Data Sufficiency

Lines N & P lie on the XY plane. Is the slope of N < the slope of P?

1) N & P intersect at (5,1)
2) Y-int. of N > y-int. of P

The first statement really doesn't communicate anything beyond what it says at face value, so it's certainly insufficient by itself.

Statement (2) by itself is not terribly helpful either -- it lets us now that line N hits the y-axis higher up than line P does, but we don't know anything about how either of the lines is moving -- falling steeply, falling slowly, rising steeply, rising slowly, etc.

Now let's combine the statements. Remember that slope is rise/run, so we'll consider each line's slope by focusing on its trip from the point where it hits the y-axis to the point (5,1). The "run" in both of these cases is just 5. The "rise" for line N is (1 - N's y-intercept), and the "rise" for line P is (1 - P's y-intercept). But since we KNOW that N's y-intercept > P's y-intercept, we can conclude that we are subtracting a greater number from 1 in our expression for the rise of line N than we are in our expression for the rise of line P. Therefore the rise of line N will turn out to be a smaller number. So, returning to rise/run form, we wind up comparing N's slope "less"/5 to P's slope "greater"/5, so P's slope is definitively greater, i.e. the answer to the question is definitively yes. So the combination of (1) and (2) is sufficient.

Great stuff thank you.

how do you know that x coordinate of the point is 0

Y2-Y1/X2-X1>Y3-Y1/X3-X1

Given Y1,X1 also Y2>Y1,but no information is given X2,X3

Well, by definition, a y-intercept occurs when a line crosses the y-axis, and a line crosses the y-axis at the point whose x-coordinate is 0. Whenever you have an intercept, that means the other coordinate must be 0, or else the line would not be crossing an axis.

@Ashley, isn't the answer to this question 'Yes' and c

thanks

Ashley@VeritasPrep wrote:... the answer to the question is definitively no. So the combination of (1) and (2) is sufficient.

Ah, yes! I misread the direction of the inequality sign in the question. I will edit above--thanks! Fortunately on data sufficiency, this wouldn't have cost me a point , since definitively no and definitively yes both mean sufficient! But you're right, it should say "definitively yes."

pemdas wrote:@Ashley, isn't the answer to this question 'Yes' and c

thanks

Ashley@VeritasPrep wrote:... the answer to the question is definitively no. So the combination of (1) and (2) is sufficient.

Ashley@VeritasPrep wrote:Ah, yes! I misread the direction of the inequality sign in the question. I will edit above--thanks! Fortunately on data sufficiency, this wouldn't have cost me a point , since definitively no and definitively yes both mean sufficient! But you're right, it should say "definitively yes."

pemdas wrote:@Ashley, isn't the answer to this question 'Yes' and c

thanks

Ashley@VeritasPrep wrote:... the answer to the question is definitively no. So the combination of (1) and (2) is sufficient.

I have a general question regarding calculating slope. If you have two points on a line, how do you determine which is (x1, y1) and which is (x2, y2)? I think this is a vital step because getting this backwards would produce a very different result depending on the points.

Bens4vcobra wrote: I have a general question regarding calculating slope. If you have two points on a line, how do you determine which is (x1, y1) and which is (x2, y2)? I think this is a vital step because getting this backwards would produce a very different result depending on the points.

Actually, good news: you can use either point for point#1 and either for point#2, as long you remain consistent throughout the calculation. Here's why it works: Let's say (arbitrarily) you've got a line sloping downwards. If you call say the point on the left point#1 and the point on the right point#2, then your slope -- (y2-y1)/(x2-x1) -- will be a negative over a positive. Had you named the points flipped -- point#2 on the left, point#1 on the right -- then (y2-y1)/(x2-x1) would have been a positive over a negative. But notice that a negative over a positive comes out to the same thing as a positive over a negative. You get the same answer either way. For a line sloping upwards, you'll either wind up with a positive over a positive (if you call the point on the right point#2) or a negative over a negative (if you call the point on the left point#2), but again, a positive over a positive is the same as a negative over a negative.

Here's a more concrete example: Consider the points (-2,6) and (3,5).
Calculated with (-2,6) as point#1 and (3,5) as point#2, my (y2-y1)/(x2-x1) = (5-6)/(3-(-2)) = -1/5.
Calculated with (3,5) as point#1 and (-2,6) as point#2, my (y2-y1)/(x2-x1) = (6-5)/(-2-3) = 1/-5.
Same answer whichever way you do it.

On a semi-related note, it's always a good idea to run a visual check on slope. Think about moving in the same direction you read (left to right). If the line is moving upwards, its slope is guaranteed positive. If it's moving downwards, guaranteed negative. Moving straight across, slop is 0. The more steeply upwards a line is slanting, the greater a slope it has, and the more steeply downwards a line is slanting, the more deeply negative a slope it has. So if that ever doesn't jibe with a calculation you've done as to slope, go back and check your calculation!

PS - If you have an example of points for which you're worried that getting this backwards would produce a very different result (or a different result at all ), check them again. If you're still getting a different result depending on which you name which, post the example and we'll see why it's misbehaving, because it shouldn't be!

Ashley@VeritasPrep wrote:The first statement really doesn't communicate anything beyond what it says at face value, so it's certainly insufficient by itself.

Statement (2) by itself is not terribly helpful either -- it lets us now that line N hits the y-axis higher up than line P does, but we don't know anything about how either of the lines is moving -- falling steeply, falling slowly, rising steeply, rising slowly, etc.

Now let's combine the statements. Remember that slope is rise/run, so we'll consider each line's slope by focusing on its trip from the point where it hits the y-axis to the point (5,1). The "run" in both of these cases is just 5. The "rise" for line N is (1 - N's y-intercept), and the "rise" for line P is (1 - P's y-intercept). But since we KNOW that N's y-intercept > P's y-intercept, we can conclude that we are subtracting a greater number from 1 in our expression for the rise of line N than we are in our expression for the rise of line P. Therefore the rise of line N will turn out to be a smaller number. So, returning to rise/run form, we wind up comparing N's slope "less"/5 to P's slope "greater"/5, so P's slope is definitively greater, i.e. the answer to the question is definitively yes. So the combination of (1) and (2) is sufficient.

Lines N & P lie on the XY plane. Is the slope of N < the slope of P?

1) N & P intersect at (5,1)
2) Y-int. of N > y-int. of P

There is a very simple way of doing this without getting bogged down by all these details...

Every line in a 2D plane is of the form : y = mx + c ,
where (x,y) are the X-Y co-ordinates , m = slope of the line, c = y-intercept

Let n be the slope of line N, a = y-intercept of line N
p be the slope of line P, b = y-intercept of line P

Here as per
1)
Line N : 1 = 5n+a
Line P : 1 = 5p+b

We cannot say if n>p or n=p or n<p unless we know the relation between a and b
So 1 is not sufficient

2)
Y-int. of N > y-int. of P
or
a > b

This is not at all sufficient

If you combine 1 and 2 you know
Line N : 1 = 5n+a
Line P : 1 = 5p+b
and
a > b
So the relation b/w slopes of Lines N and P (n and p) can be found unambiguously.....

So C is the OA


You can also solve this if you want....But I will never recommend you to solve if you can get the answer without solving...After it is a DATA SUFFICIENCY question and not a PROBLEM SOLVING question.

krishp84 described an algebraic approach that I find can be efficient in many of the complicated coordinate geometry data sufficiency questions.

Here is a video explanation to this problem(graphical and algebraic approach):
https://www.gmatquantum.com/shared-posts ... try18.html

Dabral