Question 175901
For  both,
{{{d = r*t}}}
Each has their own rates, times and distances, so
{{{d[t] = r[t]*t[t]}}} Tony's equation
{{{d[g] = r[g]*t[g]}}} Gerry's equation
Each bikes to the others house and they leave at the same time, so
For Tony,
{{{d[t] = r[t]*t[t]}}}
{{{11 = 30*t[t]}}}
{{{t[t] = 11/30}}}hrs
and, for Gerry,
{{{d[g] = r[g]*t[g]}}}
{{{11 = 25*t[g]}}}
{{{t[g] = 11/25}}}hr
Comparing these, with LCD 
{{{t[t] = 55/150}}}
{{{t[g] = 66/150}}}
So, Tony got to Gerry's house {{{11/150}}}hrs quicker
than Gerry got to Tony's house.
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Now, they turn around and head back toward eachother.
If they left at the exact same time, the elapsed
time for each until they met would be the same
{{{t[t] = t[g]}}}, but Tony gets a head start.
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Assume I have a stopwatch and I'm in a helicopter overhead
so I can see both of them. I'll start the stopwatch when
Gerry turns around {{{11/150}}} hr after Tony has turned around.
If Gerry's time to where they meet from Tony's house is {{{t[g]}}} 
then Tony's time from Gerry's house is {{{t[g] + (11/150)}}}
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Now my equations are
{{{d[t] + d[g] = 11}}}km
{{{d[t] = 11 - d[g]}}}  
where these are the distances each travel to where they meet
(1) {{{d[t] = 30*(t[g] + (11/150))}}}
(2) {{{d[g] = 25t[g]}}}
rewriting (1)
(1) {{{11 - d[g] = 30*(t[g] + (11/150))}}}
Substitute (2) in (1)
{{{11 - 25t[g] = 30t[g] + 30*(11/150)}}}
{{{55t[g] = 11 - 11/5}}}
{{{275t[g] = 55 - 11}}}
{{{t[g] = 44/275}}}hr
{{{t[g] = .16}}}hrs or {{{9.6}}} min
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The distance Tony travels is the distance from Gerry's
house when they meet
(1) {{{d[t] = 30*(t[g] + (11/150))}}}
{{{d[t] = 30*(.160 + .0733)}}}
{{{d[t] = 30*.2333}}}
{{{d[t] = 7}}}km
They meet 7 km from Gerry's house
check answer:
(2) {{{d[g] = 25t[g]}}}
{{{d[g] = 25*.16}}}
{{{d[g] = 4}}}
and
{{{d[t] + d[g] = 11}}}km
{{{7 + 4 = 11}}}
OK