Question 1047181
Let {{{ c }}} = the rate of the current
Let {{{ t }}} = time in hrs for both trips
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Equation for going upstream:
(1) {{{ 10 = ( 9 - c )*t }}}
Equation for going downstream:
(2) {{{ 20 = ( 9 + c )*t }}}
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(1) {{{ t = 10/( 9 - c ) }}}
Plug (1) into (2)
(2) {{{ 20 = ( 9 + c )*( 10/( 9 - c ) ) }}}
(2) {{{ 20*( 9 - c ) = 10*( 9 + c ) }}}
(2) {{{ 180 - 20c = 90 + 10c }}}
(2) {{{ 30c = 90 }}}
92) {{{ c = 3 }}}
The rate of the current is 3 mi/hr
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check:
(1) {{{ 10 = ( 9 - c )*t }}}
(1) {{{ 10 = ( 9 - 3 )*t }}}
(1) {{{ 10 = 6t }}}
(1) {{{ t = 5/3 }}}
and
(2) {{{ 20 = ( 9 + c )*t }}}
(2) {{{ 20 = ( 9 + 3 )*t }}}
(2) {{{ 20 = 12t }}}
(2) {{{ t = 5/3 }}}
OK