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# M24-12

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Joined: 02 Sep 2009
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M24-12  [#permalink]

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16 Sep 2014, 01:21
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85% (hard)

Question Stats:

42% (01:49) correct 58% (01:31) wrong based on 122 sessions

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If $$ac \ne 0$$, do graphs $$y=ax^2+b$$ and $$y=cx^2+d$$ intersect?

(1) $$a = -c$$

(2) $$b \gt d$$

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Re M24-12  [#permalink]

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16 Sep 2014, 01:21
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1
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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17 Aug 2016, 23:11
Is Parabola concept tested in Gmat?..if yes then how often?
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Re: M24-12  [#permalink]

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18 Aug 2016, 07:50
subhajit1 wrote:
Is Parabola concept tested in Gmat?..if yes then how often?

Yes but not very often.
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Re M24-12  [#permalink]

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23 Aug 2016, 05:36
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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28 Aug 2016, 06:38
Good question. Could be solver quickly through graphical approach.
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Re: M24-12  [#permalink]

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20 Jul 2017, 02:57
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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26 Sep 2017, 06:21
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]

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15 Jan 2018, 07:58
1
An alternate solution =

quadratic equations make a parabola.

(1) a = -c --- for a moment ignore b and d. the parabolas will be the exact mirror image of each other, one with a trough and the other with a peak, and both will intersect at the tip of the trough and peak. adding a number b or d will make these parabolas rise or fall along the y axis depending on what these numbers are. it is possible that the parabola with the trough will rise while that with a peak will fall and thus the curves will have no intersection. or the exact opposite could happen and more than one intersection would occur. therefore INSUFFICIENT.

(2) b>d this only mentions the shift on the y axis and nothing else about the curves, therefore clearly INSUFFICIENT

(1 &2) b > d. this means that the curve with b will rise more or fall less than the one with d which will rise less or fall more, depending on the signs or values of b and d.
Now from the stem we know that b is paired with the x coefficient -c. therefore this is the curve with the peak. and d is paired with the curve with c which is positive and therefore the curve with a trough. for all possible values of b or d, there will be an intersection.
BUT BEWARE! we do not know if -c is in fact positive i.e. c could be a negative number. therefore b could be paired with the curve with trough and d would be paired with the curve with the peak. in this case for all values of b and d where b>d, the curves will not intersect. therefore INSUFFICIENT.
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Re: M24-12  [#permalink]

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12 Mar 2018, 13:29
This can easily be solved visually but it is very easy to make the mistake that you assume A is positive, which makes (i) + (ii) sufficient - when it isn't
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Re: M24-12  [#permalink]

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23 Apr 2018, 07:02
+1 for option E.

Since the two have a common solution, x=sqrt[(d-b)/(a-c)]. For x to have a valid value, the qty inside sqrt root must be positive.

St 1 - NS
St 2 - NS

Together - Be careful, you do not know the sign of c. Hence insufficient.

Hence option E.
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Re: M24-12  [#permalink]

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01 Apr 2019, 10:03
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Top Contributor
Bunuel wrote:
If $$ac \ne 0$$, do graphs $$y=ax^2+b$$ and $$y=cx^2+d$$ intersect?

(1) $$a = -c$$

(2) $$b \gt d$$

Target question: Do the graphs y=ax²+b and y= cx²+d intersect?

This is a good candidate for rephrasing the target question.

Key concept: If two lines (or curves) intersect at a point, then the COORDINATES of that intersection point must satisfy BOTH equations.
Since BOTH of the given equations are set equal to y, we can look for a value of x that yields the same value of y
In other words, if there's a solution to the equation ax²+b = cx²+d, then the two graphs intersect.

REPHRASED target question: Is there a value of x that satisfies the equation ax²+b = cx²+ d?

Head straight to . . .

Statements 1 and 2 combined
There are several values of a, b, c and d that satisfy BOTH statements. Here are two:
Case a: a = -1, c = 1, b = 4 and d = 2. Our equation becomes (-1)x² + 4 = (1)x²+ 2
Rearrange to get: 2 = 2x²
One solution is x = 1
In this case, the answer to the REPHRASED target question is YES, there is a value of x that satisfies the equation ax²+b = cx²+ d

Case b: a = 1, c = -1, b = 4 and d = 2. Our equation becomes (1)x² + 4 = (-1)x²+ 2
Rearrange to get: 2x² = -2
This means x² = -1, and there is no solution to this equation
In this case, the answer to the REPHRASED target question is NO, there is no value of x that satisfies the equation ax²+b = cx²+ d

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Re: M24-12   [#permalink] 01 Apr 2019, 10:03
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