Question 804041
Let {{{ t }}} = the time in hours going both
upstream and downstream
Let {{{ s }}} = the speed of the boat in still water
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Going downstream:
(1) {{{ 20 = ( s + 7 )*t }}}
Going upstream:
(2) {{{ 10 = ( s - 7 )*t }}}
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(1) {{{ 20 = s*t + 7t }}}
and
(2) {{{ 10 = s*t - 7t }}}
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Subtract (2) from (1)
(1) {{{ 20 = s*t + 7t }}}
(2) {{{ -10 = -s*t + 7t }}}
{{{ 10 = 14t }}}
{{{ t = 5/7 }}}
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(1) {{{ 20 = ( s + 7 )*t }}}
(1) {{{ 20 = ( s + 7 )*(5/7) }}}
(1) {{{ 140 = 5s + 35 }}}
(1) {{{ 5s = 105 }}}
(1) {{{ s = 21 }}}
The speed of the boat in still water is 21 mi/hr
check:
(2) {{{ 10 = ( s - 7 )*t }}}
(2) {{{ 10 = ( 21 - 7 )*(5/7) }}}
(2) {{{ 10 = 14*(5/7) }}}
(2) {{{ 70 = 70 }}}
OK