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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Is |p|^2<|p| ? [#permalink]
18 Apr 2013, 14:52

1

This post received 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

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

The Question asks....

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.

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

2

This post received 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<1AND \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<1AND \neq{0}? Statement 1: -1\leq{p}\leq{1} Statement 2 p\neq{-,+1}

1+2 -1<p<1But no one says anything aboutp=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 1465 times ]

_________________

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

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

1+2 -1<p<1But no one says anything aboutp=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.

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!

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. _________________