Question 270049
{{{b}}} = speed of boat in still water
{{{c}}} = speed of current
With the current:
{{{d = (b + c)*t[1]}}}
Against the current:
{{{d = (b - c)*t[2]}}}
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given:
{{{d = 28}}} mi
{{{t[1] = 4}}} hrs
{{{t[2] = 7}}} hrs
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{{{d = (b + c)*t[1]}}}
{{{28 = (b + c)*4}}}
(1) {{{4b + 4c = 28}}}
and
{{{d = (b - c)*t[2]}}}
{{{28 = (b - c)*7}}}
(2) {{{7b - 7c = 28}}}
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(1) {{{4b + 4c = 28}}}
(1) {{{b + c = 7}}}
and
(2) {{{7b - 7c = 28}}}
(2) {{{b - c = 4}}}
and
(2) {{{b + c = 7}}}
(1) {{{b - c = 4}}}
{{{2b = 11}}}
{{{b = 5.5}}}
And, from (2),
{{{b - c = 4}}}
{{{5.5 - c = 4}}}
{{{c = 5.5 - 4}}}
{{{c = 1.5}}}
The speed of the river current is 1.5 mi/hr
check:
{{{28 = (b + c)*4}}}
{{{28 = (5.5 + 1.5)*4}}}
{{{28 = 7*4}}}
{{{28 = 28}}}
and
{{{28 = (b - c)*7}}}
{{{28 = (5.5 - 1.5)*7}}}
{{{28 = 4*7}}}
{{{28 = 28}}}
OK