74 (1008). A tourist who set out from $A$ to $B$ walked the first half of the distance at a speed of $v_{1}$, and the second half at a speed of $v_{2}\left(v_{1} \neq v_{2}\right)$. On the return trip, he walked half of the total time at a speed of $v_{1}$, and the other half at a speed of $v_{2}$. In which case did the tourist spend more time: when walking from $A$ to $B$ or when walking from $B$ to $A$?
Solution. I method. Let the time spent by the tourist on the journey from $B$ to $A$ be $2 t$. Then half of the time ($t$) he walked at a speed of $v_{1}$, and the second half of the time at a speed of $v_{2}$. The entire distance, therefore, is $v_{1} t + v_{2} t$, and half the distance is $\frac{v_{1} t + v_{2} t}{2}$.

The tourist spent $\frac{v_{1} t + v_{2} t}{2 v_{1}} = \frac{1}{2} + \frac{v_{2} t}{2 v_{1}}$ units of time on the first half of the distance at a speed of $v_{1}$, and $\frac{v_{1} t + v_{2} t}{2 v_{2}} = \frac{1}{2} + \frac{v_{1} t}{2 v_{2}}$ units of time on the second half of the distance at a speed of $v_{2}$. The total time spent on the journey from $A$ to $B$ is

$$
\frac{t}{2} + \frac{v_{2} t}{2 v_{1}} + \frac{t}{2} + \frac{v_{1} t}{2 v_{2}} = t + \frac{t}{2} \left( \frac{v_{2}}{v_{1}} + \frac{v_{1}}{v_{2}} \right)
$$

which is more than $2 t$, since $\frac{v_{2}}{v_{1}} + \frac{v_{1}}{v_{2}} > 2$ when $v_{1} \neq v_{2}$

$$
\left( \frac{v_{2}^{2} - 2 v_{1} v_{2} + v_{1}^{2}}{v_{1} v_{2}} = \frac{(v_{2} - v_{1})^{2}}{v_{1} v_{2}} > 0 \text{ when } v_{1} > 0, v_{2} > 0, v_{1} \neq v_{2} \right)
$$

Thus, the tourist spent more time on the journey from $A$ to $B$ than on the journey from $B$ to $A$.

II method. Let the distance $A B$ be denoted by $s$. Then the time spent by the tourist on the journey from $A$ to $B$ is $t_{1} = \frac{s}{2 v_{1}} + \frac{s}{2 v_{2}} = \frac{s}{2} \cdot \frac{v_{1} + v_{2}}{v_{1} \cdot v_{2}}$. If the time spent by the tourist on the return journey is denoted by $t_{2}$, then $s = v_{1} \frac{t_{2}}{2} + v_{2} \frac{t_{2}}{2}$, from which $t_{2} = \frac{2 s}{v_{1} + v_{2}}$. Let's compare $t_{1}$ and $t_{2}$.

$$
\frac{s}{2} \cdot \frac{v_{1} + v_{2}}{v_{1} v_{2}} \vee \frac{2 s}{v_{1} + v_{2}}
$$

Since $s > 0$, by canceling $s$, we get:

$$
\begin{gathered}
\frac{v_{1} + v_{2}}{2 v_{1} v_{2}} \vee \frac{2}{v_{1} + v_{2}} \\
(v_{1} + v_{2})^{2} \bigvee 4 v_{1} v_{2}, \\
(v_{1} - v_{2})^{2} > 0
\end{gathered}
$$

Therefore, $t_{1} > t_{2}$.

Answer: the tourist spent more time on the journey from $A$ to $B$ than on the journey from $B$ to $A$.

Remark. This problem can be given in a more general formulation: which is faster - to walk half the distance at a speed of $v_{1}$ and the other half at a speed of $v_{2} (v_{1} \neq v_{2})$ or to walk half the total time at a speed of $v_{1}$ and the other half at a speed of $v_{2}$?

$75(1009)$. One dump truck can transport construction materials 24 hours faster than the other. If the first dump truck transports two-thirds of all materials and then the second transports the remaining part, it will take 33 hours longer than if both dump trucks work simultaneously. How many hours can each dump truck transport the construction materials?

Solution. Let the first dump truck transport the construction materials in $t$ hours, then the second in $(t + 24)$ hours. In 1 hour, the first dump truck can transport $\frac{1}{t}$ of the materials, and the second in 1 hour $\frac{1}{t + 24}$ of the materials. When working simultaneously, in 1 hour the dump trucks can transport $\frac{2 t + 24}{t(t + 24)}$ of the materials $\left(\frac{1}{t} + \frac{1}{t + 24}\right)$, spending $\frac{t(t + 24)}{2 t + 24}$ hours on the entire work $\left(1 : \frac{2 t + 24}{t(t + 24)} = \frac{t(t + 24)}{2 t + 24}\right)$.

The first dump truck can transport $\frac{2}{3}$ of all materials in $\frac{2 t}{3}$ hours $\left(\frac{2}{3} : \frac{1}{t} = \frac{2}{3} t\right)$, and the remaining part $\left(\frac{1}{3}\right)$ the second dump truck can transport in $\frac{t + 24}{3}$ hours $\left(\frac{1}{3} : \frac{1}{t + 24} = \frac{t + 24}{3}\right)$. According to the problem, we form the equation

$$
\frac{2 t}{3} + \frac{t + 24}{3} - 33 = \frac{t(t + 24)}{2 t + 24}
$$

from which

$$
\begin{gathered}
t + 8 - 33 = \frac{t(t + 24)}{2 t + 24} \\
(t - 25)(2 t + 24) = t(t + 24) \\
t^{2} - 50 t - 600 = 0 \\
t_{1} = 60, t_{2} = -10
\end{gathered}
$$

136

The only positive root of the equation that satisfies the problem is $t = 60$. Therefore, the first dump truck can transport the construction materials in 60 hours, and the second in 84 hours $(60 + 24 = 84)$.

Answer: 60 hours, 84 hours.