Question

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of the journey.

Answer 1

Speed= $\frac{distance}{time}$

let the speed of the train be x and time taken y

Therefore if the train goes at 5km/hr faster;
its speed is (x+5)km/hr
and the time taken is (y+3)hours

On the other hand, if the train goes at 4km/hr slower;
its speed will be (x-4)km/hr
and time taken will be (y-3)hours

After establishing these facts now you can form equations, using speed, time and distance formula:

Speed= $\frac{distance}{time}$ and distance = time $\times$ speed

Since distance is the same for all the three instances, you can equate the equations for distance.

equation 1:

distance will be given by: x $\times$ y = xy

equation 2:

distance will be given by: (x+5)(y-3) = xy-3x+5y-15

equation 3:

distance will be given by: (x-4)(y+3) =xy+3x-4y-12

All these three equations are equal as they are equations for distance. Therefore equating any two equations will give you a linear equation you can use to solve for time, distance and speed.

hence xy = xy - 3x+5y - 15......therefore 3x - 5y = -15
3x = 5y -15

and xy = xy +3x - 4y-12 .......therefore 3x-4y =12
substitute for 3x >>>5y-15-4y =12
y=27

substitute for x: 3x=5y - 15
3x= 135-15 3x=120
x =40

therefore time taken = 27 hours
while speed = 40km

distance= speed $\times$ time
=27 $\times$ 40
= 1080km
Therefore length of journey = 1080 km