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Bunuel
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 04:43
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65% (hard)

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51% (01:35) correct 49% (01:48) wrong based on 97 sessions

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If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0


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Bunuel
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 04:44
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Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0
Show more

Similar question to practice: https://gmatclub.com/forum/if-p-is-a-po ... 27163.html
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 05:00
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Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0
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Very important -- p is a positive integer.

(1) \(p^4 > 256\)
Since \(4^4=256.....p^4>4^4.....p>4\)

(2) (\(p^2 -1)(p - 4) > 0\)
\((p-1)(p+1)(p-4)>0\)
Now p+1 is always >0, so p-1 and p-4 will have same sign as (p-1)(p-4)>0
Two cases -
a) both positive..
p-1>0 and p-4>0....p>1 and p>4...so p>4
b) both negative.
p-1<0 and p-4<0, so p<1, but that is not possible as \(p\geq{1}\)
Thus both p-1 and p-4 are positive and p>4.


D
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 05:06
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Can you elaborate how p>4 for statement 1.
When P= 4, how can I say p>4.

Would you please explain, Chetan2U Sir!
chetan2u
Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0


Very important -- p is a positive integer.

(1) \(p^4 > 256\)
Since \(4^4=256, p>4\)

(2) (\(p^2 -1)(p - 4) > 0\)
\((p-1)(p+1)(p-4)>0\)
Now p+1 is always >0, so p-1 and p-4 will have same sign as (p-1)(p-4)>0
Two cases -
a) both positive..
p-1>0 and p-4>0....p>1 and p>4...so p>4
b) both negative.
p-1<0 and p-4<0, so p<1, but that is not possible as \(p\geq{1}\)
Thus both p-1 and p-4 are positive and p>4.


D
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 05:13
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Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0
Show more

Solution


Step 1: Analyse Question Stem


    • p is a positive integer.
    • We need to find out if \(p > 4\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(p^4 > 256\)
    • \(p^4 – 4^4 > 0\)
    \(⟹ (p^2 + 4^2)(p^2 – 4^2) > 0\)
    \(⟹ (p^2 + 4^2) ( p- 4)(p+ 4) > 0\)

    • Thus, either \(p > 4 \space or \space p < -4\).
      o Since p is a positive integer. So p is always greater than 4.
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.

Statement 2: \((p^2 – 1)(p-4) > 0\)

    • \((p – 1)( p+1)(p – 4) > 0\)



    • So, -1 < p < 1 or p > 4
      o Since, there is no positive integer, between – 1 to 1 , both excluded, p cannot lie in this region.
         Thus, p > 4

Hence, statement 2 is also sufficient and the correct answer is Option D.
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 05:18
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gvij2017
Can you elaborate how p>4 for statement 1.
When P= 4, how can I say p>4.

Would you please explain, Chetan2U Sir!
chetan2u
Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0


Very important -- p is a positive integer.

(1) \(p^4 > 256\)
Since \(4^4=256, p>4\)

(2) (\(p^2 -1)(p - 4) > 0\)
\((p-1)(p+1)(p-4)>0\)
Now p+1 is always >0, so p-1 and p-4 will have same sign as (p-1)(p-4)>0
Two cases -
a) both positive..
p-1>0 and p-4>0....p>1 and p>4...so p>4
b) both negative.
p-1<0 and p-4<0, so p<1, but that is not possible as \(p\geq{1}\)
Thus both p-1 and p-4 are positive and p>4.


D
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Now \(p^4>256\) or \(p^4>4^4\), hence \(p>4\)
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If p is a positive integer, is p > 4 ? 20 Jun 2020, 20:28
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Bunuel
If p is a positive integer, is p > 4 ?

(1) p^4 > 256
(2) (p^2 -1)(p - 4) > 0
Show more

S1: In other words, p^4 is greater than 4^4. Since we know p is positive, it must be greater than 4. SUFFICIENT.

S2: The first part of the equation is a difference of squares. This can be rewritten as (p+1)(p-1)(p-4)>0. Since p>4 is a solution and the other two don't contradict it, the original equation must be true. SUFFICIENT.

ANSWER: D
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