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

 It is currently 03 Aug 2015, 05:54

### 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

# Is |p|^2<|p| ?

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 147

Kudos [?]: 1458 [8] , given: 219

Is |p|^2<|p| ? [#permalink]  18 Apr 2013, 12:32
8
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

34% (01:54) correct 66% (00:50) wrong based on 177 sessions
Is $$|p|^2<|p|$$ ?

1.$$p^2\leq{1}$$

2.$$p^2-1\neq{0}$$

Hi guys! I created this DS, if you want give it a try!

This is my first creation, I appreciate any feedback
Scroll down for OE.
[Reveal] Spoiler: OA

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Last edited by Zarrolou on 18 Apr 2013, 14:51, edited 1 time in total.
Added OA and OE
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 134 [1] , given: 39

Re: Is |p|^2<|p| ? [#permalink]  18 Apr 2013, 13:29
1
KUDOS
Zarrolou wrote:
Is $$|p|^2<|p|$$ ?

1.$$p^2\leq{1}$$
2.$$p^2-1\neq{0}$$

Hi guys! I created this DS, if you want give it a try!

I'll post the solution after some discussion
This is my first creation, I appreciate any feedback

E it is ... my bad

Last edited by yezz on 18 Apr 2013, 15:31, edited 1 time in total.
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
Followers: 62

Kudos [?]: 750 [1] , given: 135

Re: Is |p|^2<|p| ? [#permalink]  18 Apr 2013, 14:34
1
KUDOS
Expert's post
Zarrolou wrote:
Is $$|p|^2<|p|$$ ?

1.$$p^2\leq{1}$$
2.$$p^2-1\neq{0}$$

Hi guys! I created this DS, if you want give it a try!

I'll post the solution after some discussion
This is my first creation, I appreciate any feedback

From F.S 1, for p=0, we have a NO. For p=-1/2, we have a YES. Insufficient.

From F.S 2, Just as above. Insufficient.

Both together, for p=0, a NO. For p=-1/2, a YES. Insufficient.

E.
_________________
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 147

Kudos [?]: 1458 [3] , given: 219

Re: Is |p|^2<|p| ? [#permalink]  18 Apr 2013, 14:48
3
KUDOS
Official Explanation

Is $$|p|^2<|p|$$ ?

Splitting the equation into two scenarios $$p>0$$ and $$p<0$$

I)$$p>0$$
$$p^2-p>0$$
$$0<p<1$$
II)$$p<{}0$$
$$(-p)^2+p<0$$
$$-1<p<0$$

We can rewrite the question as: is $$-1<p<1$$ and $$p\neq{0}$$?

1.$$p^2\leq{1}$$
$$-1\leq{}p\leq{}1$$
We cannot say if given this interval p will have one of those values: $$-1<p<1$$ and $$p\neq{0}$$
Not Sufficient

2.$$p^2-1\neq{0}$$
$$p\neq{}+,-1$$
Clearly not Sufficient

1+2. Using both 1 and 2 we can conclude that $$-1<p<1$$ but p could equal 0. Not Sufficient

OA:
[Reveal] Spoiler: OA
E

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Intern
Joined: 16 Dec 2012
Posts: 1
Followers: 0

Kudos [?]: 1 [1] , given: 1

Re: Is |p|^2<|p| ? [#permalink]  18 Apr 2013, 14:52
1
KUDOS
/p/(/p/-1)<0 we cannot divide the inequality with /p/ since we don't know if /p/=0.
from statement 1 ,we have that -1<=p<=1 insuff
from statement 2, p<>+-1. insuff

combined p lies in(-1 U +1). p=0.5 ,we have YES,p=0 we have NO. insuff
IMO E
Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 92
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 66 [1] , given: 11

Re: Is |p|^2<|p| ? [#permalink]  19 Apr 2013, 10:36
1
KUDOS
Zarrolou wrote:
Official Explanation

Is $$|p|^2<|p|$$ ?

Splitting the equation into two scenarios $$p>0$$ and $$p<0$$

I)$$p>0$$
$$p^2-p>0$$
$$0<p<1$$
II)$$p<{}0$$
$$(-p)^2+p<0$$
$$-1<p<0$$

We can rewrite the question as: is $$-1<p<1$$ and $$p\neq{0}$$?

1.$$p^2\leq{1}$$
$$-1\leq{}p\leq{}1$$
We cannot say if given this interval p will have one of those values: $$-1<p<1$$ and $$p\neq{0}$$
Not Sufficient

2.$$p^2-1\neq{0}$$
$$p\neq{}+,-1$$
Clearly not Sufficient

1+2. Using both 1 and 2 we can conclude that $$-1<p<1$$ but p could equal 0. Not Sufficient

OA:
[Reveal] Spoiler: OA
E

----------------------------------------------------------------

Hey, Good Q ..... I'm bit confused ... Correct me if I'm wrong ....

Is |p|^2<|p| ? , one must always keep in mind that Is |p| is always positive, no matter the sign of P ..Okay & second thing that p^2 will always remain +ve. Now, |p|^2<|p| will only be possible if |p| must be a proper fraction okay. So, we have to find if |p|
is a proper fraction or not. okay.

now statement 1 says .......... p^2 =<1 ...... this means that P can be equal to +1, -1 or p can be equal to any proper fraction as P^2 must always be less than 1 only if P is a proper fraction. so from this we have three values of P... 1, -1 & any proper fraction... however this is insufficient because of multiple possibilities.

Now statement 2 says ....... p^2-1 is not equal to Zero. that means that p is not equal to +,-1 , therefore it is clearly insufficient. .....

Now, 1+2 ........ as statement 1 says that p can be 1 or -1 or any proper fraction & statement 2 says that P is not equal to +,-1 , therefore we can conclude that p = Proper fraction & if P is a Proper fraction whether +ve Proper fraction or -ve Proper fraction, ..... |p|^2<|p| will always remain true. Hence , C.

Please correct me if I'm wrong. Pls......
_________________

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

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 147

Kudos [?]: 1458 [2] , given: 219

Re: Is |p|^2<|p| ? [#permalink]  19 Apr 2013, 10:54
2
KUDOS
Hi manishuol,

lets see if I can remove your doubts. The graph will help a lot, scroll down it's $$|p|^2-|p|<0$$ we have to look where it is negative!

$$|p|^2<|p|$$, remember that both |p| and |p|^2 don't have to be positive, they can also equal 0.

"So, we have to find if |p| is a proper fraction or not. okay." Not correct, we must find out if p is in the interval $$-1<p<1$$ AND $$\neq{0}$$.

"this means that P can be equal to +1, -1 or p can be equal to any proper fraction" AND 0 I would add, p ranges in the interval $$-1\leq{p}\leq{1}$$
"however this is insufficient because of multiple possibilities." Correct

"Now statement 2 is clearly insufficient" Correct

Question: is p $$-1<p<1$$ AND $$\neq{0}$$?
Statement 1: $$-1\leq{p}\leq{1}$$
Statement 2 $$p\neq{-,+1}$$

1+2 $$-1<p<1$$ But no one says anything about $$p=0$$

The point is all here: p could still equal 0, and both statement don't give us info about this possibility.
Attachments

mia.png [ 8.74 KiB | Viewed 1799 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 92
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 66 [0], given: 11

Re: Is |p|^2<|p| ? [#permalink]  19 Apr 2013, 11:06
Zarrolou wrote:
Hi manishuol,

lets see if I can remove your doubts. The graph will help a lot, scroll down it's $$|p|^2-|p|<0$$ we have to look where it is negative!

$$|p|^2<|p|$$, remember that both |p| and |p|^2 don't have to be positive, they can also equal 0.

"So, we have to find if |p| is a proper fraction or not. okay." Not correct, we must find out if p is in the interval $$-1<p<1$$ AND $$\neq{0}$$.

"this means that P can be equal to +1, -1 or p can be equal to any proper fraction" AND 0 I would add, p ranges in the interval $$-1\leq{p}\leq{1}$$
"however this is insufficient because of multiple possibilities." Correct

"Now statement 2 is clearly insufficient" Correct

Question: is p $$-1<p<1$$ AND $$\neq{0}$$?
Statement 1: $$-1\leq{p}\leq{1}$$
Statement 2 $$p\neq{-,+1}$$

1+2 $$-1<p<1$$ But no one says anything about $$p=0$$

The point is all here: p could still equal 0, and both statement don't give us info about this possibility.

-----------------------------------

Yeah You're certainly right. I missed out 0 as one of the possibility. .......I really appreciate your quick help.Thanks !! Brother !!
_________________

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

Senior Manager
Joined: 13 May 2013
Posts: 475
Followers: 1

Kudos [?]: 87 [0], given: 134

Re: Is |p|^2<|p| ? [#permalink]  25 Jun 2013, 15:59
Is |p|^2<|p| ?

1.p^2≤1

2.p^2-1≠0

Is |p|^2<|p|? ==> is -1<p<1

1.) insufficient as p could be 1 or -1 meaning p|^2 could equal |p| or p|^2 could be less than |p| Insufficient.

2.) p^2-1≠0 insufficient. p could = 1/2 which when squared = 1/4 which doesn't = 0 and p^2 would be less than P. On the other hand P could = 10 and 10^2-1 > 10.

1+2) we know that -1<p<1 and that p^2-1≠0 so the range of values to test is between -1 and 1. Fine. For ALL values of p except -1, 0 and 1 is |p|^2<|p|. However, while -1^2 - 1 and 1^2 - 1 = 0 (which rules out p being -1 or 1) p could be 0 In which case p^2 = p or it could be a fraction in which p^2 is less than p!

Great question Zarrolou!
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 5724
Followers: 323

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

Re: Is |p|^2<|p| ? [#permalink]  24 Jul 2014, 04:16
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Is |p|^2<|p| ?   [#permalink] 24 Jul 2014, 04:16
Similar topics Replies Last post
Similar
Topics:
6 Is p^2 - 1 divisible by 12? 7 15 Aug 2014, 07:13
3 Is p^2 an odd integer? 4 06 Feb 2014, 00:22
12 If n= p^2-p+17, is n prime? 7 13 May 2013, 12:53
1 Does p^2 = q if p is a prime number? 4 08 Apr 2012, 13:08
1 Is p<0? a. p+6<|p| p+6<p 6<0; not possible. 1 21 Feb 2011, 08:03
Display posts from previous: Sort by

# Is |p|^2<|p| ?

 Question banks Downloads My Bookmarks Reviews Important topics

 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®.