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

Question

Is |p + 2| - p > 5?

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

MikeScarn · Answer

|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 0 < p < 1 NO. Sufficient. Answer: D

Princ · Answer

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}

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 5 : No Statement 1 alone is sufficient (2) p > p^2 This means 0 5 : No Statement 2 alone is sufficient

PKN · Answer

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, -2p^2 Or, p^2

sumit411 · Answer

Answer is D Consider kudos if that helped 1534955144452.jpg Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

fskilnik · Answer

\left| {p + 2} ight| - p = \left\{ \begin{gathered} \left( {p + 2} ight) - p = 2\,\,\,\,\,if\,\,\,p \geqslant 2 \hfill \ \left( {-p - 2} ight) - p = -2p - 2\,\,\,\,\,if\,\,\,p < 2 \hfill \ \end{gathered} ight. ?\,\,\,\,\,:\,\,\,\,\,\,\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} ight. (1) -2 {p^2}\,\,\, \Leftrightarrow \,\,\,p\left( {p - 1} ight) < 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.

DavidTutorexamPAL · Answer

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!

Archit3110 · Answer

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

KanishkM · Answer

from 1, we can get values as -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

CareerGeek · Answer

Is |p+2| - p > 5 —> lp+2l > p + 5 ? (1) lpl < 2 —> -2 -2+5 —> NO l2+2l > 2+5 —> NO Sufficient (2) P > P^2 —> p^2 - p < 0 —> p(p - 1) < 0 —> 0 0 + 5 —> NO l1+2l > 1 + 5 —> NO Sufficient IMO Option D Pls Hit kudos if you like the solution Posted from my mobile device

CrackverbalGMAT · Answer

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!

bumpbot · Answer

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

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