Question 1082577
Let {{{ c }}} = the rate of the current in mi/hr
{{{ 18 + c }}} = the rate of the boat going downstream in mi/hr
{{{ 18 - c }}} = the rate of the boat going upstream in mi/hr
Let {{{ t }}} = time in hrs for both upstream and downstream trips
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Going downstream:
(1) {{{ 10 = ( 18 + c )*t }}}
Going upstream:
(2) {{{ 5 = ( 18 - c )*t }}}
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(1) {{{ t = 10 / ( 18 + c ) }}}
Plug (1) into (2)
(2) {{{ 5 = ( 18 - c )*( 10 / ( 18 + c ) ) }}}
(2) {{{ 5*( 18 + c ) = 10*( 18 - c ) }}}
(2) {{{ 90 + 5c = 180 - 10c }}}
(2) {{{ 15c = 90 }}}
(2) {{{ c = 6 }}}
The rate of thr current is 6 mi/hr
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check:
(1) {{{ t = 10 / ( 18 + c ) }}}
(1) {{{ t = 10 / ( 18 + 6 ) }}}
(1) {{{ t = 10/24 }}} 
(1) {{{ t = 5/12 }}} hrs
and
(2) {{{ t = 5 / ( 18 - c ) }}}
(2) {{{ t = 5 / ( 18 - 6 ) }}}
(2) {{{ t = 5/12 }}} hrs
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Note:
{{{ 5/12*60 = 25 }}} min
OK