Answer:
a) 37
b) 2
c) 17
d) 8
Step-by-step explanation:
90 Students went to the zoo. 3 had hamburger, milk and cake; 5 had milk and hamburger, 10 had cake and milk; 8 had cake and hamburger; 24 had hamburger; 38 had cake; 20 had milk. How many had a. nothing b. cake only c. milk only d. hamburger only
Solution:
Let h represent students that ate hamburger, m represent students that had milk and c represent students that had cake.
Given that:
n(h ∩ m ∩ c) = 3, n(m ∩ h) = 5, n(c ∩ m) = 10, n(c ∩ h) = 8, n(h) = 24, n(c) = 38, n(m) = 20
The number of students that had nothing = n(h ∪ m ∪ C)'
The number of students that had only milk = n(m ∩ h' ∩ C')
The number of students that had only cake = n(m' ∩ h' ∩ C)
The number of students that had only hamburger = n(m' ∩ h ∩ C')
a) n(m ∩ h' ∩ C') = n(m) - n(m ∩ h) - n(c ∩ m) - n(h ∩ m ∩ c) = 20 - 5 - 10 - 3 = 2
n(m' ∩ h ∩ C') = n(h) - n(m ∩ h) - n(c ∩ h) - n(h ∩ m ∩ c) = 24 - 5 - 8 - 3 = 8
n(m' ∩ h' ∩ C) = n(m) - n(m ∩ c) - n(c ∩ h) - n(h ∩ m ∩ c) = 38 - 10 - 8 - 3 = 17
n(m ∩ h' ∩ C') + n(m' ∩ h ∩ C') + n(m' ∩ h' ∩ C) + n(h ∪ m ∪ C)' + n(h ∩ m ∩ c) + n(m ∩ h) + n(c ∩ m) + n(c ∩ h) = 90
2 + 8 + 17 + 5 + 10 + 8 + 3 + n(h ∪ m ∪ C)' = 90
53 + n(h ∪ m ∪ C)' = 90
n(h ∪ m ∪ C)' = 37
b) n(m' ∩ h' ∩ C) = 17
c) n(m ∩ h' ∩ C') = 2
d) n(m' ∩ h ∩ C') = 8
