Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 02:04 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The function y=px^2-4x+q in the x-y plane attains a

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7599
GMAT 1: 760 Q51 V42 GPA: 3.82
The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

11 00:00

Difficulty:   55% (hard)

Question Stats: 51% (01:18) correct 49% (01:30) wrong based on 143 sessions

### HideShow timer Statistics [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) $$q = 5$$

_________________

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.
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1197
Location: India
GPA: 3.82
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

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 V
Joined: 16 Aug 2015
Posts: 7599
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

=>

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.
_________________
Intern  B
Joined: 29 May 2016
Posts: 9
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

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.
Senior Manager  G
Joined: 09 Aug 2017
Posts: 359
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

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
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1197
Location: India
GPA: 3.82
The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

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/
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1197
Location: India
GPA: 3.82
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

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  B
Joined: 10 Jul 2017
Posts: 31
Schools: ISB '20
Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

### Show Tags

2
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   [#permalink] 20 Aug 2018, 00:34
Display posts from previous: Sort by

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