"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)
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.
5 answers
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
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
oops, copy error !!
I used 20 instead of 25 in line #2
Make the necessary changes, or else solve oobleck's equation
I used 20 instead of 25 in line #2
Make the necessary changes, or else solve oobleck's equation
good
A family has two cars. The first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of 30 miles per gallon of gas. During one particular week, the two cars went a combined total of 1300 miles, for a total gas consumption of 50 gallons. How many gallons were consumed by each of the two cars that week?