Is slope of line L1 less than the slope of line L2, where L1 and L2 are non parallel?
I. The two lines intersect each other in the first quadrant.
II. Y-intercept of L1 is less that Y-intercept of L2.
Thank you in advance! =)
Data Sufficiency - Coordinate Geometry
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i hope the slope here is taking consideration of signs and not just considering absolute values
i feel the answer sud be C for with both statement i can conclude that slope of line L1 is greater than slope of line L2
i feel the answer sud be C for with both statement i can conclude that slope of line L1 is greater than slope of line L2
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i am not sure how to draw graphs here on this site but i can help u as how to do it : draw the 4 quadrants .i guess u might have come to a position where u know that individually u cannot solve this question .so taking them together u will observe that no matter how u draw ,as long as u meet both the condition , the two lines will make the slopes such that line L1 has greater slope than line L2
this is just an outline as how to solve it and trust me this method doesn't take much time its just that until and unless u draw these random lines its little difficult to comprehend
this is just an outline as how to solve it and trust me this method doesn't take much time its just that until and unless u draw these random lines its little difficult to comprehend
Last edited by neha24 on Sun Mar 10, 2013 2:02 am, edited 1 time in total.
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Drawing the lines as neha24 described is the easiest method to solve this problem. But if you are having problem with that approach then you can follow this mathematical approach but this is way more time consuming.
Let us assume the equations of L1 and L2 are y = ax + b and y = px + q, respectively.
Hence, slopes of the lines are a and p and y-intercept of the lines are b and q, respectively.
As the lines are non-parallel, we know that a ≠b.
We need to determine whether, a < p or not.
Statement 1: Let us assume that the intersection point is (X, Y), where X > 0 and Y > 0.
Hence, Y = aX + b and Y = pX + q
Solving these two equations, X = (q - b)/(a - p) and Y = (aq - bp)/(a - p)
Hence, (q - b)/(a - p) > 0 and (aq - bp)/(a - p) > 0
In other words, (q - b), (a - p) and (aq - bp) are of same sign.
Clearly this is not enough to conclude whether a < p or not.
Not sufficient
Statement 2: b < q
Clearly this is not enough to conclude whether a < p or not.
Not sufficient
1 & 2 Together: Now we know that, (q - b) > 0
Hence, (a - p) is positive too.
Hence, a > p
Sufficient
The correct answer is C.
Let us assume the equations of L1 and L2 are y = ax + b and y = px + q, respectively.
Hence, slopes of the lines are a and p and y-intercept of the lines are b and q, respectively.
As the lines are non-parallel, we know that a ≠b.
We need to determine whether, a < p or not.
Statement 1: Let us assume that the intersection point is (X, Y), where X > 0 and Y > 0.
Hence, Y = aX + b and Y = pX + q
Solving these two equations, X = (q - b)/(a - p) and Y = (aq - bp)/(a - p)
Hence, (q - b)/(a - p) > 0 and (aq - bp)/(a - p) > 0
In other words, (q - b), (a - p) and (aq - bp) are of same sign.
Clearly this is not enough to conclude whether a < p or not.
Not sufficient
Statement 2: b < q
Clearly this is not enough to conclude whether a < p or not.
Not sufficient
1 & 2 Together: Now we know that, (q - b) > 0
Hence, (a - p) is positive too.
Hence, a > p
Sufficient
The correct answer is C.
Anju Agarwal
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i received a message about this thread. just a follow-up comment:
see, the deal is this:
the vast majority of coordinate geometry problems on the exam -- especially in DS -- are specifically designed to test your ability to visualize the concepts.
most of them are much easier (and much faster) to solve if you approach them by trying to visualize the situation. you should only use y = mx + b type stuff as a last resort, in cases when you just can't use visualization (either because it's impossible, or because you just can't imagine what the situation would look like).
while anju's solution is clearly correct, i would give different advice in case of trouble. my advice would be, "if you have trouble with that approach, then try to develop a better visual understanding".Anju@Gurome wrote:Drawing the lines as neha24 described is the easiest method to solve this problem. But if you are having problem with that approach then you can follow this mathematical approach but this is way more time consuming.
...
see, the deal is this:
the vast majority of coordinate geometry problems on the exam -- especially in DS -- are specifically designed to test your ability to visualize the concepts.
most of them are much easier (and much faster) to solve if you approach them by trying to visualize the situation. you should only use y = mx + b type stuff as a last resort, in cases when you just can't use visualization (either because it's impossible, or because you just can't imagine what the situation would look like).
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron