Challenge me on my method to solving this problem

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Hi everyone,

I came across a tough problem that I (eventually) figured out, but would like someone to verify that my logic makes sense, or propose a better way of doing the problem.

(OG 12 Diagnostic Test Prob #39)

In the xy-plane, if line k has negative slop and passes through the point (-5, r), is the x-intercept of line k positive?

(1) The slop of the line k is -5
(2) r>0


My way of doing the problem:

1. we know that x intercept would be (x,0)
2. we know that slope of line = (Y1-Y2) / (X1-X2)
3. The equation we're solving becomes slope = (r-0)/(-5-x)

(1) -5 = (r-0)/(-5-x) --> doesn't help, already know that slope is negative, and still have two variables to solve

(2) - slope = (r>0) / (-5-x) --> this tells us that the numerator is positive, so the denominator must be negative.

If x>0, then numerator is negative.
If x=0, then numerator is still negative
If -4<x<0, then numerator still negative

So, since x could be positive or negative or 0, (2) is no good.

Combining the two doesn't tell us anything new.

-5 = (r>0) / (-5-x)

x could still be positive or negative.

thoughts?
-v
Source: — Data Sufficiency |

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by DanaJ » Thu Aug 20, 2009 11:06 pm
I think you've got right. I don't really know if there are any faster methods, but my guess would be no.

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by tohellandback » Thu Aug 20, 2009 11:20 pm
IMO yes there is a faster way
the x-intercept is 0 when the line passes through the origin. so you can imagine in the x-y plane that x- intercept will be positive when the Y intercept is positive
Y=mx+C. so C should be positive.
none of the statements guarantee that.
1) Nothing new. we already know that slope is negative
2)r doesn't matter. we need to know about C
The powers of two are bloody impolite!!

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thanks

by vladig172 » Thu Aug 27, 2009 8:07 am
for the responses guys

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by vidhya16 » Sun Jan 08, 2012 1:37 pm
tohellandback wrote:IMO yes there is a faster way
the x-intercept is 0 when the line passes through the origin. so you can imagine in the x-y plane that x- intercept will be positive when the Y intercept is positive
Y=mx+C. so C should be positive.C
none of the statements guarantee that.
1) Nothing new. we already know that slope is negative
2)r doesn't matter. we need to know about

Can some one clarify how the tohellandback has made an assumption that x-intercept is positive when Y intercept is positive?

Vidhya

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by pemdas » Sun Jan 08, 2012 3:59 pm
if you continue applying algebra to such kind of problems, and in general to most of the coordinate geometry questions on GMAT, you risk to spend 4+ minutes on single question and come with unsecured answer choice.

Don't do that

Draw a pic and prove anything you want
the only way for +ve slope (line directed from the upper left to the lower right) to cross the positive side of x -abscess is to cross the positive side of y-abscess first

Otherwise you either have straight line x=0,y=0 or -ve slope when the line is directed from the lower left to the upper right
vidhya16 wrote:
tohellandback wrote:IMO yes there is a faster way
the x-intercept is 0 when the line passes through the origin. so you can imagine in the x-y plane that x- intercept will be positive when the Y intercept is positive
Y=mx+C. so C should be positive.C
none of the statements guarantee that.
1) Nothing new. we already know that slope is negative
2)r doesn't matter. we need to know about

Can some one clarify how the tohellandback has made an assumption that x-intercept is positive when Y intercept is positive?

Vidhya
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by GMATGuruNY » Mon Jan 09, 2012 3:55 am
Many slope questions are best solved by DRAWING. Check my posts at the link below. The first solves by drawing (recommended); the second offers an algebraic solution.

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