College Algebra Word Problem

Question

You are choosing between two long distance telephone planes. Plan A has monthly fee of $15 with a charge of 8 cents per minute for long distance calls. Plan B has a monthly fee of $3 with a charge of 12 cents per minute for long distance calls. How many minutes of long distance calls in a month make plan A the better deal? Use algebra to solve.

Answer 1

Hi Destiney, let x represent the number of minutes used. Set up an equation and solve for x.

 

3 + .12x = 15 + .08x

 

Solve for x:

 

.04x = 12, 

 

or x = 300

 

If you use less than 300 minutes, say 299, you would be better with the $3 monthly fee plus $.12 per minute used.

 

At 300 minutes, it does not matter which plan use, the result is the same amount.

 

Once you pass 300 minutes, the plan that costs $15 per month with $.08 per minute is the better plan.

 

You can check this by changing the number of minutes above and below 300 and see what plan is better.

 

Note: you can also solve this using the following inequality:

 

3 + .12x > 15 + .08x

 

Solve for x, which shows for x > 300, the plan costing $3 per month and $.12 per minute will cost more if x is greater than 300.

 

Questions?