Question 13.

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

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

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

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

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

Solving the quadratic equation, we get:

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

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

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

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

$= \frac{12}{3 + \sqrt{5}} = \frac{(12)(3 - \sqrt{5})}{4}$

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