Reaction of glucose acetal with acetic anhydride
I think there's nothing to discussed about this further as all comments directing to the correct answer: $x = 4, \ y = 6,$ and $ z = 5.$ I think it's better show in the structures:

Glucose is naturally occurring in pyranose hemiacetal form. When subjected to react with $\ce{H+/R-OH}$, the remaining acetal hydroxy group (anomeric hydroxyl group) will undergoes alkylation. Thus, the first scheme is showing alkylation of this anomeric hydroxyl group by methanol (Ref.1). The resultant methylglucose has four free hydroxy group, hence, $x = 4$.
When subjected to reduction with suitable conditions (e.g., $\ce{Ni/H2}$, $\ce{Ru/C/H2}$, etc.), the hemiacetal group would be reduced to a $\ce{-CH2OH}$ group (second scheme in the image). The resultant product is sorbitol (Ref.2), which has six free hydroxy groups, hence, $y = 6$.
Finally, pyranose glucose structure has five free hydroxy groups, hence, $z = 5$.

References:

B. Helferich, W. Schäfer (Checked by H. T. Clarke and M. R. Brethen), &quot;$\alpha$-methyl d-Glucoside,&quot; Org. Synth. 1926, 6, 64 (DOI: 10.15227/orgsyn.006.0064).
J. Wildschut, J. Arentz, C.B. Rasrendra, R.H. Venderbosch, H.J. Heeres, &quot;Catalytic Hydrotreatment of Fast Pyrolysis Oil: Model Studies on Reaction Pathways for the Carbohydrate Fraction,&quot; Environmental Progress &amp; Sustainable Energy 2009, 28(3), 450-460 (https://doi.org/10.1002/ep.10390).