XY Plane- GMATPrep

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XY Plane- GMATPrep

by shahab03 » Mon Feb 22, 2010 6:50 pm
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by papgust » Mon Feb 22, 2010 7:08 pm
Must be C

The question really asks whether lines l and l are perpendicular? The product of slopes of Perpendicular lines are always -1.

1. No info about line k. Insufficient.
2. No info about line l. Insufficient.

Combined,
Line K has a slope of 2 (m1) [Points are (0,0) and (1,2). Slope = 2-0 / 1-2 = 2]

Line L has a slope of -1/2 (m2) [Equation of the line is 2x+4y=8. Slope = -2/4 = -1/2]

m1 * m2 = 2 * -1/2 = -1 (Lines are perpendicular)

Sufficient.

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by lunarpower » Tue Feb 23, 2010 6:16 am
papgust wrote:Must be C
yeah.

what a ... strange question.
The question really asks whether lines l and l are perpendicular? The product of slopes of Perpendicular lines are always -1.

1. No info about line k. Insufficient.
2. No info about line l. Insufficient.

so far so good.
Combined,
Line K has a slope of 2 (m1) [Points are (0,0) and (1,2). Slope = 2-0 / 1-2 = 2]

Line L has a slope of -1/2 (m2) [Equation of the line is 2x+4y=8. Slope = -2/4 = -1/2]

m1 * m2 = 2 * -1/2 = -1 (Lines are perpendicular)

Sufficient.
totally unnecessary work. don't waste your time.

here's all you have to do:
* statement (1): there is EXACTLY ONE LINE that does this.
* statement (2): there is EXACTLY ONE LINE that does this.
* so, if you have them together, IT'S EITHER "YES" OR "NO".
sufficient.
done.

TAKEAWAY:
DO NOT GET YOUR HEAD STUCK IN "PROBLEM SOLVING MODE" WHEN YOU WORK DATA SUFFICIENCY PROBLEMS.

remember - a lot of computation is unnecessary when you're doing a DS problem. if you know that you CAN get an answer, that's all you need - there is no reason at all to spend time going through the steps.
Ron has been teaching various standardized tests for 20 years.

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by papgust » Tue Feb 23, 2010 6:21 pm
Hey Ron,

Even i'm aware that the computations are unnecessary. I just wanted to detail the concept behind it for the question poster. Probably, i should have mentioned in the end clearly that these steps are unnecessary and i just intended to explain the concept.

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by shahab03 » Tue Feb 23, 2010 7:25 pm
papgust wrote:Must be C

The question really asks whether lines l and l are perpendicular? The product of slopes of Perpendicular lines are always -1.

1. No info about line k. Insufficient.
2. No info about line l. Insufficient.

Combined,
Line K has a slope of 2 (m1) [Points are (0,0) and (1,2). Slope = 2-0 / 1-2 = 2]

Line L has a slope of -1/2 (m2) [Equation of the line is 2x+4y=8. Slope = -2/4 = -1/2]

m1 * m2 = 2 * -1/2 = -1 (Lines are perpendicular)

Sufficient.

can you please tell me how did you figure out that for line L has slope = -1/2? we are given x-intercept and y-intercept. how did you derive slope from the two intercepts?

thanks

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by thephoenix » Tue Feb 23, 2010 7:35 pm
shahab03 wrote:
papgust wrote:Must be C

The question really asks whether lines l and l are perpendicular? The product of slopes of Perpendicular lines are always -1.

1. No info about line k. Insufficient.
2. No info about line l. Insufficient.

Combined,
Line K has a slope of 2 (m1) [Points are (0,0) and (1,2). Slope = 2-0 / 1-2 = 2]

Line L has a slope of -1/2 (m2) [Equation of the line is 2x+4y=8. Slope = -2/4 = -1/2]

m1 * m2 = 2 * -1/2 = -1 (Lines are perpendicular)

Sufficient.

can you please tell me how did you figure out that for line L has slope = -1/2? we are given x-intercept and y-intercept. how did you derive slope from the two intercepts?

thanks
x inter cept is 4 means its coordinates are 4,0(x1,y1)

and y intercept is 2 means its coorniates are 0,2(x2,y2)

slope m=(y2-y1)/(x2-x1)----->(2-0)/(0-4)=-1/2

hth