ghostdude wrote:

Greg and Brian are both at Point A (above). Starting at the same time, Greg drives to point B while Brian

drives to point C. Who arrives at his destination first?

(1) Greg's average speed is 2/3 that of Brian's.

(2) Brian's average speed is 20 miles per hour greater than Greg's.

I am not sure of the OA

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"

_________________

~fluke

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