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Sagot :
To solve the problem of determining after how many days persons A, B, and C will meet again at the starting point, we need to analyze their running speeds and find when their paths coincide.
### Step-by-Step Solution:
1. Calculate the Time Each Person Takes to Complete One Circle:
- The length of the circular road is 325 li.
- Person A runs at a speed of 150 li per day.
[tex]\[ \text{Time taken by A to complete one circle} = \frac{325 \text{ li}}{150 \text{ li/day}} = \frac{325}{150} \text{ days} = \frac{13}{6} \text{ days} \][/tex]
- Person B runs at a speed of 120 li per day.
[tex]\[ \text{Time taken by B to complete one circle} = \frac{325 \text{ li}}{120 \text{ li/day}} = \frac{325}{120} \text{ days} = \frac{65}{24} \text{ days} \][/tex]
- Person C runs at a speed of 90 li per day.
[tex]\[ \text{Time taken by C to complete one circle} = \frac{325 \text{ li}}{90 \text{ li/day}} = \frac{325}{90} \text{ days} = \frac{65}{18} \text{ days} \][/tex]
2. Convert These Fractions to a Common Format:
To simplify finding the least common multiple (LCM), it's easier to work with these fractions by scaling them to a suitable common base:
[tex]\[ \frac{13}{6}, \frac{65}{24}, \frac{65}{18} \][/tex]
3. Find the LCM of the Numerators and the Denominators Separately:
Firstly, note that:
- The least common multiple (LCM) of the denominators 6, 24, and 18 must be identified.
- Factorize these numbers:
[tex]\[ 6 = 2 \times 3 \][/tex]
[tex]\[ 24 = 2^3 \times 3 \][/tex]
[tex]\[ 18 = 2 \times 3^2 \][/tex]
- LCM involves taking the highest powers of all prime factors:
[tex]\[ \text{LCM}(6, 24, 18) = 2^3 \times 3^2 = 8 \times 9 = 72 \][/tex]
- Convert all fractions to have the denominator 72:
[tex]\[ \frac{13}{6} = \frac{13 \times 12}{6 \times 12} = \frac{156}{72} \][/tex]
[tex]\[ \frac{65}{24} = \frac{65 \times 3}{24 \times 3} = \frac{195}{72} \][/tex]
[tex]\[ \frac{65}{18} = \frac{65 \times 4}{18 \times 4} = \frac{260}{72} \][/tex]
4. Find the LCM of the Numerators 156, 195, and 260:
- Factorize these numbers:
[tex]\[ 156 = 2^2 \times 3 \times 13 \][/tex]
[tex]\[ 195 = 3 \times 5 \times 13 \][/tex]
[tex]\[ 260 = 2^2 \times 5 \times 13 \][/tex]
- LCM involves taking the highest powers of all prime factors:
[tex]\[ \text{LCM}(156, 195, 260) = 2^2 \times 3 \times 5 \times 13 = 4 \times 3 \times 5 \times 13 = 4 \times 3 \times 65 = 12 \times 65 = 780 \][/tex]
5. Converting to a Whole Number of Days:
Since the least common multiple of the fractions' numerators (scaled to 72) is the smallest multiple of 72 that is a common multiple of 156, 195, and 260, the time when they will meet again at the starting point is:
[tex]\[ \frac{780}{72} \text{ days} \approx 10.83 \text{ days} \][/tex]
How many days will they meet again? Simplify the fraction:
[tex]\[ \text{LCM of all relevant days is} \frac{780}{72} = 10.83 \text{ days} \][/tex]
So they will meet again at the starting point after 780/72 days, or approximately 10.83 days. Converting this to a whole number for better understanding, it is approximately 11 days as they would meet again at the starting point.
### Step-by-Step Solution:
1. Calculate the Time Each Person Takes to Complete One Circle:
- The length of the circular road is 325 li.
- Person A runs at a speed of 150 li per day.
[tex]\[ \text{Time taken by A to complete one circle} = \frac{325 \text{ li}}{150 \text{ li/day}} = \frac{325}{150} \text{ days} = \frac{13}{6} \text{ days} \][/tex]
- Person B runs at a speed of 120 li per day.
[tex]\[ \text{Time taken by B to complete one circle} = \frac{325 \text{ li}}{120 \text{ li/day}} = \frac{325}{120} \text{ days} = \frac{65}{24} \text{ days} \][/tex]
- Person C runs at a speed of 90 li per day.
[tex]\[ \text{Time taken by C to complete one circle} = \frac{325 \text{ li}}{90 \text{ li/day}} = \frac{325}{90} \text{ days} = \frac{65}{18} \text{ days} \][/tex]
2. Convert These Fractions to a Common Format:
To simplify finding the least common multiple (LCM), it's easier to work with these fractions by scaling them to a suitable common base:
[tex]\[ \frac{13}{6}, \frac{65}{24}, \frac{65}{18} \][/tex]
3. Find the LCM of the Numerators and the Denominators Separately:
Firstly, note that:
- The least common multiple (LCM) of the denominators 6, 24, and 18 must be identified.
- Factorize these numbers:
[tex]\[ 6 = 2 \times 3 \][/tex]
[tex]\[ 24 = 2^3 \times 3 \][/tex]
[tex]\[ 18 = 2 \times 3^2 \][/tex]
- LCM involves taking the highest powers of all prime factors:
[tex]\[ \text{LCM}(6, 24, 18) = 2^3 \times 3^2 = 8 \times 9 = 72 \][/tex]
- Convert all fractions to have the denominator 72:
[tex]\[ \frac{13}{6} = \frac{13 \times 12}{6 \times 12} = \frac{156}{72} \][/tex]
[tex]\[ \frac{65}{24} = \frac{65 \times 3}{24 \times 3} = \frac{195}{72} \][/tex]
[tex]\[ \frac{65}{18} = \frac{65 \times 4}{18 \times 4} = \frac{260}{72} \][/tex]
4. Find the LCM of the Numerators 156, 195, and 260:
- Factorize these numbers:
[tex]\[ 156 = 2^2 \times 3 \times 13 \][/tex]
[tex]\[ 195 = 3 \times 5 \times 13 \][/tex]
[tex]\[ 260 = 2^2 \times 5 \times 13 \][/tex]
- LCM involves taking the highest powers of all prime factors:
[tex]\[ \text{LCM}(156, 195, 260) = 2^2 \times 3 \times 5 \times 13 = 4 \times 3 \times 5 \times 13 = 4 \times 3 \times 65 = 12 \times 65 = 780 \][/tex]
5. Converting to a Whole Number of Days:
Since the least common multiple of the fractions' numerators (scaled to 72) is the smallest multiple of 72 that is a common multiple of 156, 195, and 260, the time when they will meet again at the starting point is:
[tex]\[ \frac{780}{72} \text{ days} \approx 10.83 \text{ days} \][/tex]
How many days will they meet again? Simplify the fraction:
[tex]\[ \text{LCM of all relevant days is} \frac{780}{72} = 10.83 \text{ days} \][/tex]
So they will meet again at the starting point after 780/72 days, or approximately 10.83 days. Converting this to a whole number for better understanding, it is approximately 11 days as they would meet again at the starting point.
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