SOLUTION: A fast train takes three hours less than the slow train for a journey of 600 km.If the speed of slow train is 10km/hr less than that of fast train.Find the speed of the two trains.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: A fast train takes three hours less than the slow train for a journey of 600 km.If the speed of slow train is 10km/hr less than that of fast train.Find the speed of the two trains.      Log On


   

Question 198348: A fast train takes three hours less than the slow train for a journey of 600 km.If the speed of slow train is 10km/hr less than that of fast train.Find the speed of the two trains.
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A fast train takes three hours less than the slow train for a journey of 600
km. If the speed of slow train is 10km/hr less than that of fast train.
Find the speed of the two trains.
:
Let s = slow train speed
then
(s+10) = fast train
:
Write a time equation: Time dist%2Fspeed
;
Slow train time - Fast train speed = 3 hrs
600%2Fs - 600%2F%28%28s%2B10%29%29 = 3
Multiply each term by s(s+10); results
600(s+10) - 600s = 3s(s+10)
:
600s + 6000 - 600s = 3s^2 + 30s
:
Arrange as a quadratic equation, (600s's cancel)
3s^2 + 30s - 6000 = 0
Simplify, divide by 3
s^2 + 10s - 2000 = 0
Factors to
(s = 50)(s - 40) = 0
positive solution
s = 40 km/hr speed of the slow train
then
50 km/hr speed of the fast train
:
:
Check solution by finding the time difference
600/40 = 15 hrs
600/50 = 12 hrs
--------------
difference 3 hrs

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let f= the speed of the fast train in km/hr
Let s= the speed of the slow train in km/hr
Let t= the time for the fast train to go 600 km
given:
s+=+f+-+10 km/hr
600+=+ft km
600+=+s%2A%28t+%2B+3%29 km
substituting:
600+=+%28f+-+10%29%2A%28t+%2B+3%29
----------------------------
600+=+ft+-+10t+%2B+3f+-+30
and since 600+=+ft,
600+=+600+-+10t+%2B+3f+-+30
0+=+3f+-+10t+-+30
3f+=+10t+%2B+30
f+=+10%2A%28t%2F3+%2B+1%29
substitute back into
600+=+ft
600+=+10%2A%28t%2F3+%2B+1%29%2At
600+=+%2810%2F3%29%2At%5E2+%2B+10t
1800+=+10t%5E2+%2B+30t
t%5E2+%2B+3t+-+180+=+0
Use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+3
c+=+-180
t+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A1%2A%28-180%29+%29%29%2F%282%2A1%29+
t+=+%28-3+%2B-+sqrt%28+9+%2B+720+%29%29%2F2+
t+=+%28-3+%2B-+27%29%2F2+
t+=+%2827+-+3%29%2F2 (ignore negative answer)
t+=+12
and, since
600+=+ft
600+=+12f
f+=+50 km/hr
also,
600+=+s%2A%28t+%2B+3%29
600+=+15s
s+=+40 km/hr
The speed of the fast train is 50 km/hr and
the speed of the slow train is 40 km/hr