Question 957496
a train covers a distance of 720 km at a speed of x km/h. On the return journey, it travels at an average speed of y km/h and completes the whole journey in an hour less. if the speed during the return journey was 10 km/h more, what were the speeds in m/s? 
distance=720km; original speed=x km/hr; return speed=y km/hr=x+10 km/hr
{{{(720km/x)-1hr=720km/(x+10(km/hr))}}} Multiply by (x)(x+10 km/hr)
{{{(720km)(x+10(km/hr))-(1hr)(x)(x+10(km/hr))=720xkm}}}
{{{720xkm+7200km^2/hr-(x^2hr+10xkm)=720xkm}}} Subtract 720xkm from each side.
{{{7200km^2/hr-x^2hr-10xkm=0}}}
-x^2-10x+7200=0*[invoke quadratic "x", -1, -10, 7200 ]
x=80 Original speed was 80 km/hr.
80km/hr+10km/hr=90km/hr Return speed was 90 km/hr.
80km/hr(1000m/km)(1hr/3600sec)=22.22 m/sec Original speed was 22.22 meters per second.
90km(1000m/1km)(1hr/3600sec)=25 m/sec Return speed was 25 meters per second.
CHECK:
(720km/80km/hr)-1hr=720km/90km/hr
9hrs-1hrs=8hrs
8hrs=8hrs