Boats and Streams

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$$

=> $$\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$$

=> $$2ds=6\left(s^2-\dfrac{d^2}{144}\right)$$

=> $$144s^2-48ds-d^2=0$$

Solving the quadratic equation, we get:

$$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$$

$$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$$

=> Required value of $$\dfrac{d}{s+r}$$

= $$\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$$

= $$\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$$

= $$3\left(3-\sqrt{\ 5}\right)$$

Let the river speed be v km/h. For Sneha still-water speed = 6 km/h:

Upstream speed = (6-v), downstream speed = (6+v).

Given $$\dfrac{14}{(6-v)}-\dfrac{14}{(6+v)}=48$$minutes =0.8 hours. 

So, $$\dfrac{14(6+v-6+v)}{36-v^2}=0.8 \Rightarrow \dfrac{28v}{36-v^2}=0.8$$

$$28v=0.8(36-v^2)=28.8-0.8v^2 \Rightarrow 0.8v^2+28v-28.8=0$$

Multiply by 5: $$4v^2+140v-144=0$$

$$v^2+35v-36=0$$

We get v = 1

Thus, the river speed is 1 km/h. For Rita, the still water speed is 5 km/h:
Upstream speed = 5-1=4 km/h, Downstream Speed = (5+1=6) km/h.

If she rows d km upstream and returns d km downstream, the total time

$$ \dfrac{d}{4}+\dfrac{d}{6}=d\Big(\dfrac{1}{4}+\dfrac{1}{6}\Big)=d\cdot\dfrac{5}{12}\ \text{hours}$$

This equals 100 minutes $$=100/60=\dfrac{5}{3}$$ hours. 

So, $$d\cdot\dfrac{5}{12}=\dfrac{5}{3}\ \Rightarrow\ d=4\ \text{km}$$

Total distance covered 2d=8 km.

The boat's speed in still water= 10km/hr. Speed upstream is 5 Km/Hr and downstream is 15 km/hr. So, a/5 + a/15 = 4. So, a = 15 Km