1a) To find the HCF (Highest Common Factor) and LCM (Least Common Multiple) of 224 and 336, we can use the prime factorization method.
Prime factorization of 224:
224 = 2 * 2 * 2 * 2 * 2 * 7 = 2^5 * 7
Prime factorization of 336:
336 = 2 * 2 * 2 * 2 * 3 * 7 = 2^4 * 3 * 7
HCF: We take the common prime factors with the lowest exponents, which are 2^4 and 7.
HCF of 224 and 336 = 2^4 * 7 = 112
LCM: We take all the prime factors with the highest exponents.
LCM of 224 and 336 = 2^5 * 3 * 7 = 672
Therefore, the HCF of 224 and 336 is 112, and the LCM is 672.
1b) Prime factorization of 18:
18 = 2 * 3 * 3 = 2 * 3^2
Prime factorization of 42:
42 = 2 * 3 * 7
HCF: The common prime factors are 2 and 3.
HCF of 18 and 42 = 2 * 3 = 6
LCM: We take all the prime factors with the highest exponents.
LCM of 18 and 42 = 2 * 3^2 * 7 = 126
Therefore, the HCF of 18 and 42 is 6, and the LCM is 126.
1c) Prime factorization of 45:
45 = 3 * 3 * 5 = 3^2 * 5
Prime factorization of 150:
150 = 2 * 3 * 5 * 5 = 2 * 3 * 5^2
HCF: The common prime factors are 3 and 5.
HCF of 45 and 150 = 3 * 5 = 15
LCM: We take all the prime factors with the highest exponents.
LCM of 45 and 150 = 2 * 3^2 * 5^2 = 450
Therefore, the HCF of 45 and 150 is 15, and the LCM is 450.
The interval at which the three patients visit the doctor can be found by taking the LCM of the intervals: 8, 15, and 24.
LCM of 8, 15, and 24 = 120
Therefore, they will next all go to the doctor on the same day after 120 days from 1 March. The date will be 29 June.
The bell rings every 15 minutes and the whistle is blown every 18 minutes. To find the time when they will be rung and blown at the same time again, we need to find the LCM of 15 and 18.
LCM of 15 and 18 = 90
Therefore, the bell will be rung and the whistle will be blown at the same time again after 90 minutes. The time will be 9:30 am.
To find the time when Anthony and Joseph will be at the same point on the circular track again, we need to find the LCM of their running times: 65 seconds and 75 seconds.
LCM of 65 and 75 = 975
Therefore, it will take 975 seconds for Anthony and Joseph to be at the same point on the circular track again.
swimmers question:
To find the time when the swimmers are side by side at one end of the pool again, we need to find the least common multiple (LCM) of their lap times.
The lap times of the three swimmers are 28 seconds, 44 seconds, and 68 seconds.
To find the LCM, we can list the multiples of each lap time until we find a common multiple:
Multiples of 28: 28, 56, 84, 112, 140, 168, 196, 224, 252, 280, 308, ...
Multiples of 44: 44, 88, 132, 176, 220, 264, 308, ...
Multiples of 68: 68, 136, 204, 272, 340, 408, ...
From the lists above, we can see that the least common multiple of 28, 44, and 68 is 308 seconds.
Therefore, it will take 308 seconds (or 5 minutes and 8 seconds) for the swimmers to be side by side at one end of the pool again.