DS:line

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DS:line

by 7806 » Mon Sep 06, 2010 12:19 am
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Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p?
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p

i am down with C/E...?? pl explain. thanks.

{ i think on -ve axis( say, -ve y axis)... if, y1=-4 and y2=-3, then y1 has greater y intercept than does y2). am i correct?.}
Source: — Data Sufficiency |

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by Rahul@gurome » Mon Sep 06, 2010 3:49 am
(1) We are given the point of intersection, but this does not give the slope of any lines.
So, (1) is NOT SUFFICIENT.

(2) y-intercept of line n > y intercept of line p, but this does not give us any indication on the slopes of two lines.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), if we take some points so that y-intercept of line n > y intercept of line p, which means higher line should be marked as n and lower should be p. So, the slope of n will always be less than slope of p.

The correct answer is [spoiler](C)[/spoiler].
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by 7806 » Mon Sep 06, 2010 4:47 am
thanks rahul...
my 2 Q-

1st: i think on -ve axis( say, -ve y axis)... if, y1=-4 and y2=-3, then y1 has greater y intercept than does y2). am i correct?.}

2nd-
ok.. say, intercept for line n=5, line p=3, and they passes from point(5,1). so, as i draw on rough paper a1(n)<a2(p); wherea1(n) angle for line n....

now, on -ve y axis, y=-5, p=-3 ... ( my 1st Q is related to this part)-i am considering that this time also y intercept is greater...considering absolute value.

so, when i draw line, a1(n)>a2(p).

..sorry, i am not able to draw fig.

..is considering absolute value is wrong...?? hope i may be..that why i am getting E.

pl confirm...!!! thanks.

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by Rahul@gurome » Mon Sep 06, 2010 5:01 am
7806 wrote:thanks rahul...
my 2 Q-

1st: i think on -ve axis( say, -ve y axis)... if, y1=-4 and y2=-3, then y1 has greater y intercept than does y2). am i correct?.}

2nd-
ok.. say, intercept for line n=5, line p=3, and they passes from point(5,1). so, as i draw on rough paper a1(n)<a2(p); wherea1(n) angle for line n....

now, on -ve y axis, y=-5, p=-3 ... ( my 1st Q is related to this part)-i am considering that this time also y intercept is greater...considering absolute value.

so, when i draw line, a1(n)>a2(p).

..sorry, i am not able to draw fig.

..is considering absolute value is wrong...?? hope i may be..that why i am getting E.

pl confirm...!!! thanks.
When you take y1=-4 and y2=-3, then intercept of y1 is lesser than intercept of y2, since -4 < -3.
Rahul Lakhani
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Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)