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The function y=px^2-4x+q in the x-y plane attains a

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The function y=px^2-4x+q in the x-y plane attains a [#permalink]

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New post Updated on: 09 Jan 2018, 18:53
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[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(xy\)-plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)

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Originally posted by MathRevolution on 08 Jan 2018, 01:26.
Last edited by MathRevolution on 09 Jan 2018, 18:53, edited 2 times in total.
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Re: The function y=px^2-4x+q in the x-y plane attains a [#permalink]

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New post 08 Jan 2018, 01:39
MathRevolution wrote:
[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(x-y\) plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)


Minimum value of a quadratic equation is at \(\frac{-b}{2a}\)

here \(b=-4\) and \(a=p\), so minimum \(x=\frac{4}{2p}\). We need the value of \(p\) to find the minimum value.

Statement 1: \(p=2\). Sufficient

Statement 2: value of \(p\) is not mentioned. Insufficient

Option A
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Re: The function y=px^2-4x+q in the x-y plane attains a [#permalink]

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New post 10 Jan 2018, 01:00
=>

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 question.
\(y=px^2-4x+q\) has a minimum value when \(x = \frac{-(-4)}{2p} = \frac{2}{p}.\)
Thus, the question asks for the value of \(p\).

Since only condition 1) gives us information about p, only condition 1) is sufficient.
Therefore, A is the answer.
Answer: A
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Re: The function y=px^2-4x+q in the x-y plane attains a   [#permalink] 10 Jan 2018, 01:00
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