Greg and Brian are both at Point A (above). Starting at the

Greg's speed: \(s_g\)
Distance covered by Greg: \(D\)
Time taken by Greg: \(t_g=\frac{D}{s_g}\)

Brian's speed \(s_b\)
Distance covered by Greg \(D\sqrt{2}\) (Hypotenuse of right isosceles triangle = root(2) times base)
Time taken by Brian \(t_b=\frac{D\sqrt{2}}{s_b}\)

1.
\(s_g=\frac{2}{3}s_b\)
\(t_g=\frac{D}{s_g}=\frac{D}{\frac{2}{3}s_b}=\frac{3D}{2s_b}\)-------------1
\(t_b=\frac{\sqrt{2}D}{s_b}\)----------------2

Comparing 1 and 2:
\(\frac{3}{2} > \sqrt{2}\). Thus, Greg took relatively more time to reach his destination.
Sufficient.

2.
\(s_b=s_g+20\)
\(t_b=\frac{\sqrt{2}D}{s_b}=\frac{\sqrt{2}D}{s_g+20}\)
\(t_g=\frac{D}{s_g}\)

Let's prove opposite of st1:
\(t_b>t_g\)
\(\frac{\sqrt{2}D}{s_g+20}>\frac{D}{s_g}\)
\(\frac{\sqrt{2}}{s_g+20}>\frac{1}{s_g}\)
\(s_g\sqrt{2}>s_g+20\)
\(0.4s_g>20\)
\(s_g>50\)

Thus, if greg's speed is approx more than 50, brian would take more time, otherwise greg would take more time.

Not Sufficient.

Ans: "A"