In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are

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Accepted Answer

We are tasked with determining the minimum and maximum number of students who opted for both swimming and running in a group of 250 students. Here's the step-by-step solution:

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### Step 1: Total number of boys and girls
The percentage of girls is given as at least 44% and at most 60%.  
- **At least 44% girls:**  
  Number of girls = $$ 250 	imes \frac{44}{100} = 110 $$  
  Number of boys = $$ 250 - 110 = 140 $$

- **At most 60% girls:**  
  Number of girls = $$ 250 	imes \frac{60}{100} = 150 $$  
  Number of boys = $$ 250 - 150 = 100 $$

Thus, the number of girls ranges from **110 to 150**, and the number of boys ranges from **140 to 100**.

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### Step 2: Students opting for swimming and running
- **Swimming:**  
  - $$50\%$$ of boys:  
    - For $$140$$ boys: $$140 	imes \frac{50}{100} = 70$$  
    - For $$100$$ boys: $$100 	imes \frac{50}{100} = 50$$  
  - $$80\%$$ of girls:  
    - For $$110$$ girls: $$110 	imes \frac{80}{100} = 88$$  
    - For $$150$$ girls: $$150 	imes \frac{80}{100} = 120$$

- **Running:**  
  - $$70\%$$ of boys:  
    - For $$140$$ boys: $$140 	imes \frac{70}{100} = 98$$  
    - For $$100$$ boys: $$100 	imes \frac{70}{100} = 70$$  
  - $$60\%$$ of girls:  
    - For $$110$$ girls: $$110 	imes \frac{60}{100} = 66$$  
    - For $$150$$ girls: $$150 	imes \frac{60}{100} = 90$$

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### Step 3: Students opting for both swimming and running
Using the principle of inclusion-exclusion, the percentage of students opting for both activities is determined by the overlap:
- Boys: $$20\%$$
- Girls: $$40\%$$

#### Minimum case:
- Girls = $$110 	imes \frac{40}{100} = 44$$, Boys = $$140 	imes \frac{20}{100} = 28$$
- Total = $$44 + 28 = 72$$

#### Maximum case:
- Girls = $$150 	imes \frac{40}{100} = 60$$, Boys = $$100 	imes \frac{20}{100} = 20$$
- Total = $$60 + 20 = 80$$

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### Final Answer:
The minimum and maximum number of students who opted for both swimming and running are **72 and 80**, respectively.

**Correct Option:** A

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