SOLUTION: A long-distance phone company charges $0.50 for the first minute and $0.15 for each additional minute. If Mark budgets $4 to call his sister once a month, how long can they talk?

Question 29517: A long-distance phone company charges $0.50 for the first minute and $0.15 for each additional minute. If Mark budgets $4 to call his sister once a month, how long can they talk? How much will he be charged?
Answer by Evy(14) (Show Source): You can put this solution on YOUR website!
Hello, I can help you!
I will solve this problem by working backwards meaning i will start with the total of $4 and work down.

First you should:
subtract .50 from $4 for the 1st minute of the call.

Next step: divide 3.50 by .15

round 23.33 minutes to 23 minutes

Now we have 23 minutes at .15(15 cents) and 1 minute at .50(50 cents):

Mark and his sister can talk for 24 minutes.

to find out how much he will get charged simply plug in the numbers:

Multiply: Now you have

So Mark will be Charged $3.95.

If you have any questions e-mail me at Southqueensb@yahoo.com
~EVY~

RELATED QUESTIONS
Customers of a phone company can choose between two service plans for long distance... (answered by ikleyn)

Customers of a phone company can choose between two service plans for long distance... (answered by felizgu)

A long distance phone call is 50 cents for the first minute and 25 cents for each... (answered by JBarnum)

Customers of a phone company can choose between two service plans for long distance... (answered by jim_thompson5910)

A phone company charges $1.00 for the first three minutes of a long-distance call and... (answered by checkley71)

I wondering how to change story problems into a linear equations such as: A phone... (answered by solver91311)

Customers of a phone company can choose between two service plans for long distance... (answered by rwm)

Customers of a phone company can choose between two service plans for long distance... (answered by drcole)

Customers of a phone company can choose between two service plans for long distance... (answered by fractalier)