Real Numbers - Case Study 2

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CASE STUDY 2: 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.

Prime factorization

• 60 = 2² × 3 × 5


• 84 = 2² × 3 × 7


• 108 = 2² × 3³

HCF : Take minimum powers of common primes:

• 2² is common, 3¹ is common, 5 and 7 are not common


• HCF = 2² × 3 = 4 × 3 = 12

Ans: b) 12

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

Divide participants by 12 (HCF) for each subject:

• 60 ÷ 12 = 5 rooms


• 84 ÷ 12 = 7 rooms


• Mathematics: 108 ÷ 12 = 9 rooms

Total rooms = 5 + 7 + 9 = 21

Ans: d) 21

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

Use prime factorizations:

• 60 = 2² × 3 × 5


• 84 = 2² × 3 × 7


• 108 = 2² × 3³

LCM = take maximum powers of each prime

• 2² (max), 3³ (max), 5¹, 7¹ → LCM = 2² × 3³ × 5 × 7


• Step by step: 2² = 4, 3³ = 27


• 4 × 27 = 108


• 108 × 5 = 540


• 540 × 7 = 3780

Ans: a) 3780

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

        HCF = 12, LCM = 3780

• Product = 12 × 3780 = 45360

Ans: d) 45360

5. 108 can be expressed as a product of its primes as a) b) c) d)

• Divide by 2 repeatedly: 108 ÷ 2 = 54, ÷2 = 27


• Divide by 3 repeatedly: 27 ÷ 3 = 9, ÷3 = 3, ÷3 = 1


• So 108 = 2² × 3³

Ans: b) 2² × 3³