Question 2

In a business school, only three specializations are offered: marketing, finance and operations. A student can opt for either one, or two, or no specializations. In the current batch of 120 students, 60 are specializing in marketing, 50 are specializing in finance and 30 are specializing in operations. During the placement process, all students specializing either in marketing or in finance are shortlisted for consulting job interviews.
If 45 students are not shortlisted for consulting job interviews, what is the MINIMUM possible number of students specializing in both marketing and finance?

Solution

In the above figure, let the number of students specializing in :

Only marketing = a, Only finance = b, Only operations = c

Both marketing and finance but not operations = d, Both marketing and operations but not finance = e, Both finance and operations but not marketing = f

All three subjects = g, None of the subjects = h

Now it is given that only students having an specialization in marketing or finance gets a consulting shortlist. So, we can say that students doing specialization only in operations and students not doing specialization in any of the 3 subjects will not get a shortlist.

Hence, c + h = 45 (given) $$\longrightarrow\ i$$

Number of students doing specialization in marketing = a + d + e + g = 60 (given) $$\longrightarrow\ ii$$

Number of students doing specialization in finance = b + d + g + f = 50 (given) $$\longrightarrow\ iii$$

Number of students doing specialization in operations = c + e + g + f = 30 (given) $$\longrightarrow\ iv$$

Number of students who received the consulting shortlist = a + b + d + e + f + g = 75 $$\longrightarrow\ v$$

By using equation iv and v, we get,

a + b + d + 30 - c = 75

a + b + d = 45 + c $$\longrightarrow\ vi$$

Add equation ii and iii,

a + b + d + (d + g) + e + f + g = 110 $$\longrightarrow\ vii$$

Now, using equation iv, vi and vii, we get,

45 + c + (d + g) + 30 - c = 110