Math problem with time

[kate]

1. A pedestrian left from point a to points B. Two hours later from A to B left a cyclist that after one hour was 1 km behind the pedestrian. After one more hour the cyclist was three times closer to B that the pedestrian. Find their speeds and the distance between A and B, if the speed of the speed of the pedestrian is 5/14 the speed of the cyclist

[shyamjayakannan]

let the distance from A to B be [tex]d[/tex] km and the cyclists speed be [tex]v[/tex] km/hr.

1) after 3hr, the pedestrian is at [tex]3\times\frac{5v}{14}=\frac{15v}{14}[/tex] km and the cyclist is at [tex]v[/tex] km because he only traveled for 1hr. It is given that [tex]\frac{15v}{14}-1=v\Rightarrow v=14[/tex] km/hr.

2) after 4hr, the pedestrian is at [tex]4\times\frac{5v}{14}=\frac{10v}{7}[/tex] km and the cyclist is at [tex]2v[/tex] km because he only traveled for 2hr. It is given that [tex]d-\frac{10v}{7}=3(d-2v)\Rightarrow 2d=\frac{32v}{7}=\frac{32\times14}{7}\Rightarrow\boxed{d=32km}[/tex].