RRB Assistant Loco Pilot and Technician Model Paper 6 with solutions

$\frac{{\mathit{x}}_1+{\mathit{x}}_2}{2}=1$ ⇒ ${\mathit{x}}_1+{\mathit{x}}_2=2$ ...(i)

$\frac{{\mathit{y}}_1+{\mathit{y}}_2}{2}=1$
⇒ ${\mathit{y}}_1+{\mathit{y}}_2=2$   ...(ii)
$\frac{{\mathit{x}}_2+{\mathit{x}}_3}{2}=3$
⇒ ${\mathit{x}}_2+{\mathit{x}}_3=6$   ...(iii)
$\frac{{\mathit{y}}_2+{\mathit{y}}_3}{2}=-3$
⇒ ${\mathit{y}}_2+{\mathit{y}}_3=-6$   ...(iv)
$\frac{{\mathit{x}}_3+{\mathit{x}}_1}{2}=4$
⇒ ${\mathit{x}}_1+{\mathit{x}}_3=8$   ...(v)
$\frac{{\mathit{y}}_3+{\mathit{y}}_1}{2}=5$
⇒ ${\mathit{y}}_1+{\mathit{y}}_3=10$   ...(vi)
Adding (i), (iii), (v), we get
$2\left({\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3\right)=16$
∴ ${\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3=8$
∴ $\frac{{\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3}{3}=\frac{8}{3}$
Similarly, adding (ii), (iv) and (vi)
$2\left({\mathit{y}}_1+{\mathit{y}}_2+{\mathit{y}}_3\right)=6$
∴ ${\mathit{y}}_1+{\mathit{y}}_2+{\mathit{y}}_3=3$
∴ $\frac{{\mathit{y}}_1+{\mathit{y}}_2+{\mathit{y}}_3}{3}=\frac{3}{3}=1$
Hence, the co-ordinates of centroid is $\left(\frac{8}{3},1\right)$