Bunuel wrote:
During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entiretrip?
\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);
\(t_2=\frac{40}{x}\) ;
Q:\(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\) . So we see that the value of
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