Last visit was: 23 Apr 2024, 22:21 It is currently 23 Apr 2024, 22:21

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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
User avatar
Director
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 859
Own Kudos [?]: 4889 [21]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Send PM
Most Helpful Reply
User avatar
Director
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 859
Own Kudos [?]: 4889 [6]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Send PM
General Discussion
Retired Moderator
Joined: 05 Jul 2006
Posts: 849
Own Kudos [?]: 1562 [1]
Given Kudos: 49
Send PM
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 485
Own Kudos [?]: 3092 [2]
Given Kudos: 141
Send PM
Re: Is |p|^2<|p| ? [#permalink]
2
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


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.
avatar
Intern
Intern
Joined: 16 Dec 2012
Posts: 1
Own Kudos [?]: 1 [1]
Given Kudos: 1
Send PM
Re: Is |p|^2<|p| ? [#permalink]
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
User avatar
Intern
Intern
Joined: 30 Sep 2012
Status:Pushing Hard
Affiliations: GNGO2, SSCRB
Posts: 44
Own Kudos [?]: 203 [1]
Given Kudos: 11
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE:Analyst (Health Care)
Send PM
Re: Is |p|^2<|p| ? [#permalink]
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:

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

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......
User avatar
Director
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 859
Own Kudos [?]: 4889 [2]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Send PM
Re: Is |p|^2<|p| ? [#permalink]
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
mia.png [ 8.74 KiB | Viewed 6087 times ]

User avatar
Intern
Intern
Joined: 30 Sep 2012
Status:Pushing Hard
Affiliations: GNGO2, SSCRB
Posts: 44
Own Kudos [?]: 203 [0]
Given Kudos: 11
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE:Analyst (Health Care)
Send PM
Re: Is |p|^2<|p| ? [#permalink]
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 !!
User avatar
Senior Manager
Senior Manager
Joined: 13 May 2013
Posts: 314
Own Kudos [?]: 565 [1]
Given Kudos: 134
Send PM
Re: Is |p|^2<|p| ? [#permalink]
1
Kudos
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!

Answer = E.

Great question Zarrolou!
Intern
Intern
Joined: 31 Jul 2017
Posts: 5
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
Re: Is |p|^2<|p| ? [#permalink]
If this question stated that P didn't equal 0, would the answer be B?
When I worked this problem out, I forgot to take into account for the number 0.

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:
Senior Manager
Senior Manager
Joined: 05 Feb 2018
Posts: 312
Own Kudos [?]: 794 [0]
Given Kudos: 325
Send PM
Re: Is |p|^2<|p| ? [#permalink]
For equation |p|² < |p|, testing values:


1) p² ≤ 1
If p = 1/2 then 1/4 < 1/2 (negative fractions will be the same), YES
If p = 1 then 1 < 1 (-1 and any value above 1 will do the same), NO
This is the trick, if p = 0 it satisfies our constraint and it is also NO in the original equation
Insufficient.

2) p² - 1 ≠ 0
(p-1)(p+1) ≠ 0 so p can't be -1 or 1, same as above and we can get even more values that work.

1&2) Even taking both constraints into account we still can't eliminate p = 0 so it is E
Manager
Manager
Joined: 04 Jun 2010
Posts: 100
Own Kudos [?]: 33 [0]
Given Kudos: 264
Location: India
GMAT 1: 660 Q49 V31
GPA: 3.22
Send PM
Re: Is |p|^2<|p| ? [#permalink]
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 \(p\neq{0}\)?


Good question..........obviously missed the p could be zero part

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:
GMAT Club Legend
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 8018
Own Kudos [?]: 4095 [0]
Given Kudos: 242
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Send PM
Re: Is |p|^2<|p| ? [#permalink]
we know that
lxl= √x^2
so
\(|p|^2<|p|\)
(√p^2)^2 <√p^2
p^2<p
need to determine whether p^2-p<0

1.\(p^2\leq{1}\)

possible when p = 1/4 or 0
we get yes & no to target
insufficient

2.\(p^2-1\neq{0}\)

p = 0 or 1/4
insufficient
from 1 &2
nothing can be determined
OPTION E is correct


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!

This is my first creation, I appreciate any feedback
Scroll down for OE.
GMAT Club Bot
Re: Is |p|^2<|p| ? [#permalink]
Moderator:
Math Expert
92883 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne