SOLUTION: this is is kinda confusing, need your help guys. alan and dave leave from the same point driving in opposite directions, alan driving at 55 miles pero hour and dave at 65 miles

Question 157645: this is is kinda confusing, need your help guys.
alan and dave leave from the same point driving in opposite directions, alan driving at 55 miles pero hour and dave at 65 miles per hour. alan has a one hour head start. how long will they be able to talk on their phones if the phones have a 250 mile range?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
OK
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
So, before Dave gets started, Alan has already driven 55*1 or 55 miles (and they have talked an hour on their phones)
Let t=time it takes them to get 250 mi apart after Dave leaves
Total distance Alan drives 55+55t
Total distance Dave drives=65t
Now when the above two distances add up to 250 mi, they will be at the outer limits of phone reception, so:
55+55t+65t=250 subtract 55 from each side
55-55+65t+55t=250-55 collect like terms
120t=195 divide each side by 120
t= 1 5/8 hr or 1hr 37.5 min--------time it takes to get 250 mi apart after Dave leaves
Time that they can talk on the phone is the 1 hour that Alan was driving alone plus the 1 5/8 hour that the were both driving and this equals 2 5/8 hrs

Hope this helps---ptaylor
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