gmatt1476 wrote:
A certain truck traveling at 55 miles per hour gets 4.5 miles per gallon of diesel fuel consumed. Traveling at 60 miles per hour, the truck gets only 3.5 miles per gallon. On a 500-mile trip, if the truck used a total of 120 gallons of diesel fuel and traveled part of the trip at 55 miles per hour and the rest at 60 miles per hour, how many miles did it travel at 55 miles per hour?
A. 140
B. 200
C. 250
D. 300
E. 360
This is a MIXTURE PROBLEM.
4.5 miles per gallon at a lower speed is combined with
3.5 miles per gallon at a higher speed to yield a MIXTURE with
500 miles per 120 gallons.
The following approach is called ALLIGATION -- a great way to handle mixture problems.
Let L = the lower speed and H = the higher speed.
Step 1: Convert to blue values to fractions over a common denominator
L = 4.5 miles per gallon \(= \frac{9}{2} = \frac{27}{6}\)
H = 3.5 miles per gallon \(= \frac{7}{2} = \frac{21}{6}\)
Whole trip = 500 miles per 120 gallons \(= \frac{500}{120} = \frac{25}{6}\)
Step 2: Plot the 3 numerators on a number line, with the numerators for L and H on the ends and the numerator for the whole trip in the middle
L 27---------25---------21 H
Step 3: Calculate the distances between the numerators.
L 27----
2----25----
4----21 H
Step 4: Determine the ratio in the mixture.
The ratio of L to H is equal to the RECIPROCAL of the distances in red.
L:H = 4:2 = 2:1
Implication of the resulting ratio:
Of every 3 gallons, 2 gallons are used at the lower speed, while 1 gallon is used at the higher speed.
Thus, 2/3 of the 120 gallons are used at the lower speed:
\(\frac{2}{3}*120 = 80\) gallons
Distance traveled at the lower speed = (4.5 miles per gallon)(80 gallons) = 360 miles
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