there are 1500 students who appeared in the cma examination under the icmab.out of this students, 450 failed in accounting, 500 failed in business mathematics and 475 are failed in costing. 200 failed in both accounting and business mathematics were 300,those who failed in both business mathematics and costing were 320, and those who failed in both accounting and costing were 350. the students who failed in all the three subject were 250.find
(1)how many students failed in at least any one of the subjects?
(2)How many students failed in no subjects?
(3)How many students failed in both accounting and business mathematics only?
Answers
Step-by-step explanation:
1) failed in any one of the subject
n(AUBUC)=
n(A)+n(B)+n(C)-n(AnB)-n(AnC)-n(BnC)+n(AnBnC)
=705
2)no subjects =total - n(AUBUC)
1500 -705 = 795
3) only A and B = n(AnB)-n(AnBnC)=300-250=50
Given:
Total students = 1500
Students failed in accounting, A = 450
Students failed in mathematics, B = 500
Students failed in costing, C = 475
Students failing in both accounting and mathematics = 300
Students failing in both mathematics and costing = 320
Students failing in both accounting and costing = 350
Students failing in all three subjects = 250
To Find:
- How many students failed in at least any one of the subjects?
- How many students failed in no subjects?
- How many students failed in both accounting and business mathematics only?
Solution:
The answers of each of these questions will be given by Set theory.
AUB = A union B
A∩B = A intersection B
1) Failed in any one of the subject,
n(AUBUC)=
n(A)+n(B)+n(C)-n(AnB)-n(AnC)-n(BnC)+n(AnBnC)
=705
2)Failed in no subjects = total - n(AUBUC)
1500 -705 = 795
3) Failed in only A and B = n(AnB)-n(AnBnC)
=300-250= 50
Hence the no. of students who failed in atleast one subjects is 705
The no of students who failed in no subject is 795
The no. of students who failed in both accounting and mathematics is 50