Bunuel wrote:

Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

\(Average \ speed = \frac{(Total \ distance)}{(Total \ time)}\)

(1) The distance from the return trip was 4/5 of the entire distance. Say the entire distance is d, then the distance from Y to X is 4d/5 and the distance from X to Y is d/5.

\(Average \ speed = \frac{(Total \ distance)}{(Total \ time)}=\frac{d}{\frac{(\frac{d}{5})}{40}+\frac{(\frac{4d}{5})}{50}}=\frac{1}{\frac{(\frac{1}{5})}{40}+\frac{(\frac{4}{5})}{50}}\) (by reducing d). Sufficient.

(2) The distance from the return trip was 160 miles. Not sufficient.

Answer: A.

Hope it's clear.

Hello, I was wondering about this part - "the distance from Y to X is 4d/5 and the distance from X to Y is d/5."

I understood why the distance from Y to X is 4d/5, but how did we conclude that the distance from X to Y is d/5?

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