Three fine bars in length L are arranged in the shape of inverted U as the figure below. The two sidebars have mass M and the central bar mass 3M. What is the location of the center of mass of the set?

The CM of the system will be the CM of the three separate pieces. So set up a coordinate system (for reference) at the lower left corner of the diagram. the CM of the left leg is at (0, L/2), the CM of the top bar is at (L/2, L), and the CM of the right leg is at (L, L/2). Now use the CM formula:

\(\displaystyle r_{CM} = \frac{\sum_{n = 1}^N r_im_i}{\sum_{n = 0}^N m_i}\)

in both coordinate directions.

So I'm getting

\(\displaystyle x_{CM} = \frac{0 \cdot M + \frac{L}{2} \cdot 3M + L \cdot M}{M + 3M + M} = \frac{L}{2}\)

as you might have expected and

\(\displaystyle y_{CM} = \frac{\frac{L}{2} \cdot M + L \cdot 3M + \frac{L}{2} \cdot M}{M + 3M + M} = \frac{4L}{5}\)

This is what arbolis told you.

-Dan