Question

In a survey of 200 students of a school it was found that 120 study mathematics 90 study physics and 70 study chemistry 40 study mathematics and physics 30 study physics and chemistry 50 study chemistry and mathematics and20 study none of these subjects find the number of students who study 1) all the 3 subjects 2) only one subject 3) at most one subject 4) at least two subjects
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Answer 1

Answer:

1)  20

2)  100

3)  120

4)  80

Step-by-step explanation:

Let M, P, C be the sets of students studying maths, physics and chemistry, respectively, and let Ω be the set of all students.  Then the given information is:

|Ω| = 200,  |M| = 120,  |P| = 90,  |C| = 70,

|M∩P| = 40,  |P∩C| = 30,  |C∩M| = 50,

| Ω - (M∪P∪C) | = 20

From the last of these, we have |M∪P∪C| = 200 - 20 = 180.

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1) By the Principle of Inclusion-Exclusion, we have:

|M∪P∪C| = |M| + |P| + |C| - |M∩P| - |P∩C| - |C∩M| + |M∩P∩C|

=> 180 = 120 + 90 + 70 - 40 - 30 - 50 + |M∩P∩C|

=> |M∩P∩C| = 180 - 120 - 90 - 70 + 40 + 30 + 50 = 20

So there are 20 students who study all three subjects.

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2) Maths only:  |M - (P∪C)| = |M| - |M∩P| - |C∩M| + |M∩P∩C| = 120 - 40 - 50 + 20 = 50

Physics only:  |P - (C∪M)| = |P| - |P∩C| - |M∩P| + |M∩P∩C| = 90 - 30 - 40 + 20 = 40

Chemistry only:  |C - (M∪P)| = |C| - |C∩M| - |P∩C| + |M∩P∩C| = 70 - 50 - 30 + 20 = 10

So the number of students who study only one subjects is 50+40+10 = 100.

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3) # who study at most one subject

= ( # who study exactly one subject ) + ( # who study no subjects )

= 100 + 20

= 120

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4) # who study at least two subjects

= ( total # of students )  -  ( # who study at most one subject )

= 200 - 120

= 80

Answer 2

Answer:

M = Mathematics ; P = Physics and C = Chemistry 

n(M) = 120 n(P) = 90 n (C) = 70 n ( M ∩ P) = 40 

n ( P ∩ C ) = 30 n ( C ∩ M ) = 50 n ( M ∪ P ∪ C )’ = 20 

Now n(M ∪ P ∪ C)’ = n(U) – n(M ∪ P ∪ C) 

20 = 200 – n (M ∪ P ∪ C) 

Therefore, n(M ∪ P ∪ C) = 200 – 20 = 180 

n(M ∪ P ∪ C) 

= n(M) + n(P) + n(C) – n(M ∩ P) – n(P ∩ C) – n(C ∩ M) + n(M ∩ P ∩ C)

180 = 120 + 90 + 70 - 40 - 30 - 50 + n(M ∩ P ∩ C)

⇒ n(M ∩ P ∩ C) =180 - 120 - 90 - 70 + 40 + 30 + 50 

⇒ n(M ∩ P ∩ C) = 20.

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Step-by-step explanation: