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10 Nov 2023, 10:44
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B
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Difficulty:
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Question Stats:
72% (01:53) correct
28%
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A certain car averaged 20 miles per gallon of gasoline on a trip from City N to City P and averaged 25 miles per gallon on the return trip, which followed a different route. How many miles per gallon did the car average for the round trip?
(1) The round trip was 175 miles.
(2) The trip from City P was 3/4 as many miles as the trip to City P.
(1) The round trip was 175 miles.
(2) The trip from City P was 3/4 as many miles as the trip to City P.
10 Nov 2023, 11:35
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averagedude23
Trip From City N to City P
- Distance = \(d_1\)
- Mileage = \(20\) miles / gallon
- Gasoline Used = \(\frac{d_1}{20}\) gallons
Trip From City P to City N
- Distance = \(d_2\)
- Mileage = \(25\) miles / gallon
- Gasoline Used = \(\frac{d_2}{25}\) gallons
Question - How many miles per gallon did the car average for the round trip?
Average miles per gallon = \(\frac{\text{Total Miles Travelled}}{\text{Total Gasoline Consumed}} = \frac{d_1+d_2}{\frac{d_1}{20}+\frac{d_2}{25}} = \frac{500(d_1+d_2)}{25d_1 + 20d_2}\)
Statement 1 =
(1) The round trip was 175 miles.
\(d_1 + d_2 = 175\)
We don't know the value of \(25d_1 + 20d_2\). Hence, the statement alone is not sufficient. We can eliminate A and D.
Statement 2
(2) The trip from City P was 3/4 as many miles as the trip to City P.
\(d_2 = \frac{3}{4}* d_1\)
Average miles per gallon = \(\frac{\text{Total Miles Travelled}}{\text{Total Gasoline Consumed}} = \frac{500(d_1+d_2)}{25d_1 + 20d_2}\)
As we have a multiplicative relationship between \(d_1\) and \(d_2\), we can express one term in terms of another. The common term will cancel out, leaving us with a definite answer.
Average miles per gallon = \(\frac{\text{Total Miles Travelled}}{\text{Total Gasoline Consumed}} = \frac{500(d_1+d_2)}{25d_1 + 20d_2}=\frac{500(d_1+0.75d_1)}{25d_1 + (20*0.75d_1)} \)
Taking \(d_1\) common from the numerator and denominator we get
Average miles per gallon =\(\frac{d_1* 500(1+0.75)}{d_1*((25 + (20*0.75))} \)
Average miles per gallon =\(\frac{500(1+0.75)}{25 + (20*0.75)} = 21.875\)
Taking \(d_1\) common from the numerator and denominator we get
Average miles per gallon =\(\frac{d_1* 500(1+0.75)}{d_1*((25 + (20*0.75))} \)
Average miles per gallon =\(\frac{500(1+0.75)}{25 + (20*0.75)} = 21.875\)
Hence, statement 2 is sufficient to find the solution.
Option B
General Discussion
19 Mar 2024, 19:52
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gmatophobia
Hey gmatophobia was wondering how we're able to take 20mpg and flip it to 20 gallons/mile like you did in your answer.
