Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Aug 2016, 10:31

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m13 q23

Author Message
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 95

Kudos [?]: 875 [0], given: 334

### Show Tags

17 Oct 2010, 10:03
Sure I can change it, but I don't see a difference. Why is "any X" wording better? Sorry if I'm missing something obvious .
Pkit wrote:
Bunuel's explanations are clear enough.
Thanks

Dzyubam, do you consider to replace "for all x" to "any X"...

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 11 Apr 2011
Posts: 32
Location: London
WE 1: 3 yrs f/t manager in aerospace (IT - Commercial - Int'l Mktg)
WE 2: 5 yrs freelance consultant to Gov't
Followers: 0

Kudos [?]: 3 [0], given: 3

### Show Tags

25 Apr 2011, 14:44
Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive...
inserting x=3....
(3^2) - (3*2) + A must be positive...
9 - 6 + A must be positive
3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated
_________________

Blogging my way through the applications/study process!!
http://mbahorizons.blogspot.com/

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 157

Kudos [?]: 1558 [0], given: 376

### Show Tags

25 Apr 2011, 15:27
bleemgame wrote:
Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive...
inserting x=3....
(3^2) - (3*2) + A must be positive...
9 - 6 + A must be positive
3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated

Question is correct IMO.
x is the variable and A a constant.
St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

Say, A=-2
For x=3; the expression is +ve. Great!!!
for x=1; the expression is -ve(falsifies the statement)
Means; A must not be -2.
_________________
Intern
Joined: 11 Apr 2011
Posts: 32
Location: London
WE 1: 3 yrs f/t manager in aerospace (IT - Commercial - Int'l Mktg)
WE 2: 5 yrs freelance consultant to Gov't
Followers: 0

Kudos [?]: 3 [0], given: 3

### Show Tags

25 Apr 2011, 17:55
fluke wrote:
x is the variable and A a constant.

St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

thanks - error in interpretation (treating 'a' as variable that will alter to satisfy the equation accordingly and not a constant).
_________________

Blogging my way through the applications/study process!!
http://mbahorizons.blogspot.com/

Intern
Joined: 17 Dec 2010
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

26 Apr 2011, 13:29
I will go with option A .

As I understood...minimum value of A which makes the expression xsq2-2*x+A for all positive. If it's given that x2-x*x+a> 0 for all x , then we must be sure that is such that the inequality holds true.

If we know that a>1 , we can answer the question ( is positive).
Manager
Joined: 03 Aug 2010
Posts: 106
GMAT Date: 08-08-2011
Followers: 1

Kudos [?]: 58 [0], given: 63

### Show Tags

06 Jul 2011, 04:58
After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not.
However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

Bunuel wrote:
CrushTheGMAT wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0.

*****
I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

Manager
Joined: 13 May 2010
Posts: 124
Followers: 0

Kudos [?]: 11 [0], given: 4

### Show Tags

23 Jan 2012, 18:45
In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?
Math Expert
Joined: 02 Sep 2009
Posts: 34438
Followers: 6257

Kudos [?]: 79467 [0], given: 10018

### Show Tags

23 Jan 2012, 19:11
teal wrote:
In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?

Sure you can find SOME x for which A will be negative, BUT (1) says that x^2-2x+A is positive FOR ALL x-es. This expression to be positive for ALL x-es A must be more than 1. Refer to my post here to see why: m13-q23-70269.html#p716027

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 34438
Followers: 6257

Kudos [?]: 79467 [0], given: 10018

### Show Tags

23 Jan 2012, 19:24
Yalephd wrote:
Bunuel wrote:
Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not.
However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

You are right: if we use the same approach for (2) then we'll get A>0 BUT if A=0 then Ax^2+1 won't be a quadratic function anymore. So this approach will work only if A doesn't equal to zero, but we can not eliminate this case and if A=0 then Ax^2+1=1 is also always positive. Hence Ax^2+1 is positive for A>0 (if we use quadratic function approach) as well as for A=0, so for $$A\geq0$$.

Hope it's clear.
_________________
Manager
Joined: 13 May 2010
Posts: 124
Followers: 0

Kudos [?]: 11 [0], given: 4

### Show Tags

02 Aug 2012, 01:22
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?
Math Expert
Joined: 02 Sep 2009
Posts: 34438
Followers: 6257

Kudos [?]: 79467 [0], given: 10018

### Show Tags

02 Aug 2012, 01:26
teal wrote:
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

_________________
Intern
Joined: 11 Mar 2013
Posts: 8
Followers: 0

Kudos [?]: 6 [0], given: 11

### Show Tags

30 Apr 2013, 06:03
It is A for the simple reason that we have to find a value for A that works for every value of x. So A has to be > 1.

Posted from my mobile device
Senior Manager
Joined: 13 Jan 2012
Posts: 309
Weight: 170lbs
GMAT 1: 740 Q48 V42
GMAT 2: 760 Q50 V42
WE: Analyst (Other)
Followers: 15

Kudos [?]: 127 [0], given: 38

### Show Tags

30 Apr 2013, 10:44
Bunuel wrote:
CrushTheGMAT wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0.

*****
I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

Your explanations are always so elegant.
Intern
Joined: 12 Apr 2013
Posts: 9
Location: Germany
Concentration: Finance, Accounting
GMAT Date: 05-28-2013
GPA: 3.6
Followers: 0

Kudos [?]: 11 [0], given: 20

### Show Tags

30 Apr 2013, 15:13
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

Basically the question is, when $$x^2 - 2x + A$$ is positive for all $$x$$ --> can we say under that prerequisitions that $$A$$ is positive?
_________________

........................................................................................
See it big and keep it simple.

Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 89
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 77 [0], given: 11

### Show Tags

30 Apr 2013, 21:15
Ultimor wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

Basically the question is, when $$x^2 - 2x + A$$ is positive for all $$x$$ --> can we say under that prerequisitions that $$A$$ is positive?

Look for the Bunuel's Post $$\Rightarrow$$ m13-q23-70269.html#p716027 . Your all doubts will be cleared completely.
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Director
Joined: 14 Dec 2012
Posts: 841
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Followers: 54

Kudos [?]: 1115 [0], given: 197

### Show Tags

01 May 2013, 00:54
teal wrote:
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

hi,

a general quadratic equation is like this:

y=ax^2+bx+c

now your doubt is "why can't we apply the negative discriminant logic in statement 2".

you can apply the logic of negative discriminant only when you are sure that coefficient of x^2 is positive...i.e "a" is positive.

now in our question equation is AX^2+1

here we dont know whether A is positive or negative now if we use the discriminant rule in this then we are already assuming that "A" is positive...

and we are neglecting the possibility of A < 0...

so TAKEAWAYS:

use the discriminant concept only when coefficient of X^2 is positive.

hope it helps.

SKM
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment

Re: m13 q23   [#permalink] 01 May 2013, 00:54

Go to page   Previous    1   2   [ 36 posts ]

Similar topics Replies Last post
Similar
Topics:
M19 Q23 2 22 Jul 2010, 09:35
v02 q23 2 18 Apr 2010, 21:28
M09 Q23 1 31 Mar 2010, 13:23
m13/q23....wrong ans 2 03 Jan 2010, 09:27
3 m04 Q23 12 14 Sep 2008, 06:12
Display posts from previous: Sort by

# m13 q23

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.