A FAST TRAIN TAKES 3 HOURS LESS THAN A SLOW TRAIN IS TRAVELLING 600 KM. IF THE SPEED OF THE FAST TRAIN IS 10 KMH MORE THAN THE SPEED OF SLOW TRAIN . FIND THE SPEED OF BOTH THE TRAINS.
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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
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
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