Question 1064928
At a speed of {{{x}}} kph (kilometers per hour),
the second cyclist cover {{{100}}}km in
{{{100/x}}} hours.
The second cyclist, at {{{x+10}}} kph,
would cover the same distance in
{{{100/(x+10)}}} hours,
but takes a break lasting {{{50/6=5/6}}} hours,
so his/her trip takes {{{100/(x+10)+5/6}}} hours.
The problem says the first cyclist trip took less time,
so {{{100/(x+10)+5/6<100/x}}} is our inequality.
Multiplying both sides of the inequality times
the positive number {{{6x(x+10)}}} , we get
{{{600x+5x(x+10)<600(x+10)}}}
{{{600x+5x^2+50x<600x+6000}}
{{{5x^2+50x<6000}}}
{{{5x^2+50x-6000<0}}}
Dividing by 5 both sides of the inequality, we get
{{{x^2+10x-1200<0}}}
You know that the quadratic polynomial
{{{x^2+10x-1200}}} is negative only between its zeros.
The solutions to
{{{x^2+10x-1200=0}}} <---> {{{(x-30)(x+40)=0}}}
are {{{x=30}}}, and {{{x=-40}}} ,
and since speeds should not be negative,
{{{0<=x<30}}} ---> {{{10<=x+10<40}}} is the solution.
The second cyclist may be moving at 0 kph (sitting still),
and then the first one would be going at 10kph,
or they both may be moving,
but the first cyclist is moving at
10 kph or more, but less than 40 kph.