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M24-12

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M24-12  [#permalink]

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New post 16 Sep 2014, 01:21
2
11
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

42% (01:48) correct 58% (01:30) wrong based on 120 sessions

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Re M24-12  [#permalink]

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New post 16 Sep 2014, 01:21
2
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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New post 15 Jan 2018, 07:58
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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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New post 01 Apr 2019, 10:03
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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

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Re: M24-12  [#permalink]

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New post 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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Re M24-12  [#permalink]

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New post 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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New post 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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New post 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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New post 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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New post 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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New post 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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M24-12  [#permalink]

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New post 31 Oct 2019, 22:43
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How do i draw a diagram for these cases?
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Re: M24-12  [#permalink]

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New post 01 Nov 2019, 00:12
Bunuel wrote:
If \(ac \ne 0\), do graphs \(y=ax^2+b\) and \(y=cx^2+d\) intersect?


(1) \(a = -c\)

(2) \(b \gt d\)


Graphs of quadratics are parabolas (upward opening or downward opening).

Attachment:
IMG_7392.jpg

The second figure shows the various ways in which a graph of y = ax^2 + b and y = cx^2 + d can look like.

(1) \(a = -c\)

One graph is upward opening and the other downward. We don't know which is which.
If a is positive, its graph is upward opening and c is negative so the graph is downward opening.
If a is negative, its graph is downward opening and c is positive so its graph is upward opening.
They may or may not intersect. I and IV do not intersect but I and II do.


(2) \(b \gt d\)
So the first graph lies higher up on the y axis than the second graph. But they may or may not intersect.
I and IV do not intersect while II and III do.

Using both, we still do not know if they intersect. I an IV do not intersect but II and III do.

Answer (E)
>> !!!

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Re: M24-12  [#permalink]

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New post 01 Nov 2019, 00:16
axezcole wrote:


Graph I could be y = x^2 + 1 (assuming the point where the graph intersects the y axis is y = 1)

y = -x^2 + 1 would just be the same graph at the same point (y = 1) but downward opening.
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Re: M24-12  [#permalink]

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New post 11 Nov 2019, 17:19
Bunuel wrote:
If \(ac \ne 0\), do graphs \(y=ax^2+b\) and \(y=cx^2+d\) intersect?


(1) \(a = -c\)

(2) \(b \gt d\)


We see that the two graphs are parabolas, and they may or may not intersect.

Statement One Only:

a = -c

This means we have two parabolas; one opens upward and the other downward. An upward parabola and a downward parabola may or may not intersect. Without further information, statement one alone is not sufficient.

Statement Two Only:

b > d

This means we have two parabolas, one with a y-intercept higher than the other. Furthermore, the y-intercept of each parabola is its vertex. However, two such parabolas may or may not intersect. Without further information, statement two alone is not sufficient.

Statements One and Two Together:

We see that the first parabola y = ax^2 + b has a higher vertex than the second parabola y = cx^2 + d, and the two parabolas open in opposite directions. If the first parabola opens downward and the second upward, then the two parabolas intersect. However, if the first parabola opens upward and the second downward, then they don’t intersect. The two statements together are still not sufficient to answer the question.

Answer: E
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Re: M24-12   [#permalink] 11 Nov 2019, 17:19
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