Answer :
Let's solve these two problems step-by-step:
a) To find out after what length of time in hours the three bells that ring at intervals of 15, 18, and 30 minutes will ring together again, we need to calculate the least common multiple (LCM) of the intervals at which they ring.
First, we list the prime factors of each number:
- The prime factors of 15 are 3 and 5 (15 = 3 × 5).
- The prime factors of 18 are 2, 3 and 3 (18 = 2 × 3^2).
- The prime factors of 30 are 2, 3 and 5 (30 = 2 × 3 × 5).
To find the LCM, we take the highest powers of all prime factors that appear in any of the numbers:
- For the prime factor 2, the highest power is 2^1 (from the number 18 and 30).
- For the prime factor 3, the highest power is 3^2 (from the number 18).
- For the prime factor 5, we have 5^1 (from the numbers 15 and 30).
The LCM is therefore 2^1 × 3^2 × 5^1 = 2 × 9 × 5 = 90.
The three bells will ring together again after 90 minutes. To convert this period into hours, we divide by 60:
90 minutes / 60 = 1.5 hours.
So the three bells will ring together again after 1.5 hours.
b) The task is to find the least number of learners in the school that can be equally divided into groups of 24, 30, and 32. This is again an LCM problem.
We will list the prime factors of each number:
- The prime factors of 24 are 2, 2, 2, and 3 (24 = 2^3 × 3).
- The prime factors of 30 are 2, 3, and 5 (30 = 2 × 3 × 5).
- The prime factors of 32 are 2, 2, 2, 2, and 2 (32 = 2^5).
We take the highest powers of all prime factors:
- For the prime factor 2, the highest power is 2^5 (from the number 32).
- For the prime factor 3, the highest power is 3^1 (from the numbers 24 and 30).
- For the prime factor 5, we have 5^1 (from the number 30).
The LCM is therefore 2^5 × 3^1 × 5^1 = 32 × 3 × 5 = 960.
The least number of learners in the school that can be divided into groups of 24, 30, and 32 without any learner being excluded is 960.
