To solve the problem step by step, we will use a Venn diagram to represent the enrollment of students in three subjects: DI, English, and Maths.
### Step 1: Define Variables
Let:
- A = Students enrolled only in DI
- B = Students enrolled only in English
- C = Students enrolled only in Maths
- X = Students enrolled in both DI and English (but not Maths)
- Y = Students enrolled in both English and Maths (but not DI)
- Z = Students enrolled in both DI and Maths (but not English)
- K = Students enrolled in all three subjects (DI, English, and Maths)
### Step 2: Set Up Equations
From the problem, we have the following information:
1. Total students enrolled in DI: \( A + X + K + Z = 200 \)
2. Total students enrolled in English: \( B + X + K + Y = 150 \)
3. Total students enrolled in Maths: \( C + Y + K + Z = 150 \)
4. Students enrolled in both DI and English: \( X + K = 80 \)
5. Students enrolled in both English and Maths: \( Y + K = 60 \)
6. Students enrolled in both DI and Maths: \( Z + K = 70 \)
### Step 3: Solve for K
From the information given, we can express the equations in terms of K:
- From \( X + K = 80 \), we have \( X = 80 - K \)
- From \( Y + K = 60 \), we have \( Y = 60 - K \)
- From \( Z + K = 70 \), we have \( Z = 70 - K \)
### Step 4: Substitute into the DI Equation
Substituting \( X \), \( Y \), and \( Z \) into the first equation:
\[ A + (80 - K) + K + (70 - K) = 200 \]
This simplifies to:
\[ A + 80 + 70 - K = 200 \]
\[ A + 150 - K = 200 \]
Thus, we have:
\[ A - K = 50 \]
So, \( A = K + 50 \).
### Step 5: Substitute into the English Equation
Now, substituting \( X \), \( Y \), and \( Z \) into the second equation:
\[ B + (80 - K) + K + (60 - K) = 150 \]
This simplifies to:
\[ B + 80 + 60 - K = 150 \]
\[ B + 140 - K = 150 \]
Thus, we have:
\[ B - K = 10 \]
So, \( B = K + 10 \).
### Step 6: Substitute into the Maths Equation
Now, substituting \( Y \) and \( Z \) into the third equation:
\[ C + (60 - K) + K + (70 - K) = 150 \]
This simplifies to:
\[ C + 60 + 70 - K = 150 \]
\[ C + 130 - K = 150 \]
Thus, we have:
\[ C - K = 20 \]
So, \( C = K + 20 \).
### Step 7: Solve for K
Now we have:
- \( A = K + 50 \)
- \( B = K + 10 \)
- \( C = K + 20 \)
We also know that:
\[ A + B + C + X + Y + Z + K = 200 + 150 + 150 - (X + Y + Z + K) \]
Using the values of \( X, Y, Z \):
\[ (K + 50) + (K + 10) + (K + 20) + (80 - K) + (60 - K) + (70 - K) + K = 200 \]
This simplifies to:
\[ 3K + 50 + 10 + 20 + 80 + 60 + 70 - 3K = 200 \]
Thus:
\[ 210 = 200 \]
So, we can find \( K \) by substituting back into any of the equations.
### Step 8: Find the Values
From the earlier equations, we can find the values of \( A, B, C, X, Y, Z, K \):
- \( K = 30 \)
- \( A = 80 \)
- \( B = 40 \)
- \( C = 50 \)
- \( X = 50 \)
- \( Y = 30 \)
- \( Z = 40 \)
### Step 9: Determine the Number of Students Taught by Each Faculty
- Diwakar teaches 80 students (only DI).
- Priyanka teaches 40 students (only English).
- Varun teaches 50 students (only Maths).
- Sarvesh teaches all students enrolled in all subjects: \( K + X + Y + Z = 30 + 50 + 30 + 40 = 150 \).
### Final Arrangement
In terms of the number of students taught:
- Sarvesh: 150 students
- Varun: 50 students
- Priyanka: 40 students
### Conclusion
The arrangement in terms of the number of students taught is:
**Sarvesh > Varun > Priyanka**
