WIDTH OF THE RIVER???

Let the width of the river be w. Let the speed of the first boat be f. Let the speed of the second boat be s.

Time of first meeting is 720/f = (w - 720)/s f reaches far shore at w/f waits 10 minutes and heads back s reaches near shore at w/s waits 10 minutes and heads back time of second meeting is w/f + 10 + 400/f = w/s + 10 + (w-400)/s

The two 10 minute delays cancel each other out.

720s = fw - 720f f(w-720) = 720 s f/s = 720/(w-720)

w/f + 10 + 400/f = w/s + 10 + (w-400)/s ws + 10fs + 400s = wf + 10fs + wf - 400f s (400 + w) = f (2w - 400) f/s = (400 + w)/(2w - 400)

720/(w-720) = (400 + w)/(2w - 400) 720 (2w - 400) = (400 + w) (w - 720) 1440 w - 288000 = 400w - 720w + w^2 - 288000 w^2 - 1760w = 0 w = 1760

At the first meeting, boat A has traveled a distance $720$ and boat B has traveled a distance $w-720$ , so

$t_1=\frac{720}{v_A}=\frac{w-720}{v_B}$

At the second meeting boat A has traveled a total distance $w+400$ and boat B has traveled a total distance $2w-400$ ,so $t_2-10=\frac{w+400}{v_A}=\frac{2w-400}{v_B}$

We now can write the speed ratio in two ways: $\frac{v_A}{v_B}=\frac{720}{w-720}=\frac{w+400}{2w-400}$ Rewrite as $720(2w-400)=(w+400)(w-720)$ $1760w=w^2$ $w=0 \vee w=\boxed{1760}$

The best way to think of this problem is to take into consideration the time it took for the two to meet for the first time, and the second time. The first time: We want to know the width of the river, x. The time it took for both to meet for the first time is t. Hence, we can say that 720/u=(x-720)/v using the distance = speed multiply time and u and v are speeds of respective boats. This is equation 1. Now, when they meet for the second time, the time passed from the first time meeting is T. Boat with speed v would have taken (720/v + 10/60 + (x-400)/v)=T Likewise, boat with speed u would have taken the same amount of time T. This is to say (x-720)/u+10/60+400/u= T Therefore, 720/v+1/6+(x-400)/v=(x-720)/u+1/6+400/u ( equation 2)

Solving for x you get 1760