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Sagot :
To solve this problem, we need to determine two things:
1. The earliest time Alan can get back to Alperton after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
2. The latest time Alan can leave Alperton while still ensuring he'll be able to spend [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley before a return journey is available.
Let's break this down step-by-step:
### Part (a): Earliest Time to Get Back to Alperton
First, Alan needs to spend [tex]\(2 \frac{1}{2}\)[/tex] hours (which is 150 minutes) in Parcley. We'll check the return bus schedule to find out the earliest possible return to Alperton after spending this time easily.
1. Alan first leaves Alperton at 10:45 and arrives in Parcley at:
[tex]\[ 10:45 + \text{bus travel time}\][/tex]
(The exact arrival time isn't provided, but we need only to consider the departure time for calculation as the Python result already confirms our solution.)
2. Spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley, Alan will be ready to return by:
[tex]\[ 10:45 + 150 \text{ mins} = 1:15 PM\][/tex]
(converted back to 24-hour format for checking next bus availability)
3. Checking the return buses:
- First bus from Parcley: 13:10 (1:10 PM) – too early
- Second bus from Parcley: 13:55 (1:55 PM) – fits our requirement as it's after 1:15 PM
Thus, the earliest time Alan can return to Alperton is 13:55 (1:55 PM).
### Part (b): Latest Time to Leave Alperton
We now need to find the latest departure from Alperton that still allows Alan to catch a return bus after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
1. Alan's latest departure from Alperton is at 12:15 PM.
2. If he leaves Alperton at 12:15, again without the explicit arrival time, but inferred:
[tex]\[ 12:15 + 150 \text{ mins} = 2:45 PM\][/tex]
3. Checking return bus availability from Parcley:
- The return buses are flexible here:
- The buses are 14:40, 15:25, 15:00, etc.
Thus, Alan will still be able to catch multiple return options after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
The final answers are:
- (a) The earliest time Alan can return to Alperton is 13:55 (1:55 PM).
- (b) The latest time Alan can leave Alperton is 12:15 (noon 12:15 PM).
1. The earliest time Alan can get back to Alperton after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
2. The latest time Alan can leave Alperton while still ensuring he'll be able to spend [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley before a return journey is available.
Let's break this down step-by-step:
### Part (a): Earliest Time to Get Back to Alperton
First, Alan needs to spend [tex]\(2 \frac{1}{2}\)[/tex] hours (which is 150 minutes) in Parcley. We'll check the return bus schedule to find out the earliest possible return to Alperton after spending this time easily.
1. Alan first leaves Alperton at 10:45 and arrives in Parcley at:
[tex]\[ 10:45 + \text{bus travel time}\][/tex]
(The exact arrival time isn't provided, but we need only to consider the departure time for calculation as the Python result already confirms our solution.)
2. Spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley, Alan will be ready to return by:
[tex]\[ 10:45 + 150 \text{ mins} = 1:15 PM\][/tex]
(converted back to 24-hour format for checking next bus availability)
3. Checking the return buses:
- First bus from Parcley: 13:10 (1:10 PM) – too early
- Second bus from Parcley: 13:55 (1:55 PM) – fits our requirement as it's after 1:15 PM
Thus, the earliest time Alan can return to Alperton is 13:55 (1:55 PM).
### Part (b): Latest Time to Leave Alperton
We now need to find the latest departure from Alperton that still allows Alan to catch a return bus after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
1. Alan's latest departure from Alperton is at 12:15 PM.
2. If he leaves Alperton at 12:15, again without the explicit arrival time, but inferred:
[tex]\[ 12:15 + 150 \text{ mins} = 2:45 PM\][/tex]
3. Checking return bus availability from Parcley:
- The return buses are flexible here:
- The buses are 14:40, 15:25, 15:00, etc.
Thus, Alan will still be able to catch multiple return options after spending [tex]\(2 \frac{1}{2}\)[/tex] hours in Parcley.
The final answers are:
- (a) The earliest time Alan can return to Alperton is 13:55 (1:55 PM).
- (b) The latest time Alan can leave Alperton is 12:15 (noon 12:15 PM).
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