FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.13or, 43 = 15 + 20 + 20 – 3 – 6 – 5 + n (A ∪ B ∪ C)
or, n (A ∩ B ∩ C) = 2.
Example 16 : In a survey of 1000 families it is found that 454 use electricity, 502 use gas, 448 use kerosene, 158
use gas and electricity, 160 use gas and kerosene and 134 use electricity and kerosene for cooking. If all of
them use at least one of the three, find how many use all the three fuels.
Solution:
Let us take E for electricity, G for gas, K for kerosene.
Now n (E) = 454, n (G) = 502, n (K) = 448.
n ( G ∩ E) = 158. n ( G ∩ K) = 160, n (E ∩ K) = 134, n (E ∩ G ∩ K) =?
n (E ∪ G ∪ K) = 1000
Again n (E ∪ G ∪ K) = n (E) + n (G) + n (K) – n (E ∩ G)
–n (G ∩ K) – n (K ∩ E) + (E ∩ G ∩ K)
or, 1000 = 454 + 502 + 448 – 158 – 160 –134 + n (E ∩ G ∩ K)
= 952 + n (E ∩ G ∩ K)
or, n (E ∩ G ∩ K) = 1000 – 952 = 48.
Example 17 : In a class of 50 students appearing for an examination of ICWA, from a centre, 20 failed in
Accounts, 21 failed in Mathematics and 27 failed in Costing, 10 failed both in Accounts and Costing, 13
failed both in Mathematics and Costing and 7 failed both in Accounts and Mathematics. If 4 failed in all
the three, find the number of
(i) Failures in Accounts only.
(ii) Students who passed in all the three subjects.
Solution:
A = Accounts, M = Mathematics, C = Costing [No. of students failed (say)]
Now n (A) = 20, n(M) = 21, n (C) = 27, n (A ∩ C) = 10, n (M ∩ C) = 13, n (A ∩ M) = 7
n (A ∩ M ∩ C) = 4.
(i) n(A M C∩ ∩ ) = n (A) – n (A ∩ M) – n (A ∩ C) + n (A ∩ M ∩ C) = 20 –7 – 10 + 4 = 7
(ii) Total no of students failed
= n (A) + n (M) + n(C) – n (A ∩ M) – n (M ∩ C) – n (A ∩ C) + n (A ∩ M ∩ C)
= 20 + 21 + 27 – 7 – 13 – 10 – 4 = 42
∴ reqd. no. of pass = 50 – 42 = 8.
SELF EXAMINATION QUESTIONS
- (i) If n (A) = 20, n(B) = 12, n (A ∩ B) = 4, find n (A ∪ B) [Ans. 28]
(ii) H n (A) = 41, n(B) = 19, n(A ∩ B) = 10, find n (A ∪ B) [Ans. 50]
(iii) If n (A) = 12, n (B) = 20, and A χ B, Find n (A ∪ B) [Ans. 20]
(iv) If n (A) = 24, n (B) = 18 and B c A, find n (A ∪ B) [Ans. 24]
[Hints. n(A ∩ B) = n (B) as B ⊆ A
n (A ∪ B) = n (A) + n (B) - n (A ∩ B) = n (A) + n (B) - n (B) = n (A) = etc.] - In a class 60 students took mathematics and 30 took Physics. If 17 students were enrolled in both the