Question

There are 40 students in a chemistry class and 60 students in a physics class. Find the number of students, which are either in physics class or chemistry class in the following case: The two classes meet at different hours and 20 students are enrolled in both the subjects.

Answer 1

Here, we will assume that A be the students in chemistry class, B be the students in physic class and then use the property
$n\left( {{\text{A}} \cup {\text{B}}} \right) = n\left( {\text{A}} \right) + n\left( {\text{B}} \right) - n\left( {{\text{A}} \cap {\text{B}}} \right)$, where $n\left( {{\text{A}} \cup {\text{B}}} \right)$, which are either in physics class or chemistry class.

Complete step-by-step answer:
Let us assume that A be the students in chemistry class and B be the students in physics class.
Since we are given that there are 40 students in a chemistry class, 60 students in a physics class and two classes meet at different timings then there can be 20 students in both the classes.
We know the property, $n\left( {{\text{A}} \cup {\text{B}}} \right) = n\left( {\text{A}} \right) + n\left( {\text{B}} \right) - n\left( {{\text{A}} \cap {\text{B}}} \right)$.
Finding the value of $n\left( {\text{A}} \right)$, $n\left( {\text{B}} \right)$ and $n\left( {{\text{A}} \cap {\text{B}}} \right)$ from the given conditions, we get
$n\left( {\text{A}} \right) = 40$
$n\left( {\text{B}} \right) = 60$
$n\left( {{\text{A}} \cap {\text{B}}} \right) = 20$
Substituting the above values in the above property to find the number of students, which are either in physics class or chemistry class, $n\left( {{\text{A}} \cup {\text{B}}} \right)$, we get

$$\Rightarrow n\left( {{\text{A}} \cup {\text{B}}} \right) = 40 + 60 - 20 \\\ \Rightarrow n\left( {{\text{A}} \cup {\text{B}}} \right) = 80 \\\$$

Thus, there are 80 students, which is either physics class or chemistry class.

Note: In solving these types of questions, students need to know the meaning of cardinality and their properties and then this problem is very simple.