In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are....
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Answers
Answer:
The Number of room will be minimum if each room accomodates maximum
number of participants. Since in each room the same number of participants
are to be seated and all of them must be of the same subject. Therefore, 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 as under.
60=2
2
×3×5,84=2
2
×3×7 and 108=2
2
×3
3
∴ HCF of 60,84 and 108 is 2
2
×3=12
Therefore, in each room 12 participants can be seated.
∴ Number of rooms required=
12
Total number of participants
=
12
60+84+108
=
12
252
=21
Answer:
HCF (60,84,108)
Step-by-step explanation:
A.T.Q
using prime factorisation method
- 60=2×2×3×5=2²×3×5
- 84=2×2×3×7=2²×3×7
- 108=2×2×3×3×3=2²×3³
Maximum number of participants that can be accommodated in each room HCF (60,84,108)
=product of the smallest power of each
common prime factor in the number .