Question 299982
Let {{{r[s]}}} = rate of boat in still water
Let {{{r[c]}}} = rate of the current
Write 2 equations, one for upstream and one for downstream
given:
{{{t[u] = 4}}} hrs
{{{t[d] = 2}}} hrs
{{{d= 320}}} mi
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upstream:
{{{d = r[u]*t[u]}}}
{{{d = (r[s] - r[c])*t[u]}}}
(1) {{{320 = (r[s] - r[c])*4}}}
downstream:
{{{d = r[d]*t[d]}}}
{{{d = (r[s] + r[c])*t[d]}}}
(2) {{{320 = (r[s] + r[c])*2}}}
-----------------
(1) {{{r[s] - r[c] = 80}}}
(2) {{{r[s] + r[c] = 160}}}
Add the equations
{{{2r[s] = 240}}}
{{{r[s] = 120}}} mi/hr
And, since
(1) {{{r[s] - r[c] = 80}}}
{{{120 - r[c] = 80}}}
{{{r[c] = 40}}} mi/hr
The rate of the boat in still water is120 mi/hr
The rate of the current is 40 mi/hr
check:
(1) {{{320 = (r[s] - r[c])*4}}}
{{{320 = (120 - 40)*4}}}
{{{320 = 320}}}
and
(2) {{{320 = (r[s] + r[c])*2}}}
{{{320 = (120 + 40)*2}}}
{{{320 = 320}}}
OK