Let students in X, Y, Z be $a, b, c$. From (X+Y): $\frac{83a + 76b}{a+b} = 79 \Rightarrow 83a + 76b = 79a + 79b \Rightarrow 4a = 3b \Rightarrow b = \frac{4}{3}a$.
From (Y+Z): $\frac{76b + 85c}{b+c} = 81 \Rightarrow 76b + 85c = 81b + 81c \Rightarrow -5b + 4c = 0 \Rightarrow c = \frac{5}{4}b = \frac{5}{4} \times \frac{4}{3}a = \frac{5}{3}a$.
Total average = $\frac{83a + 76(\frac43 a) + 85(\frac53 a)}{a + \frac43 a + \frac53 a} = \frac{83a + \frac{304}{3}a + \frac{425}{3}a}{a + \frac43 a + \frac53 a}$. Numerator = $\frac{249 + 304 + 425}{3}a = \frac{978}{3}a = 326a$. Denominator = $a + 1.333a + 1.667a = 4a$. Average = $\frac{326a}{4a} = 81.5$.