SOLUTION: A train was late by 3 hours.It had to cover a distance of 900 km. To cover the time, the driver increased the speed by 15 kmph.Find the increased speed of the train

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Question 698362: A train was late by 3 hours.It had to cover a distance of 900 km. To cover the time, the driver increased the speed by 15 kmph.Find the increased speed of the train
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the increased speed +s+%2B+15+
With speed +s+, time is +t+%2B+3+
where +t+ is on time
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Equation being late by 3 hrs:
(1) +900+=+s%2A%28+t+%2B+3+%29+
Equation for being on time:
(2) +900+=+%28+s+%2B+15+%29%2At+
-----------------------
(1) +900+=+s%2At+%2B+3s+
(1) +s%2At+=+900+-+3s+
(1) +t+=+%28+900+-+3s+%29+%2F+s+
Substitute (1) into (2)
(2) +900+=+%28+s+%2B+15+%29%2A%28+900+-+3s+%29+%2F+s+
(2) +900s+=+%28+s+%2B+15+%29%2A%28+900+-+3s+%29+
(2) +900s+=+900s+%2B+13500+-+3s%5E2+-+45s+
(2) +3s%5E2+%2B+45s+-+13500+=+0+
(2) +s%5E2++%2B+15s+-+4500+=+0+
Use quadratic formula
+s+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+1+
+b+=+15+
+c+=+-4500+
+s+=+%28+-15+%2B-+sqrt%28+15%5E2+-+4%2A1%2A%28-4500%29+%29%29+%2F+%282%2A1%29+
+s+=+%28+-15+%2B-+sqrt%28+225+%2B+18000+%29%29+%2F+2+
+s+=+%28+-15+%2B-+sqrt%28+18225+%29%29+%2F+2+ ( ignore the (-) square root )
+s+=+%28+-15+%2B+135%29+%2F+2+
+s+=+120%2F2+
+s+=+60+
+s+%2B+15+=+75+
The increased speed of the train is 75 km/hr
check:
(1) +900+=+60%2A%28+t+%2B+3+%29+
(1) +t+%2B+3+=+15+
(1) +t+=+12+
and
(2) +900+=+%28+60+%2B+15+%29%2A12+
(2) +900+=+75%2A12+
(2) +900+=+900+
OK