Question 262931
I have a stopwatch and I'll start it when the 2nd 
cyclist leaves. At that time the 1st cyclist has gone
{{{r[1]*3}}}
{{{6*3 = 18}}} mi
Now they will each go for the same time on my stopwatch, so
{{{t[1] = t[2]}}}, which I'll just call {{{t}}}
{{{d[1] = r[1]*t}}}
{{{d[2] = r[2]*t}}}
From these,
{{{d[1]/r[1] = d[2]/r[2]}}}
{{{d[1]*r[2] = d[2]*r[1]}}}
{{{10d[1] = 6d[2]}}}
{{{d[1] = .6d[2]}}}
I know 1 more thing about the distances:
{{{d[2] = d[1] + 18}}}, since the 2nd cyclist has to make up 
the headstart the 1st one got
Substituting:
{{{d[2] = .6d[2] + 18}}}
{{{.4d[2] = 18}}}
{{{d[2] = 45}}} mi
and
{{{d[1] = .6*d[2]}}}
{{{d[1] = .6*45}}}
{{{d[1] = 27}}} mi
Now, I need to find {{{t}}}
(1) {{{d[1] = r[1]*t}}}
(2) {{{d[2] = r[2]*t}}}
-------------------
(1) {{{27 = 6t}}}
{{{t = 4.5}}} hrs
and, also
(2) {{{45 = 10t}}}
{{{t = 4.5}}} hrs
How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? 
This means that the 3 hr headstart is not to be included in answer
so, 4.5 hrs will pass 
The answer is B