Question

A passenger train takes 3 hours less for a journey of 360km . if its speed increased by 10km/hr from its usual speed . what is its usual speed?

Answer 1

360 = v t
360 = (v+10)(t-3) = vt -3v + 10t -30

so
vt = vt -3v + 10 t - 30

10 t = 3 (v+10)
but t = 360/v

3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30

Answer 2

360 = v t 
360 = (v+10)(t-3) = vt -3v + 10t -30 

so 
vt = vt -3v + 10 t - 30 

10 t = 3 (v+10) 
but t = 360/v 

3600/v = 3 v + 30 
3 v^2 +30 v - 3600 = 0 
v^2 + 10 v - 1200 = 0 
v+40)(v-30) = 0 
v = 30