<p>Can't figure out how do this one.</p>
<p>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?</p>
<p>Thanks.</p>
<p>i will show you 2 methods that can be used to solve this problem</p>
<p>METHOD 1
after a certain point, carrier A costs 0.07 a minute
carrier B costs 0.06 a minute (always)</p>
<p>carrier A costs 1.00 for the first 20 minutes combined. this (realistically) means that you save a bit of money – it only costs 0.05 a minute
carrier A: 0.05 a minute ----> 0.07 a minute
carrier B: 0.06 a minute</p>
<p>as you can see, in relation to carrier B, carrier A goes from 0.01 cheaper to 0.01 more expensive.</p>
<p>Since you get the 0.01 cheaper rate for 20 minutes, you would need to get the 0.01 more expensive rate for 20 MORE minutes if you want to balance it out at 0.06 (0.00 cheaper, or exactly the same rate)</p>
<p>20+20=40</p>
<p>here’s proof:
carrier A = $1.00 (20 minutes) + 0.07<em>20 (20 minutes) = 1 + 1.4 = 2.4
carrier B = 0.06</em>40 (40 minutes) = 2.4</p>
<p>answer is 40 minutes</p>
<p>METHOD 2
carrier A: 1.00+0.07(t-20)
t = 21 minutes = 20 minutes (1.00) + 1 minute (0.07<em>1)
t = 22 minutes = 20 minutes (1.00) + 2 minutes (0.07</em>2)
etc.
carrier B: 0.06t</p>
<p>1+0.07(t-20)=0.06t
1+0.07t-1.4=0.06t
0.01t=0.4
t=40</p>