Three particles of masses 100 g, 150 g and 200 g are placed at the vertices O(0, 0), A(0.5, 0) and B(0.25, 0.25√3) of an equilateral triangle of side 0.5 m. Find the x-coordinate of the centre of mass of the system.

X = (m₁x₁ + m₂x₂ + m₃x₃)/(m₁+m₂+m₃) = [100(0) + 150(0.5) + 200(0.25)] g·m / (450 g) = (75 + 50) / 450 m = 125/450 m = **5/18 m** ≈ 0.278 m. Distractor 1/6 m drops the 200 g contribution; 1/4 m is the geometric centroid x-coordinate (equal masses); 5/9 m comes from mistakenly using 250 in the numerator instead of 125.