Is p^2q > pq^2? (1) pq < 0 (2) p < 0

Rephrase the question as:

p * p * q > p * q * q

This simplifies to

is p > q ?

I) only tells us that p & q have opposite signage
II) specifically tells us p < 0 , but tells us nothing about q

using I & II:

we know that p < 0 and p&q have opp signage, so q must be > 0 and so p<q
This is sufficient, so the right answer is C

From st. 1&2 we know that pq < 0 & p < 0 therefore q>0 and p<q
but what if : p= -1/2 & q is 1/4 then we have p^ 2q =(-1/2) ^1/2 (square root of a negative number) & pq^2 = -1/2 *1/16= -1/32 , how do we decide?

If \(p=-\frac{1}{2}\) and \(q=\frac{1}{4}\), then \(p^2q=\frac{1}{16} >-\frac{1}{32}=pq^2\)

Is p^2q > pq^2?

The question asks whether \(pq(p-q) >0\).

(1) pq < 0. The question becomes whether \(p-q<0\) or whether \(p<q\). We don't know that. Not sufficient.

(2) p < 0. Clearly insufficient.

(1)+(2) Since from (2) \(p < 0\), then from (1) we'd have that \(q>0\), thus \(p=negative<q=positive\). Sufficient.

Answer: C.

Is p^2q > pq^2?

(1) pq < 0
(2) p < 0

Here is my method for this one:
1). p x q < 0. This means, we have two cases. Either p<0 & q>0 or q<0 & p>0. Going ahead with the first case. Is (-ve)^2 x (+ve) > (-ve) x (+ve)^2 ? --> +ve x +ve > -Ve. So YES. Going ahead with second case. Is (+ve)^2 x (-ve) > (+ve) x (-ve)^2 ? --> -ve > +Ve. This cannot be true. We get two answers. Insufficient.

2). p < 0. This doesnt tell us anything about q. Insufficient.

Together => We identified two cases from statement 1. Considering statement 2, We can only use the first case from statement 1. So, YES. Sufficient.

Alternate Soln

hi karishma,

i wanted a clarification here

p^2 .q > p. q ^2
now p^2 > 0 always and so is q^2 > 0
hence, the question boils down to

q > p ??
is this approach correct ??

Is \(p^2 .q > p. q ^2\)?
does not reduce to 'Is \(q >p\)?'

Note that we could have ignored \(p^2\) and \(q^2\) had we known that they are both equal in magnitude and non zero.

Say, if you divide the inequality by \(q^2\) (assuming q is not 0), you get

Is \(\frac{p^2}{q^2} * q > p\)?

Is this the same as 'Is \(q>p\)?'

No, to make this same, we are assuming that \(\frac{p^2}{q^2} = 1\).

Hi Bunuel - I have a fundamental problem when reading these kind of questions, I never get what the questions is i.e. p^2q means (p^2)q or p^(2q)? Since parenthesis are rarely used, is there a convention in the Gmat exam? Thanks

Good news is that on the exam you'll see formulas so you won't be confused.

To answer your question p^2q ALWAYS means p^2*q if it were p^(2q) it would be written that way.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

\(p^2q > pq^2\)
\(⇔ p^2q - pq^2 > 0\)
\(⇔ pq(p-q) > 0\)

Since we have 2 variables (p and a) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.

Condition 1) & 2):
Since \(pq < 0\) and \(p < 0\), we have \(q > 0\).
\(p - q < 0\) since \(p < 0 < q\) or \(p < q\).
Thus pq(p-q) > 0 since p < 0, q > 0 and p - q < 0.
Both conditions together 1) & 2) are sufficient.

Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
We don't know that \(p - q\) is positive or negative from the condition 1) only.
The condition 1) is not sufficient.

Condition 2)
We don't have any information about \(q\) from the condition 2)
The condition 2) is not sufficient.

Therefore, the answer is C.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Couldn't you further reduce this? You stopped before reaching where arnabs got to.

\(p^2 *q > p* q ^2\)

\(=\frac{q}{q^2} * p^2 > p\)

\(=\frac{q}{q^2} > \frac{p}{p^2}\)

\(=\frac{1}{q} > \frac{1}{p}\)

\(=q > p\)

Since P^2 and Q^2 are squares, you know they are positive. You can raise each side to the -1 to get the variables out of the denominator.

Could you let us know if you disagree with this approach?

Thanks

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

\(p^2q > pq^2\)
\(⇔ p^2q - pq^2 > 0\)
\(⇔ pq(p-q) >0\)

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since \(pq < 0\) and \(p < 0\), we have \(q > 0\) and \(p-q < 0\).
Then \(pq(p-q) > 0\) and we have a unique answer "yes".
Thus both conditions together are sufficient.

Obviously, each of conditions is not sufficient alone.

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

We cannot simplify this statement as we cannot tell if p or q are positive, we shall proceed with the original question.

(1) This tells us p and q have opposite signs. If q were positive then q > p, and we would have \(p^2q > q^2p\) due to positive > negative. The reverse is true if p were positive instead. Therefore insufficient.
(2) Let p = -1 for convenience, then with q = 1 the answer to the question is “yes” but with q = -1 the answer would be “no”. Insufficient.
(3) Combining the information we are stuck with the q > p case. Then the inequality is always positive > negative, sufficient.

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