`1//4th` of number of boys and `3//8th` of numbers of girls participated in annual sports of the school. What fraction part of total number of students participated ?

Options

A. 0.32
B. 0.2
C. 0.36
D. Data inadequate

Answer

To solve the problem step by step, let's denote the total number of boys as \( B \) and the total number of girls as \( G \).

### Step 1: Determine the number of boys and girls participating
According to the problem:
- The fraction of boys participating is \( \frac{1}{4} \) of \( B \), which can be expressed as:
\[ \text{Boys participating} = \frac{1}{4}B \]
- The fraction of girls participating is \( \frac{3}{8} \) of \( G \), which can be expressed as:
\[ \text{Girls participating} = \frac{3}{8}G \]

### Step 2: Calculate the total number of participants
Now, we can find the total number of students participating in the sports event by adding the number of boys and girls participating:
\[ \text{Total participants} = \frac{1}{4}B + \frac{3}{8}G \]

### Step 3: Calculate the total number of students
The total number of students in the school is the sum of the total number of boys and girls:
\[ \text{Total students} = B + G \]

### Step 4: Determine the fraction of total students that participated
To find out what fraction of the total number of students participated, we will divide the total number of participants by the total number of students:
\[ \text{Fraction of students participating} = \frac{\frac{1}{4}B + \frac{3}{8}G}{B + G} \]

### Step 5: Simplify the expression
To simplify the expression, we can find a common denominator for the fractions in the numerator. The common denominator for \( 4 \) and \( 8 \) is \( 8 \):
\[ \frac{1}{4}B = \frac{2}{8}B \]
Thus, we can rewrite the numerator:
\[ \frac{2}{8}B + \frac{3}{8}G = \frac{2B + 3G}{8} \]
Now, substituting this back into our fraction:
\[ \text{Fraction of students participating} = \frac{\frac{2B + 3G}{8}}{B + G} \]
This can be simplified further:
\[ = \frac{2B + 3G}{8(B + G)} \]

### Final Answer
Thus, the fraction of the total number of students that participated in the annual sports is:
\[ \frac{2B + 3G}{8(B + G)} \]