Question 389197
Stephen meets Jason in {{{t}}} min
Stephen went {{{d[1]}}} m and
Jason went {{{d[2]}}} m
I can write 2 equations:
{{{d[1] = 100t}}} m
{{{d[2] = 80t}}} m
Now add the equations:
{{{d[1] + d[2] = 180t}}} m
Let {{{d[1] + d[2] = d}}}, the distance between the towns
{{{d = 180t}}}
In 8 more min, stephen goes {{{d[3] = 100*8}}} m further
{{{d[3] = 800}}} m , so he went {{{d[1] + d[3] = 100t + 800}}}
total distance to meet Melvin
Now I want to know how far Melvin went in {{{t + 8}}} min
He went {{{d[4] = 75*(t + 8)}}} m
{{{d[4] = 75t + 600}}} m
Note that {{{d[1] + d[3] + d[4] = d}}} the distance between the towns
{{{d = 100t + 800 + 75t + 600}}}
and, from before, I said {{{d = 180t}}}, so
{{{180t = 175t + 1400}}}
{{{5t = 1400}}}
{{{t = 280}}}
and, if {{{d = 180t}}}
{{{d = 180*280}}}
{{{d = 50400}}} m
The distance between towns A and B is 50.4 km
check answer:
{{{d[1] = 100t}}}
{{{d[1] = 100*280}}}
{{{d[1] = 28000}}} m
and
{{{d[2] = 80t}}}
{{{d[2] = 80*280}}}
{{{d[2] = 22400}}}
{{{d[1] + d[2] = 28000 + 22400}}} m
{{{d = 50400}}} m
OK
{{{d[3] = 800}}} m
{{{d[1] + d[3] + d[4] = 28000 + 800 + 75t + 600}}}
{{{d = 28000 + 800 + 75*280 + 600}}}
{{{d = 28800 + 21000 + 600}}}
{{{d = 50400}}}
OK