Question 518231
Let {{{ s }}} = rate in still water
Let {{{ c }}} = rate of current
going downstream:
(1) {{{ d[1] = ( s + c )*t }}}
Going upstream:
(2) {{{ d[2] = ( s - c )*t }}}
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given:
{{{ d[1] = 30 }}} mi
{{{ d[2] = 18 }}} mi
{{{ s = 8 }}} mi/hr
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(1) {{{ 30 = ( 8 + c )*t }}}
(2) {{{ 18 = ( 8 - c )*t }}}
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(1) {{{ 30 = 8t + c*t }}}
(2) {{{ 18 = 8t - c*t }}}
Add the equations
{{{ 48 = 16t }}}
{{{ t = 3 }}} hrs
and, since
(1) {{{ 30 = 8t + c*t }}}
(1) {{{ 30 = 24 + 3c }}}
(1) {{{ 3c = 6 }}}
(1) {{{ c = 2 }}}
The rate of the current is 2 mi/hr
check:
(1) {{{ 30 = ( 8 + c )*t }}}
(1) {{{ 30 = ( 8 + 2 )*3 }}}
(1) {{{ 30 = 10*3 }}}
OK