Write down: the prime factors of 132 . the HCF of 47 and 97 . the I.CM of 24 and 96 . Three boys take steps of \( 54 \mathrm{~cm}, 60 \mathrm{~cm} \) and 72 cm respectively. They start walking together in step from the same starting point and walk in the same direction. What is the shortest distance they must walk before they are in step again? Divide R 6500 in the ratio of \( 3: 4: 6 \). 15 pears cost R 5,60 . How much will 27 of the same pears cost? 7 men build a house in 300 days. How many men, working at the same rate, can build the house in 70 days?

Alright, I need to tackle these mathematical problems step by step. Let's start with the first one: finding the prime factors of 132. First, I'll recall that prime factors are the prime numbers that multiply together to give the original number. So, I'll begin by dividing 132 by the smallest prime number, which is 2. 132 ÷ 2 = 66 Now, 66 is still even, so I'll divide by 2 again. 66 ÷ 2 = 33 33 is not divisible by 2, so I'll move to the next prime number, which is 3. 33 ÷ 3 = 11 11 is a prime number, so I'll stop here. Therefore, the prime factors of 132 are 2, 2, 3, and 11. Next, I need to find the Highest Common Factor (HCF) of 47 and 97. Since both numbers are prime, their HCF should be 1, as they don't share any common factors other than 1. Moving on to the Least Common Multiple (LCM) of 24 and 96. To find the LCM, I'll first find the prime factors of both numbers. For 24: 24 ÷ 2 = 12 12 ÷ 2 = 6 6 ÷ 2 = 3 3 ÷ 3 = 1 So, the prime factors of 24 are 2, 2, 2, and 3. For 96: 96 ÷ 2 = 48 48 ÷ 2 = 24 24 ÷ 2 = 12 12 ÷ 2 = 6 6 ÷ 2 = 3 3 ÷ 3 = 1 So, the prime factors of 96 are 2, 2, 2, 2, 2, and 3. To find the LCM, I'll take the highest power of each prime number present in the factorizations: - The highest power of 2 is 2^5 (from 96). - The highest power of 3 is 3^1 (from both 24 and 96). Therefore, LCM = 2^5 * 3^1 = 32 * 3 = 96. Now, for the next problem: Three boys take steps of 54 cm, 60 cm, and 72 cm respectively. They start walking together from the same starting point and walk in the same direction. I need to find the shortest distance they must walk before they are in step again. This is essentially finding the Least Common Multiple (LCM) of their step lengths. Let's find the LCM of 54, 60, and 72. First, find the prime factors: - 54: 2 * 3^3 - 60: 2^2 * 3 * 5 - 72: 2^3 * 3^2 The LCM will be the product of the highest powers of all primes present: - 2^3 (from 72) - 3^3 (from 54) - 5^1 (from 60) So, LCM = 2^3 * 3^3 * 5 = 8 * 27 * 5 = 1080 cm. Therefore, the shortest distance they must walk before they are in step again is 1080 cm. Next, I need to divide R 6500 in the ratio of 3:4:6. This means the total parts are 3 + 4 + 6 = 13 parts. Each part is worth R 6500 / 13 ≈ R 500. So, the amounts are: - First part: 3 * 500 = R 1500 - Second part: 4 * 500 = R 2000 - Third part: 6 * 500 = R 3000 Now, for the next problem: 15 pears cost R 5.60. I need to find out how much 27 pears will cost. First, find the cost per pear: R 5.60 / 15 pears = R 0.3733 per pear. Then, multiply by 27 pears: 0.3733 * 27 ≈ R 10.08. So, 27 pears will cost approximately R 10.08. Finally, the last problem: 7 men build a house in 300 days. I need to find out how many men are needed to build the house in 70 days, working at the same rate. First, determine the total man-days required to build the house: 7 men * 300 days = 2100 man-days. Now, to build the house in 70 days, the number of men needed is: 2100 man-days / 70 days = 30 men. Therefore, 30 men are needed to build the house in 70 days.