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Is p^2<p ? [#permalink]
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18 Apr 2013, 12:32
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Is \(p^2<p\) ? 1.\(p^2\leq{1}\) 2.\(p^21\neq{0}\) 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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Last edited by Zarrolou on 18 Apr 2013, 14:51, edited 1 time in total.
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Re: Is p^2<p ? [#permalink]
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18 Apr 2013, 13:29
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Zarrolou wrote: Is \(p^2<p\) ?
1.\(p^2\leq{1}\) 2.\(p^21\neq{0}\)
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.



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Re: Is p^2<p ? [#permalink]
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18 Apr 2013, 14:34
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Zarrolou wrote: Is \(p^2<p\) ?
1.\(p^2\leq{1}\) 2.\(p^21\neq{0}\)
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, 14:48
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Official ExplanationIs \(p^2<p\) ? Splitting the equation into two scenarios \(p>0\) and \(p<0\) I)\(p>0\) \(p^2p>0\) \(0<p<1\) 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.\(p^21\neq{0}\) \(p\neq{}+,1\) 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:
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Re: Is p^2<p ? [#permalink]
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/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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Zarrolou wrote: Official ExplanationIs \(p^2<p\) ? Splitting the equation into two scenarios \(p>0\) and \(p<0\) I)\(p>0\) \(p^2p>0\) \(0<p<1\) 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.\(p^21\neq{0}\) \(p\neq{}+,1\) 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^21 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, 10:54
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Hi manishuol, lets see if I can remove your doubts. The graph will help a lot, scroll down it's \(p^2p<0\) 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, 11:06
Zarrolou wrote: Hi manishuol,
lets see if I can remove your doubts. The graph will help a lot, scroll down it's \(p^2p<0\) 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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Is p^2<p ?
1.p^2≤1
2.p^21≠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^21≠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 10^21 > 10.
1+2) we know that 1<p<1 and that p^21≠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, 19:14
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 ExplanationIs \(p^2<p\) ? Splitting the equation into two scenarios \(p>0\) and \(p<0\) I)\(p>0\) \(p^2p>0\) \(0<p<1\) 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.\(p^21\neq{0}\) \(p\neq{}+,1\) 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:




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