1. **Problem:** Three bells ring at intervals of 6, 8, and 10 minutes. They ring together at 9:00 a.m. When will they ring together again? Step 1: Find the Least Common Multiple (LCM) of 6, 8, and 10. Step 2: Factorize each number: 6 = 2 \times 3 8 = 2^3 10 = 2 \times 5 Step 3: LCM is the product of the highest powers of all prime factors: $$\text{LCM} = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120$$ Step 4: So they will ring together every 120 minutes. Step 5: Adding 120 minutes to 9:00 a.m. gives 11:00 a.m. **Answer:** They will ring together again at 11:00 a.m. 2. **Problem:** A gardener has 72 roses, 96 lilies, and 120 tulips and wants to arrange them in the maximum number of rows with the same number of each flower per row. Step 1: Find the Highest Common Factor (HCF) of 72, 96, and 120. Step 2: Factorize each number: 72 = 2^3 \times 3^2 96 = 2^5 \times 3 120 = 2^3 \times 3 \times 5 Step 3: HCF is the product of the lowest powers of common prime factors: $$\text{HCF} = 2^3 \times 3 = 8 \times 3 = 24$$ Step 4: Maximum number of rows = 24. Step 5: Each row has: Roses = 72 \div 24 = 3 Lilies = 96 \div 24 = 4 Tulips = 120 \div 24 = 5 **Answer:** Maximum 24 rows, each with 3 roses, 4 lilies, and 5 tulips. 3. **Problem:** A fruit seller has 240 apples, 300 oranges, and 360 mangoes. He wants to pack them in boxes with equal number of fruits of the same kind per box. (a) Find the greatest number of boxes he can use. Step 1: Find HCF of 240, 300, and 360. Step 2: Factorize: 240 = 2^4 \times 3 \times 5 300 = 2^2 \times 3 \times 5^2 360 = 2^3 \times 3^2 \times 5 Step 3: HCF = lowest powers of common primes = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 Step 4: Greatest number of boxes = 60. (b) Number of fruits per box: Apples: 240 \div 60 = 4 Oranges: 300 \div 60 = 5 Mangoes: 360 \div 60 = 6 **Answer:** (a) 60 boxes; (b) each box has 4 apples, 5 oranges, 6 mangoes. 4. **Problem:** Find the minimum possible length of rope that can be cut into pieces of 75 cm, 90 cm, and 120 cm without any leftover. Step 1: Find LCM of 75, 90, and 120. Step 2: Factorize: 75 = 3 \times 5^2 90 = 2 \times 3^2 \times 5 120 = 2^3 \times 3 \times 5 Step 3: LCM = highest powers of all primes: $$2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$$ Step 4: Minimum rope length = 1800 cm = 18 meters. **Answer:** The rope must be at least 1800 cm long. 5. **Problem:** Two traffic lights change every 36 seconds and 54 seconds respectively. If they turn green together at 9:00 a.m., when will they turn green together again? Step 1: Find LCM of 36 and 54. Step 2: Factorize: 36 = 2^2 \times 3^2 54 = 2 \times 3^3 Step 3: LCM = 2^2 \times 3^3 = 4 \times 27 = 108 Step 4: They turn green together every 108 seconds. Step 5: 108 seconds = 1 minute 48 seconds. Step 6: Add 1 minute 48 seconds to 9:00:00 a.m. to get 9:01:48 a.m. **Answer:** The lights will turn green together again at 9:01:48 a.m.