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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2

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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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Is |p + 2| - p > 5?

(1) |p| < 2
(2) p > p^2

Originally posted by a70 on 03 Aug 2018, 09:01.
Last edited by Bunuel on 03 Aug 2018, 09:14, edited 1 time in total.
Renamed the topic and edited the question.
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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 03 Aug 2018, 10:35
ankit7055 wrote:
Is |p + 2| - p > 5?

(1) |p| < 2
(2) p > p^2


|p + 2| > 5 + p

Answer is YES for negative numbers with high magnitude

Answer is NO for any number greater than -4

(1) |p| < 2

-2 < p < 2 NO. Sufficient.

(2) p > p^2

0 < p < 1 NO. Sufficient.

Answer: D
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 00:39
1
Can someone please explain this better?
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 02:30
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surabhikhandelwal13 wrote:
Can someone please explain this better?


Quote:
Is \(|p + 2| - p > 5\)?
(1) \(|p| < 2\)
(2) \(p > p^2\)


OA: D

Reducing Question stem, we get

Case 1 : If \(p+2 ≥0\) , \(p≥-2\)
\(p+2-p>5\)
\(2>5\) (Not possible)
So For \(p≥-2\), Answer for Question : Is \(|p + 2| - p > 5\) : No

Case 2 :If \(p+2 <0 , p<-2\)
\(-(p+2)-p>5\)
\(-p-2-p>5\)
\(-2p-2>5\)
\(-2p>7\)
\(p<-\frac{7}{2}\)

for \(-\frac{7}{2}≤p<-2\)
Answer for Question : Is \(|p + 2| - p > 5\) : No

For \(-\frac{7}{2}<p\)
Answer for Question : Is \(|p + 2| - p > 5\) : Yes

Summarising
\(p≥-\frac{7}{2}\) ; Is \(|p + 2| - p > 5\) : No
\(p<-\frac{7}{2}\) ; Is \(|p + 2| - p > 5\) : Yes


(1) \(|p| < 2\)
This means \(-2<p<2\)
As \(p≥-\frac{7}{2}\), Is \(|p + 2| - p > 5\) : No
Statement 1 alone is sufficient

(2) \(p > p^2\)
This means \(0<p<1\)
As \(p≥-\frac{7}{2}\), Is \(|p + 2| - p > 5\) : No
Statement 2 alone is sufficient
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 07:21
surabhikhandelwal13 wrote:
Can someone please explain this better?

Is \(|p+2|−p>5?\)
(1) \(|p|<2\)
(2) \(p>p^2\)

Simplifying question stem:-
\(|p+2|−p>5\)

Case-1 (when \((p+2)\geq{0}\) or \(p\geq{-2}\), |p+2|=p+2)
p+2-p>5
2>5
Invalid.

Case-2 (when \((p+2)<{0}\) or p< -2, |p+2|=-(p+2))
-(p+2)-p>5
Or, -2p-2>5
Or, -2p>7
Or, \(p < \frac{-7}{2}\)

Combining the ranges: a) p>-2 and invalid OR b) p<-2 and \(p < \frac{-7}{2}\), we have
\(p < \frac{-7}{2}\)

Re-phrased question stem:- Is \(p < \frac{-7}{2}(=-3.5)?\)

St1:- \(|p|<2\)
Or, \(-2<p<2\)
Answer to question stem is always NO.
Sufficient.

St2:- \(p>p^2\)
Or, \(p^2<p\) (Switching sides)
Or, \(p^2-p<0\)
Or, \(p(p-1)<0\)
Or, \(0<p<1\)
Answer to question stem is always NO.
Sufficient.

Ans. (D)
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 08:24
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Answer is D

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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 09:16
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Quote:
Is |p + 2| - p > 5?

(1) |p| < 2
(2) p > p^2


\(\left| {p + 2} \right| - p = \left\{ \begin{gathered}
\left( {p + 2} \right) - p = 2\,\,\,\,\,if\,\,\,p \geqslant 2 \hfill \\
\left( {-p - 2} \right) - p = -2p - 2\,\,\,\,\,if\,\,\,p < 2 \hfill \\
\end{gathered} \right.\)

\(?\,\,\,\,\,:\,\,\,\,\,\,\left\{ \begin{gathered}
2\,\,\,\,\mathop > \limits^? \,\,\,5\,\,\,\,\,\,\,\,if\,\,\,p \geqslant 2\, \hfill \\
-2p - 2\,\,\,\,\mathop > \limits^? \,\,\,5\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,p\,\,\mathop < \limits^? - \frac{7}{2}\,\,\,\,\,\,\,if\,\,\,p < 2\, \hfill \\
\end{gathered} \right.\)

(1) -2 < p < 2 , hence we are dealing with the second expression in the last two possibilities. And NO, p is not less than -3.5 , for sure. Sufficient.

(2) \(p > {p^2}\,\,\, \Leftrightarrow \,\,\,p\left( {p - 1} \right) < 0\,\,\, \Leftrightarrow \,\,\,0 < p < 1\)
We are again dealing with the second expression in the last two possibilities. And NO, p is not less that -3.5, for sure. Sufficient.

The solution above follows the notations and rationale taught in the GMATH method.
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Is |p + 2| − p > 5?  [#permalink]

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New post 02 Jan 2019, 09:50
Is |p + 2| − p > 5?

(1) |p| < 2

(2) p > p^2
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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post Updated on: 03 Jan 2019, 02:46
1
The answer is D.
We can answer with a mixture of Alternative and Logical methods
1) let's try the edge cases, max and min:
if p=2, then |p + 2| − p = 4-2 = 2 - no!
if p=-2, then |p + 2| − p = |-2 + 2| − (-2)=|0|+2=2 - no!
two edge cases, same answer - sufficient!

2) this condition (p > p^2) is only true for positive fractions (0<p<1), for whom |p + 2| − p is definitely smaller than 5 - sufficient!
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Originally posted by DavidTutorexamPAL on 02 Jan 2019, 10:38.
Last edited by DavidTutorexamPAL on 03 Jan 2019, 02:46, edited 1 time in total.
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Re: Is |p + 2| − p > 5?  [#permalink]

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New post 02 Jan 2019, 11:18
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please correct OA to B
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Re: Is |p + 2| − p > 5?  [#permalink]

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New post 02 Jan 2019, 11:28
Shouldn't OA be D since both statements are sufficient to answer the prompt? DavidTutorexamPAL barryseal



DavidTutorexamPAL wrote:
The answer is B.
We can answer with a mixture of Alternative and Logical methods
1) let's try the edge cases, max and min:
if p=2, then |p + 2| − p = 4-2 = 2 - no!
if p=-2, then |p + 2| − p = |-2 + 2| − (-2)=|0|+2=2 - no!
two edge cases, same answer - sufficient!

2) this condition (p > p^2) is only true for positive fractions (0<p<1), for whom |p + 2| − p is definitely smaller than 5 - sufficient!
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Re: Is |p + 2| − p > 5?  [#permalink]

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New post 02 Jan 2019, 11:48
Yes you're right. Sorry

funsogu wrote:
Shouldn't OA be D since both statements are sufficient to answer the prompt? DavidTutorexamPAL barryseal



DavidTutorexamPAL wrote:
The answer is B.
We can answer with a mixture of Alternative and Logical methods
1) let's try the edge cases, max and min:
if p=2, then |p + 2| − p = 4-2 = 2 - no!
if p=-2, then |p + 2| − p = |-2 + 2| − (-2)=|0|+2=2 - no!
two edge cases, same answer - sufficient!

2) this condition (p > p^2) is only true for positive fractions (0<p<1), for whom |p + 2| − p is definitely smaller than 5 - sufficient!
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 03 Jan 2019, 02:38
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 03 Jan 2019, 02:46
funsogu wrote:
Shouldn't OA be D since both statements are sufficient to answer the prompt? DavidTutorexamPAL barryseal



DavidTutorexamPAL wrote:
The answer is B.
We can answer with a mixture of Alternative and Logical methods
1) let's try the edge cases, max and min:
if p=2, then |p + 2| − p = 4-2 = 2 - no!
if p=-2, then |p + 2| − p = |-2 + 2| − (-2)=|0|+2=2 - no!
two edge cases, same answer - sufficient!

2) this condition (p > p^2) is only true for positive fractions (0<p<1), for whom |p + 2| − p is definitely smaller than 5 - sufficient!

Yes, corrected it thanks!
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 03 Jan 2019, 08:37
sumit411 wrote:
Answer is D

Consider kudos if that helped
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Appreciations to your detailed explanation, thanks bro..
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 06 Jan 2019, 02:41
a70 wrote:
Is |p + 2| - p > 5?

(1) |p| < 2
(2) p > p^2


from the given info we can say;
p>7/2 or 3.5

#1:
lpl<2

sufficient since p would be less thean 3.5

#2
p > p^2

1>p

since 1>p so p wont be >3.5 sufficeint

IMO D
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Re: Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2 &nbs [#permalink] 06 Jan 2019, 02:41
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