A boat travels 60 kilometres downstream and 20 kilometres upstream in 4 hours. The same boat travels 40 kilometres downstream and 40 kilometres upstream in 6 hours. What is the speed (in km./hr.) of the stream?

Let the rate downstream of boat be u kmph. Its rate upstream = v kmph. According to the question, $\frac{60}{u}+\frac{20}{v}=4$ and, $\frac{40}{u}+\frac{40}{v}=6$ $\Rightarrow \frac{20}{u}+\frac{20}{v}=3$ By equation (i) - (ii), $\frac{60}{u}-\frac{20}{u}=4-3$ $\Rightarrow \frac{40}{u}=1$ $\Rightarrow u=40 \mathrm{kmph}$ From equation (i), $\frac{60}{40}+\frac{20}{v}=4$ $\Rightarrow \frac{20}{v}=4-\frac{3}{2}=\frac{8-3}{2}$ $\Rightarrow \frac{20}{v}=\frac{5}{2} \Rightarrow v=\frac{20 \times 2}{5}$ $=8 \mathrm{kmph}$ $\therefore$ Speed of current $=\frac{1}{2}(u-v)$ $=\frac{1}{2}(40-8) \mathrm{kmph}=16 \mathrm{kmph}$

Let

Speed of boat = $b$ kmph

Speed of current = $w$ kmph

According to the question,

$\frac{60}{b+w}+\frac{2}{b-w}=4$

.....(i)

and $\frac{40}{b+w}+\frac{40}{b-w}=6$

$\frac{20}{b+w}+\frac{20}{b-w}=3$

From equation (i) – (ii),

$\frac{60}{b+w}-\frac{20}{b+w}=1$

$\frac{40}{b+w}=1$

$b+w=40$

Now from equation (i),

$\frac{60}{40}+\frac{20}{b-w}=4$

$\frac{20}{b-w}=4-\frac{3}{2}$

$b-w=8$

Solving equations (iii) & (iv)

b = 16 kmph