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

It is currently 23 Oct 2014, 00:09

Close

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
Your Progress

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

ds:

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 02 Feb 2004
Posts: 345
Followers: 1

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

ds: [#permalink] New post 22 Apr 2005, 06:33
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
please show workings
Attachments

0023.jpg
0023.jpg [ 5.76 KiB | Viewed 526 times ]

Current Student
avatar
Joined: 28 Dec 2004
Posts: 3403
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 164 [0], given: 2

 [#permalink] New post 22 Apr 2005, 06:56
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...
Intern
Intern
avatar
Joined: 17 Mar 2005
Posts: 14
Location: Philadelphia
Followers: 0

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

 [#permalink] New post 22 Apr 2005, 07:54
C for me

for I if x = 5, then A can be negative or positive for the equation to be greater than zero. Because 15+A>0 hence A can be -14 or greater.
So A and D are out.

for II if x=1/2 then A/4+1>0. Hence A can be -3 or greater to satisfy the equation. So not B

Therefore either C or E. No time left, For me Joe bloggs would think E but i would think there are two equations two variables, i should be able to find the value therefore C
Intern
Intern
avatar
Joined: 15 Feb 2005
Posts: 23
Followers: 0

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

 [#permalink] New post 22 Apr 2005, 12:07
coolcal wrote:
C for me

for I if x = 5, then A can be negative or positive for the equation to be greater than zero. Because 15+A>0 hence A can be -14 or greater.
So A and D are out.

for II if x=1/2 then A/4+1>0. Hence A can be -3 or greater to satisfy the equation. So not B

Therefore either C or E. No time left, For me Joe bloggs would think E but i would think there are two equations two variables, i should be able to find the value therefore C


Joe Bloggs may be right. My answer is E.

Agree with your individual explanations. So the answer has to be either C or E.

Combining I and II,

If x=3,
From I) 9 - 6 +a >0 ==> A > -3
From II) 9A + 1 > 0 ===> A > -(1/9)

As the problem does not say A is an integer, A could be any number greater than -(1/9). So A could be +ve or -ve.

Answer E.
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1493
Location: Germany
Followers: 4

Kudos [?]: 52 [0], given: 0

 [#permalink] New post 23 Apr 2005, 04:30
D)...

1) to be true for ALL x. when you pick x=16 you can choose if A is +ve or -ve. when you pick x=1 you MUST choose for A a +ve number => sufficient

2) same reasonig as above. try x=16 and try x=1 => sufficient
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Director
Director
avatar
Joined: 27 Dec 2004
Posts: 908
Followers: 1

Kudos [?]: 12 [0], given: 0

 [#permalink] New post 23 Apr 2005, 04:38
D..for all X means when X is either negative , positive or non integer. I stand corrected on this one.
Intern
Intern
avatar
Joined: 17 Mar 2005
Posts: 14
Location: Philadelphia
Followers: 0

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

 [#permalink] New post 23 Apr 2005, 06:17
Christoph/Folaa - please let me know how you think D. I showed with examples earlier that we cannot prove A is either positive or negative conclusively when x = 5 or x = 1/2 so your solution of D is not ringing a bell with me.
Director
Director
User avatar
Joined: 07 Jun 2004
Posts: 618
Location: PA
Followers: 3

Kudos [?]: 191 [0], given: 22

 [#permalink] New post 23 Apr 2005, 13:47
My take on this is

From I

x^2 - 2x + A = Positve

lets take sample value of x
x = 10

100 - 20 + A = Positve
or

80 + A = +ve so A can be - 3.14 or 1547888 it does not matter

I is insufficient

From II

Ax^2 + 1 = +ve

Let x = 1/2

then

A/4 + 1 = +ve Again A can be - 1/8 or 20 it does not matter

II is insufficient

so its C / E

from I and II we cannot co-relate

My answer to this will be E
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3403
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 164 [0], given: 2

 [#permalink] New post 23 Apr 2005, 17:12
guys
the key is the word in the problem stem "for all x"...which included + ve and -ve values...A has to be positive...

if for all X was not mentioned I would have picked E as my answer...
Manager
Manager
avatar
Joined: 24 Jan 2005
Posts: 217
Location: Boston
Followers: 1

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

 [#permalink] New post 24 Apr 2005, 06:00
ax^2 + bx+c = y (Say)

The sign of y is same as a if Determinant D <= 0

If on the other hand D > 0 then there are two distinct real roots say p and q (p<q). so for all x <p, y has same sign as a and also for all x>q, y has same sign as a. But for p<x<q the sign of y is opposite to that of a.

Now D = 4-4A in our case. Depending on this value of D we can decide on the sign of the expression.

Neither of the statements 1 or 2 clearly let us decide on the sign of D

So if 4-4A <=0 or if A >=1 then y is +ve, but for 0<A<1 D > 0 so different possibilities.

E
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1493
Location: Germany
Followers: 4

Kudos [?]: 52 [0], given: 0

 [#permalink] New post 24 Apr 2005, 12:07
OA pls :-D
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

 [#permalink] New post 24 Apr 2005, 13:16
"A"


state 1...(x-1)^2 + A-1 > 0.....as (x-1)^2 is always > 0...then A-1 > 0...
A> 1...suff

state 2...insuff....A can be < 0 or > 0 and still be able to satisfy.
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1126
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 25 [0], given: 0

 [#permalink] New post 24 Apr 2005, 18:20
it's E

the critical values are reached when -1<x<0 , there can be different solutions, just take the value (-1/2):

1) X^2-2X+A >0 -> Let's use (-1/2)
1/4 + 1 + A > 0 ; 5/4 + A > 0

A could be any +ive number or any -ive number from 0 to -5/4 excluded, for example if A=-1 -> 5/4 -1 = 1/4 which is correct (+ive)

2) AX^2+1>0 -> Let's use (-1/2)
A*1/4 + 1 > 0

A could be any +ive number or any -ive number from 0 to -4 excluded, for example if A=-2 , -2*1/4 + 1 = 1/2 which is correct (+ive)

Finally, when X is between -1 and 0, A can be +ive or -ive, hence E
Senior Manager
Senior Manager
User avatar
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC
Followers: 2

Kudos [?]: 9 [0], given: 0

 [#permalink] New post 24 Apr 2005, 18:31
banerjeea_98 wrote:
"A"


state 1...(x-1)^2 + A-1 > 0.....as (x-1)^2 is always > 0...then A-1 > 0...
A> 1...suff

state 2...insuff....A can be < 0 or > 0 and still be able to satisfy.



I almost fell for the explanation for A being sufficient ;-)

1. If (x-1)^2 + a-1 > 0 , a can still be +ve or -ve ,
eg. let (x-1)^2 = 5 , a can be -1 ,-3 or 21 and the expression will still be >0
so A is insufficient.

2. B is also insufficient. ax^2 + 1 > 0

a> -1/x^2 ; let x^2 = 5 implies a > -1/5
so a could be -1/2 or a could be +3.
B is insufficient too.

Ans is C or E. To eliminate one of these, lets add the equations 1 and 2.

x^2 -2x +a >0 - I
ax^2 + 1>0 - II

adding and solving for a we get.
x^2 - 2x + 1 + a + ax^2 > 0
or (x-1)^2 + a (1 + x^2) > 0
or a > - (1+x^2)/(x-1)^2

for any value of X , you will get something like
a > - 3/4
so a could be still +ve or -ve.

Finally ans is E
_________________

ash
________________________
I'm crossing the bridge.........

VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1497
Followers: 6

Kudos [?]: 31 [0], given: 0

Re: ds: [#permalink] New post 24 Apr 2005, 20:45
mirhaque wrote:
please show workings

it is definitely a difficult problem. but i go with E.
from i, x^2-2x+a=+ve . it doesnot mean that x must be positive but the whole equation must have positive value. suppose x=1, and the value of the eq is 1 (at least +ve).
x^2-2x+a=1
a =1............ the equation is satisfied.

if x=3, and the value of the eq is 1 (at least +ve).
x^2-2x+a =1
3+a =1
a =-2 ............ the equation is still satisfied but the information given st i is insufficient.

from ii, it is even insufficientsuppose x =1, and ax^2+1=2
a+1=2
a = 1

if x = 3, but ax^2+1=1/5
9a+1=1/5
a = -4/45 so insufficient.

combining also it doesnot give any solution (because i have used the same values in st i and ii). therefore it should be E.
VP
VP
avatar
Joined: 26 Apr 2004
Posts: 1226
Location: Taiwan
Followers: 2

Kudos [?]: 101 [0], given: 0

Re: ds: [#permalink] New post 25 Apr 2005, 03:55
mirhaque wrote:
please show workings


E it is.

A) if A = (-1), then we still have a positive answer.

B) if A = (-0.5), x^2 =1, then we still have a positive answer.
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1493
Location: Germany
Followers: 4

Kudos [?]: 52 [0], given: 0

 [#permalink] New post 25 Apr 2005, 04:11
i change from D) to A)...

1) there is no chance that the eq is +ve for ALL x (or am i missing something?). how is the eq +ve for x=1 ? => sufficient

2) A can be +ve or -ve for the eq to be +ve => insufficient
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.


Last edited by christoph on 25 Apr 2005, 12:09, edited 1 time in total.
Senior Manager
Senior Manager
avatar
Joined: 02 Feb 2004
Posts: 345
Followers: 1

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

 [#permalink] New post 25 Apr 2005, 12:01
fresinha12 wrote:
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...


Thanks Fresh answer and explanation. OA is D!
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1126
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 25 [0], given: 0

 [#permalink] New post 25 Apr 2005, 17:46
mirhaque wrote:
fresinha12 wrote:
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...


Thanks Fresh answer and explanation. OA is D!


SO TRICKY, I understand ythe meaning of "for all X", it is definitely a greart problem. For example in the first statement it means that you will never consider values of X that will lead to a -ive value when added to A...
Terrific problem :P
VP
VP
avatar
Joined: 26 Apr 2004
Posts: 1226
Location: Taiwan
Followers: 2

Kudos [?]: 101 [0], given: 0

Re: ds: [#permalink] New post 26 Apr 2005, 06:15
(1) x = (2+ (4-4A)^1/2) / 2 and (2 - (4-4A)^1/2) / 2

x =(2 - (4-4A)^1/2) / 2 must be positive, so 0 <A <1

but I disagree with the second condition.

(2)AX^2 + 1 must be positive for all x.

if A=0, AX^2 + 1 will be always positive for all x.

How could D?
Re: ds:   [#permalink] 26 Apr 2005, 06:15
    Similar topics Author Replies Last post
Similar
Topics:
DS jpv 3 24 Apr 2005, 07:55
DS: mirhaque 2 15 Apr 2005, 06:17
ds: mirhaque 3 25 Mar 2005, 06:44
ds mirhaque 5 23 Mar 2005, 05:54
ds sakshi 2 16 Jan 2005, 21:38
Display posts from previous: Sort by

ds:

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 22 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.