Row, row, row your boat

Let c be the current rate in mph and let m be Mia's still water rowing rate. In 20 minutes of rowing upstream Mia travels $(1/3)(m-c)$ miles. After turning around, Maria must row for $\frac{(1/3)(m-c)+2.4}{m+c}$ hours to get the cap. The cap floated downstream for $\frac{(1/3)(m-c)+2.4}{m+c}+\frac{1}{3}$ hours at a rate of c mph. Thus $2.4=c[\frac{1}{3}+\frac{(1/3)(m-c)+2.4}{m+c}]$ . The expression in brackets reduces to $\frac{(2/3)m+2.4}{m+c}$ , so we get $2.4(m+c)=c[(2/3)m+2.4]$ , which further reduces to $2.4m=(2/3)cm$ , so $c=3.6$ .

Let $M$ be Mia's rowing rate in still water and let $c$ be the rate of the current. Let $t$ be the amount of time it takes for Mia to retrieve the cap once she turns around. I was able to get the following two equations:

$c(t+\frac{1}{3})=2.4,$

$(M+c)t-\frac{1}{3}(M-c)=2.4.$

With some simple manipulation, the second equation can also be written as:

$M(t-\frac{1}{3})+c(t+\frac{1}{3})=2.4$

Substituting $2.4$ in for $c(t+\frac{1}{3}$ :

$M(t-\frac{1}{3}) + 2.4 = 2.4$

$\Rightarrow M(t-\frac{1}{3})=0$

$\Rightarrow t=\frac{1}{3}$

So, $c(\frac{1}{3}+\frac{1}{3}) = 2.4$

$\Rightarrow c = (2.4)\frac{3}{2} = 3.6$

let say that cap is blew 0ff at p0int "X" <<<initial P.>>> l00K at this graph

X(blew 0ff)---------------->----------20 min------------------>------------------>>(she n0ticed) ================ ========X<<--<<----------<<---------------20 min-----------------------------<<<(turned)

(((((((she didn't find the cap at p0int X but it m0ves abt 2.4 miles away taken the time """"frm 1stly blew 0ff >>>> n0ticed >>>turned >>arrived again ))) s0 .....................

cap m0ved 2.4 Miles >>>>>> at 40 Minutes ========= ???????>>>>>>>at 60 Minutes ( 1 hr )

????????????????= 60 2.4 / 40 = 3.6 M/hr ;) : :D :v