solviong algebraic word problem

Question

You are debating between buying 1 of 2 trucks with the following specs. (buy outright)

Truck #1 - 35/miles gal retail =$22,000

Truck #2 - 30/miles gal retail $18,000

Cost of gas is $2.00 gal

 

Write 2 expressions for the cost of each truck (let x = # gallons of gas.

 

 

Also for what values of x is truck #1 more practical?

Also for what values of x is truck #2 more practical?

Answer 1

Hi Cortnee.

 

This question is asking you how many miles you'll have to drive for the more expensive but better mileage car to eventually become a better deal than the cheaper but worse mileage car.

 

Truck 1 gets 35 miles/ gallon, and it will cost 2.00 dollars/ gallon, so Truck 1 will pay:

 

(2 dollars/ gallon) / (35 miles/ gallon) = $1 / 17.5 miles

 

Truck 2 gets 30 miles/ gallon, and it will cost 2.00 dollars, gallon, so Truck 2 will pay:

 

(2 dollars/ gallon) / (30 miles/ gallon) = $1 / 15 miles

 

Let's call C1 the cost of driving Truck 1.

 

C1 = 22,000 + 1/17.5 * x

 

This is the starting cost of the car + $1 / 17.5 miles * x miles.

 

Let's call C2 the cost of driving Truck 2.

 

C2 = 18,000 + 1/15 * x

 

This is the starting cost of the car + $1 / 15 miles * x miles.

 

I know your question called x the number of gallons, but I think it's easier to evaluate by miles and then convert it back to gallons afterward based on the mileage.

 

You want to find out how many miles need to be driven for C1 and C2 to be equal.

 

So:

 

22000 + 1/17.5 * x = 18000 + 1/15*x

 

4000 = x*(1/15 - 1/17.5)

 

x = 420,000 miles

 

Truck 1 gets 35 miles/ gal.

 

420,000 miles / 35 miles/gal = 12,000 gallons.

 

Truck 2 gets 30 miles/ gal.

 

420,000 miles / 30 miles/gal = 14,000 gallons.

 

Since Truck 1 is more expensive upfront, anything after 12,000 gallons will make Truck 1 more practical since its better mileage will catch up to its start-up cost.

 

Since Truck 2 is less expensive upfront, anything before 14,000 gallons will make Truck 2 more practical since at that point its worse mileage will overwhelm the savings at retail.

 

Those answers are strictly from a math standpoint. Honestly, though, if you can drag a car out to 420,000 miles worth of engine life you'll be paying a lot more than just retail and gas. It'll probably be a cube in some scrapyard before it reaches that point. Probably better just to go with Truck 2 since Truck 1 won't last long enough to be practical. But I wouldn't put that on your homework.

 

Hope this helps.