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805+ (Hard)|   Inequalities|      
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rajthakkar
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If p > q, is p a negative number? Updated on: 10 Jun 2015, 10:50
Originally posted by rajthakkar on 08 Feb 2015, 01:12.
Last edited by Bunuel on 10 Jun 2015, 10:50, edited 1 time in total.
Renamed the topic and edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

41% (02:24) correct 59% (02:59) wrong based on 61 sessions

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If p > q, is p a negative number?

(1) (p^2)(q) + (p)(q^2) is negative
(2) (p^2)(q) - (p)(q^2) is positive
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Official Answer
C
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If p > q, is p a negative number? 10 Jun 2015, 11:28
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rajthakkar
If p > q, is p a negative number?

(1) (p^2)(q) + (p)(q^2) is negative
(2) (p^2)(q) - (p)(q^2) is positive
Show more

If p > q, is p a negative number?

(1) (p^2)(q) + (p)(q^2) is negative

\((p^2)(q) + (p)(q^2) < 0\);
\(pq(p + q) < 0\).

If p = 2 and q = -1, then the answer is NO.
If p = -1 and q = -2, then the answer is YES.

Not sufficient.

(2) (p^2)(q) - (p)(q^2) is positive

\((p^2)(q) - (p)(q^2) > 0\);
\(pq(p - q) > 0\).

Since from the stem p - q > 0, then it must be true that pq > 0 (p and q have the same sign).
If p = 2 and q = 1, then the answer is NO.
If p = -1 and q = -2, then the answer is YES.

Not sufficient.

(1)+(2) From (2) we have that p and q have the same sign. Both of them cannot be positive because in this case (p^2)(q) + (p)(q^2) would be positive, not negative as stated in (1). Therefore both p and q must be negative. Sufficient.

Answer: C.
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If p > q, is p a negative number? 10 Jun 2015, 11:29
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Bunuel
rajthakkar
If p > q, is p a negative number?

(1) (p^2)(q) + (p)(q^2) is negative
(2) (p^2)(q) - (p)(q^2) is positive

If p > q, is p a negative number?

(1) (p^2)(q) + (p)(q^2) is negative

\((p^2)(q) + (p)(q^2) < 0\);
\(pq(p + q) < 0\).

If p = 2 and q = -1, then the answer is NO.
If p = -1 and q = -2, then the answer is YES.

Not sufficient.

(2) (p^2)(q) - (p)(q^2) is positive

\((p^2)(q) - (p)(q^2) > 0\);
\(pq(p - q) > 0\).

Since from the stem p - q > 0, then it must be true that pq > 0 (p and q have the same sign).
If p = 2 and q = 1, then the answer is NO.
If p = -1 and q = -2, then the answer is YES.

Not sufficient.

(1)+(2) From (2) we have that p and q have the same sign. Both of them cannot be positive because in this case (p^2)(q) + (p)(q^2) would be positive, not negative as stated in (1). Therefore both p and q must be negative. Sufficient.

Answer: C.
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Similar question to practice: is-p-2q-pq-153912.html
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If p > q, is p a negative number? 10 Jun 2015, 11:41
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IMO answer should be B

Option A
P^2/q + P/q^2 is negative

=((P^2*q)+p)/q^2 should be -ive
as denominator is q^2 it needs too be positive
so we are left with
P^2*q +p
for this to be negative
Either p& q both need to negative
or only q can be negative
Eg p=-1 q=-2
-1^2 * -2 + _1
-3 negative
or p=1 q=-2
1^2 * -2 +1
-1

so p can and cannot be negative

Now Choice 2

P^2/q-p/q^2 is positive

=((P^2 )* q - p)/q^2
again q^2 is always positive
numerator needs to decide
p^2 * q- p positive
so this only possible when both p and q are positive
EG p=2 q=1
4*1-2
2
q cannot be negative
Eg q=-2 p=1
1*-2 - 1
-3 but question states it should be positive
so only one option left both p & q needs to be positive

So B gives us the answer
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If p > q, is p a negative number? 10 Jun 2015, 11:45
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kanigmat011
IMO answer should be B

Option A
P^2/q + P/q^2 is negative

=((P^2*q)+p)/q^2 should be -ive
as denominator is q^2 it needs too be positive
so we are left with
P^2*q +p
for this to be negative
Either p& q both need to negative
or only q can be negative
Eg p=-1 q=-2
-1^2 * -2 + _1
-3 negative
or p=1 q=-2
1^2 * -2 +1
-1

so p can and cannot be negative

Now Choice 2

P^2/q-p/q^2 is positive

=((P^2 )* q - p)/q^2
again q^2 is always positive
numerator needs to decide
p^2 * q- p positive
so this only possible when both p and q are positive
EG p=2 q=1
4*1-2
2
q cannot be negative
Eg q=-2 p=1
1*-2 - 1
-3 but question states it should be positive
so only one option left both p & q needs to be positive

So B gives us the answer
Show more

It's p^2*q - p*q^2 > 0 not p^2/q-p/q^2.
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If p > q, is p a negative number? 10 Jun 2015, 20:58
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Bunuel
kanigmat011
IMO answer should be B

Option A
P^2/q + P/q^2 is negative

=((P^2*q)+p)/q^2 should be -ive
as denominator is q^2 it needs too be positive
so we are left with
P^2*q +p
for this to be negative
Either p& q both need to negative
or only q can be negative
Eg p=-1 q=-2
-1^2 * -2 + _1
-3 negative
or p=1 q=-2
1^2 * -2 +1
-1

so p can and cannot be negative

Now Choice 2

P^2/q-p/q^2 is positive

=((P^2 )* q - p)/q^2
again q^2 is always positive
numerator needs to decide
p^2 * q- p positive
so this only possible when both p and q are positive
EG p=2 q=1
4*1-2
2
q cannot be negative
Eg q=-2 p=1
1*-2 - 1
-3 but question states it should be positive
so only one option left both p & q needs to be positive

So B gives us the answer

It's p^2*q - p*q^2 > 0 not p^2/q-p/q^2.
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Thankx Bunuel
Even I jst realized that
Although just on lighter node was my approach correct if it would had been p^2/q-p/q^2
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If p > q, is p a negative number? 11 Jan 2022, 20:16
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