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M24-12

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M24-12 [#permalink]

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New post 16 Sep 2014, 00:21
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E

Difficulty:

  55% (hard)

Question Stats:

46% (01:32) correct 54% (01:05) wrong based on 28 sessions

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Re M24-12 [#permalink]

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New post 16 Sep 2014, 00:21
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Official Solution:


Notice that graphs of \(y=ax^2+b\) and \(y=cx^2+d\) are parabolas.

Algebraic approach:

(1) \(a = -c\). Given: \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). Now, if these two parabolas 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 \(\ge 0\). So, the question becomes: is \(discriminant=0-8a(b-d) \ge 0\)? Or, is \(discriminant=-8a(b-d) \ge 0\)? Now can we determine whether this is true? We know nothing about \(a\), \(b\), and \(d\), hence no. Not sufficient.

(2) \(b \gt d\). 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 \(\ge 0\). So, the question becomes: is \(discriminant=0-4(a-c)(b-d) \ge 0\)? Or, is \(discriminant=-4(a-c)(b-d) \ge 0\)? Now can we determine whether this is true? We know that \(b-d \gt 0\) but what about \(a-c\)? Hence no. Not sufficient.

(1)+(2) We have that \(a=-c\) and \(b \gt d\), so \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). The same steps as above: \(2ax^2+(b-d)=0\) and the same question remains: is \(discriminant=-8a(b-d) \ge 0\) true? \(b-d \gt 0\) but what about \(a\)? Not sufficient.

Else consider two cases.

First case: \(y=-x^2+1\) and \(y=x^2+0\) (upward and downward parabolas). Notice that these parabolas satisfy both statements and they cross each other;

Second case: \(y=x^2+1\) and \(y=-x^2+0\) (also upward and downward parabolas). Notice that these parabolas satisfy both statements and they do not cross each other.


Answer: E
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Re: M24-12 [#permalink]

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New post 17 Aug 2016, 22:11
Is Parabola concept tested in Gmat?..if yes then how often?

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Re M24-12 [#permalink]

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New post 23 Aug 2016, 04:36
I think this is a high-quality question and I agree with explanation.

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Re: M24-12 [#permalink]

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New post 28 Aug 2016, 05:38
Good question. Could be solver quickly through graphical approach.

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Re: M24-12 [#permalink]

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New post 20 Jul 2017, 01:57
hi brunel

i struggle with parabola questions..may be because of lack of clarity in the concepts.
In the algebra approach i understood why both the equations have been equated but how you break to get the dicriminants i.e from 2ax2+(b−d)=0 to discriminant=0−8a(b−d)≥0 ?

also, can you help via diagram as i could not understand second method

thanks in advance

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Re: M24-12 [#permalink]

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New post 26 Sep 2017, 05:21
Bunnuel, just to check if my graphical solution is right:

If they would say that a is positive, and c is negative, would the answer be (C) ? (Given that one parabola opens upward and the ther one downwards?

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Re: M24-12   [#permalink] 26 Sep 2017, 05:21
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