Question 567816: the train covers 90km in uniform speed.if speed is raised 15km/hr then train covers distance in half hour less.find the original speed. Found 2 solutions by mananth, josmiceli:Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! let speed be x km/h
increase by 15
=x+15
original time - time with increased speed = 1/2 hour
d=90 km
d/r = t
90/x-(90/(x+15)=1/2
multiply the equation by 2x(x+15) to eliminate the denominator
180(x+15)-180x=x(x+15)
180x+2700-180x=x^2+15x
90x cancels off
x^2+15x-2700=0
x^2+60x-45x-2700=0
x(x+60)-45(x+60)=0
(x+60)(x-45)=0
x= 45 km/h the original speed. Ignore negative value
You can put this solution on YOUR website! Let = original speed in km/hr
Let = time in hrs to cover km at speed
given:
(1)
(2)
------------------------
(1)
Substitute (1) into (2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
Use quadratic formula ( I can't use the negative square root )
The original speed is 45 km/hr
check answer:
(1)
(1)
(1) hrs
and
(2)
(2)
(2)
(2)
OK
