Question

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 she 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

• A. $\frac{t_1 t_2}{t_2 - t_1}$
• B. $\frac{t_1 t_2}{t_2 + t_1}$
• C. $t_1 - t_2$
• D. $ \frac{t_1 + t_2}{2} $

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between Preeti's walking speed on the escalator, the speed of the escalator, and the resultant speed when both are combined.

1. Let the length of the escalator be L.
2. When Preeti walks up the stationary escalator, she covers the distance L in t_1 time. Therefore, her speed v_p is given by: v_p = \frac{L}{t_1}.
3. If she remains stationary and the escalator is moving, it covers the same distance L in t_2 time. Thus, the speed of the escalator v_e is: v_e = \frac{L}{t_2}.
4. When Preeti walks on the moving escalator, both the speeds add up. Therefore, the resultant speed is: v = v_p + v_e = \frac{L}{t_1} + \frac{L}{t_2}.
5. The time taken t to travel the length L with this resultant speed is given by: t = \frac{L}{v} = \frac{L}{\frac{L}{t_1} + \frac{L}{t_2}}.
6. Simplifying the expression, we have: t = \frac{1}{\frac{1}{t_1} + \frac{1}{t_2}} = \frac{t_1 t_2}{t_1 + t_2}.

Thus, the time taken by Preeti to walk up on the moving escalator is \frac{t_1 t_2}{t_1 + t_2}, which corresponds to the correct answer choice.