CAT 2003 Slot 2 — QA Question 36

Let x be the distance from the later position of the car and the tower (i.e. when the angle of elevation was 60°).

Since the triangle formed (i.e. ∆ABD) is a 30°-60°-90° triangle, we have,

height of the tower, h = x$\sqrt{3}$

Now, since the triangle formed by the initial position of the car (i.e. ∆ABC) is an isosceles triangle, AB = BC

i.e. BC = $x\sqrt{3}$

∴ DC = x$\sqrt{3}$ - x = x($\sqrt{3}$ - 1)

Time taken to travel distance DC is 10 minutes, thus,

Speed $s=\frac{x(\sqrt{3}-1)}{10}$

Time taken to travel distance x = $\frac{x}{{\displaystyle \frac{x(\sqrt{3}-1)}{10}}}=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

Hence, option (a).