A person climbs up a stalled escalator in 60s. If standing on the same but escalator running with constant velocity, he takes 40 s. How much time is taken by the person to walk up the moving escalator?

Options

A. 37 s
B. 27 s
C. 24 s
D. 45 s

Explanation

To solve the problem, we need to find out how much time the person takes to walk up the moving escalator. Let's break it down step by step.

### Step 1: Define Variables
- Let \( L \) be the length of the escalator.
- Let \( V_m \) be the speed of the person when climbing the escalator.
- Let \( V_e \) be the speed of the escalator.
- Let \( t' \) be the time taken by the person to walk up the moving escalator.

### Step 2: Calculate Speeds
1. When the escalator is stalled (not moving), the person takes 60 seconds to climb it:
\[ V_m = \frac{L}{60} \]

2. When the escalator is moving, the person takes 40 seconds to reach the top:
\[ V_e = \frac{L}{40} \]

### Step 3: Determine the Total Speed on the Moving Escalator
When the person walks up the moving escalator, the effective speed is the sum of the person's speed and the escalator's speed:
\[ V_{total} = V_m + V_e \]

### Step 4: Substitute the Speeds
Substituting the expressions for \( V_m \) and \( V_e \):
\[ V_{total} = \frac{L}{60} + \frac{L}{40} \]

### Step 5: Find a Common Denominator
To add these fractions, we need a common denominator. The least common multiple of 60 and 40 is 120:
\[ V_{total} = \frac{2L}{120} + \frac{3L}{120} = \frac{5L}{120} = \frac{L}{24} \]

### Step 6: Calculate the Time Taken on the Moving Escalator
Using the formula for speed, we can find the time taken \( t' \) to walk up the moving escalator:
\[ t' = \frac{L}{V_{total}} = \frac{L}{\frac{L}{24}} = 24 \text{ seconds} \]

### Final Answer
The time taken by the person to walk up the moving escalator is **24 seconds**.

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