Question

CASE STUDY: A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects, the number of participants in Mathematics, Science and English are 72, 60, 96 respectively Give the answer of the following questions(1,2 and3) 1. In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can be accommodated in each room are: 1 point 16 18 14 12 Clear selection 2. Number of rooms required during the event are: 1 point 19 20 16 18 3. The product of HCF and LCM of three numbers = Product of three numbers 1 point False True​

Answer 1

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

Step-by-step explanation:

hope it helps you

Answer 2

Answer:

Number of participants in each room

= HCF of 36, 60, 84 36

= 2 x 2 x 3 x 3 60

= 2 x 2 x 3 x 5 84

= 2 x 2 x 3 x 7

HCF = 2 x 2 x 3

= 12 So, the number of students in each class

= 12 Total number of students

= 36 + 60+ 84 = 180

Number of rooms required

= 180 / 12

= 15.