$A$ cubic solid is made of two elements $P$ and $Q$. Atoms of $Q$ are at the corners of the cube and $P$ at the body-centre. What is the formula of the compound? What are the coordination numbers of $P$ and $Q$?

(N/A) It is given that the atoms of $Q$ are present at the corners of the cube.
Therefore,the number of atoms of $Q$ in one unit cell $= 8 \times \frac{1}{8} = 1$.
It is also given that the atoms of $P$ are present at the body-centre.
Therefore,the number of atoms of $P$ in one unit cell $= 1$.
This means that the ratio of the number of $P$ atoms to the number of $Q$ atoms is $P:Q = 1:1$.
Hence,the formula of the compound is $PQ$.
In this structure (body-centered cubic),each $P$ atom at the body-centre is surrounded by $8$ $Q$ atoms at the corners,and each $Q$ atom at a corner is surrounded by $8$ $P$ atoms at the body-centres of adjacent unit cells.
Thus,the coordination number of both $P$ and $Q$ is $8$.