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mathematics

A certain car model cost Birr 20,000 with a gasoline engine and Br 25,000 with a diesel engine. The number of miles per gallon of fuel for cars with these two engines is 25 and 30, respectively. Assume that the price of both types of fuel is Birr 1.50 per gallon.
Required:
a) Drive the equation for the cost of driving a gasoline powered car.
b) Drive the equation for the cost of driving a diesel powered car.
c) Find the break-even point, that is, find the mileage at which the diesel-powered car becomes more economical than the gasoline powered car.

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  1. "derive" not "drive"
    cost in $/mi is just mi/(mi/gal) * $/gal
    so, the cost of driving x miles is
    (a) 20000 + x/25 * 1.50 = 0.06x + 20000
    do the same for (b) and then solve for (c)

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  2. number of miles driven = x
    cost to drive the gas car = 20000 + x/20 * 1.5 = 3x/40 + 20000
    cost to drive the diesel - 25000 + x/30*1.5 = x/20 + 25000

    3x/40 + 20000 = x/20 + 25000
    3x/40 - x/20 = 5000
    x/40 = 5000
    x = 20,000

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  3. oops, copy error !!
    I used 20 instead of 25 in line #2

    Make the necessary changes, or else solve oobleck's equation

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  4. good

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