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18 Mar 2026, 05:32
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Difficulty:
95%
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Question Stats:
45% (03:01) correct
55%
(03:14)
wrong
based on 42
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Olek drove yesterday at 40 miles per hour in the city and 60 miles per hour on the highway. His car averaged 30 miles per gallon in the city and 50 miles per gallon on the highway. If Olek spent a fraction f of his total driving time on the highway, what was his car’s overall fuel efficiency, in miles per gallon?
A. 150(2 + f) / (10 - f)
B. 30 + 20f
C. (40 + 20f) / (1 - f)
D. 150(2 + f) / (10 + f)
E. 30(2 + 3f) / (2 + f)
A. 150(2 + f) / (10 - f)
B. 30 + 20f
C. (40 + 20f) / (1 - f)
D. 150(2 + f) / (10 + f)
E. 30(2 + 3f) / (2 + f)
18 Mar 2026, 07:44
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Here's how i did it, we need to calculate the total miles covered and the total gallons consumed, which gives us the overall mpg
So the miles are \(60f\) on the highway and \(40(1-f)\) in the city
The gallons are \(\frac{60f}{50}\) on the highway and \(\frac{40(1-f)}{30}\) in the city
The total distance is therefore \(40 + 20f\) or \(20(f + 2)\) miles
The total fuel used is \(\frac{6f}{5} + \frac{4(1-f)}{3} = \frac{18f + 20(1-f)}{15} = \frac{20 - 2f}{15}\)
Now, when we divide the divide the two, we get \(\frac{20(f+2)*15}{2(10-f)}\), or \(\frac{150(2+f)}{(10-f)}\), which is is in our answer choices.
So the miles are \(60f\) on the highway and \(40(1-f)\) in the city
The gallons are \(\frac{60f}{50}\) on the highway and \(\frac{40(1-f)}{30}\) in the city
The total distance is therefore \(40 + 20f\) or \(20(f + 2)\) miles
The total fuel used is \(\frac{6f}{5} + \frac{4(1-f)}{3} = \frac{18f + 20(1-f)}{15} = \frac{20 - 2f}{15}\)
Now, when we divide the divide the two, we get \(\frac{20(f+2)*15}{2(10-f)}\), or \(\frac{150(2+f)}{(10-f)}\), which is is in our answer choices.
20 Mar 2026, 03:04
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The miles are 60*f on the highway and 40*(1−f) in the city.
The gallons are 60*f/50 on the highway and 40*(1−f)/30 in the city
The total distance is therefore 40+20f miles
Thus the total fuel used is 6f/5+4(1−f)/3=[18f+20(1−f)]/15 = (20-2f)/15
Now, when we divide the divide the two, we get 150(2+f)/(10-f)
Therefore, A
The gallons are 60*f/50 on the highway and 40*(1−f)/30 in the city
The total distance is therefore 40+20f miles
Thus the total fuel used is 6f/5+4(1−f)/3=[18f+20(1−f)]/15 = (20-2f)/15
Now, when we divide the divide the two, we get 150(2+f)/(10-f)
Therefore, A
