SOLUTION: Two trains travel a 160 km track each day. The express travels 10 km h-1 faster and takes 30 minutes less time than the normal train. Find the speed of the express

Algebra ->  Equations -> SOLUTION: Two trains travel a 160 km track each day. The express travels 10 km h-1 faster and takes 30 minutes less time than the normal train. Find the speed of the express      Log On


   

Question 641310: Two trains travel a 160 km track each day. The express travels 10 km h-1 faster and takes 30 minutes less time than the normal train. Find the speed of the express
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the slower train in km/hr
+s+%2B+10+ = the speed of the faster train in km/hr
Let +t+ = time in hrs for the slower train to go 160 km
+t+-+.5+ = time in hrs for faster train to go 160 km
---------------
Slower train:
(1) +160+=+s%2At+
Faster train:
(2) +160+=+%28+s+%2B+10+%29%2A%28+t+-+.5+%29+
--------------------------
(2) +160+=+s%2At+%2B+10t+-+.5s+-+5+
From (1)
(1) +t+=+160%2Fs+
and
(2) +160+=+s%2A%28+160%2Fs%29+%2B+10%2A%28+160%2Fs+%29+-+.5s+-+5+
(2) +160s+=+160s+%2B+1600+-+.5s%5E2+-+5s+
(2) +.5s%5E2+%2B+5s+-+1600+=+0+
(2) +5s%5E2+%2B+50s+-+16000+=+0+
(2) +s%5E2+%2B+10s+-+3200+=+0+
+s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+10+
+c+=+-3200+
+s+=+%28-10+%2B-+sqrt%28+10%5E2+-+4%2A1%2A%28-3200%29+%29%29%2F%282%2A1%29+
+s+=+%28-10+%2B-+sqrt%28+100+%2B+12800+%29%29+%2F+2+
+s+=+%28-10+%2B-+sqrt%28+12900+%29%29+%2F+2+
Check my work and then check the results
for +s+, the slower train and +s%2B10+,
the faster train