I rented a car for $105.00 for 3 days and 300 miles. Tom was charged $195.00 for 5 days and 600 miles. What was charged per day and per mile?

Question

My fee was $105.00 for 3 days and 300 miles driven.  Tom's fee was $195.00 for 5 days and 600 miles driven. What were the charges per day and per mile?

Answer 1

Basically, we have two variables, the charge per day and the charge per miles. If we just had your fee or Tom's fee, we wouldn't be able to work it out. We need to use both formulas to help us find the answer.

My fee: 3d + 300m = 105
Tom's fee: 5d + 600m = 195

Next we are going to isolate one of the variables by rearranging the equation.

My fee: 3d + 300m = 105
Isolate d: 3d = 105 + -300m
Simplify: d = 35 + -100m

Now, we can substitute this new equation into Tom's original fee equation, because we know what d equals.

Tom's fee: 5d + 600m = 195
Substitute: 5(35 + -100m) + 600m = 195
Distribute 5: 175 + -500m + 600m = 195
Subtract: 175 + 100m = 195
Isolate m: 100m = 20
Simplify: m = .2

Great! Now we have the value of m, the charge $0.20 per mile. Now that we've narrowed it down to one unknown variable, we can plug our value of m into Tom's fee equation and figure out the daily charge.

Tom's fee: 5d + 600m = 195
Insert value of m: 5d + 600(.2) = 195
Multiply: 5d + 120 = 195
Isolate d: 5d = 75
Simplify: d = 15

Now we have our daily fee charge of $15/day. Now, while we know both of our variables, we should double check. We can insert those values into the first fee equation, and if it works, we know we've got it right.

My fee: 3d + 300m = 105
Insert values: 3(15) + 300(.2) = 105
Multiply: 45 + 60 = 105

We got it! The charge per day is $15, and the charge per mile is $0.20.

Answer 2

Hey Judy -- "eyeballing" differences can help ... Tom pays $90 more for 2 added days and 300 more miles ... you pay $105 for 3 days, 300 miles ... meaning you pay $15 for another day => day rate $15, 20 cents/mile