<PRE><B>Question 290060</B>
Two cyclists drive a distance of 15 kilometres.
 One of the cyclists drives 10 km h faster than the other one and arrives at
 their destination 20 minutes earlier.
 What are the speeds of the cyclists?
:
Let s = speed of the slow cyclist
then
(s+10) = speed of the faster one
:
Change 20 min to {{{1/3}}} hr
:
Write a time equation: Time = dist/speed
:
slow time - fast time = {{{1/3}}} hr
{{{15/s}}} - {{{15/((s+10))}}} = {{{1/3}}}
Multiply by 3s(s+10), results:
3(15)(s-10) - 3(15)s = s(s+10)
:
45(s+10) - 45s = s(s+10)
:
45s + 450 - 45s = s^2 + 10s
:
0 = s^2 + 10s - 450
:
Use the quadratic formula to find s:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this equation, x=s; a=1, b=10; c=-450
{{{s = (-10 +- sqrt(10^2-4*1*-450 ))/(2*1) }}}
:
{{{s = (-10 +- sqrt(100 -(-1800) ))/2 }}} 
;
{{{s = (-10 +- sqrt(100 + 1800))/2 }}}
:
{{{s = (-10 +- sqrt(1900 ))/2 }}}
Positive solution
{{{s = (-10 + 43.589)/2 }}}
s = {{{33.589/2}}}
s = 16.7945 ~ 16.8km/h is the slow cyclist
and
26.8 km/h is the fast cyclist
;
:
Check solution by finding the times
Slow: 15/16.8 = .89 hrs
Fast: 15/26.8 = .56 hrs
------------------------
difference is:  .33 hrs which is about 20 min

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