To solve the problem, let’s break it down step-by-step.
Given:
- Total students = 400
- Students offering Chemistry = 300
- Students offering neither Biology nor Chemistry = 70
Step 1: Calculate Students Offering Either Biology or Chemistry
First, we calculate the total number of students that are offering either Biology or Chemistry, which can be found by subtracting the students who are offering neither from the total students.
\[ \text{Students offering either Biology or Chemistry} = \text{Total students} - \text{Students offering neither} \]
\[ = 400 - 70 = 330 \]
Step 2: Use the Venn Diagram Formula
Let:
- \( A \) = number of students offering Biology
- \( B \) = number of students offering Chemistry
- \( |B| = 300 \) (as given)
- \( |A \cup B| = 330 \) (students offering at least one of Biology or Chemistry)
- \( |A \cap B| \) = number of students offering both Biology and Chemistry
From the formula for the union of two sets, we have:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Substituting the known values, we can express it as:
\[ 330 = |A| + 300 - |A \cap B| \]
Step 3: Determine Students Offering Only Biology
From this equation, we can isolate \( |A| \):
\[ |A| = 330 + |A \cap B| - 300 \]
\[ |A| = 30 + |A \cap B| \]
Step 4: Analyze Another Equation
We now also know that the number of students offering Chemistry is 300:
\[ |A \cap B| + (\text{Only Chemistry}) = 300 \]
Let's denote the "Only Chemistry" students as \( C \). Therefore:
\[ |A \cap B| + C = 300 \]
Step 5: Solve the Equations
To find \( |A \cap B| \), we observe that \( |A| \) is also expressed as \( |A| + C = 330 \).
We can substitute \( |A| = 30 + |A \cap B| \) into \( |A| + C = 330 \):
\[ (30 + |A \cap B|) + C = 330 \]
From \( C = 300 - |A \cap B| \), substitute:
\[ 30 + |A \cap B| + (300 - |A \cap B|) = 330 \]
This simplifies to:
\[ 30 + 300 = 330 \]
This equation is true for any \( |A \cap B| \), thus we can assign values. If we suppose \( |A \cap B| = x \), then we have:
\[ |A| = 30 + x \] \[ C = 300 - x \]
Conclusion
To meet the conditions specified:
- For simplicity, we can assign \( |A \cap B| = 30 \) (assuming 30 students like both subjects):
- Subsequently, \( |A| = 30 + 30 = 60 \) (for Biology)
- Thus \( C = 300 - 30 = 270 \) (for only Chemistry)
Final Count
- Students offering both Biology and Chemistry: 30
- Students offering at least one of Biology or Chemistry: 330
Venn Diagram Representation
You would typically represent this using circles where:
- Circle for Biology \( |A| = 60 \)
- Circle for Chemistry \( |B| = 300 \)
- The intersection \( |A \cap B| = 30 \)
- Exclusive Biology students \( = 60 - 30 = 30 \)
- Exclusive Chemistry students \( = 300 - 30 = 270 \)
- Outside both circles representing neither \( = 70 \)
This helps visualize the numbers effectively!