3.In a survey of 25 students, it was found that 15 had taken mathematics,12 had taken Physics and 11 had taken chemistry.5 had takenMathematics and Chemistry,9 had taken mathematics and Physics,4 had taken Chemistry and Physics and 3 had taken all the three subjects. Find the number of students that had i)only Chemistry ii)only Physics iii)only Mathematics iv)Physics and Maths but not Chemistry v)Mathematics and Chemistry but not Physics vi)only one of the subjects vii)atleast one of the subjects viii)none of the subjects
Answers
Step-by-step explanation:
Venn diagrams are used to solve problems pertaining to relations among two or more set of items. Overlapping of circles are used to identify the common things among the sets.
Please find the figure below to understand various sets and relations between them.
The following things are given in the problem:
N(MUCUP) = 25 (M-Mathematics; P-Physics; C-Chemistry)
N(M) = 15
N(P) = 12
N(C) = 11
N(M∩C) = 5
N(M∩P) = 9
N(C∩P) = 4
N(M∩P∩C) = 3
i) Only Chemistry: Number of students who take only chemistry but not maths or physics.
Method I: From the first venn diagram, number of students who take only chemistry is 5.
Refer image 1.
Method II:
Number of students who read only Chemistry =n(C∩P′∩M′)
= n((C∩(PUM)′)
= n(C)-n(C∩(PUM))
= n(C)-n((C∩P)U(C∩M))
= n(C) - [n(C∩P)+n(C∩M)-n(C∩P∩M)]
= 11 - (5 + 4 - 3) = 5
ii) Only Mathematics: Number of students who study only maths but not physics or chemistry.
Method I:
We can see that 2 read maths and chemistry, 3 read all the three, 6 read maths and chemistry, only 4 read just mathematics.
Therefore 4 read only maths.
Refer Image 2.
Method II:
Number of students who read only maths = n(M∩P′∩C′)
= n((M∩(PUC)′)
=n(M)-n(M∩(PUC))
=n(M)-n((M∩P)U(M∩C))
=n(M) - [n(M∩P)+n(M∩C)-n(M∩P∩C)]
= 15 - (9 + 5 - 3) = 4
iii) Only Physics: Number of students who take only physics but not maths or chemistry.
Method I: From the venn diagram, we can say the number of students who take physics is 2.
Refer image 3.
Method II:
Number of students who take only Physics =n(P∩M′∩C′)= n((P∩(MUC)′)
=n(P)-n(P∩(MUC))
=n(P)-n((P∩M)U(P∩C))
=n(P) - [n(P∩M)+n(P∩C)-n(P∩M∩C)]
= 12 – (9+4-3) = 2
iv) Physics and chemistry but not mathematics
Number of students = n(P∩C∩M′)
= n(P∩C) - n(P∩C∩M) = 4 - 3 = 1
v) Maths and Physics but not chemistry
Number of students = n(P∩M∩C′)
= n(P∩M) – n(P∩C∩M) = 9 - 3 = 6
vi) Only one of the subjects
Number of students = n(P∩M′∩C′) + n(M∩P′∩C′)+ n(C∩M′∩P′)
From the first venn diagram, only one of the subjects = only maths + only physics + only chemistry
= 4 + 5 + 2 = 11
vii) Atleast one of the subjects : n(MUPUC)
All the students who take one, two or three.
n(MUPUC) = n(M) + n(P) + n(C) - n(P∩M) - n(P∩C) - n(C∩M) + n(M∩C∩P)
= 15+12+11-9-4-5+3 = 23
viii) None of the subjects:
n(M′∩P′∩C′) = n(MUPUC)′ = n(U) - n(MUPUC) = 25 - 23 = 2
(Where U is the total number of students)
