Word problem with speed and trains

Question

By traveling 10km/hr faster than its normal speed, a train can complete a journey of 300km in 1 hour less. Determine the train's normal speed

Answer 1

distance=rate*time

d=rt

300=rt

300=(r+10)(t-1)

from 300=rt we get 300/r=t

substitute

300=(r+10)([300/r]-1)

300=(r+10)([300-r]/r)

multiply both sides by r

300r=(r+10)(300-r)

300r=300r+3000-r^2-10r

subtract 300r from both sides

0=3000-r^2-10r

r^2+10r-3000=0

factor (using the quadratic formula will be easier but I like to factor)

3000=3*2*2*2*5*5*5=(3*2*2*5)*(2*5*5)=60*50 and 60-50=10

(r+60)(r-50)=0

r-50=0

r=50 kph is the train's normal speed

check:

300=50*6

300=60*5