Group QuestionAnswer the following question based on the information g...
Let the number of candidates who got placed in only Infrastructure be x. Then, the number of candidates who got placed in Services as well as Consultancy, but not in Infrastructure is also x.
Let y be the number of candidates who got placed in all the three industries, then we have y < 5 ... (i)
Let z be the number of candidates who got placed in only Consultancy. Then, we have the following graphical representation:
Now, the number of candidates who got placed in Services as well as Consultancy was one more than those who got placed in Infrastructure as well as Services, but was one less than those who got placed in Infrastructure as well as Consultancy. Since the number of candidates who got placed in Services as well as Consultancy (but not Infrastructure) is jc, hence the number of candidates who got placed in Infrastructure as well as Services (but not Consultancy) is (x - 1) and the number of candidates who got placed in Infrastructure and Consultancy (but not services) would be (x + 1).
Let p be the number of candidates who got placed in only Services. Since the number of candidates who did not get placed in Services is the same as those who did not get placed in Consultancy, we have:
x + (x - 1) +p = x + (x + 1) + z; implying that p = z + 2 Hence, the final distribution would be as shown below:
Now, since the total number of candidates in this institute was less than 40, we have;
4x + y + 2z + 2 < 40
Implying that 4x + y + 2z < 3 8 ... (ii)
Further, since the number of candidates who got placement from exactly one industry was 15, we have:
x + 2z + 2 = 15, implying x + 2z = 13 ... (iii)
From (ii) and (iii), we get, 3x + y < 25 ... (iv)
Now, since y cannot be negative, we have x < 8 ...(v)
The number of candidates who got placed in at least two industries, one of which is Services, is more than those who got placed in Infrastructure as well as Consultancy. Hence, we have,
2x+y—>x+y+9 implying x > 2
Thus, we have, 2 < x < 8 and further from (iii) x can take only odd values, hence x = 3 or 5 or 7
Case 1: If x = 3, then z = 5 Case 2: If x = 5, then z = 4 Case 3: Ifx = 7, thenz = 3
However, the number of candidates who did not get placed in Services was different from those who did not get placed in Infrastructure; i.e. 2x + z + 1 ^ x + 2z + 2 i.e. x^z + 1
Thus, Case 2 gets eliminated.
Hence, now we have two possibilities:
Since we are supposed to maximize the number of placements, this would be possible in Case 2, where the total number of candidates is 36 + y. Now since the maximum number of candidates in this institute is less than 40, we have 36 +y = 39, implying thaty = 3. Note that in Case 1, even though we get the total number of candidates as 24 +y = 39, implying thaty =15. However, since the maximum possible value ofy can be 5, the total number of placements can be (3 + 5 + 7) + 2(2 + 4 + 3) + 3(5) = 48.
Hence, from Case 2, the maximum number of placements would be: (7 + 5 + 3) + 2(6 + 8 + 7) + 3(3) = 15 + 42 + 9 = 66
Hence, option 4.