Question 761851
Let {{{ c }}} = the speed of the current
{{{ 7 - c }}} = her speed going against the current
{{{ 7 + c }}} = her speed going with the current
Let {{{ t }}} = her time going with the current
{{{ t + 10 }}} = her time going against the current
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Against the current:
(1) {{{ 4 = ( 7 - c )*( t + 10 ) }}}
With the current:
(2) {{{ 4 = ( 7 + c )*t }}}
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(1) {{{ 4 = 7t - c*t + 70 - 10c }}}
(1) {{{ 7t - c*t = 10c - 66 }}}
(1) {{{ t*( 7 - c ) = 10c - 66 }}}
(1) {{{ t = ( 10c - 66 ) / ( 7-c ) }}}
and
(2) {{{ 4 = ( 7 + c )*t }}}
(2) {{{ t = 4/( 7 + c ) }}}
By substitution:
(1) {{{ 4/( 7 + c ) = ( 10c - 66 ) / ( 7-c ) }}}
Multiply both sides by {{{ ( 7+c )*( 7-c ) }}}
(1) {{{ 4*( 7-c ) = ( 10c - 66 )*( 7+c ) }}}
(1) {{{ 4*( 7-c ) = 2*( 5c - 33 )*( 7+c ) }}}
(1) {{{ 14 - 2c = 35c - 231 + 5c^2 - 33c }}}
(1) {{{ 5c^2 + 4c - 245 = 0 }}}
Use quadratic formula
{{{ c = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 5 }}}
{{{ b = 4 }}}
{{{ c = -245 }}} ( not the same c as current )
{{{ c = (-4 +- sqrt( 4^2 - 4*5*(-245) )) / (2*5) }}}
{{{ c = (-4 +- sqrt( 16 + 4900)) / 10 }}}
{{{ c = (-4 +- sqrt( 4916)) / 10 }}}
{{{ c = (-4 + 70.114 )/ 10 }}}
{{{ c = 66.114 / 10 }}}
{{{ c = 6.611 }}}
The speed of the current is 6.611 mi/hr
check:
(2) {{{ 4 = ( 7 + c )*t }}}
(2) {{{ 4 = ( 7 + 6.611 )*t }}}
(2) {{{ 4 = 13.611t }}}
(2) {{{ t = .294 }}}
and
(1) {{{ 4 = ( 7 - 6.611 )*( t + 10 ) }}}
(1) {{{ 4 = .389t + 3.89 }}}
(1) {{{ .389t = .11 }}}
(1) {{{ t = .283 }}}
This method should be OK, but it doesn,t
exactly check- maybe you can find error
Hope this helps