Is t^2 t/3?

Question

Is t^2 > t/3?

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

Bunuel · Answer

Welcome to Gmat Club. Is t^2>\frac{t}{3} ? --> Is 3t^2>t ? --> Is t(3t-1)>0 ? The given inequality holds true in 2 cases: A. t>0 and 3t-1>0 , t>\frac{1}{3} --> which gives us the range t>\frac{1}{3} ; B. t<0 and 3t-1<0 , t<\frac{1}{3} --> which gives us the range t<0 . So if t is in either in ranges above t^2>\frac{t}{3} holds true. (1) t>0 . Not sufficient. (2) |t|>\frac{1}{3} --> t>\frac{1}{3} and t<-\frac{1}{3} . Any t from these ranges falls in the ranges defined for t^2>\frac{t}{3} to hold true. Sufficient. Answer: B. Hope it helps.

Igor010 · Answer

IMO the answer is D. :wink:

Bunuel · Answer

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\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.

Igor010 · Answer

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

sidhu4u · Answer

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.

tarun · Answer

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

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

bangalorian2000 · Answer

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.

TheMechanic · Answer

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.

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Bunuel · Answer

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?

MathRevolution · Answer

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, 1/30 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

TheMechanic · Answer

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

minustark · Answer

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.

exc4libur · Answer

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)

Is t^2 > t/3?

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