Question 259736: The cost of a telephone call using long-distance carrier A is $1.00 for any time up to and including 20 minutes and $0.07 per minute thereafter. The cost using long-distance carrier B is $0.06 per minute for any amount of time. For a call that lasts t minutes, the cost using carrier A is the same as the cost using carrier B. If t is a positive integer greater than 20, what is the value of t?
Found 2 solutions by richwmiller, drk:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! $1 20 minutes +.07
.06 per minute
20 minutes each
$1 for a
1.20 for b
now to find where they are the same after those 20 minutes
7x=6x+20
x=20
after 20 MORE minutes they are equal
1+20*.07=1.20+.06*20
2.40=1.20+1.20
2.40=2.40
A 40 minute call will cost the same on both carriers.
Answer by drk(1908)  (Show Source):
You can put this solution on YOUR website! PLAN A:
If t <= 20, then 1$
If t> 20, then .07*(x-20)
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PLAN B
if t > 0, then .06t
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To be the same cost we set them equal and get
.07*(t-20) + 1 = .06t
using distributive property, we get
,07t -1.4 + 1 = .06t
combining like terms, we get
.07t -.4 = .06t
subtracting .07t, we get
-.4 = -.01t
dividing we get
t = 40
So, we have a 40 minute call being the same cost.
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