Question

. All the students in SS3 of a named school take either Mathematics (M), or Physics (P) or Chemistry (C). 40 take Mathematics, 42 take physics, 38 take Chemistry, 20 take Mathematics and Physics, 28 take Physics and Chemistry while 25 take mathematics and chemistry.
How many take
(a) All the three subject:
(b) Mathematics, but neither Physics nor Chemistry
(c) Physics, but neither Mathematics nor Chemistry

Answer 1

Answer:

(a) = 13

(b) = 8

(c) = 5

Step-by-step explanation:

Addition theorems on sets are

Theorem 1 :

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

Theorem 2 :

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

Total number of students in the school is not given

so let there are 60 students in the school

using theorem 2

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

let n(A) = Mathematics, n(B) =Physics and n(C) = Chemistry

so putting values,

   60             =    40 + 42 + 38 - 20 - 28 - 25 + n(AnBnC)

60 +73 -120 = n(AnBnC)

         13         = n(AnBnC)

therefore, there are total 13 students who take all three subjects

Number of students who had taken only Mathematics  =
n(A) - n(AnB) - n(AnC) + n(AnBnC)

40 - 20 - 25 + 13

53 - 45 = 8 students

Number of students who had taken only Physics  =

n(B) - n(BnA) - n(BnC) + n(AnBnC)

42 - 20 - 28 + 13

53 - 48 = 5 students

learn more about sets and vein diagrams at

https://brainly.com/question/2332158

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