Summer3 wrote:
A car averages 40 miles per hour for the first 6 hours of a trip and averages 60 miles per hour for each additional hour of travel time. If the average speed for the entire trip is 55 miles per hour, how many hours long is the trip?
A. 8
B. 12
C. 16
D. 18
E. 24
This is a MIXTURE problem.
Two speeds (40mph and 60mph) are combined to form a mixture with an average speed of 55mph.
To determine how much time must be spent at each speed, we can use ALLIGATION.
Let S = the slower speed and F = the faster speed.
Step 1: Plot the 3 speeds on a number line, with F and S on the ends and the speed for the mixture (55mph) in the middle.S 40-----------55----------60 F
Step 2: Calculate the distances between the values on the number line.S 40----
15----55----
5-----60 F
Step 3: Determine the ratio of the two given speeds.The ratio of S to F is equal to the RECIPROCAL of the distances in red.
S:F = 5:15 = 1:3
The resulting ratio implies the following:
Of every 4 hours of travel, 1 hour must be traveled at the slower speed (40mph), and 3 hours must be traveled at the faster speed (60mph).
Thus, the 6 hours traveled at 40mph must be equal to \(\frac{1}{4}\) of the total travel time:
\(6 = \frac{1}{4}t\)
\(24 = t\)
The reasoning behind the ratio of the two speeds for every 4 hours of travel time:
Since the average speed (55) is closer to the faster speed (60) than to the slower speed (40), more time must be spent at the faster speed (3 hours) than at the slower speed (1 hour).
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