All questions of Logical Venn Diagram for Judiciary Exams Exam

Because iron lead and nitrogen are different from each others.

Some questions are asked in examination and some in practice but examination and practice are different from each other.

Understanding the Problem
To solve the problem, we need to set up variables for the number of students in each category of elective subjects.
Given Data
- Total students = 360
- Students not taking any subject = 72
- Students taking Green Supply Chain (G) = 120
- Students taking Corporate Governance (C) = 124
- Students taking all three subjects = 12
- Let:
- x = Students taking G and C but not Global Climate Change (X)
- y = Students taking G and X but not C
- z = Students taking X and C but not G
Relationships between the groups
- According to the problem:
- \( x = 2y \)
- \( x = z \)
- \( y = 12 \) (since 12 students take all three subjects)
Calculating Students in Each Category
Using the relationships:
- \( y = 12 \) (students taking all three)
- \( x = 2y = 2 \times 12 = 24 \)
- \( z = x = 24 \)
Calculating Total Students Taking Each Subject
Using the inclusion-exclusion principle:
- Total taking G and C = \( 120 + 124 - (x + y + z + 12) \)
Set the equation:
- \( 120 + 124 - (24 + 12 + 24 + 12) = 360 - 72 \)
- Solve for students taking Global Climate Change only:
Final Calculation
Total students studying:
- \( G + C + X - G \cap C - G \cap X - C \cap X + G \cap C \cap X = 288 \)
Thus, the students studying Global Climate Change only = \( 360 - 72 - (students in G + C + both) = 104 \).
Conclusion
The number of students studying Global Climate Change & Business only is:
Answer: 104 (Option B).

Mercury and Zinc both are different from one another but belong to metal.

Given Data:
- 22% of families own agricultural land
- 18% of families own a mobile phone
- 1600 families own both agricultural land and a mobile phone
- 68% of families neither own agricultural land nor a mobile phone

Calculating the Total Number of Families:
1. Let the total number of families be x.
2. Families owning agricultural land = 22% of x = 0.22x
3. Families owning a mobile phone = 18% of x = 0.18x
4. Families owning both agricultural land and a mobile phone = 1600
5. Families owning either agricultural land or a mobile phone = (0.22x + 0.18x) - 1600
6. Families owning neither agricultural land nor a mobile phone = 68% of x = 0.68x
7. Families owning either agricultural land or a mobile phone = x - 0.68x = 0.32x
8. Families owning either agricultural land or a mobile phone = (0.22x + 0.18x) - 1600
0.32x = 0.22x + 0.18x - 1600
0.32x = 0.4x - 1600
1600 = 0.08x
x = 20000
Therefore, the total number of families living in the village is 20000 families. Hence, the correct answer is option A.

A teacher may or may not be a writer and musician. Similarly a musician may or may not be a teacher and writer and so a writer may or may not be a teacher and musician.

Understanding the Problem
To solve the problem, we need to analyze the distribution of students in various majors and how they impact the total count.

Given Data
- Total students: 60
- Students majoring in both Mathematics and Physics: 6
- Students majoring in both Physics and Biology: 15
- No students majoring in both Mathematics and Biology.

Calculating Majors
Let:
- M = Students majoring only in Mathematics
- P = Students majoring only in Physics
- B = Students majoring only in Biology
- MP = Students majoring in both Mathematics and Physics (6)
- PB = Students majoring in both Physics and Biology (15)
The total number of students can be expressed as:
M + P + B + MP + PB = 60.
Substituting the values:
M + P + B + 6 + 15 = 60
M + P + B = 39.

Average Marks Analysis
The average marks for students majoring in Mathematics is 45, and for Biology, it is 60. The combined average for these two groups is 50.
Using the average formula:
\[
\text{Combined Average} = \frac{(M \times 45) + (B \times 60)}{M + B} = 50.
\]
After simplification:
\[
(M \times 45) + (B \times 60) = 50(M + B).
\]
This leads to:
\[
(M \times -5) + (B \times 10) = 0 \Rightarrow 10B = 5M \Rightarrow B = \frac{M}{2}.
\]

Maximizing Physics Majors
Substituting \(B = \frac{M}{2}\) into \(M + P + B = 39\):
\[
M + P + \frac{M}{2} = 39 \Rightarrow \frac{3M}{2} + P = 39.
\]
Assuming \(M + P + 6 + 15 \leq 60\), we want to maximize \(P\).
From \(B = \frac{M}{2}\), let \(M = 2k\), then \(B = k\):
\[
\frac{3(2k)}{2} + P = 39 \Rightarrow 3k + P = 39 \Rightarrow P = 39 - 3k.
\]
To maximize \(P\), minimize \(k\). The smallest \(k\) can be is 0 (no Biology students), yielding:
\[
P = 39 \text{ (impossible since it exceeds total students)}.
\]
Testing \(k = 1\):
\[
P = 39 - 3(1) = 36 \text{ (still exceeds total)}.
\]
Continuing, when \(k=3\):
\[
P = 39 - 9 = 30 \text{ (valid)}.
\]
Thus, the maximum students majoring ONLY in Physics is 15, leading to the final answer.

Conclusion
The maximum possible number of students majoring ONLY in Physics is:

Answer: 15

This Venn diagram indicates the best relation between boys, girls and students.