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)
