Is t^2 > t/3?

\(t^2>\frac{t}{3}\) holds true when \(t\) is in the range \(t>\frac{1}{3}\) or \(t<0\).

(1) is not sufficient because \(t>0\) means \(t\) can be in the range \(0<t\leq{\frac{1}{3}}\) too and for \(t\) from this range \(t^2>\frac{t}{3}\) doesn't hold true. For example \(t=\frac{1}{9}\) --> \(t^2=\frac{1}{81}<\frac{t}{3}=\frac{1}{27}\).

Hope it's clear.

Now I see!!! You're right! Thanks for your expl.!

t^2 > t/3 => 3t^2 - t>0=> t (3t -1) >0 => t>1/3 and t <0 we have to prove
stmt1: t>0 cannot say anything from this so insuff.
stmt2: |t|>1/3 => t>1/3 or t<-1/3
definitely if t<-1/3 => t<0 so we are proving both the stmts
hence B is the answer.

or you can plug in numbers, fractions and integers and find the result.

It's so nice to read the way Bunuel explains.

Its B.
Solving question stmt
t^2-t/3>0
t(t-1/3)>0

stmt 1: t>0 but t<1/3 and in tht case equation above is false hence insufficient.

stmt 2: |t|>1/3 then t >1/3 or t<-1/3 hence equation in both cases >0 so sufficient.

Hi Bunuel,

While I understand your approach I tried one of mine too and I couldn't understand why It's wrong.

Here, why can't we simplify the actual question to:

\(t^2\)>\frac{t}{3},
or,\(t/3<0\) since,\(t^2\) is always > 0.
or t<0?

So the answer to the above question becomes A. But that's not the OA answer.

What is wrong with this approach? If this approach is wrong, I do find its relevant uses in the other questions. But it seems I can't connect the links where this approach is fine and where it is not. A little background will help solidify the approach. Thanks.

The highlighted part is not clear. Yes, t/3 is less than some non-negative quantity (t^2) but why does this mean that t/3 must be negative? Why cannot it be positive but less than t^2?

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 t^2 > t/3?

(1) t > 0
(2) |t| > 1/3

in case of inequality questions we're not going to simply substitute all the values, but we'll be comparing the range of que and the range of con and see if the range of que includes that of con. If it does, then the condition is sufficient.

Transforming the original condition, we have t^2-t/3>0?, t(t-1/3)>0? gives us t<0, 1/3<t?. Since there is 1 variable, we need 1 equation and there is 1 each in 1) and 2). therefore D has high probability of being the answer.

In case of 1), t>0 does not make the range of que include the range of con. Therefore the condition is not sufficient.
In case of 2), t<-1/3, 1/3<t and thus the range of que includes the range of con. Therefore the condition is sufficient and the answer is B.

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.

This is very interesting!!! Needlessto say unheard and unseen approach :D

I started solving this by taking t > 1/3 (dividing both side of the inequality by t), which leads me to answer C. Thanks Bunuel for your awesome explanation.
As t^2 is given, we have to keep in mind both the +ve and -ve t for such kinds of question. Really helpful question.

t^2>t/3, 3t^2-t>0, t(3t-1)>0
t≥0: 3t-1>0, 3t>1, t>1/3;
t<0: 3t-1<0, 3t<1, t<1/3;
Is t>1/3 or t<0?

(1) t > 0 insufic

(2) |t| > 1/3 sufic
t>1/3 or t<-1/3 (t<0)

Ans (B)