SOLUTION: I subscribe to the Comic-of-the- Month Club. Each month I can buy any number of the 48 titles offered by the club. The first month I bought five comics for $3.07. The second month
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: I subscribe to the Comic-of-the- Month Club. Each month I can buy any number of the 48 titles offered by the club. The first month I bought five comics for $3.07. The second month
Log On
Question 278456: I subscribe to the Comic-of-the- Month Club. Each month I can buy any number of the 48 titles offered by the club. The first month I bought five comics for $3.07. The second month I bought two comics for $1.72. The next month I bought six of the club offerings for $3.52. In May I bought three more for $2.17. The club charges a fee for each comic and a handling fee for the entire order. How much would it have cost to buy all 48 titles at the same time?
I need to solve using a guess and check table. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! I subscribe to the Comic-of-the- Month Club. Each month I can buy any number of the 48 titles offered by the club. The first month I bought five comics for $3.07. The second month I bought two comics for $1.72. The next month I bought six of the club offerings for $3.52. In May I bought three more for $2.17. The club charges a fee for each comic and a handling fee for the entire order. How much would it have cost to buy all 48 titles at the same time?
:
Let c = cost of each comic including the fee
Let h = handling fee per order
:
Write equation for the 1st two transactions
5c + h = 3.07
2c + h = 1.72
----------------subtraction eliminates h, find c
3c = 1.35
c =
c = .45 for each comic
:
Find h
5(.45) + h = 3.07
2.25 + h = 3.07
h = 3.07 - 2.25
h = .82 is the handling fee
:
Confirm this by checking it in the 3rd transaction
6(.45) + .82 = 3.52
2.70 + .82 = 3.52
:
All 48 comics
48(.45) + .82 = $22.42
