Simultaneous Equation

A train running from A to B meets with an accident 75 km from A, after which it travels with 3/5th of its original velocity and arrives 3 hours late at B. If the accident had occurred 75 km further on, it would have been only 2 hours late. Find the distance from A to B and the original velocity of the train.


Answer:

let the Distance be "y" and the original velocity be "v"

if there were no accident, then the train would reach B in = y/v hours

time taken by train to cover the first 75 km = 75/v

velocity after the accident = 3/5 of v = 3v/5

therefore, time taken to reach B 
= 75/v + (y - 75) / (3v/5) 
= 75/v + 5(y - 75) / 3v 

= (225 + 5y -375) / 3v

= (5y - 150)/ 3v
 
But actual time taken is y/v + 3

So, equating the two, we get 

(5y - 150)/ 3v = y/v + 3

or, (5y - 150) = 3y + 9v

or, 2y - 9v = 150  -------------------(1)



in the second case, the time taken to reach B is,

150/v + (y - 150) / (3v/5) 

= 150/v + 5(y - 150)/3v

= (450 + 5y -750) / 3v

= (5y - 300) / 3v 

But the actual time taken is y/v + 2

So, equating the two we get

(5y - 300) / 3v  = y/v + 2

or, 5y - 300 = 3y + 6v

or, 2y - 6v = 300 -----------(2)

Solving equation (1) and (2), we get

6v - 9v = 150 - 300

or, v = 50 km/hr

o, y = 300 km

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