word problem?
Posted in Homework Help on 10/21/2008 12:53 am by admin
Pam K asked:
A discount airline has fixed charge for processing tickets plus mileage fee. A ticket for a 200-mile trip costs $48 while a ticket for a 300 mile trip costs $61. find the fixed charge and charge per mile
NELSON
A discount airline has fixed charge for processing tickets plus mileage fee. A ticket for a 200-mile trip costs $48 while a ticket for a 300 mile trip costs $61. find the fixed charge and charge per mile
NELSON
10/21/2008 at 9:49 pm
ROBIN
Ok, here’s one way to do this problem.
You’re basic equation is:
Fixed Charge + (Fee x Mileage) = Cost of Ticket
Let’s shorten that to:
C + ( F x M ) = T
You know that:
1. C + ( F x 200 ) = 48
2. C + ( F x 300 ) = 61
You’ve got two unknown variables (C and F) in the equation; makes it very difficult to solve. So, try to find a way to re-state things to one variable. If you re-write the second equation, you get: C = 61 – ( F x 300 ). You can substitute this in place of C in the first equation and only have to solve for F, like this:
[ 61 - ( F x 300 ) ] + ( F x 200 ) = 48
Now you can solve as follows:
61 – F300 + F 200 = 48
61 – F100 = 48
- F100 = – 13
F = 0.13
So, the mileage fee is $0.13 per mile and the fixed processing charge is $22.00.
You could also simply subtract one equation from the other. Like subtract the second equation from the first equation. The C’s would knock each other out and you’d be left with:
- F100 = – 13
F = 0.13