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For which value of x will y=ax^2+100x+b have a maximum in the xy-pla

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For which value of x will y=ax^2+100x+b have a maximum in the xy-pla [#permalink]

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New post 28 Nov 2017, 00:55
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  55% (hard)

Question Stats:

52% (00:47) correct 48% (01:04) wrong based on 113 sessions

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[GMAT math practice question]

For which value of x will \(y=ax^2+100x+b\) have a maximum in the \(xy\)-plane?

1) \(b=100\)
2) \(a=-4\)

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For which value of x will y=ax^2+100x+b have a maximum in the xy-pla [#permalink]

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New post 28 Nov 2017, 09:56
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MathRevolution wrote:
[GMAT math practice question]

For which value of x will \(y=ax^2+100x+b\) have a maximum in the \(xy\)-plane?

1) \(b=100\)
2) \(a=-4\)


The maximum or minimum value of a quadratic function \(ax^2+bx+c\) occurs at

\(x=-\frac{b}{2a}\)

so if \(a>0\) then minima will occur and the parabola will open upwards

and if \(a<0\) then maxima will occur and the parabola will open downwards

so for \(y=ax^2+100x+b\)

maxima will be at \(x=-\frac{100}{2a}\) where \(a<0\)

Statement 1: no information about \(a\). Insufficient

Statement 2: provides the value of \(a\). Sufficient

Option B

Alternatively, we can use derivative (out of scope for GMAT)

\(y=ax^2+100x+b\)

so \(\frac{dy}{dx} = 2ax+100\)

maximum point will be \(\frac{dy}{dx}=0 => 2ax+100=0\)

so \(x=-\frac{100}{2a}\). We need the value of \(a\) to get the maximum point.
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Re: For which value of x will y=ax^2+100x+b have a maximum in the xy-pla [#permalink]

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New post 30 Nov 2017, 01:19
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the numbers of variables and equations.

We can modify the original condition and question as follows:

If \(a < 0\), the function will have a maximum at \(x = \frac{(-100)}{(2a)} = \frac{(-50)}{a}\).
If \(a > 0\), the function has no maximum. So, to answer the question, we need to find the value of a.
Thus, condition 2) is sufficient.

Note: condition 1) cannot be sufficient as it provides no information about the value of a.

Therefore, the answer is B.

Answer: B
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Re: For which value of x will y=ax^2+100x+b have a maximum in the xy-pla   [#permalink] 30 Nov 2017, 01:19
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