Solution:

Let’s break this problem down carefully. We know:

1. Sharat takes 40 steps to reach the bottom.
2. Chandra takes two steps for every one step taken by Sharat.
3. The time taken by both to complete their respective journeys is the same.

Let’s analyze the situation to figure out how many steps more Chandra takes before they cross each other.

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Key Variables:

• Let the speed of the escalator be $E$ (in steps per second).
• Sharat’s step rate = $S$ (steps per second).
• Chandra’s step rate = $2S$ (since she takes 2 steps for every 1 step Sharat takes).

The total steps on the escalator (without either walking) would be the same for both travelers, since the escalator is running at the same speed for both.

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Total Time for the Journey:

• Sharat’s total time to reach the bottom:
  $T = \frac{40}{S + E}$
• Chandra’s total time to reach the top:
  $T = \frac{40}{2S - E}$ Since the time is the same for both, we equate the two equations: $\frac{40}{S + E} = \frac{40}{2S - E}$

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Solving for the Escalator's Speed:

We can cancel 40 from both sides: $\frac{1}{S + E} = \frac{1}{2S - E}$ Cross-multiplying: $2S - E = S + E$ Rearrange the equation: $2S - S = E + E \quad \Rightarrow \quad S = 2E$ So, the speed of the escalator is half of Sharat’s walking speed:
$E = \frac{S}{2}$

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Finding the Meeting Point:

Let’s assume they meet after t seconds. In these $t$ seconds:

• Sharat will have walked: $S \cdot t$ steps.
• Chandra will have walked: $2S \cdot t$ steps.
• The escalator will have moved: $E \cdot t = \frac{S}{2} \cdot t$ steps.

So, the total distance traveled on the escalator by both at the meeting point is: $\text{Total steps covered by Sharat} = S \cdot t + \frac{S}{2} \cdot t = \frac{3S}{2} \cdot t$ $\text{Total steps covered by Chandra} = 2S \cdot t + \frac{S}{2} \cdot t = \frac{5S}{2} \cdot t$

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Difference in Steps:

The difference in steps taken by Chandra and Sharat before they meet: $\text{Difference} = \left(\frac{5S}{2} \cdot t\right) - \left(\frac{3S}{2} \cdot t\right) = S \cdot t$ This means Chandra takes exactly $S \cdot t$ more steps than Sharat before they meet.

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Conclusion:

Since the time $t$ is the same for both travelers, and $S \cdot t$ represents the extra steps taken by Chandra, Chandra takes exactly as many extra steps as Sharat’s step rate in the time they cross each other.

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Would you like more details or a specific part explained further?

Here are 5 follow-up questions:

1. What if Sharat walked at twice his original speed?
2. How would the problem change if the escalator was stationary?
3. What happens if both travelers have identical walking speeds?
4. How can this problem be visualized graphically?
5. What is the effect of the escalator’s speed on their meeting point?

Tip: Problems involving moving walkways or escalators can often be solved using relative velocity concepts.