[Algebra]
Problem: Let $T=10$. Ann spends 80 seconds climbing up a $T$ meter rope at a constant speed, and she spends 70 seconds climbing down the same rope at a constant speed (different from her upward speed). Ann begins climbing up and down the rope repeatedly, and she does not pause after climbing the length of the rope. After $T$ minutes, how many meters will Ann have climbed in either direction?

Solution: In 150 seconds (or 2.5 minutes), Ann climbs up and down the entire rope. Thus in $T$ minutes, she makes $\left\lfloor\frac{T}{2.5}\right\rfloor$ round trips, and therefore climbs $2 T\left\lfloor\frac{T}{2.5}\right\rfloor$ meters. After making all her round trips, there are $t=60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)$ seconds remaining. If $t \leq 80$, then the remaining distance climbed is $T \cdot \frac{t}{80}$ meters, and if $t>80$, then the distance climbed is $T+T \cdot\left(\frac{t-80}{70}\right)$ meters. In general, the total distance in meters that Ann climbs is

$$
2 T\left\lfloor\frac{T}{2.5}\right\rfloor+T \cdot \min \left(1, \frac{60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)}{80}\right)+T \cdot \max \left(0, \frac{60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)-80}{70}\right) .
$$

With $T=10$, Ann makes exactly 4 round trips, and therefore climbs a total of $4 \cdot 2 \cdot 10=\mathbf{8 0}$ meters.

Final answer: 80