SOLUTION: I am stuck on this problem that has to do with an application problem dealing with inequalities. Consumeration: During a weekday, to call a city 40 miles away from a certain pa

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Question 339849: I am stuck on this problem that has to do with an application problem dealing with inequalities.
Consumeration: During a weekday, to call a city 40 miles away from a certain pay phone costs $.70 for the first 3 min. and $.15 for aeach additional minute. If you use a calling card, there is a $.35 fee, and then the rates are $.196 for the first min. and $.126 for each addition minute. How many minutes must a call last if it is to be cheaper to pay with coins rather than a calling card?
I came out with 7 min or less. I had to do it by making columns and then adding for each minute.
I then set up a problem:
70
___ (3) + .015x > .035 = .196 x 1 + .126
3
I'm sure these are not right. I did come up with the right answer when I added in columns for each minute.
Thanks

3
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Let x = no. of minutes
:
Write an expression for each statement:
:
" During a weekday, to call a city 40 miles away from a certain pay phone costs
$.70 for the first 3 min. and $.15 for each additional minute."
.15(x-3) + .70
:
"If you use a calling card, there is a $.35 fee, and then the rates are $.196
for the first min. and $.126 for each addition minute."
.126(x-1) + .196 + .35
.126(x-1) + .546
:
How many minutes must a call last if it is to be cheaper to pay with coins
rather than a calling card?
:
Coin cost less than Tel card
.15(x-3) + .70 < .126(x-1) + .546
.15x - .45 + .70 < .126x - .126 + .546
.15x + .25 < .126x + .42
.15x - .126x < .42 - .25
.024x < .17
x <
x < 7.0833
x = 7
we can say a 7 min call would be cheaper with coins
:
:
Prove that
.15(4) + .70 = 1.30 with coins
.126(6) + .546 = 1.302, slightly more
:
but for 8 min
.15(5) + .70 = 1.45
.126(7) + .546 = 1.43, the card is cheaper