In an university, out of 100 students, 16 opted mathematics only, 24 opted physics only, 30 opted for chemisty only and 8 opted chemistry and mathematics, 14 opted physics and chemistry, 4 opted mathematics and physics, and 48 opted chemistry. By drawing Venn diagram, find the number of students who i) opted mathematics ii)opted physics iii)did not opted any of the above three subjects

Explanation

To solve the problem step by step, we will use a Venn diagram to represent the students opting for Mathematics, Physics, and Chemistry. Let's denote:

- A: Students who opted for Mathematics only
- B: Students who opted for Physics only
- C: Students who opted for Chemistry only
- D: Students who opted for both Physics and Mathematics
- E: Students who opted for both Chemistry and Mathematics
- F: Students who opted for both Physics and Chemistry
- G: Students who opted for all three subjects

### Step 1: Set up the equations based on the information provided

1. **Mathematics only (A)**: 16 students
2. **Physics only (B)**: 24 students
3. **Chemistry only (C)**: 30 students
4. **Chemistry and Mathematics (E + G)**: 8 students
5. **Physics and Chemistry (F + G)**: 14 students
6. **Mathematics and Physics (D + G)**: 4 students
7. **Total students in Chemistry (C + E + F + G)**: 48 students

### Step 2: Write down the equations

From the information above, we can derive the following equations:

1. \( A = 16 \)
2. \( B = 24 \)
3. \( C = 30 \)
4. \( E + G = 8 \)
5. \( F + G = 14 \)
6. \( D + G = 4 \)
7. \( C + E + F + G = 48 \)

### Step 3: Substitute known values into the equations

We know \( C = 30 \), so substituting into the seventh equation:

\[ 30 + E + F + G = 48 \]

This simplifies to:

\[ E + F + G = 18 \quad \text{(Equation 1)} \]

### Step 4: Solve for E, F, and G

From \( E + G = 8 \) (Equation 4) and \( F + G = 14 \) (Equation 5), we can express \( E \) and \( F \) in terms of \( G \):

1. From \( E + G = 8 \), we have \( E = 8 - G \)
2. From \( F + G = 14 \), we have \( F = 14 - G \)

Substituting these into Equation 1:

\[ (8 - G) + (14 - G) + G = 18 \]

This simplifies to:

\[ 22 - G = 18 \]

Thus, solving for \( G \):

\[ G = 4 \]

### Step 5: Find values for E and F

Substituting \( G = 4 \) back into the equations for \( E \) and \( F \):

1. \( E = 8 - 4 = 4 \)
2. \( F = 14 - 4 = 10 \)

### Step 6: Find value for D

Using \( D + G = 4 \):

\[ D + 4 = 4 \implies D = 0 \]

### Step 7: Summarize the values

Now we have:

- \( A = 16 \)
- \( B = 24 \)
- \( C = 30 \)
- \( D = 0 \)
- \( E = 4 \)
- \( F = 10 \)
- \( G = 4 \)

### Step 8: Calculate the total number of students who opted for each subject

i) **Students who opted for Mathematics**:

\[ \text{Total for Mathematics} = A + D + E + G = 16 + 0 + 4 + 4 = 24 \]

ii) **Students who opted for Physics**:

\[ \text{Total for Physics} = B + D + F + G = 24 + 0 + 10 + 4 = 38 \]

iii) **Students who did not opt for any subject**:

Total students = 100

Students who opted for at least one subject:

\[ \text{Total} = A + B + C + D + E + F + G = 16 + 24 + 30 + 0 + 4 + 10 + 4 = 88 \]

Thus, students who did not opt for any subject:

\[ 100 - 88 = 12 \]

### Final Answers:

i) 24 students opted for Mathematics.
ii) 38 students opted for Physics.
iii) 12 students did not opt for any of the above three subjects.