Let the total no of students doing CA be n(A), those doing actuarial be n(B) and those doing MBA be n(C). Now, ‘CA’ is a set of all even numbered students, thus n(A) = = 75 ‘Actuarial’ is a set of all the students whose number are divisible by 5, thus n(B) = = 30 ‘MBA’ is a set of all the students whose number are divisible by 7, thus n(C) = = 21 The 10th, 20th, 30th…… numbered students would be doing both CA and Actuarial ∴ n(A∩B) = = 15 The 14th, 28th, 42nd…… numbered students would be doing both CA and MBA ∴ n(A∩C) = = 10 The 35th, 70th, …… numbered students would be doing both Actuarial and MBA ∴ n(B∩C) = = 4 And the 70th and 140th students must be doing all the three. ∴n(A∩B∩C) = 2 Now, n(A∪B∪C) = n(A)+n(B)+n(C) – n(A∩B) – n(A∩C) - n(B∩C) + n(A∩B∩C) = 99 ∴ Number of students doing nothing = 150 – 99 = 51 = (c) Answer - C Correct Answer: 51.
All features of the online course, including the classes, discussion board, quizes and more, on a mobile platform.
Download videos onto your mobile so you can learn on the fly, even when the network gets choppy!