Correct Answer: 27. (3)

Solution: 27. (3) $d$ : downstream,

$u$ : upstream

$\mathrm{V}_{\mathrm{u}}=\frac{15 \mathrm{km}}{\left(\frac{5}{4}\right) h r}=12 \mathrm{km} / \mathrm{h}$ $\quad$ ……..(i)

$\frac{t_{d}}{t_{u}}=\frac{4}{5}$

$\therefore \frac{V_{d}}{V_{u}}=\frac{\frac{S_{d}}{t_{d}}}{\frac{S_{u}}{t_{u}}}=\frac{\frac{S_{d}}{S_{u}}}{\frac{t_{d}}{t_{u}}}=\frac{\frac{4}{3}}{\frac{4}{5}}$

$=\frac{5}{3}$

Let $V_{\mathrm{s}}$ be the speed of the boatman in still water and $V_{c}$ the speed of the current

$\Rightarrow \frac{V_{s}+V_{c}}{V_{s}-V_{c}}=\frac{5}{3}$

Using reverse componendo - dividendo, we get

$\frac{V_{s}}{V_{c}}=\frac{4}{1} \Rightarrow V_{\mathrm{s}}=4 V_{\mathrm{c}}$

Now $V_{\mathrm{u}}=V_{\mathrm{s}}-V_{\mathrm{c}}=4 V_{\mathrm{c}}-V_{\mathrm{c}}$

$=3 \mathrm{V}_{\mathrm{c}}=12 \mathrm{km} / \mathrm{hr} \quad $ [from (i)]

$\therefore V_{\mathrm{c}}=4 \mathrm{km} / \mathrm{hr}$