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- kuvyojhmoob

A salesperson purchased an automobile that was advertised as averaging 27 mi/gal in the city and 38 mi/gal on the highway. A recent sales trip that covered 1978 miles required 59 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city?

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- kuvyojhmoob

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- anonymous

Assume the number of miles driven in the city is \(x\). Then the amount of gas used in the city is \(\frac{x}{27}\). The number of miles driven on the highway is therefore \(1978-x\) and the amount of gas used on the highway is \(\frac{1978-x}{38}\). The total amount of gas used is then\[\frac{ x }{ 27 } + \frac{ 1978-x }{ 38 } = 59\]Solve for \(x\).

- kuvyojhmoob

×=648

- mathmate

Yes, x=648 is correct
x/27+1978/38-x/38=59
x/27-x/38=59-1978/38
multiply by 27*38=1026 to eliminate the denominators
38x-27x =60354-1978=7128
x=7128/12=648

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