Is p^2q pq^2? (1) pq

Question

If  pq ≠ 0 , is  p^2q > pq^2 ?

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

got confused in C & E

and guessed C

Please explain how to eliminate this doubt, as I always get wrong in these two options.

Zarrolou · Accepted Answer

Is p^2q>pq^2? Rewrite as p^2q-pq^2>0 or pq(p-q)>0 ? (1) pq<0 The first term is negative, no info about the sign of (p-q) . Not sufficient (2) p<0 Clearly not sufficient 1+2)Now we know that pq<0 and that p<0 , so q>0 . From those we get q>p so p-q<0 . The first term pq is negative, the second is negative => their product is positive. Sufficient

KarishmaB · Answer

There is a problem in this approach. 'Is p^2q > pq^2 ?' does not get simplified to 'Is p > q ?' You cannot reduce the expression by p and q without knowing the signs of p and q. If a variable is negative, the inequality sign flips when you divide both sides by it. Note Zarrolou's method above: Is pq( p - q) > 0? This is how you should proceed. The final answer was not compromised here because it is a DS question. Since you know that pq < 0, actually the question is now 'Is p < q?' so whether you answer with yes or no, it doesn't matter in DS as long as you have a clear cut answer. But in some other question, this distinction could be the difference between the correct and incorrect answer.

labheshr · Answer

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

dentobizz · Answer

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?

Bunuel · Answer

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 p0 , thus p=negative

taransambi · Answer

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.

samsonfred76 · Answer

Alternate Soln

arnabs · Answer

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 ??

KarishmaB · Answer

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 .

bazu · Answer

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

Bunuel · Answer

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.

MathRevolution · Answer

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.

CupOfWater · Answer

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

MathRevolution · Answer

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.

TestPrepUnlimited · Answer

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

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