Question

a fast train takes three hrs less than slow train for journey of 600km. if the speed of the slow train is 10km/hr less than that of the fast train. find the speeds of the two train

Answer 1

Let the speed of fast train be x 
then speed of slow train will be x-10 
d = 600 
let time taken by slow train be y 
so time taken by fast train will be y-3
d = speed × time 
so 
x × (y-3) = 600           eq. 1 
(x-10) × y = 600           eq. 2
⇒
xy - 3x = 600
xy - 10y = 600

equating both, we get 
xy-3x = xy-10y 
-3x = -10y ⇒ x = 10/3 y 
putting the value of x in eq. 1 
10/3 y into (y-3) = 600
10/3 y² - 10 y = 600  ⇒ 10 y² - 30 y = 1800
solving the eq. by splitting the middle term, we get 
10 y² - 150 y + 120 y - 1800 = 0 
10y (y-15) + 120(y-15) = 0 
y= 15 and y = -120/10= -12 
but as y cannot be negative, we take y = 15 
so, x = 10/3 y = 10/3 × 15 = 50 

so, the speed of the fast train is 50 and of the slow train is 40 km/hr