SOLUTION: Hi! I have another question on applications of linear equations. That is, word problems. A fast train was help up by a red-light signal for 16 minutes and made up for the lost t

Algebra ->  Linear-equations -> SOLUTION: Hi! I have another question on applications of linear equations. That is, word problems. A fast train was help up by a red-light signal for 16 minutes and made up for the lost t      Log On


   

Question 628316: Hi! I have another question on applications of linear equations. That is, word problems.
A fast train was help up by a red-light signal for 16 minutes and made up for the lost time on a 80-km stretch travelling at 10 kph faster than called by schedule. What is the scheduled speed of the train?
Thank you so much!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A fast train was held up by a red-light signal for 16 minutes and made up
for the lost time on a 80-km stretch traveling at 10 kph faster than
called by schedule.
What is the scheduled speed of the train?
:
Change 16 min to hrs: 16/60 = 4%2F15 hrs
:
Let s = the scheduled speed of the train
then
(s+10) = catch up speed of the train
:
Write a time equation time = dist/speed
:
Sched time - faster time = 16 min
80%2Fs - 80%2F%28%28s%2B10%29%29 = 4%2F15
:
Multiply by 15s(s+10)
15s(s+10)*80%2Fs - 15s(s+10)*80%2F%28%28s%2B10%29%29 = 15s(s+10)*4%2F15
:
Cancel the denominators, leaving
15(s+10)*80 - 15s(80) = 4s(s+10)
:
80(15s+150) - 1200s = 4s^2 + 40s
:
A quadratic equation
1200s + 12000 - 1200s = 4s^2 + 40s
0 = 4s^2 + 40s - 12000
:
simplify, divide by 4
s^2 + 10s - 3000 = 0
:
Factors to
(s+60)(s-50) = 0
s = 50 mph is the scheduled speed of the train
:
:
Check this, find the time of each trip
80/50 = 1.6 hrs
80/60 = 1.33 hrs
----------------
diff: .27 hr ~ 16 min