This problem involves two people, A and B, using an escalator where A is descending and B is ascending simultaneously. The speeds of A and B are given relative to each other, and we need to determine the number of visible steps on the escalator when it is stationary.

Step-by-step solution:

Step 1: Define the variables

Let:

• $N$ be the total number of steps on the escalator when it is stationary.
• $v$ be the speed of the escalator in terms of steps per unit time.
• $s_A$ be the speed at which person A descends (in steps per unit time).
• $s_B$ be the speed at which person B ascends (in steps per unit time).
• $t_A$ be the time taken by A to reach the ground floor.
• $t_B$ be the time taken by B to reach the first floor.

From the problem statement:

• The speed of B is twice the speed of A: $s_B = 2s_A$.
• A takes 30 steps, and B takes 120 steps.

Step 2: Express the relationship between time, steps, and speed

For person A (who is descending):

• In time $t_A$, A takes 30 steps and the escalator moves downward by $vt_A$ steps.
• The total number of steps $N$ is covered by A's 30 steps plus the steps moved by the escalator: $N = 30 + vt_A$

For person B (who is ascending):

• In time $t_B$, B takes 120 steps and the escalator moves downward by $vt_B$ steps.
• Since B is moving against the escalator, the total number of steps $N$ is covered by B's 120 steps minus the steps moved by the escalator: $N = 120 - vt_B$

Step 3: Relate the times for A and B

Since the speeds of A and B are given relative to each other: $t_B = \frac{t_A}{2}$ because B is moving twice as fast as A, they reach their destinations in half the time.

Step 4: Solve the system of equations

From the two expressions for $N$: $N = 30 + vt_A$ and $N = 120 - vt_B.$ Substitute $t_B = \frac{t_A}{2}$ into the second equation: $N = 120 - v\left(\frac{t_A}{2}\right).$ Now we have two equations:

1. $N = 30 + vt_A$
2. $N = 120 - \frac{vt_A}{2}$

Equating these two: $30 + vt_A = 120 - \frac{vt_A}{2}.$

Step 5: Simplify and solve for $vt_A$

Multiply through by 2 to eliminate the fraction: $60 + 2vt_A = 240 - vt_A.$ Combine like terms: $3vt_A = 180.$ Solve for $vt_A$: $vt_A = 60.$

Step 6: Find $N$

Substitute $vt_A = 60$ into the first equation for $N$: $N = 30 + 60 = 90.$

Thus, the total number of visible steps on the escalator when stationary is $\boxed{90}$.

Conclusion:

The possible number of steps visible on the escalator when it is stationary is 90.

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Do you need more details on any of the steps? Here are some additional questions you might explore:

1. What happens if A and B move at the same speed instead of different speeds?
2. How would the time taken by A and B change if the escalator's speed was doubled?
3. What if the escalator were stationary—how many steps would A and B take then?
4. Can we generalize the relationship between the number of steps and the speeds for different ratios between A and B's speeds?
5. How would the result change if A's speed were thrice that of B instead of half?

Tip:

Whenever solving such problems, focus on clearly defining relationships between speed, time, and distance/steps to simplify the process of setting up equations.