x-intercept say, (7.0001, 0 ) will always be -ve [in order to intersect (7,2)] and the slope of the line with a smaller x-intercept say (6.9999, 0 ) will always be +ve [in order to intersect (7,2)]
Suff.
Following is the reasoning:
stmt 1: we just have two points on the x-axis. we can draw many lines via these two points and having different slopes. Insuff.
stmt2: we have the intersection point in the first quadrant. Now, draw a line perpendicular from (7,2) to the x-axis. This will be a vertical line. Now select two points on the axis just to right and left of (7,0) on x-axis. Let these points be the x-intercepts of the lines. Now the slope of line with a greater x-intercept say, (7.0001, 0 ) will always be -ve [in order to intersect (7,2)] and the slope of the line with a smaller x-intercept say (6.9999, 0 ) will always be +ve [in order to intersect (7,2)]
Suff.
. Thanks for explanation Bunuel! 
