Math, asked by nishoobajwa, 6 hours ago

A school has a total enrollment of 150 students. There are 63 students taking Physics, 48 taking chemistry, and 21 taking both
How many students are taking neither Physics nor chemistry?

Answers

Answered by mintdew
1
Answer:

number of students for each subject:
total = 150
physics = 63
chemistry = 48
both = 21

to find the number of students that took neither:
150 - (63 + 48 + 21)
63 + 48 + 21
= 132
= 150 - 132
= 18
thus, the number of students that took neither are 18


I hoped it helped you :)
Answered by tennetiraj86
1

Step-by-step explanation:

Given :-

A school has a total enrollment of 150 students. There are 63 students taking Physics, 48 taking chemistry, and 21 taking both.

To find :-

How many students are taking neither Physics nor chemistry?

Solution :-

Given that

Total enrollment of a school = 150 students

Let P represents Physics and C represents Chemistry

Number of students who takes Physics = 63

=> n(P) = 63

Number of students who takes Chemistry = 48

=> n(C) = 48

Number of students who takes both subjects = 21

=> n(PnC) = 21

We know that

n(AUB) = n(A)+n(B)-n(AnB)

Now,

n(PUC) = n(P) + n(C) - n(PnC)

=> n(PUC) = 63+48-21

=> n(PUC) = 111-21

=> n(PUC) = 90

Therefore, Number of students who takes either Physics or Chemistry = 90

Now,

Number of students are taking neither Physics nor Chemistry

=> Total number of students-Number of students who takes either Physics or Chemistry

=> 150-90

=> 60

Answer:-

Number of students are taking neither Physics nor Chemistry = 60

Used formulae:-

Fundamental Theorem on Sets :-

n(AUB) = n(A)+n(B)-n(AnB)

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