Question 1046114
Let {{{ c }}} = the speed of the curent in mi/hr
{{{ 6 - c }}} = her speed going upstream
{{{ 6 + c }}} = her speed going downstream
Let {{{ t }}} = her time in hrs for both trips
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Going upstream:
(1) {{{ 10 = ( 6-c )*t }}}
Going downstream:
(2) {{{ 30 = ( 6+c )*t }}}
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(1) {{{ t = 10/( 6-c ) }}}
Plug (1) into (2)
(2) {{{ 30 = ( 6+c )*( 10 / ( 6 - c ) ) }}}
(2) {{{ 30*( 6-c ) = 10*( 6 + c ) }}}
(2) {{{ 180 - 30c = 60 + 10c }}}
(2) {{{ 40c = 120 }}}
(2) {{{ c = 3 }}}
The rate of the current is 3 mi/hr
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check:
(1) {{{ 10 = ( 6-c )*t }}}
(1) {{{ 10 = ( 6-3 )*t }}}
(1) {{{ 10 = 3t }}}
(1) {{{ t = 10/3 }}}
and
(2) {{{ 30 = ( 6+c )*t }}}
(2) {{{ 30 = ( 6+3 )*t }}}
(2) {{{ 30 = 9t }}}
(2) {{{ t = 10/3 }}}
OK