Do the lines cross is another way of asking if they are parallel.

Statement 1 is sufficient. We know that if a = 1 and -c = a, then c = -1. These are not parallel. Also keep in mind the forumula for a line. Is y = [slope]x + [y-intercept] (or minus y-intercept).

Statement 2 is telling us information about the y-intercept. It is saying that no matter what d is, b is greater than d. This doesn't tell us any information about a or c, which are the most important because these two variables determine the slope of the line. Without knowing the slope of the lines, we cannot know if the lines cross.

Answer is A, and the anwer with the info in A is "Yes, the lines do cross."


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You guys might have forgotten that we are dealing with quadratic equations here. The upward and downward parabolas may not cross even if the "slope" is different for the two equations. See the OE for more info.

The right answer is E.


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Answer should be E. We know (given S1) that one of them is upward parabola, the other is inverted. However, we don't know which one is inverted and which one is upward. I.e. - we dont know the sign of "a", it can be a=2 or a=-2

This question is a bit different than a normal line equation. A normal line equation is y = [slope]x + [y intercept], but here the X is actually squared. This will lead to a curve similar to a parabola rather than a straight light. I don't think this is avery good question, but that's just me. I have not seen this before becuase I only did the first 12 or 13 math tests before my GMAT.

As for your questions prior to the DS question. Yes, generally, all you need to know for lines that cross is the slope. Remember that a Cartesian plane extends to infinity, so if the slopes are not IDENTICAL, then the lines will cross at some point. The y-intercept doesn't matter because at some point, the lines will cross unless they are EXACTLY parallel. To determine perpendicular lines, you are correct, the opposite sign (positive or negative) and reciprocal will be the test for perpendicular.

If slope is -\frac{1}{C} then the perpendicular slope will be C because we take the reciprocal so \frac{C}{1} and it's positive because the other was negative.


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J Allen Morris
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Both statements required to answer this question.

Explanation:
Lets equate these two equations first.
We have no idea about the signs of a,b,c or d. so we cant say anything about the nature of the graph whether it is concave or convex.
After equating these equation we can get x^2 = (d-b)/(a-c).
For getting real values of x, the above result must satisfy that a!=c!=0.
Also, (d-b)/(a-c) must always be positive.
To find this we must have both the conditions. i.e, if a=-c and b>d then x^2 will be equal to a negative value so we can't have real values of x. Thus we can say that these two lines are not crossing each other.

1. a = -c
2. b > d

1) If a = -c then first equation will become as follow,

1) -cx2 –y + b = 0
2) Cx2 –y + b = 0
As per rule two lines are parallel (if a1/a2 = b1/b2) but here a1/a2 = -1 and b1/b2 = 1 and so they are not equal and so line must intercept.

2) b > d, does not give enough information about a and c and so it is not sufficient.

Answer is A.

Hi abhishekik,
If we dont have real values then how do we say that lines dont intersect !! Imaginary numbers can also be plotted on plane and they can also be an intersection point??!!
http://en.wikipedia.org/wiki/Imaginary_number

Or my understanding is not correct? Please explain.

Statement 1: It states that each parabola has a different sign (or they both equal 0). If one slope is positive with an intercept that is positive and one is negative with an intercept that is negative the lines will never cross. (There is nothing defining which line has a positive or negative slope).

Statement 2: Obviously not sufficient says nothing about the slopes.

Together: Since we don't know which line is positive and which is negative or rather if there both not zero than the y intercept can either mean they cross or don't so still not definitive.

E

I get E

only if product of two slops =-1 then the lines are perpendicular to each other so they cross.

I'd like to comment on this one:

First of all: equations given ARE NOT linear equations. We know that if two different lines do not cross each other they are parallel. How can we tell, based on the equations, whether the lines are parallel? We can check the slopes of these lines: parallel lines will have the same slopes. NOT that the slopes of lines must be negative reciprocals of each other (as it was mentioned in the earlier posts): in this case they are perpendicular not parallel.

Second of all: we have quadratic equations. These equations when drawn give parabolas. The question is: do they cross? This CAN NOT be transformed to the question: "are they parallel?" In the wast majority of cases the word "parallel" is used for the lines. Well we can say that concentric circles are parallel, BUT GMAT, as far as I know, uses this word ONLY about the lines (tutors may correct me if I'm wrong). Next, the word "parallel" when used for curves (lines, ...) means that these curves remain a constant distance apart. So strictly speaking two parabolas to be parallel they need not only not to intersect but also to remain constant distance apart. In this case, I must say that this can not happen. If a curve is parallel (as we defined) to the parabola it won't be quadratic: so curve parallel to a parabola is not a parabola. So I think that at this point we can stop considering this concept in regard to the original question.

So in which cases parabolas do not cross? There are number of possibilities: We can shift the vertex: the parabolas y=x^2 and y=x^2+1 will never intersect (note that they won't be exactly parallel but they will never intersect). We can consider downward and upward parabolas and in some cases they also never intersect... Of course there can be other cases as well.

As for the solution. We can follow the way dzyubam proposed (and I think it's the fastest way, provided we can identify correct examples) and consider two cases. First case: y=-x^2+1 and y=x^2+0 (upward and downward parabolas), which satisfies both statements, and see that in this case answer is YES, they cross each other; and the second case: y=x^2+1 and y=-x^2+0 (also upward and downward parabolas), which also satisfies both statements, and see that in this case answer is NO, they do not cross each other. Two different answers to the question, hence not sufficient.

Answer: E.

We can solve the question algebraically as well:

Do lines y = ax^2 + b and y = cx^2 + d cross?

(1) a = -c --> y_1= ax^2 + b and y_2=-ax^2 + d, now if they cross then for some x, ax^2+b=-ax^2 + d should be true --> which means that equation 2ax^2+(b-d)=0 must have a solution, some real root(s), or in other words discriminant of this quadratic equation must be \geq0 --> d=0-8a(b-d)\geq0? --> d=-8a(b-d)\geq0? Now can we determine whether this is true? We know nothing about a, b, and d, hence no. Not sufficient.

(2) b>d --> the same steps: if y_1= ax^2 + b and y_2= cx^2 + d cross then for some x, ax^2 +b=cx^2+d should be true --> which means that equation (a-c)x^2+(b-d)=0 must have a solution or in other words discriminant of this quadratic equation must be \geq0 --> d=0-4(a-c)(b-d)\geq0? --> d=-4(a-c)(b-d)\geq0? Now can we determine whether this is true? We know that b-d>0 but what about a-c? Hence no. Not sufficient.

(1)+(2) a=-c and b>d --> y_1= ax^2 + b and y_2=-ax^2 + d --> same steps as above --> 2ax^2+(b-d)=0 --> and the same question remains: is d=-8a(b-d)\geq0 true? b-d>0 but what about a? Not sufficient.

Answer: E.

Hope it helps.
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