111 parabola intersection

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111 parabola intersection

by ern5231 » Mon Oct 05, 2009 4:19 pm
Q1 y = x ^ 2, px + qy +1 = 0, are two equations in the XY axis. Is there a point of intersection?
a. p ^ 2 = 4q
b. q ^ 2 = 4p


Q2 Parabola y = x ^ 2, and qx + py +1 = 0, Is there a point of intersection?
(1) p = q ^ 2 / 4 (2) q = 4p ^ 2

OA later
Source: — Data Sufficiency |

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Re: 111 parabola intersection

by umaa » Mon Oct 05, 2009 5:55 pm
ern5231 wrote:Q1 y = x ^ 2, px + qy +1 = 0, are two equations in the XY axis. Is there a point of intersection?
a. p ^ 2 = 4q
b. q ^ 2 = 4p


Q2 Parabola y = x ^ 2, and qx + py +1 = 0, Is there a point of intersection?
(1) p = q ^ 2 / 4 (2) q = 4p ^ 2

OA later
IMO C for both the question.

First question can be plotted by using both the equation. There is no intersection point.

But in the second q, the p value is sqrrot(4).

If we know the values of p and q, we can definitely find whether both the graphs intersect. So, I assume the answer is C for the second question.

Source of the question and OA pls.
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Re: 111 parabola intersection

by schumi_gmat » Mon Oct 05, 2009 7:22 pm
ern5231 wrote:Q1 y = x ^ 2, px + qy +1 = 0, are two equations in the XY axis. Is there a point of intersection?
a. p ^ 2 = 4q
b. q ^ 2 = 4p


Q2 Parabola y = x ^ 2, and qx + py +1 = 0, Is there a point of intersection?
(1) p = q ^ 2 / 4 (2) q = 4p ^ 2

OA later
IMO E

p and Q are not defined as constants. So We can have multiple values of p and q

hence E

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by ern5231 » Fri Oct 09, 2009 12:11 am
OA for 1st one is C and second one is A. Can anyone explain?

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by Katrusya » Fri Oct 09, 2009 1:10 pm
I got A for both of them.
OA is strange.

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by umaa » Fri Oct 09, 2009 8:06 pm
IMO A can't be the answer for the second question. If Statement 1 is sufficint, statement 2 should be sufficient.

Let me know what is the source of the question..
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by ern5231 » Tue Oct 13, 2009 12:37 am
Guys this is a question from Gmat mock test at my prep center. It is a real toughie. I am not understanding anything from your solutions. Hope an instructor can help!