Question 129937
There are 2 equations:
{{{d = (b + c)*t}}} boat going with the current
and
{{{d = (b - c)*t}}} boat going against the current
where {{{d}}}=distance
{{{b}}}= boat's speed in still water
{{{c}}}= current's speed
{{{20 = (b + c)*3}}}
{{{20 = (b - c)*4}}}
{{{3b + 3c = 20}}}
{{{4b - 4c = 20}}}
divide both sides of the 2nd equation by 4
{{{b - c = 5}}}
{{{b = c + 5}}}
substitute this value of {{{b}}} in the 1st equation
{{{3*(c + 5) + 3c = 20}}}
{{{3c + 15 + 3c = 20}}}
{{{6c = 5}}}
{{{c = 5/6}}}
{{{b = c + 5}}}
{{{b = 5/6 + 30/6}}}
{{{b = 35/6}}}
The speed of the boat in still water is 35/6 mi/hr
The speed of the current is 5/6 mi/hr
check answer:
{{{20 = (b + c)*3}}}
{{{20 = (40/6)*3}}}
{{{20 = 20}}}
{{{20 = (b - c)*4}}}
{{{20 = (30/6)*4}}}
{{{20 = 20}}}
OK