7.3. In the entrance, Lera rides the elevator, while Lada walks down (the speeds of the elevator and Lada are constant). Once, when they were descending from the 4th floor to the 1st, Lada was faster and waited for Lera at the bottom for some time. Another time, the girls were descending from the 8th floor, and Lada doubled her speed, as a result of which she waited at the bottom three times longer than the first time. How many times is Lada's initial speed greater than the speed of the elevator?
Answer: In $11 / 4$ times.

Solution: Let $U, V$ be the speeds of Lada and the elevator, respectively, and $S$ be the distance between two floors. Then, the first time the girls covered a distance of 3S, so Lada waited at the bottom for $(3 S / V) - (3 S / U)$. Similarly, the second time it was $(7S/V) - (7S/2U)$. Since the second number is three times the first, we can set up the equation:

$$
\begin{gathered}
3(3 S / V - 3 S / U) = (7 S / V - 7 S / 2 U), \\
9 S / V - 9 S / U = 7 S / V - 7 S / 2 U, \\
18 / V - 18 / U = 14 / V - 7 / U, \\
18 U - 18 V = 14 U - 7 V, \\
4 U = 11 V \\
U = 11 / 4 V
\end{gathered}
$$

That is, Lada's speed is $11 / 4$ times greater.

Criteria: Answer, answer with verification - 0 points. Correctly set up equation, nothing else done - 3 points.