Question 214879
I'll write 2 equations, 1 for upriver
and 1 for downriver.
{{{d[u] = r[u]*t[u]}}}
and
{{{d[d] = r[d]*t[d]}}}
Let the rate of the river's current = {{{r[c]}}}
Let {{{r[b]}}} = the rate of the boat in still water 
given:
{{{d[u] = 24}}} mi
{{{d[d] = 24}}} mi
{{{r[u] = 10}}} mi/hr
{{{r[d] = 12}}} mi/hr
{{{t[u] + t[d] = 5.5}}} hr
--------------------------
{{{d[u] = r[u]*t[u]}}}
{{{24 = 10*t[u]}}}
and
{{{24 = 12*t[d]}}}
From these,
{{{t[u] = 2.4}}}
{{{t[d] = 2}}}
Let {{{r[u] = r[b] - r[c]}}}
Let {{{r[d] = r[b] + r[c]}}}
and
{{{r[b] = r[u] + r[c]}}}
{{{r[b] = r[d] - r[c]}}}
{{{r[b] = 10 + r[c]}}}
{{{r[b] = 12 - r[c]}}}
So,
{{{10 + r[c] = 12 - r[c]}}}
{{{2r[c] = 2}}}
{{{r[c] = 1}}} mi/hr
The rate of the current is 1 mi/hr
check answer:
{{{d[u] = r[u]*t[u]}}}
 {{{d[u] = (r[b] - r[c])*t[u]}}}
{{{24 = (r[b] - 1)*2.4}}}
{{{24 = 2.4r[b] - 2.4}}}
{{{2.4r[b] = 26.4}}}
{{{r[b] = 11}}} mi/hr
and
{{{d[d] = r[d]*t[d]}}}
{{{d[d] = (r[b] + r[c])*t[d]}}}
{{{24 = (r[b] + 1)*2}}}
{{{24 = 2r[b] + 2}}}
{{{2r[b] = 22}}}
{{{r[b] = 11}}} mi/hr
OK