```Question 182034
A Fed ex truck leaves San Diego to deliver a package to egg ville. One hour out
of San Diego on the road it develops mechanical problems. It now slows its
speed down to 3/5th of the average speed up to the time of the mechanical
trouble. It hobbles along and gets to egg ville 2 hours late. If the mechanical
troubles occurred 50 miles beyond where it had, the delay to egg ville would
have been reduced by 40 minutes. Find the distance from San Diego to to egg ville.
:
use .6 for {{{3/5}}}
:
Let s = normal average speed
Let t = normal time for the trip
Then
st = distance of the trip
1s = distance traveled before problem developed (1 hr)
Then
(st - 1s) = distance traveled at .6s average speed
:
write a time equation
:
1 hr at normal speed + time at slow speed = normal time + 2 hrs
1 + {{{((st-1s))/(.6s)}}} = t + 2
factor out s
1 + {{{(s(t-1))/(.6s)}}} = t + 2
Cancel s
1 + {{{((t-1))/.6}}} = t + 2
multiply equation by .6
.6 + t - 1 = .6(t+2)
.6 - 1 + t = .6t + 1.2
-.4 + t = .6t + 1.2
t - .6t = 1.2 + .4
.4t = 1.6
t = {{{1.6/.4}}}
t = 4 hrs normal time for the trip
then
4s = distance of the trip
and
trip with breakdown: 4 + 2 = 6 hrs (1hr at normal speed & 5hrs at .6 speed)
:
"If the mechanical troubles occurred 50 miles beyond where it had, the delay
to egg ville would have been reduced by 40 minutes. (2/3 hr)
:
Distance at slow speed now = 4s - 1s - 50 = (3s-50)
total Time now:  6 - {{{2/3}}} = 5.33 hrs (40 min earlier)
:
Another time equation
1 + {{{50/s}}} + {{{(3s-50)/(.6s)}}} = 5.33
Multiply equation by .6s
.6s + .6(50) + 3s - 50 = .6s(5.33)
.6s + 30 + 3s - 50 = 3.2s  (got rid of the repeating decimal)
.6s + 3s - 20 = 3.2s
3.6s - 3.2s = 20
.4s = 20
s = {{{20/.4}}}
s = 50 mph is normal speed
and
.6(50) = 30 mph is breakdown speed
then
4 * 50 = 200 mi is the distance, which is what they want
:
:
Check solution, find the distance using these values:
1(50)  + 5(30) =
50 + 150 = 200 mi
and find the time (dist/speed) for the  2nd scenario
50/50 + 50/50 + 100/30 =
1 + 1 + 3.33 = 5.33 hrs
: