Data Sufficiency

Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p

ans b

Statement 1 tells us where the lines intersect, but tells us nothing about either of the lines' slopes. If you draw a picture of two lines intersecting at (5,1), you can see that with the information given, we could label either one n. We could make the one with greater slope n, or the one with less slope n. So, we don't have enough information from statement 1.

Now let's consider statement 2, that says that the y intercept of n is greater than that of the y intercept of p. The slopes could be unequal (think about intersecting lines), or we could have parallel lines, in which case the slopes are equal. So, statement 2 is not sufficient on its own.

Now let's consider the statements together. The lines intersect at (5,1), and n has the higher y intercept. Let's look at 3 cases:

1) Both have y intercepts above y=1
Since n intersects higher, then we know n had further to descend, so its slope is steeper (but more negative) than p's. Thus, p has a greater slope.

2) n has intercept above y=1, p has intercept below
n would have a negative slope and p a positive, so p has a greater slope

3) Both have y intercepts below y=1
Both have positive slopes, but p has further to ascend. Thus, p has a greater slope.

Since combining the information tells us that p always has a greater slope, we have sufficient information with both statements and the answer is C, NOT B.

I agree with C.

The case that two lines are parellel should be considered.

VP_Tatiana wrote: Putting the two statements together, we know we have intersecting lines with n's intercept greater than p's. In that case, we know n has a greater slope than p, so that is sufficient information.

Thus, the answer is C, NOT B.

How is it possible?
Consider line n is defined by the equation y=1 and line p by y=x-4. So we have that this case satisfies both conditions: y intersection of n is greater than that of p and lines intersect each other in the point (5,1). However, the slope of line n is less than the slope of line p

Olika,

Thanks for catching my typo. I meant to say that n was steeper, not that it had a greater slope. In the case I was illustrating, n had a steeper DOWNWARD slope... thus more negative... thus smaller than p. I edited my original post to illustrate 3 distinct cases, to more explicitly show why that is true. Hope that helps.

Tatiana

Tatiana,

Can you explain this part? I'm having trouble understanding how we determined that n would have a negative slope?

Thanks!
r
-----------------------

1) Both have y intercepts above y=1
Since n intersects higher, then we know n had further to descend, so its slope is steeper (but more negative) than p's. Thus, p has a greater slope.

2) n has intercept above y=1, p has intercept below
n would have a negative slope and p a positive, so p has a greater slope

3) Both have y intercepts below y=1
Both have positive slopes, but p has further to ascend. Thus, p has a greater slope.

Got stuck with this problem in the test.

Later,while revieing --analyzed the problem.

We can arrive at the answer by using no values,just by sheer knowledge of slopes.

Hope i don't get stuck in the real thing.

Hi aj5105,

can you explain please?

thanks in advance!
r

@Montreal06
Just read the problem carefully,you too can crack it. (No values needed)

Consider Sn < Sp --- yes
Sn >=Sp ---no

Statement 1 ---it's not possible.
Sn can be less than or greater than Sp.

Statement 2 --- Imagine line n running parallel to x axis with y intercept 2.
Imagine line p cutting line n with a positive slope and y intercept 1.
Hence slope of n < slope p (because slope of n is 0)

Now consider line n and p running parallel to each satisfying the condition given in statement 2.Obviously here Slope of n > Slope of p.

Consider both-- Considering lines n and p intersect at (5,1) and n having greater y intercept, Slope of p has to be greater than slope of n.

VP_Tatiana wrote:Statement 1 tells us where the lines intersect, but tells us nothing about either of the lines' slopes. If you draw a picture of two lines intersecting at (5,1), you can see that with the information given, we could label either one n. We could make the one with greater slope n, or the one with less slope n. So, we don't have enough information from statement 1.

Now let's consider statement 2, that says that the y intercept of n is greater than that of the y intercept of p. The slopes could be unequal (think about intersecting lines), or we could have parallel lines, in which case the slopes are equal. So, statement 2 is not sufficient on its own.

Now let's consider the statements together. The lines intersect at (5,1), and n has the higher y intercept. Let's look at 3 cases:

1) Both have y intercepts above y=1
Since n intersects higher, then we know n had further to descend, so its slope is steeper (but more negative) than p's. Thus, p has a greater slope.

2) n has intercept above y=1, p has intercept below
n would have a negative slope and p a positive, so p has a greater slope

3) Both have y intercepts below y=1
Both have positive slopes, but p has further to ascend. Thus, p has a greater slope.

Since combining the information tells us that p always has a greater slope, we have sufficient information with both statements and the answer is C, NOT B.

I really liked your 3 cases. This was excellent. Thanks so much.

Why did nobody just say the easy way to figure this out?

Equation of a line is y=mx+b

Statement 1 lets us fill in a little bit of the information:

line n:
y=mx+b
1=m5+b

Line p:
y=mx+b
1=m5+b

From this we cant tell jack sh*t about the slopes of the line.

Statement 2: Fill in some numbers

Line n:
y=mx+b
y=mx+5

line p:
y=mx+b
y=mx+2

From this we cannot tell anything about the slope. Even if you set both line equations equal to m, we can not tell because we would need to then know the either 1 point on each line, or the point at which they intersect... bringing us to option C.

If you want to solve to make sure then:

Line N:
y=mx+b
1=5m+5
-4/5=m

Line P
y=mx+b
1=5m+2
-1/5

I am really confused whether the answer is C or E

https://gmatclub.com/forum/ds-geometry-3 ... ml#p647049

I would like the experts to comment

uptowngirl92 wrote:I am really confused whether the answer is C or E

https://gmatclub.com/forum/ds-geometry-3 ... ml#p647049

I would like the experts to comment

Hi uptowngirl92!

Since you would like to know whether the answer is C or E, let's proceed immediately to considering them in combination.

There are a couple of different approaches here.

First, we can take an algebraic (and/or reasoning)approach, using the line equation: y = mx + b. We know that they intersect at 5,1, so we can use these coordinates as "x" and "y" respectively for the line equations of both lines.

for line n:
(1) = m(5) + b
let's rewrite the equation using the subscript "n" to designate this as line n's equation:
1 = 5mn + bn

Similarly, for line p:
1 = 5mp + bp

The left-hand sides of both of these equations is 1. Because the right hand sides of both equations equal to 1, and because 1 is equal to 1, the right hand sides of both equations are equal to each other:

5mp + bp = 5mn + bn

But statement two tells us that bn>bp. bn being larger than bp would make the right hand side of our equation larger than the left hand side. But this can't happen because the two sides of the equation must equal each other. So, to compensate for this, 5mp will have to be bigger than 5mn (or, 5mn will have to be smaller than 5mp), and the answer to the question is "definitely yes." Choose C

Algebraically:

bn>bp
bn-bp>0

rearranging first equation:

5mp - 5mn = bn - bp

Because bn-bp > 0, the right hand side of the above equation must be positive. Therefore, the left hand side must be positive: 5mp - 5mn > 0. Then, 5mp>5mn, and choose C.


The second approach is to plot it out. If we do it this way, there are actually four cases with one case dividing into two subcases.

Just draw a coordinate plane, and point 5,1. Go below the flat line y=1 (preferably a few units below origin), and draw two different lines going up from left to right, intersecting at 5,1. Again, don't even worry about which one is p (or which one is n) until after you've drawn these two lines. Now, label the one with the higher y intercept as line n. Clearly, line p has a greater slope. We can call this case 1.

Let's look at case 2. Do the same thing, except this time, go a few units above y=1. Don't even worry about which one is p (or which one is n.) Just draw two different falling lines intersectring at 5,1. When we have "falling" lines (ie, going downwards from left to right), then the line that "falls" more slowly has a higher slope. In this case, p is falling more slowly and so has a higher slope.

Case 3: Draw one line falling from left to right (like both lines in case 2). Draw the other line rising from left to right (like both lines in case 1). The falling line must be line n because that is the line that has the higher y intercept. But the line that rises automatically has a higher slope than the falling line. So, again, line p has a higher slope.

Case 4: One line is flat (don't worry whether it is p or n). That is, one line is just y = 1. Now, the other line can intersect in two ways: either by rising up into the intersection point or else falling down into the intersection point.

Case 4a: intersecting line rising up into the intersection: in this case, the flat line has a higher y intercept, and so the flat line is line n, and the rising line is line p. Therefore, again, line p has a higher slope.

case 4b: intersecting line falling down into the intersection: in this case, the falling line has a higher y intercept, and so it is line n, and the flat line is line p. Because line n is falling, line p, again, must have a higher slope. (Line p's slope is zero while line n's slope is negative).

The statements, although in sufficient in isolation, are sufficient in combination. (C)

How about another approach

Line N

y= m1x + b1

Line P

y=m2x + b2


Statement 1---Insufficient

1 = 5m1 + b1
m1 = (1-b1)/5---------------------------------------(1)

1=5m2 + b2
m2= (1-b2)/5---------------------------------------(2)

Since we dont know b1 or b2 ( y intercepts)-----INSUFFICIENT

Statement 2------INSUFFICIENT

b1>b2. No relationship established

Combining both,

From (1) and (2),

m1 = (1-b1)/5 and m2 = (1-b2)/5

Since, b1>b2

m1<m2

Hence Sufficient

Hope you guys find this method easier