One of the 400 students in the final year in a senior secondary school, 300 are offering chemistry.

1. How many students are offering both biology and chemistry. If only 70 students are offering neither Biology nor chemistry.
2. How many students are offering at least one of Biology or Chemistry. And it must be shown in Venn diagram

1 answer

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:

  1. 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!