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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 miles per hour 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 had traveled at an average speed of x miles per hour for the entire trip?
(1) x = 48.
(2) y = 20.
We are given that 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 at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. We are asked to determine what percentage the travel time for the 40-mile trip was of the time it would have taken her, had she traveled the entire trip at an average speed of x miles per hour.
Since time = distance/rate, the time it would have taken her to travel the entire trip at an average speed of x miles per hour is 40/x, and the actual time it took her to travel the entire trip at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip is y/x + (40 – y)/(1.25x). So the question becomes:
y/x + (40 – y)/(1.25x) is what percent of 40/x ?
[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?
[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?
[y/x + (40 – y)/(1.25x)] * (x/40) * 100 = ?
[y/40 + (40 – y)/50] * 100 = ?
2.5y + 2(40 – y) = ?
Thus, knowing the value of y will allow us to determine the above percentage.
Statement One Alone:x = 48
Since we do not have any information about y, statement one alone is not sufficient. We can eliminate answer choices A and D.
Statement Two Alone:y = 20
Since we have a value for y, statement two alone is sufficient. For practice, we can calculate the percentage as follows:
2.5(20) + 2(40 – 20) = 50 + 40 = 90
Answer: B