Circles C, B, and P overlap. Circle C contains 9, circle B contains 11, and circle P contains 7. The overlap of circles C and P contains 6, the overlap of circles C and B contains 5, and the overlap of circles B and P contains 4. The overlap of all 3 circles contains 3.

Westside High’s 2015 senior-class science enrollment is summarized by the Venn diagram. Chemistry, biology, and physics class enrollments are denoted by the sets C, B, an P, respectively.

How many students are in the 2015 senior class?

How many students are taking chemistry and biology, but not physics?

How many students are not taking science?

1 answer

To solve these problems, we can use the information about the overlaps of the circles representing the students in different science classes. Let's denote the following:

  • \( C \): Students taking Chemistry
  • \( B \): Students taking Biology
  • \( P \): Students taking Physics

Given data:

  • \( |C| = 9 \)
  • \( |B| = 11 \)
  • \( |P| = 7 \)
  • \( |C \cap P| = 6 \)
  • \( |C \cap B| = 5 \)
  • \( |B \cap P| = 4 \)
  • \( |C \cap B \cap P| = 3 \)

Step 1: Find the number of students in the 2015 senior class

We can calculate the number of students in each of the individual categories of the Venn Diagram.

  1. Students taking only Chemistry \( |C \text{ only}| \): \[ |C \text{ only}| = |C| - |C \cap B| - |C \cap P| + |C \cap B \cap P| \] \[ |C \text{ only}| = 9 - 5 - 6 + 3 = 1 \]

  2. Students taking only Biology \( |B \text{ only}| \): \[ |B \text{ only}| = |B| - |B \cap C| - |B \cap P| + |C \cap B \cap P| \] \[ |B \text{ only}| = 11 - 5 - 4 + 3 = 5 \]

  3. Students taking only Physics \( |P \text{ only}| \): \[ |P \text{ only}| = |P| - |P \cap C| - |P \cap B| + |C \cap B \cap P| \] \[ |P \text{ only}| = 7 - 6 - 4 + 3 = 0 \]

  4. Students taking Chemistry and Biology, but not Physics \( |C \cap B \text{ only}| \): \[ |C \cap B \text{ only}| = |C \cap B| - |C \cap B \cap P| = 5 - 3 = 2 \]

  5. Students taking Chemistry and Physics, but not Biology \( |C \cap P \text{ only}| \): \[ |C \cap P \text{ only}| = |C \cap P| - |C \cap B \cap P| = 6 - 3 = 3 \]

  6. Students taking Biology and Physics, but not Chemistry \( |B \cap P \text{ only}| \): \[ |B \cap P \text{ only}| = |B \cap P| - |C \cap B \cap P| = 4 - 3 = 1 \]

Now calculating total number of students:

Total number of students in the senior class: \[ |C| + |B| + |P| - |C \cap B| - |C \cap P| - |B \cap P| + |C \cap B \cap P| \] \[ = 9 + 11 + 7 - 5 - 6 - 4 + 3 = 15 \]

Answer 1: There are 15 students in the 2015 senior class.

Step 2: Students taking Chemistry and Biology, but not Physics

From the above calculation, we found that: \[ |C \cap B \text{ only}| = 2 \]

Answer 2: There are 2 students taking Chemistry and Biology, but not Physics.

Step 3: Students not taking science

To find the number of students not taking any science, we can consider the total number of students minus those that are in the science classes. Since we calculated 15 students taking at least one of the subjects, we denote the total number of seniors:

  • If we assume the total number of seniors is 15 (from previous computations and it is common to consider it for simplicity unless specified), there are no students outside of any science.

Thus, if 15 is the total: \[ \text{Students not taking science} = \text{Total students} - \text{Students taking science} = 0 \]

Answer 3: There are 0 students not taking science, assuming total students = 15.

In conclusion:

  • Total students in class: 15
  • Chemistry and Biology, but not Physics: 2
  • Students not taking science: 0