a) 8 min 17 second
b) 10 min 57 second
c) 14 min 34 second
d) 12 min 23 second

Answer: b
Explanation:

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y
$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x} \cr & \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow h = x\,……\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AB}}{{\left( {BC + CD} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{x + y}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}} \cr & x + y = \sqrt 3 \,h \cr & y = \sqrt 3 \,h – x \cr} $$
$$ y = \sqrt 3 \,h – h$$     ( Substituted the value of x from equation 1 )
$$ y = h\left( {\sqrt 3 – 1} \right)$$
Given that distance y is covered in 8 minutes.
i.e, distance $$h\left( {\sqrt 3 – 1} \right)$$   is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
$$\eqalign{ & h\left( {\sqrt 3 – 1} \right) \propto 8\,……\left( {\text{A}} \right) \cr & h \propto t\,…………..\left( {\text{B}} \right) \cr & \frac{{\left( {\text{A}} \right)}}{{\left( {\text{B}} \right)}} \Rightarrow \frac{{h\left( {\sqrt 3 – 1} \right)}}{h} = \frac{8}{t} \cr & \Rightarrow \left( {\sqrt 3 – 1} \right) = \frac{8}{t} \cr & \Rightarrow t = \frac{8}{{\left( {\sqrt 3 – 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{\left( {1.73 – 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{.73}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{800}}{{73}}\,{\text{minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 10\frac{{70}}{{73}}\,{\text{minutes}} \cr & \approx {\text{10}}\,{\text{minutes}}\,{\text{57}}\,{\text{seconds}} \cr} $$