Question 1064441
{{{ 35 }}} mi/hr is the speed without a current
The 2 trips each have the same distance, {{{ d }}} in miles
Let {{{ cv }}} = the speed of the current in mi/hr
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Boat going upstream:
(1) {{{ d = ( 35 - c )*(30/60) }}}
Boat going downstream:
(2) {{{ d = ( 35 + c )*(20/60) }}}
( note that I converted minutes to hours )
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(1) {{{ d = ( 35 - c )*(1/2) }}}
(1) {{{ 2d = 35 - c }}}
(1) {{{ 2d + c = 35 }}}
and
(2) {{{ d = ( 35 + c )*(1/3) }}}
(2) {{{ 3d = 35 + c }}}
(2) {{{ 3d - c = 35 }}}
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Add (1) and (2)
(1) {{{ 2d + c = 35 }}}
(2) {{{ 3d - c = 35 }}}
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{{{ 5d = 70 }}}
{{{ d = 14 }}}
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(1) {{{ 2d + c = 35 }}}
(1) {{{ 2*14 + c = 35 }}}
(1) {{{ 28 + c = 35 }}}
(1) {{{ c = 7 }}}
The speed of the current is 7 mi/hr
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check:
(1) {{{ d = ( 35 - c )*(1/2) }}}
(1) {{{ 14 = ( 35 - 7 )*(1/2) }}}
(1) {{{ 14 = 28/2 }}}
(1) {{{ 14 = 14 }}}
and
(2) {{{ d = ( 35 + c )*(1/3) }}}
(2) {{{ 14 = ( 35 + 7 )*(1/3) }}}
(2) {{{ 14 = 42/3 }}}
(2) {{{ 14 = 14 }}}
OK