Video Explanation

Text Explanation

See this blog for more information about regular polygons. This is a particularly challenging high level Quant problem.

First of all, you must remember that the sum of the angles in any n-sided polygon equals (n – 2)*180°. Thus, the angles in a hexagon add up to 720°, and in a regular hexagon, each full angle is 720°/6 = 120°.

Now, look at triangle LMN. LM = MN, so it's an isosceles triangle. The angle at M is 120°, so the other two angles, at L & N, must each be 30°. Thus, ∠MNQ = 30°.

Of course, triangle KLM must be identical to triangle LMN, so ∠LMQ also equals 30°. Because the big angle at M, ∠LMN, is 120°, this means that

(∠QMN) = (∠LMN) – (∠LMQ) = 120° – 30° = 90°

Thus, triangle QMN has a 30° angle and a 90° angle, so the third angle, ∠MQN, must equal 60°. Answer = (D).