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# gmat prep DS : slopes

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Senior Manager
Joined: 02 Dec 2007
Posts: 433
gmat prep DS : slopes [#permalink]

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26 Sep 2008, 12:40
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Intern
Joined: 20 Aug 2008
Posts: 27
Re: gmat prep DS : slopes [#permalink]

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30 Sep 2008, 12:47
Hi,

You actually need to draw this figure.

C is correct, The reason is from the first statement the two lines, would either cross the X axis at a point where x>0 or here x<0. In both the cases we cannot determine what would be the product of the slopes.

Coming to 2nd statement since it is negative, it means that one line crosses the y axis at a point where y>0 and the other line crosses the y axis at a point where y<0. This situation alone cant help. If we combine both, then you will find that it is possible only when x>0 - from teh inference of statement 1. best , is if you can draw a diagram.

Award points if you feel. Let me know for any further clarifications

Regards,
Max
Intern
Joined: 29 Sep 2008
Posts: 46
Re: gmat prep DS : slopes [#permalink]

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30 Sep 2008, 13:04
1
KUDOS
the basic equation of a line is y = mx + c. where the slope is m, y-intercept is c and x-intercept is -c/m
let's say:

l is y = m1x + c1 with y-intercept = c1 and x-intercept = -c1/m1
k is y = m2x + c2 with y-intercept = c2 and x-intercept = -c2/m2

statement I - c1c2/m1m2 > 0. does not say anything about c1c2 so we cannot say if m1m2 is negative or not

statement I - c1c2 < 0. does not say anything about m1 and m2 so we cannot say if m1m2 is negative or not.

I and II - if c1c2 < 0 then m1m2 has to be negative.

Thus C
VP
Joined: 05 Jul 2008
Posts: 1379
Re: gmat prep DS : slopes [#permalink]

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30 Sep 2008, 19:38
aim2010 wrote:
the basic equation of a line is y = mx + c. where the slope is m, y-intercept is c and x-intercept is -c/m
let's say:

l is y = m1x + c1 with y-intercept = c1 and x-intercept = -c1/m1
k is y = m2x + c2 with y-intercept = c2 and x-intercept = -c2/m2

statement I - c1c2/m1m2 > 0. does not say anything about c1c2 so we cannot say if m1m2 is negative or not

statement I - c1c2 < 0. does not say anything about m1 and m2 so we cannot say if m1m2 is negative or not.

I and II - if c1c2 < 0 then m1m2 has to be negative.

Thus C

This is what I did. How ever, I was wondering (4,3) were never used when I solved. Then I started thinking Did I miss some thing here?

What is the OA?
Intern
Joined: 29 Sep 2008
Posts: 46
Re: gmat prep DS : slopes [#permalink]

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30 Sep 2008, 21:52
icandy wrote:
aim2010 wrote:
the basic equation of a line is y = mx + c. where the slope is m, y-intercept is c and x-intercept is -c/m
let's say:

l is y = m1x + c1 with y-intercept = c1 and x-intercept = -c1/m1
k is y = m2x + c2 with y-intercept = c2 and x-intercept = -c2/m2

statement I - c1c2/m1m2 > 0. does not say anything about c1c2 so we cannot say if m1m2 is negative or not

statement I - c1c2 < 0. does not say anything about m1 and m2 so we cannot say if m1m2 is negative or not.

I and II - if c1c2 < 0 then m1m2 has to be negative.

Thus C

This is what I did. How ever, I was wondering (4,3) were never used when I solved. Then I started thinking Did I miss some thing here?

What is the OA?

The point is given as a trap to attract people to draw lines and form cases and waste time. without the point too the question would remain the same, just that when they find no other way, people would straightaway jump to the equation based approach. Just my feeling, I might be terribly wrong.
OA is C, see image in the first post.
Re: gmat prep DS : slopes   [#permalink] 30 Sep 2008, 21:52
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