You are simply squaring the denominator on the left and multiplying the denominator of t by 3 on the right. 1/3 is an inflection point where t^2 will be bigger if t is > 3. Absolute value works because the left will always be positive, the right will not.

Answer c

jim,
request to explain stmtnt 1 again with example.
sorry.

pappueshwar wrote:IS t^2 > t/3?

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

OA IS B. heavily confused

t² > t/3
3t² > t
3t²-t > 0
t(3t-1) > 0.

The lefthand side is equal to 0 if t=1 or t=1/3.
These are the CRITICAL POINTS: the values where t²=t/3.
When t is ANY OTHER VALUE, t²>t/3 or t²<t/3.
To determine the range where t²>t/3, test one value to the left and right of each critical point.

t<0:
Let t=-1.
(-1)² > 1/3.
1 > 1/3.
This works.
Thu, t<0 is part of the range where t² > t/3.

0<t<1/3:
Let t=1/4.
(1/4)² > (1/4)/3.
1/16 > 1/12.
Doesn't work.
Thus, 0<t<1/3 is not part of the range where t² > t/3.

t>1/3:
Let t=1.
(1)² > 1/3.
1 > 1/3.
This works.
Thus, t>1/3 is part of the range where t² > t/3.

Thus, the only range where it is NOT true that t² > t/3 is 0≤t≤1/3.

Question rephrased: Is 0≤t≤1/3?

Statement 1: t>0
No way to determine whether 0≤t≤1/3.
INSUFFICIENT.

Statement 2: |t|>1/3
Thus, it is not possible that 0≤t≤1/3.
SUFFICIENT.

The correct answer is B.

thanks mitch that was amazing