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Is q > t ?

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Is q > t ?  [#permalink]

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New post 11 Nov 2014, 05:41
4
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A
B
C
D
E

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  15% (low)

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73% (01:06) correct 27% (01:17) wrong based on 188 sessions

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Re: Is q > t ?  [#permalink]

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New post 11 Nov 2014, 07:10
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IMO A.

ST 1 p^2 (q-t) < 0, p^2 must be +ve so q-t<0, so q<t....Sufficent

ST 2 p^3(q-t) < 0, but if p^3 is negative q-t is +ve, and vice versa. Insufficient.
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Re: Is q > t ?  [#permalink]

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New post 12 Nov 2014, 05:00
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1
As per Statement 1.
qp^2 < tp^2
Get terms on one side
qp^2 - tp^2 < 0
p^2 ( q-t) < 0
If multiplication of two terms is zero then both terms should have opposite signs.
p^2 will always be +ve .
So q-t < 0
q<t Sufficient

As per Statement 2 qp^3 > tp^3
P^3 (q-t) > 0

If mulitplication of two terms are positive then both terms either will be +ve or -ve
P^3 can be +ve or -ve. So q>t or q<t.
Statement 2 is insufficient

Answer: A

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Is q > t ?  [#permalink]

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New post Updated on: 08 Apr 2015, 06:31
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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".

Originally posted by Ruchika2302 on 08 Apr 2015, 04:07.
Last edited by Ruchika2302 on 08 Apr 2015, 06:31, edited 1 time in total.
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New post 08 Apr 2015, 04:09
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Is q > t ?  [#permalink]

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New post 07 Apr 2015, 18:24
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~
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Re: Is q > t ?  [#permalink]

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New post 07 Apr 2015, 19:48
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.
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Re: Is q > t ?  [#permalink]

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New post 08 Apr 2015, 03:50
tusharacc wrote:
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....
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Re: Is q > t ?  [#permalink]

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New post 08 Apr 2015, 18:48
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
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Re: Is q > t ?  [#permalink]

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New post 11 Dec 2015, 03:02
Bunuel wrote:

Tough and Tricky questions: Inequalities.



Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3

Kudos for a correct solution.


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
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Re: Is q > t ?  [#permalink]

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New post 13 Dec 2015, 04:14
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.
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Re: Is q > t ?   [#permalink] 12 Apr 2019, 22:35
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