SOLUTION: peter leaves his house at 6:45 am and drives east. At 7:15 am, Paul leaves his house driving east. Paul's speed is 25 mph faster than half peters speed. At 7:35 am, they pass each

Click here to see ALL problems on Travel Word Problems

Question 1146228: peter leaves his house at 6:45 am and drives east. At 7:15 am, Paul leaves his house driving east. Paul's speed is 25 mph faster than half peters speed. At 7:35 am, they pass each other on the road. Find Pauls speed if they live 59.3 miles apart.
Found 2 solutions by ankor@dixie-net.com, richwmiller:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Does not really make sense as written, lets change it to Paul driving west (toward each other)
:
peter leaves his house at 6:45 am and drives east.
At 7:15 am, Paul leaves his house driving west.
Paul's speed is 25 mph faster than half peters speed.
At 7:35 am, they pass each other on the road.
Find Pauls speed if they live 59.3 miles apart.
:
let x = Peter's speed
then
(.5x+25) = Paul's speed
:
6:45 to 7:35 = 50 min is Peter's travel time (.8333 hr)
and
7:15 to 7:35 = 20 min is Paul's travel time (.3333 hr)
:
When they meet, the sum of their distances = 59.3 mi
Write a dist equation; dist = speed (mph) * time (hr)
:
Peter's dist + Paul's dist = 59.3 mi
.8333x + .333(.5x+25) = 59.3
.8333x + .1665x + 8.325 = 59.3
.9998x = 59.3 - 8.325
close enough to 1
x = 50.975 ~ 51 mph is Peter's speed
then
.5(50.975) + 25 = 50.4875 ~ 50.5 mph is Paul's speed
:
Seems the are going about the same speed, see if this checks out
.8333(51) = 42.5 mi
.3333(50.5) = 16.8 mi
---------------------
total distance 59.3 m
:
:
A rather strange problem!

Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
Paul leaves his house driving EAST!
They start 59.3 miles apart and head in the same direction (east).
There is a delay of 30 minutes before Paul starts.
Peter travels 59.3 miles +Paul's distance