The school organised a farewell party of 100 students and school management decided three types of drinks distributes in party are Milk (M), Coffee (C) and Tea (T). He reported the following 10 students had all the three drinks M,C,T, 20 had M and C; 30 had C and T, 25 had M and T. 12 had M only; 5 had C only; 8 had T only.
(i) The number of students who did not take any drinks, is
(a) 20 (b) 30 (c) 10 (d) 25
(ii) The number of students who prefer Milk is (a) 47 (b) 45 (c) 53 (d) 50
(iii) The number of students who prefer Coffee is
(a)47 (b) 53 (15 (8) 45 (d) 50
(iv) The number of student who prefer Tea is (a) 51 (b) 53 (c) 50 (d) 47
(v) The number of students who prefer Milk and Coffee but not tea is
(a) 12 (b) 10 (c) 15 (d) 20
Answers
Answer:
A)Solution
Correct option is
A
20
According to given information, the venn diagram is below.
Now, n(M∪C∪T)=12+5+8+10+15+20+10=80
Now, number of people who did not took any of the drinks is n(M
′
∩C
′
∩T
′
)=n(M∪C∪T)
′
⇒n(M∪C∪T)
′
=N−n(M∪C∪T)
⇒n(M∪C∪T)
′
=100−80=20
B) Solution
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
n(U)=600,n(T)=150,n(C)=225,n(T∩C)=100
To find: Number of student taking neither tea nor coffee i.e., we have to find n(T
′
∩C
′
).
n(T
′
∩C
′
)=n(T∪C)
′
=n(U)−n(T∪C)
=n(U)−[n(T)+n(C)−n(T∩C)]
=600−[150+225−100]
=600−275
=325
Hence, 325 students were taking neither tea nor coffee.
Step-by-step explanation:
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