Question

There 40 students in a chemistry class and 60 students in physics class. Find the number of students which are

either in Physics class or Chemistry class in the cases. (i) the two classes meet at the same hour. (ii) the two classes meet at different hours and 20 students are enrolled in both the subjects​

Answer 1

Answer:

Step-by-step explanation:

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(A∪B)=n(A)+n(B)−n(A∩B)

.

Finding the value of n(A)

, n(B)

and n(A∩B)

from the given conditions, we get

n(A)=40

n(B)=60

n(A∩B)=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(A∪B)

, we get

⇒n(A∪B)=40+60−20⇒n(A∪B)=80

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

Answer 2

Correct option is (A) 100

Let A be the set of students in chemistry class and B be the set of students in physics class. It is given that n(A)=40 and n(B)=60.

If two classes meet at the same hour, then there will not be a common student sitting in both the classes.

Therefore, n(A∩B)=0

∴n(A∪B)=n(A)+n(B)−n(A∩B)

⇒n(A∪B)=40+60−0=100