A class has total 30 students all have different weights in natural numbers. 60% of the students in the class are boys and the rest of the students are girls. The maximum and minimum weight of the boys are 82 kg and 56 kg respectively. The maximum and minimum weight of the girls are 62 kg and 48 kg respectively What is the maximum possible difference between the average weight of the boys and the average weight of the girls

Let's break down the problem step by step to find the maximum possible difference between the average weight of the boys and the girls.

### Step 1: Determine the number of boys and girls.
- Total number of students = 30
- 60% of the students are boys, so:
  \[
  \text{Number of boys} = 0.6 \times 30 = 18
  \]
- The remaining 40% are girls, so:
  \[
  \text{Number of girls} = 0.4 \times 30 = 12
  \]

### Step 2: Analyze the given weights.
- Maximum weight of boys = 82 kg
- Minimum weight of boys = 56 kg
- Maximum weight of girls = 62 kg
- Minimum weight of girls = 48 kg

### Step 3: Calculate the maximum average weight of the boys.
To maximize the average weight of the boys, we need to assign the highest possible weights to the boys. Since all weights are different natural numbers:
- The 18 boys will have weights as close to 82 kg as possible. The maximum weights could be \( 65, 66, \dots, 82 \) (i.e., 18 consecutive numbers).

Now calculate the average of these numbers:
\[
\text{Sum of weights of boys} = 65 + 66 + \dots + 82
\]
This is an arithmetic series where:
- First term \( a_1 = 65 \)
- Last term \( a_n = 82 \)
- Number of terms \( n = 18 \)

The sum \( S \) of an arithmetic series is given by:
\[
S = \frac{n}{2} \times (\text{First term} + \text{Last term}) = \frac{18}{2} \times (65 + 82) = 9 \times 147 = 1323 \text{ kg}
\]
So, the average weight of the boys is:
\[
\text{Average weight of boys} = \frac{1323}{18} = 73.5 \text{ kg}
\]

### Step 4: Calculate the minimum average weight of the girls.
To minimize the average weight of the girls, we need to assign the lowest possible weights to the girls. The 12 girls could have weights \( 48, 49, \dots, 59 \) (12 consecutive numbers).

Now calculate the average of these numbers:
\[
\text{Sum of weights of girls} = 48 + 49 + \dots + 59
\]
This is also an arithmetic series where:
- First term \( a_1 = 48 \)
- Last term \( a_n = 59 \)
- Number of terms \( n = 12 \)

The sum \( S \) of this arithmetic series is given by:
\[
S = \frac{n}{2} \times (\text{First term} + \text{Last term}) = \frac{12}{2} \times (48 + 59) = 6 \times 107 = 642 \text{ kg}
\]
So, the average weight of the girls is:
\[
\text{Average weight of girls} = \frac{642}{12} = 53.5 \text{ kg}
\]

### Step 5: Calculate the maximum possible difference.
The maximum possible difference between the average weight of the boys and the average weight of the girls is:
\[
\text{Maximum difference} = 73.5 \text{ kg} - 53.5 \text{ kg} = 20 \text{ kg}
\]

### Final Answer:
The maximum possible difference between the average weight of the boys and the average weight of the girls is **20 kg**.

Would you like more details or have any questions? Here are some related problems you might find interesting:

1. What would be the difference if the number of boys and girls were different?
2. How would the result change if the total number of students increased?
3. What if the weights were not distinct but allowed to repeat?
4. How does the distribution of weights affect the average?
5. Can the problem be generalized for other percentages of boys and girls?

**Tip:** Understanding the distribution of values is key in maximizing or minimizing averages.

Discover how to calculate the maximum possible difference between the average weight of boys and girls in a class of 30 students, where 60% are boys with weights ranging from 56 kg to 82 kg, and 40% are girls with weights ranging from 48 kg to 62 kg. Learn the arithmetic steps and formulas involved in determining this difference.

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