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

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Is |p|^2<|p| ?  [#permalink]

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Updated on: 18 Apr 2013, 15:51
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Question Stats:

37% (01:36) correct 63% (01:31) wrong based on 245 sessions

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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.

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Originally posted by Zarrolou on 18 Apr 2013, 13:32.
Last edited by Zarrolou on 18 Apr 2013, 15:51, edited 1 time in total.
Added OA and OE
##### Most Helpful Community Reply
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Re: Is |p|^2<|p| ?  [#permalink]

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18 Apr 2013, 15:48
4
1
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:
Spoiler: :: OA
E

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##### General Discussion
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Re: Is |p|^2<|p| ?  [#permalink]

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Updated on: 18 Apr 2013, 16:31
1
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

Originally posted by yezz on 18 Apr 2013, 14:29.
Last edited by yezz on 18 Apr 2013, 16:31, edited 1 time in total.
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Re: Is |p|^2<|p| ?  [#permalink]

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18 Apr 2013, 15:34
2
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.
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Re: Is |p|^2<|p| ?  [#permalink]

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18 Apr 2013, 15:52
1
/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
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Re: Is |p|^2<|p| ?  [#permalink]

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19 Apr 2013, 11:36
1
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:
Spoiler: :: OA
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......
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Re: Is |p|^2<|p| ?  [#permalink]

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19 Apr 2013, 11:54
2
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.
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Re: Is |p|^2<|p| ?  [#permalink]

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19 Apr 2013, 12:06
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 !!
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Re: Is |p|^2<|p| ?  [#permalink]

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25 Jun 2013, 16:59
1
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!
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Re: Is |p|^2<|p| ?  [#permalink]

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17 Sep 2017, 20:14
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:
Spoiler: :: OA
E
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Re: Is |p|^2<|p| ?  [#permalink]

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01 Mar 2019, 09:43
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
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Re: Is |p|^2<|p| ?  [#permalink]

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16 May 2019, 05:56
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:
Spoiler: :: OA
E
Re: Is |p|^2<|p| ?   [#permalink] 16 May 2019, 05:56
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