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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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New post Updated on: 03 Aug 2018, 10:14
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D
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Difficulty:

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Is |p + 2| - p > 5?

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

Originally posted by a70 on 03 Aug 2018, 10:01.
Last edited by Bunuel on 03 Aug 2018, 10:14, 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, 11:35
1
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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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 03:30
1
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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Is |p + 2| - p > 5? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post 22 Aug 2018, 08:21
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, 09:24
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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, 10:16
1
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? (1) |p| < 2 (2) p > p^2  [#permalink]

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New post Updated on: 03 Jan 2019, 03:46
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, 11:38.
Last edited by DavidTutorexamPAL on 03 Jan 2019, 03:46, edited 1 time in total.
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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, 03:41
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  [#permalink]

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New post 02 Feb 2019, 09:21
a70 wrote:
Is |p + 2| - p > 5?

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


from 1, we can get values as -2 < p < 2,

-1,0,1, when put back in the question will give 2 > 5 a No .
Sufficient

for 2 to be possible p > p^2

p can only be values from 0.1 <= x <= 0.3

Which when put back in the question will always give us a No
Sufficient

D
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Re: Absolute Values/Modulus  [#permalink]

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New post 11 Jun 2019, 22:41
1
TarunTilokani wrote:
Is |p+2| - p > 5 ?

A) |p| < 2

B) P > P^2


Is |p+2| - p > 5
—> lp+2l > p + 5 ?

(1) lpl < 2
—> -2 < p < 2
Try substituting -2 & 2
l-2+2l > -2+5 —> NO
l2+2l > 2+5 —> NO

Sufficient

(2) P > P^2
—> p^2 - p < 0
—> p(p - 1) < 0
—> 0 < p < 1
Try substituting 0 & 1
l0+2l > 0 + 5 —> NO
l1+2l > 1 + 5 —> NO

Sufficient

IMO Option 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 12 Jun 2019, 05:02
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This is a moderately difficult question on Inequalities and absolute values. There are multiple concepts tested in this question and therefore, you will have to have a sound grasp of these if you want to crack this question.

If |x| < a, then the range of x that satisfies the inequality is –a<x<a.

The only range where x>\(x^2\) is when 0<x<1.

Also remember that this is a DS question where a YES or a NO are expected as answers; so keep in mind that a definite NO is also an acceptable answer.

Using statement I alone, we know that |p| < 2. This means -2<p<2. For the boundary values of p = 2 and p = -2, we get the LHS of the inequality as,

|4| - 2 = 2 which is NOT More than 5 and

|0| + 2 = 2 which is also NOT more than 5.

Clearly, when we substitute other smaller values from this range, the LHS of the inequality given in the question stem is bound to be lesser than 5. So, we obtain a clear NO as the answer.
Statement I alone is therefore sufficient. Possible answer options are A and D; options B, C and E can be ruled out.

Using statement II alone, we know that p is a positive proper fraction i.e. 0<p<1. Even if we substitute the upper bound i.e. p=1, we have |1+2| - 1 = 2 which is NOT more than 5.
Therefore, for smaller values of p, LHS of the inequality WILL be smaller than 5.
Here also, we get a clear NO as an answer. Statement II alone is sufficient.

So, the correct answer option is D.

The best approach in such questions is to use a balance of concepts and values to solve the question. Use the concepts to determine the range and pick values from that range to plug in and verify.
Ideally, this question should take anywhere between 1 minute to 2 minutes for you to solve when you are well prepared.

Hope this helps!
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Re: Is |p + 2| - p > 5 ? (1) |p| < 2 (2) p > p^2   [#permalink] 12 Jun 2019, 05:02
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