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IN A CLASS OF 60 STUDENTS, 25 PLAYS BASKETBALL, 2O PLAY VOLLEYBALL, AND 10 STUDENTS PLAY BOTH OF THEM. HOW MANY STUDENTS PLAY NEITHER BASKETBALL NOR VOLLEYBALL?

Sagot :

Answer:

25

Step-by-step explanation:

We'll compare it with the study of sets, the most noticeable operation of which is:

  • n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

where,

n(A ∪ B) is the Union Set, i. e, the set that contains all the elements

n(A) is a subset of the Union Set

n(B) is another subset of the Union Set, and

n(A ∩ B) is the Intersection Set ,i.e, the set contains common elements from both A and B sets.

In the question:

  • n(A ∪ B) = ? (the total number of students in the class who are into the above mentioned sports)

Let set A contains the students who play basketball and set B, the students who play Volleyball.

  • n(A) = 25
  • n(B) = 20

10 students play both of them, i. e.,

  • n(A ∩ B) = 10 (as 10 students have common sports - Volleyball and Basketball)

Using the above operation:

=> n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

=> n(A ∪ B) = 25 + 20 - 10

=> n(A ∪ B) = 35

Final Step to the Answer:

The total number of students in the class

= students into the given sports + students who don't play any of them

  • Total number of students in the class is 60
  • Total number fo students playing basketball and volleyball is 35

The students who play neither = 60 - 35

= 25