Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time `t_1`. On other days, if the remains stationary on the moving escalator, then the escalator takes her up in time `t_2`. The time taken by her to walk up on the moving escalator will be :

Options

A. `(t_(1)+t_(2))/(2)`
B. `(t_(1)t_(2))/(t_(2)-t_(1))`
C. `(t_(1)t_(2))/(t_(2)+t_(1))`
D. `t_(1)-t_(2)`

Answer

To solve the problem, we need to determine the time taken by Preeti to walk up the moving escalator. We will break down the solution step by step.

### Step 1: Define Variables
Let:
- \( d \) = length of the escalator
- \( t_1 \) = time taken by Preeti to walk up the stationary escalator
- \( t_2 \) = time taken by the escalator to take Preeti up when she is stationary

### Step 2: Calculate Velocities
From the information given, we can calculate the velocities:
1. When Preeti walks up the stationary escalator:
\[ v_1 = \frac{d}{t_1} \]
(This is Preeti's walking speed)

2. When the escalator moves and Preeti stands still:
\[ v_2 = \frac{d}{t_2} \]
(This is the speed of the escalator)

### Step 3: Combined Velocity
When both Preeti and the escalator are moving, the combined velocity is:
\[ v_{\text{combined}} = v_1 + v_2 = \frac{d}{t_1} + \frac{d}{t_2} \]

### Step 4: Time Taken on the Moving Escalator
The time taken \( t_3 \) for Preeti to walk up the moving escalator can be calculated using the formula:
\[ t_3 = \frac{d}{v_{\text{combined}}} \]
Substituting the expression for \( v_{\text{combined}} \):
\[ t_3 = \frac{d}{\left(\frac{d}{t_1} + \frac{d}{t_2}\right)} \]

### Step 5: Simplifying the Expression
Now, we simplify the expression for \( t_3 \):
\[ t_3 = \frac{d}{d\left(\frac{1}{t_1} + \frac{1}{t_2}\right)} = \frac{1}{\left(\frac{1}{t_1} + \frac{1}{t_2}\right)} \]
This can be rewritten as:
\[ t_3 = \frac{t_1 t_2}{t_1 + t_2} \]

### Conclusion
Thus, the time taken by Preeti to walk up the moving escalator is:
\[ t_3 = \frac{t_1 t_2}{t_1 + t_2} \]