We unpack 187 books in a box, which weigh a total of $189 \mathrm{~kg}$. The average weight of large books is $2.75 \mathrm{~kg}$, medium-sized books weigh $1.5 \mathrm{~kg}$ on average, and small books weigh $\frac{1}{3} \mathrm{~kg}$ on average. How many books are there of each size, if the total weight of the large books is the greatest and the total weight of the small books is the smallest?
If we denote the number of books by $x, y, z$, then according to the problem,

$$
x+y+z=187
$$

and

$$
\frac{11}{4} x+\frac{3}{2} y+\frac{1}{3} z=-189
$$

which is equivalent to

and

$$
\begin{gathered}
33 x+18 y+4 z=2268 \\
4 x+4 y+4 z=748
\end{gathered}
$$

Subtracting $\left(1^{\prime}\right)$ from $\left(2^{\prime}\right)$

$$
\begin{gathered}
29 x+14 y=1520 \\
y=\frac{1520-29 x}{14}=108-2 x+\frac{8-x}{14}=108-2 x+t
\end{gathered}
$$

Thus,

$$
x=8-14 t, \quad y=108-16+28 t+t=92+29 t, \quad z=187-x-y=87-15 t
$$

Considering the constraints in the problem,

$$
\frac{11}{4}(8-14 t)>\frac{3}{2}(92+29 t)
$$

and

$$
\frac{3}{2}(92+29 t)>\frac{1}{3}(87-15 t)
$$

From (3),

From (4),

$$
44-77 t>276+87 t, \quad \text { which implies } \quad-\frac{232}{164}>t .
$$

Thus,

$$
828+261 t>174-30 t, \quad \text { which implies } \quad t>-\frac{654}{291}
$$

$$
-\frac{654}{291}<t<-\frac{232}{164}
$$

which implies

$$
t=-2
$$

and thus

$$
x=8-14 t=36, \quad y=92+29 t=34, \quad z=87-15 t=117
$$

József Szabados (Bp. III., Árpád g. II. o.t.)