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by Night reader » Tue Jan 11, 2011 12:56 am
In the xy-plane, if line m has negative slope and passes through the point (r; 6), is the x-intercept of line m positive?

(1) The slope of line m is - 1/2

(2) r > 0
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by 4GMAT_Mumbai » Tue Jan 11, 2011 1:35 am
Hi,

Thanks for the interesting Q ...

1. The x intercept could either be -ve or +ve as it entirely depends on the value of r. If the line meets the x-axis at (a,0);

then the slope = (6-0) / (r - a) = -0.5

12 = a-r.

Whether a is +ve or not depends on r. Hence, insuff ...

2. r > 0.

then slope = (6-0) / (r - a) = some -ve number

So, r - a < 0

a > r

a > a positive r.

So, a also has to be +ve.

Answer: B ?
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by Night reader » Tue Jan 11, 2011 2:46 am
4GMAT_Mumbai wrote:Hi,

Thanks for the interesting Q ...

1. The x intercept could either be -ve or +ve as it entirely depends on the value of r. If the line meets the x-axis at (a,0);

then the slope = (6-0) / (r - a) = -0.5

12 = a-r.

Whether a is +ve or not depends on r. Hence, insuff ...

2. r > 0.

then slope = (6-0) / (r - a) = some -ve number

So, r - a < 0

a > r

a > a positive r.

So, a also has to be +ve.

Answer: B ?
flawless reasoning and correct answer! thanks

BTW, when the e-book of 4gmat on number theory and inequalities will be available? I got one for perms, prob and combin. - awesome stuff
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by Anurag@Gurome » Tue Jan 11, 2011 3:52 am
Night reader wrote:In the xy-plane, if line m has negative slope and passes through the point (r; 6), is the x-intercept of line m positive?

(1) The slope of line m is - 1/2
(2) r > 0
Sometimes the problems on coordinate geometry can be easily solved by drawing or visualizing the scenario. This problem is a good example to learn that. See the below figure.
Image

Statement 1: two lines are shown with negative slope (I've not shown the slope as the value of the slope doesn't matter), one of them has positive x-intercept and other one has negative.

Not sufficient

Statement 2: As r is positive, the point (r, 6) lies on the first quadrant. Any line having negative slope and passing through any point on the first quadrant must have a positive x-intercept.

Sufficient.

The correct answer is B.
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by ankur.agrawal » Tue Jan 11, 2011 5:35 am
Night reader wrote:
4GMAT_Mumbai wrote:Hi,

Thanks for the interesting Q ...

1. The x intercept could either be -ve or +ve as it entirely depends on the value of r. If the line meets the x-axis at (a,0);

then the slope = (6-0) / (r - a) = -0.5

12 = a-r.

Whether a is +ve or not depends on r. Hence, insuff ...

2. r > 0.

then slope = (6-0) / (r - a) = some -ve number

So, r - a < 0

a > r

a > a positive r.

So, a also has to be +ve.

Answer: B ?
flawless reasoning and correct answer! thanks

BTW, when the e-book of 4gmat on number theory and inequalities will be available? I got one for perms, prob and combin. - awesome stuff
Hi dude. I am facing sum hiccups pn P& C & Probability. Is the E-book of 4GMAT a good buy.
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by anshumishra » Tue Jan 11, 2011 5:49 am
Anurag@Gurome wrote:
Night reader wrote:In the xy-plane, if line m has negative slope and passes through the point (r; 6), is the x-intercept of line m positive?

(1) The slope of line m is - 1/2
(2) r > 0
Sometimes the problems on coordinate geometry can be easily solved by drawing or visualizing the scenario. This problem is a good example to learn that. See the below figure.
Image

Statement 1: two lines are shown with negative slope (I've not shown the slope as the value of the slope doesn't matter), one of them has positive x-intercept and other one has negative.

Not sufficient

Statement 2: As r is positive, the point (r, 6) lies on the first quadrant. Any line having negative slope and passing through any point on the first quadrant must have a positive x-intercept.

Sufficient.

The correct answer is B.
Nicely done Anurag !
Graphical approach helps a lot. Many a times, it helps us from making silly mistakes.
Thanks
Anshu

(Every mistake is a lesson learned )
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