Q) A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

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 participants that can accommodated in each room are
a) 14                    b) 12                     c) 16                          d) 18

2. What is the minimum number of rooms required during the event?
a) 11                    b) 31                      c) 41                          d) 21

3. The LCM of 60, 84 and 108 is
a) 3780                b) 3680                 c) 4780                      d) 4680

4. The product of HCF and LCM of 60,84 and 108 is
a) 55360              b) 35360               c) 45500                     d) 45360

5. 108 can be expressed as a product of its primes as
a) 2³ x 3²             b)2³ x 3³               c) 2² x 3²                   d) 2² x 3³

Ans:

STEP BY STEP SOLUTION

1. Maximum number of participants in each room:

Here, we need to find the maximum number of participants and this number need to be same across all 3 subjects.

Therefore, that number will be a factor, common to all the three subjects’ participants and has to be highest.

Hence, we will find the HCF of 60, 84 and 108

By Prime factorisation, let’s find factors of 60, 84 and 108:

60 = 2 x 2 x 3 x 5

84 = 2 x 2 x 3 x 7

108 = 2 x 2 x 3 x 3 x 3

Now the HCF will have only the common factors among all 3 numbers

Hence, the HCF is: (2 x 2 x 3) = 12

Therefore, option b) is correct

2. Number of rooms required: 

Now a set of12 participants for each subject will occupy a room.

Therefore, total number of rooms = Rooms for Hindi (R _H ) + Rooms for English (R _E ) + Rooms for Mathematics (R _M )

Now Rooms required for Hindi =

∴ R _H = = 5

Rooms required for English =

∴ R _E = = 7

Rooms required for Mathematics =

∴ R _M = = 9

∴ Total number of rooms = R _H  + R _E  + R _M 

= 5 + 7 + 12 = 21

Therefore, option d) is correct

3. LCM of 60, 84, 108:

In part 1, By Prime factorisation method, we have already calculated factors of 60, 84 and 108

Now the LCM will have common factors among all 3 numbers, uncommon factors of 1st number, uncommon factors of 2nd number.and uncommon factors of 3nd number.

∴ LCM (60, 84, 108) = (2 x 2 x 3) x 5 x 7 x 3 x 3

∴ LCM (60, 84, 108) = 3780

Therefore, option a) is correct

4. Product of HCF and LCM:

In part 1, we calculated HCF = 12

In part 3, we calculated LCM = 3780

∴ HCF x LCM = 12 x 3780 = 45,360

Therefore, option d) is correct

5. Expression for 108: 

By Prime factorisation method, we write 108 = 2 x 2 x 3 x 3 x 3

∴ 108 = 2² x 3³

Therefore, option d) is correct

Please press the “Heart” button if you like the solution.