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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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Updated on: 09 Jan 2018, 18:53
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Difficulty:

55% (hard)

Question Stats:

52% (01:17) correct 48% (01:34) wrong based on 121 sessions

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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) $$q = 5$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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. PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 Location: India GPA: 3.82 Re: The function y=px^2-4x+q in the x-y plane attains a [#permalink] ### Show Tags 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6239 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The function y=px^2-4x+q in the x-y plane attains a [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Joined: 29 May 2016
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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13 Aug 2018, 13:10
MathRevolution wrote:
=>

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.

Can you please explain how did you derive the minimum value? Frankly, I still don't understand.
Manager
Joined: 09 Aug 2017
Posts: 198
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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13 Aug 2018, 18:25
I have one question here.
How did you assume that both roots are equal?

-b/2a is minimum value of quadratic equation when b^2-4ac is equal to 0 and a>0; Both roots are equal.

What do you say?

niks18 wrote:
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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Joined: 25 Feb 2013
Posts: 1217
Location: India
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The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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14 Aug 2018, 10:13
gvij2017 wrote:
I have one question here.
How did you assume that both roots are equal?

-b/2a is minimum value of quadratic equation when b^2-4ac is equal to 0 and a>0; Both roots are equal.

What do you say?

niks18 wrote:
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

hi gvij2017

a quadratic equation is of the form $$y=ax^2+bx+c$$, it is a parabola and the value of $$a$$ decides whether the parabola opens upwards or downwards

the maximum/minimum value of the equation will be at $$\frac{dy}{dx}=0$$, so on differentiating the quadratic equation you will get

$$\frac{dy}{dx}=2ax+b=0 =>\frac{dy}{dx}=\frac{-b}{2a}$$

Hence the equation will have minimum/maximum point at $$\frac{-b}{2a}$$

I have used calculus here but the relation can be derived using simple algebra also. check out the below link for detailed explanations -

https://brilliant.org/wiki/maximum-valu ... -equation/
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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14 Aug 2018, 10:16
1
Xin Cho wrote:
MathRevolution wrote:
=>

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.

Can you please explain how did you derive the minimum value? Frankly, I still don't understand.

hi Xin Cho

pls see my post above for mathematical derivation. But for GMAT it is not required
Intern
Joined: 10 Jul 2017
Posts: 19
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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20 Aug 2018, 00:34
1
Well if you have knowledge of differentiation, this question could be solved easily.

Given y = px^2−4x+qy
by differentiating :
dy/dx = 2px -4 = 0 (At lowermost point, the slope of the curve is Zero that's why dy/dx (slope) =0 )
x=4/2p => x= 2/p ( now you just need the value of 'p')
Re: The function y=px^2-4x+q in the x-y plane attains a &nbs [#permalink] 20 Aug 2018, 00:34
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