The ratio of close-packed atoms to tetrahedral holes in cubic close packing is

$A$ solid has an $hcp$ lattice. Atoms of $Z$ (anions) form the $hcp$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids in the lattice. Atoms of $Y$ (cations) occupy half of the tetrahedral voids. What is the molecular formula of the solid?

The number of tetrahedral voids present in $0.5 \ mole$ of $hcp$ crystal structure is:

$A$ compound is formed by $X$ and $Y$ elements. Atoms of $Y$ (anions) form $hcp$ lattice. Atoms of $X$ (cations) are in some octahedral holes. The formula of the compound is $XY_3$. What is the fraction of octahedral holes unoccupied by $X$?

$A$ compound is formed by elements $P$ and $Q$. Atoms of $Q$ are in $ccp$ arrangement. If $P$ occupies all the tetrahedral voids,what is the formula of the compound?

Consider the $bcc$ unit cells of the solid $1$ and $2$ with the position of atoms as shown below. The radius of atom $B$ is twice that of atom $A$. The unit cell edge length is $50\%$ more in solid $2$ than in $1$. What is the approximate packing efficiency in solid $2$? $........... \%$