Read the data given below and answer the questions based on it.
There were 3 sections namely A, B and C in a test. Out of three 33 students passed in section A, 34 students passed in Section B and 32 passed in Section C. 10 students passed in Section A and Section B, 9 passed in Section B and Section C, 8 passed in Section A and Section C. The number of students who passed each section alone was equal and was 21 for each section.
The ratio of the number of students passed in one or more of the sections to the number of students who passed in section A alone is :
A. 78/21
B. 3
C. 73/21
D. 75/21

Options

A. B
B. D
C. C
D. A

Answer

To solve the problem step by step, we will use a Venn diagram approach to visualize the distribution of students across the three sections (A, B, and C).

### Step 1: Set Up the Venn Diagram
We have three sections: A, B, and C. We know the following:
- 33 students passed in Section A.
- 34 students passed in Section B.
- 32 students passed in Section C.
- 10 students passed in both A and B.
- 9 students passed in both B and C.
- 8 students passed in both A and C.
- 21 students passed each section alone.

### Step 2: Define Variables
Let:
- \( x \) = number of students who passed all three sections (A, B, and C).

### Step 3: Fill in the Venn Diagram
From the information provided:
- Students who passed only A = 21
- Students who passed only B = 21
- Students who passed only C = 21

Now, we can express the number of students who passed in combinations:
- Students who passed A and B only = \( 10 - x \)
- Students who passed B and C only = \( 9 - x \)
- Students who passed A and C only = \( 8 - x \)

### Step 4: Set Up Equations
Now we can set up equations based on the total number of students who passed in each section:

1. For Section A:
\[ 21 + (10 - x) + (8 - x) + x = 33 \]
Simplifying this:
\[ 21 + 10 - x + 8 - x + x = 33 \implies 39 - x = 33 \implies x = 6 \]

2. For Section B:
\[ 21 + (10 - x) + (9 - x) + x = 34 \]
Substituting \( x = 6 \):
\[ 21 + (10 - 6) + (9 - 6) + 6 = 34 \implies 21 + 4 + 3 + 6 = 34 \text{ (This holds true)} \]

3. For Section C:
\[ 21 + (8 - x) + (9 - x) + x = 32 \]
Substituting \( x = 6 \):
\[ 21 + (8 - 6) + (9 - 6) + 6 = 32 \implies 21 + 2 + 3 + 6 = 32 \text{ (This holds true)} \]

### Step 5: Calculate Total Students Who Passed in One or More Sections
Now, we need to calculate the total number of students who passed in one or more sections:
- Students who passed only A = 21
- Students who passed only B = 21
- Students who passed only C = 21
- Students who passed A and B only = \( 10 - 6 = 4 \)
- Students who passed B and C only = \( 9 - 6 = 3 \)
- Students who passed A and C only = \( 8 - 6 = 2 \)
- Students who passed all three = 6

Adding these together:
\[ 21 + 21 + 21 + 4 + 3 + 2 + 6 = 78 \]

### Step 6: Calculate the Ratio
The ratio of the number of students who passed in one or more sections to the number of students who passed in section A alone is:
\[ \text{Ratio} = \frac{78}{21} \]

### Final Answer
Thus, the answer is \( \frac{78}{21} \).