Mr. Bean went to a shopping mall where he rode an escalator. Once he moved exactly half the number of stairs on the escalator with the help of the moving stairs only, he realised that his cat ‘Kitty’ was unable to ride the escalator and was waiting for his help at the base of the escalator. So, Mr. Bean came down at thrice the speed of the moving escalator, picked up ‘Kitty’ in no time, and reached the top of the escalator with the help of the escalator alone. He took 63 seconds in this whole process to reach the top. How much time (in seconds) would it have taken for Mr. Bean to reach the top if he had not had to come down to pick ‘Kitty’?

To solve this problem, let’s break it down step by step, using the following key variables:

- Let \( N \) be the total number of stairs on the escalator.
- Let \( v_e \) be the speed of the escalator (in stairs per second).
- Let \( v_b \) be the speed of Mr. Bean when moving by himself, i.e., when he moves without the escalator.

We are given the following information:
1. Mr. Bean covers half the escalator with the help of the moving stairs only, so he moves half the distance at the speed of the escalator, \( v_e \).
2. Mr. Bean comes back down at three times the speed of the escalator, i.e., at a speed of \( 3v_e \).
3. Mr. Bean picks up his cat and then rides the escalator all the way to the top with the help of the escalator alone (at speed \( v_e \)).
4. The entire process took 63 seconds.

We aim to determine how long it would have taken Mr. Bean to reach the top without coming down to pick up his cat.

### Step-by-Step Solution:

1. **Time to move up halfway using only the escalator**:
   Since Mr. Bean moves halfway up the escalator only using the escalator's movement, the time taken for this is:
   \[
   t_1 = \frac{N/2}{v_e} = \frac{N}{2v_e}
   \]

2. **Time to come back down at thrice the speed of the escalator**:
   When Mr. Bean comes down, he moves at a speed of \( 3v_e \). The distance he has to travel is \( N/2 \), and the time taken for this is:
   \[
   t_2 = \frac{N/2}{3v_e} = \frac{N}{6v_e}
   \]

3. **Time to move up the entire escalator with the cat**:
   After picking up his cat, Mr. Bean moves the entire length of the escalator while standing still, letting the escalator carry him. The time taken for this is:
   \[
   t_3 = \frac{N}{v_e}
   \]

4. **Total time for the entire process**:
   The total time for the entire process is the sum of these three times, which is given as 63 seconds:
   \[
   t_1 + t_2 + t_3 = 63
   \]
   Substituting the expressions for \( t_1 \), \( t_2 \), and \( t_3 \):
   \[
   \frac{N}{2v_e} + \frac{N}{6v_e} + \frac{N}{v_e} = 63
   \]

5. **Solving the equation**:
   Combine the terms on the left-hand side:
   \[
   \frac{N}{2v_e} + \frac{N}{6v_e} + \frac{N}{v_e} = \frac{3N}{6v_e} + \frac{N}{6v_e} + \frac{6N}{6v_e} = \frac{10N}{6v_e}
   \]
   Thus, the equation becomes:
   \[
   \frac{10N}{6v_e} = 63
   \]
   Simplifying:
   \[
   \frac{5N}{3v_e} = 63
   \]
   Solving for \( N / v_e \):
   \[
   \frac{N}{v_e} = \frac{63 \times 3}{5} = 37.8
   \]

6. **Time if Mr. Bean had not come down**:
   If Mr. Bean had not come down, he would have traveled the entire length of the escalator with its help. The time taken for this would simply be:
   \[
   t_{\text{no down trip}} = \frac{N}{v_e}
   \]
   From our earlier calculation, \( \frac{N}{v_e} = 37.8 \) seconds.

### Final Answer:
If Mr. Bean had not come down to pick up his cat, it would have taken him 37.8 seconds to reach the top of the escalator.

Would you like more details or have any questions?

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Here are some related questions to explore further:
1. How would the answer change if Mr. Bean's speed down the escalator were different?
2. What would happen if the escalator speed increased by 50%?
3. Can this problem be solved using a different method, like calculus?
4. How can we find Mr. Bean's speed relative to a stationary escalator?
5. What would be the time taken if the escalator were twice as long?

**Tip:** Problems involving relative motion often require analyzing movement separately and summing individual times.

Solve a relative motion problem involving Mr. Bean on an escalator. Learn how to calculate the time it would take for Mr. Bean to reach the top if he hadn’t come back down for his cat. This problem involves key algebraic concepts, including time and speed relations, and is suitable for Grades 9-11.

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