2. Find the LCM for these sets of numbers. \( \quad \) b) 18 and 48 \( \begin{array}{ll}\text { a) } 24 \text { and } 90 & \text { c) } 90,105 \text { and } 150 \\ \text { Mr Abrahams is able to divide the learners in his Mathematics class into equal } \\ \text { groups of } 2,5 \text { or } 6 \text { without leaving any learners out. What is the smallest } \\ \text { possible number of learners in the class? } \\ \text { In a village, the bell rings every } 30 \text { minutes at the Junior School and every } 40 \\ \text { minutes at the High School. Both schools start lessons at 8:00 a.m. } \\ \text { a) At what times between 8:00 a.m. and 2:30 p.m. do the bells both ring at } \\ \text { the same time? }\end{array} \) If the High School bell rings every 45 minutes, how many times during school day would this bell ring at the same time as the Junior School

To find the LCM of 18 and 48, first, we determine their prime factorization: 18 is \(2 \times 3^2\) and 48 is \(2^4 \times 3\). The LCM is found by taking the highest powers of all prime factors present, which results in \(2^4 \times 3^2 = 144\). Thus, the LCM of 18 and 48 is 144. Now, considering Mr. Abrahams' class, to find the smallest number of learners that can be divided by 2, 5, and 6, we first calculate the LCM of these numbers. The prime factors give us: 2 as \(2^1\), 5 as \(5^1\), and 6 as \(2^1 \times 3^1\). Therefore, the LCM is \(2^1 \times 3^1 \times 5^1 = 30\). Mr. Abrahams could have 30 learners in his class. As for the bells from the Junior School ringing every 30 minutes and the High School every 40 minutes, the LCM of 30 and 40 is 120 minutes or 2 hours. Therefore, the bells will ring together at 8:00 a.m., 10:00 a.m., and 12:00 p.m. Afterward, with the High School bell ringing every 45 minutes as well, both bells will ring together at 10:30 a.m. and 1:30 p.m. During the school day, this means the High School bell will ring simultaneously with the Junior School bell three times total: at 10:00 a.m., 10:30 a.m., and 1:30 p.m.