Math, asked by harshu38377, 1 year ago

7)
Out of 200 students; 35 students failed in
MHT-CET, 40 in AIEEE and 40 in IIT
entrance, 20 failed in MHT-CET and
AIEEE, 17 in AIEEE and IIT entrance, 15
in MHT-CET and IIT entrance and 5 failed
in all three examinations. Find how many
students.
i) did not fail in any examination.
ii) failed in AIEEE or IIT entrance.​

Answers

Answered by ashvin60
3

Answer:

I)37 students

II}40 students

Answered by sharonr
19

i)The number of students did not fail in any examination is 132

ii) The number of students failed in AIEEE or IIT entrance is 63

Solution:

Let n(M) be the number of students who failed in MHT-CET

n(M) = 35

Let n(A) be the number of students who failed in AIEEE,

n(A) = 40

Let n(I) be the number of students who failed in IIT,

n(I) = 40

Given 20 failed in MHT-CET and AIEEE,

n( M ∩ A) = 20

Given 17 failed in AIEEE and IIT entrance,

n(A ∩ I) = 17,

Given 15 failed in MHT-CET and IIT entrance

n(M ∩ I) = 15

Given 5 failed in all three examinations,

n(A ∩ M ∩ I) = 5

i) To find number of student who did not fail in any of the examination

= [n(A ∪ M ∪ I)']

= n(x) - n(A ∪ M ∪ I)

n(A ∪ M ∪ I) = n(A) + n(M) + n(I) - n(A∩M) - n(A∩I) - n(M∩ I) + n(A∩M∩I)

= 35 + 40 + 40 - 20 - 17 - 15 + 5

= 68

So n(A∪M∪I) = 68

Number of students who did not fail in any examination will be

200 - 68

= 132

ii) Students failed in AIEEE or IIT

= n(A∪I)

= n(A) + n(I) - n(A∩I)

= 40 + 40 - 17

= 63

Hence the students failed in AIEEE or IIT is 63

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