a) 26.28 km/hr
b) 32.42 km/hr
c) 24.22 km/hr
d) 31.25 km/hr

Answer: a
Explanation:

Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m
$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & 1 = \frac{{AB}}{{100}} \cr & AB = 100\,……\left( {eq:1} \right) \cr} $$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$ \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$     ( Substituted the value of AB from equation 1)
$$ BD = 100\sqrt 3 $$
$$\eqalign{ & CD = \left( {BD – BC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 – 100} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 – 1} \right) \cr} $$
It is given that the distance CD is covered in 10 seconds.
i.e., the distance $$100\left( {\sqrt 3 – 1} \right)$$     is covered in 10 seconds.
$$\eqalign{ & {\text{Required}}\,{\text{speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{100\left( {\sqrt 3 – 1} \right)}}{{10}} \cr & = 10\left( {1.73 – 1} \right) \cr & = 7.3\,{\text{meter/seconds}} \cr & = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = 26.28\,{\text{km/hr}} \cr} $$