Concept : 1 Mark
Application : 2 Marks
Conclusion : 1 Mark
It is given that, each room the same number of participants are to be seated and all of them must be of the same subject.
The number of rooms will be minimum if each room accomodates maximum number of participants.
∴ The number of participants in each room must be the HCF of 60, 84 and 108.
The prime factorisations of 60, 84 and 108 are,
60=22×3×5
84=22×3×7
108=22×32
∴ HCF of 60, 84 and 108 is 22×3=12
Therefore, in each room 12 participants can be seated.
∴ Number of rooms required=Total number of participants12
=60+84+10812=25212=21