In a class of 50 students, 40 students offered Physics and 30 offered Biology. How many offered both Physics and Biology?

To determine the number of students who offered both Physics and Biology, we can use the concept of set intersection. We know that there are 40 students who offered Physics and 30 students who offered Biology. If we represent the set of students who offered Physics as set P, and the set of students who offered Biology as set B, then we can find the number of students who offered both by finding the intersection of the two sets, denoted as P ∩ B. Using the formula for set intersection, we get: |P ∩ B| = |P| + |B| - |P U B| where |P| is the cardinality (number of elements) of set P, |B| is the cardinality of set B, and |P U B| is the cardinality of the union of sets P and B (i.e., the total number of students who offered either Physics or Biology, or both). Substituting the values we know, we get: |P ∩ B| = 40 + 30 - |P U B| We don't know |P U B| yet, but we can see that it represents the total number of students in the class, which is 50. So we can rewrite the equation as: |P ∩ B| = 40 + 30 - 50 Simplifying, we get: |P ∩ B| = 20 Therefore, there were 20 students who offered both Physics and Biology. : 20 is the correct answer.