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Is p > q ?

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Is p > q ?  [#permalink]

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New post Updated on: 29 Aug 2017, 11:28
2
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A
B
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D
E

Difficulty:

  25% (medium)

Question Stats:

72% (01:02) correct 28% (01:17) wrong based on 61 sessions

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Is p > q ?

(1) p^2 > q^2
(2) p^3 > q^3

Originally posted by saikiranjadav on 29 Aug 2017, 11:25.
Last edited by Bunuel on 29 Aug 2017, 11:28, edited 1 time in total.
Renamed the topic and edited the question.
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Is p > q ?  [#permalink]

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New post 29 Aug 2017, 11:31
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saikiranjadav wrote:
Is p > q ?

(1) p^2 > q^2
(2) p^3 > q^3


Is p > q ?

(1) p^2 > q^2

Take the square root: |p| > |q|. This means that p is further from 0 than q is, which is clearly insufficient to answer the question whether p > q.

(2) p^3 > q^3

Take the cube root: p > q. Sufficient.

Answer: B.

P.S. Manipulating Inequalities (adding, subtracting, squaring etc.).
P.P.S. Please name topics properly. Check rule 3 here: https://gmatclub.com/forum/rules-for-po ... 33935.html Thank you.
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Re: Algebra- inequalities  [#permalink]

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New post 29 Aug 2017, 11:36
consider negative range to find extreme scenarios
A) p^2 > q^2
Insufficient as
q can be 2 while p can be -3 still satisfying equation 9> 4 but -3 < 2
also q can be 2 and p can be 3 satisfying equation 9> 4 also 3> 2
Squares hide the original sign

B) if p =-2 and q= -3
p^3 = -8
q^3 = -27
-8 > -27 and -2 > -3 therefore p > q
Also, p=2 , q=-3 then 8> -27 therefor p>q

similarly if we choose one + one -ve to satisfy the equation, we will get p>q
similar behavior for fractions.

hence B
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Re: Is p > q ?  [#permalink]

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New post 29 Aug 2017, 18:30
saikiranjadav wrote:
Is p > q ?

(1) p^2 > q^2
(2) p^3 > q^3


Condition 1) \(p^2 - q^2 > 0\)

\((p+q)(p-q) > 0\)
\(p + q > 0\) implies \(p - q > 0\) or \(p > q\).
\(p - q < 0\) implies \(p - q < 0\) or \(p < q\).

Thus we have \(p = 2\), \(q = 1\) and \(p = -2\), \(q = -1\).

Thus this condition is not sufficient.


Condition 2) \(p^3 > q^3\)

\(p^3 - q^3 = (p-q)(p^2+pq+q^2) > 0\)

Without loss of generality, we can assume \(q \ne 0\).
\(p^2 + pq + q^2\)
\(= q^2 \{ \frac{p^2}{q^2} + \frac{p}{q} + 1 \}\)
\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + 1 \}\)
\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + \frac{1}{4} + \frac{3}{4} \}\)
\(= q^2 \{ ( \frac{p}{q} + \frac{1}{2} )^2 + \frac{3}{4} \} \ge \frac{3}{4} q^2 > 0\)
Thus we have \(p^2 + pq + q^2 > 0\).

Since \(p^3 - q^3 > 0\), \(p^2 + pq + q^2 > 0\) and \(p^3 - q^3 = (p-q)(p^2 + pq + q^2) > 0\), we have \(p - q > 0\) or \(p > q\).

This is sufficient.

Answer: B
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Re: Is p > q ?   [#permalink] 29 Aug 2017, 18:30
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