A bee flies at 20 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for 18 ​minutes, and then flies directly back to the hive at 12 feet per second. It is away from the hive for a total of 22 minutes.


To solve this problem, let's break it down step by step:

1. **Let \( d \) be the distance from the hive to the flowerbed in feet.**
  
2. **Time spent flying to the flowerbed:**
   - The bee flies to the flowerbed at 20 feet per second.
   - The time taken to reach the flowerbed is \( \frac{d}{20} \) seconds.

3. **Time spent flying back to the hive:**
   - The bee flies back at 12 feet per second.
   - The time taken to return is \( \frac{d}{12} \) seconds.

4. **Total time spent flying:**
   - The total time spent flying is the sum of the time taken to go to the flowerbed and the time taken to return:
   \[
   \text{Total flying time} = \frac{d}{20} + \frac{d}{12} \text{ seconds}
   \]

5. **Convert the total time away from the hive into seconds:**
   - The bee is away from the hive for 22 minutes.
   - Convert 22 minutes to seconds:
   \[
   22 \times 60 = 1320 \text{ seconds}
   \]

6. **Total time equation:**
   - The total time away from the hive includes the time spent flying and the time spent at the flowerbed:
   \[
   \frac{d}{20} + \frac{d}{12} + 18 \times 60 = 1320 \text{ seconds}
   \]
   \[
   \frac{d}{20} + \frac{d}{12} + 1080 = 1320
   \]

7. **Simplify the equation:**
   - Subtract 1080 from both sides:
   \[
   \frac{d}{20} + \frac{d}{12} = 240
   \]

8. **Find a common denominator to combine the fractions:**
   - The common denominator of 20 and 12 is 60:
   \[
   \frac{3d}{60} + \frac{5d}{60} = 240
   \]
   \[
   \frac{8d}{60} = 240
   \]

9. **Solve for \( d \):**
   \[
   \frac{d}{7.5} = 240
   \]
   \[
   d = 240 \times 7.5 = 1800 \text{ feet}
   \]

So, the distance from the hive to the flowerbed is **1800 feet**.

Would you like more details or have any questions?

Here are five related questions:

1. What would the distance be if the bee flew back at 10 feet per second instead?
2. How long would the bee stay at the flowerbed if it flew back at the same speed but was away for 25 minutes in total?
3. How does the total travel time change if the bee's return speed is doubled?
4. What is the average speed of the bee during the entire trip?
5. How would the problem change if the flowerbed was 2000 feet away instead?

**Tip:** When solving time-distance problems, always ensure all time units are consistent (e.g., all in seconds).

Learn how to solve a distance-speed-time problem involving a bee's flight from hive to flowerbed and back. Calculate the distance using given speeds and times with detailed step-by-step solution. Suitable for high school students.

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