Is q > t ? (1) qp^2 < tp^2 (2) qp^3 > tp^3

I think the assumption that p^2 is positive is wrong..

p^2 can also be negative... The question stem doesn't say anything about p.
How can you take it for granted that p^2 is positive?

Dear Bunuel,
Could you please clarify~

p^2 can also be negative...

navneetb --> p can be negative only if, p is a complex number, and even though inquestion it has not been specified whether p is real or imaginary. However, I think in gmat they always mean real number.

Thanks for the reply..

I agree that GMAT doesn't deal with imaginary numbers. I presume it is true when it comes to real calculations...
Somehow in case inequalities, according to my experience, I believe GMAT does allow squared numbers to be taken as negative.

I am not sure 100% though.

It will be great if some expert chips in here to confirm....

This has to be 'A' .

Is q> t ?

Statement 1 : (1) qp^2 < tp^2

=> qp^2 - tp^2 <0
p^2 ( q -t ) < 0

Now , p^2 can never be negative ( GMAT does not deal with complex numbers ) , therefore, to satisfy the above equation ,

(q-t) has to be negative , i.e (q-t) < 0 ,

Therefore , q<t

a confirmed NO . Therefore sufficient statement .

Statement 2 : qp^3 > tp^3

qp^3 -tp^3 > 0
p^3 (q - t ) > 0

Now the above equation will hold when , p^3 and (q-t) are of same signs .

So if p is positive , then p^3 is positive , therefore (q-t) is positive , therefore q>t . ----> Yes ( q>t)

But if p is negative , then p^3 is negative , therefore (q-t) is negative , therefore q<t . -----> NO ( q <t )

Conflicting answers ( both yes and no ) , so the statement is not sufficient .

Therefore the answer is "A".

On the GMAT the square of a number cannot be negative. It can be 0 or positive.

Is q > t ?

(1) qp^2 < tp^2
qp^2 < tp^2 => P is not 0 and q < t => Sufficient
(2) qp^3 > tp^3
qp^3 > tp^3 => p is not 0 and we have 2 cases

- Case1: p>0 => q>t
- Case2: p<0 => q<t
=> Insufficient

=> Chose A

Since this is an inequality, terms common(which could otherwise be cancelled, if non-zero) on both sides of the inequality cannot be 0 and hence can be cancelled.
1) qp^2 < tp^2 will always be q < p, because p^2 can never be negative, and in this case never be 0 as well (reason mentioned above). This answers the question. Sufficient!
2) qp^3 > tp^3. In this case, although p can never be 0, p^3 can definitely be both positive and negative. Thus the sign will flip. Insufficient!

Answer A

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3


-> In inequality, it is important to ignore square. That is, when it comes to multiplication in the other equation with 0, you need to ignore square. Thus, the question above is q > t ?.

In (1) qp^2-tp^2<0, (q-t)p^2<0, as q-t<0(p^2 is always positive number, even if you divide the both equations, the direction of the inequality sign is not changed. As q<t, it is (2) qp^3 > tp^3, (q-t)p^3>0, (q-t)p>0. So, the answer is A. In 1), it becomes no and is sufficient. Therefore, the answer is A.



-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

Bunuel
what if statement 1 : p=1/2, q=-2 & t=-1 i.e qp^2<tp^2 => (-2)(1/4) <(-1)(1/4)
but if p=1, q=2 & t=3 i.e qp^2<tp^2 => (2)(1) <(3)(1)

In both of your examples q is less than t. This is exactly what we get from (1): qp^2 < tp^2 is equivalent to q < t (after reducing by p^2), so we have a definite NO answer to the question.

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