Thread: unit problem

Car costs $14,500 and has a rated mileage of 28 miles/gallon. another car costs $21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is $1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
thanks a lot
it's driving me crazy
nertil

Originally Posted by nertil1

Car costs $14,500 and has a rated mileage of 28 miles/gallon. another car costs $21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is $1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
thanks a lot
it's driving me crazy
nertil

I'll tell you, but first you have to tell me what 19 kilos per liter are in miles per gallon.

44.7 mi/gal

Originally Posted by nertil1

44.7 mi/gal

Originally Posted by nertil1

Car costs $14,500 and has a rated mileage of 28 miles/gallon. another car costs $21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is $1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
thanks a lot
it's driving me crazy
nertil

Ok then.

What you need to do is write a function to compare the amount of miles gone ( ) to the amount of money paid ( ) for both cars.

For the first car, you would spend

or just 5 dollars for every 112 miles

So the equation for the cost of the first car is:

Now your job is to find the equation of the second car. And then see when they intersect...

Originally Posted by nertil1

where did 5/122 come from?

it's 1.25/28

that's the smallest fraction it could be without having a decimal place (I hate decimal places in fractions)