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Another MGmat Problem

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Source: — Data Sufficiency |
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by thirst4edu » Sun Sep 26, 2010 7:36 pm
HPengineer wrote:If pq ≠ 0, is p2q > pq2?

(1) pq < 0

(2) p < 0

Love to see some other ways to work this one..
IMO C

Rephrasing the question -
Given, pq <> 0 ,
p^2 * q > p * q^2 ?
p (pq) > q (pq) ?
Dividing both sides by pq (since we are given pq is not zero)

p > q? So this is the question

1) pq < 0
This means p and q both have opposite signs, but we dont know p < q or q < p, not sufficient.
2) p < 0
we dont know anything about q, Not sufficient.

1) & 2)
we know that p < 0, so according to 1) q has to be > 0, hence we know that p > q is not true. Hence Sufficient.
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by Jim@Grockit » Sun Sep 26, 2010 7:48 pm
HPengineer wrote:If pq ≠ 0, is p2q > pq2?

(1) pq < 0

(2) p < 0

Love to see some other ways to work this one..
Other ways to work this one other than . . . what? You didn't give us one.

Statement 1: This tells us that either p or q is negative, but not both. This means that (p^2)q and p(q^2) are a negative number and a positive number, but we don't know which is which. If p is positive and q negative, the answer is no; if q is positive and p negative, the answer is yes. INSUFFICIENT

Statement 2: We know p is negative, but know nothing about q. If q is positive, the answer is yes. If q is negative, we end up with a positive times a negative on each side and no values for the variables to answer the question. INSUFFICIENT

TOGETHER, we know that p is negative and that pq<0, making q positive. This makes the expression on the left of the inequality positive for all values and the one on the right negative for all values. SUFFICIENT
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by HPengineer » Sun Sep 26, 2010 9:16 pm
Ok i see where i went wrong.. Thanks for the help guys.
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by benjiboo » Tue Sep 11, 2012 1:45 pm
If pq ≠ 0, is p^2q > pq^2?

(1) pq < 0

(2) p < 0

Is it ok to rearrange the question prompt as such:

p^2q > pq^2 =

1/q > 1/p

(I divided both sides by q^2 and p^2... since we may not know the sign of p or q, but we know that either squared is positive)
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by mcdesty » Tue Jul 08, 2014 5:43 pm
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