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Sagot :
To find the least possible number of adult men in KZN, we start by establishing the relationship between the number of adult men and adult women based on the given information:
1. According to the problem, [tex]\(\frac{2}{3}\)[/tex] of the adult men are married to [tex]\(\frac{3}{5}\)[/tex] of the adult women.
2. Let [tex]\(m\)[/tex] represent the number of adult men and [tex]\(w\)[/tex] represent the number of adult women.
3. The relationship translates to:
[tex]\[ \frac{2}{3} \cdot m = \frac{3}{5} \cdot w \][/tex]
4. Simplifying this equation, we get:
[tex]\[ 2m = \frac{9}{5}w \][/tex]
Multiplying both sides by 5 to clear the fraction:
[tex]\[ 10m = 9w \][/tex]
5. Solving for [tex]\(w\)[/tex] in terms of [tex]\(m\)[/tex], we have:
[tex]\[ w = \frac{10}{9}m \][/tex]
We need the number of women, [tex]\(w\)[/tex], to be an integer. Starting with at least 300 men, we check if there is an integer value for [tex]\(w\)[/tex] corresponding to an integer [tex]\(m \geq 300\)[/tex]:
1. If we set [tex]\(m = 300\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 300 = \frac{3000}{9} \approx 333.33 \quad \text{(not an integer)} \][/tex]
Since [tex]\(333.33\)[/tex] is not an integer, [tex]\(m = 300\)[/tex] is not a solution. We increment [tex]\(m\)[/tex] until we find the smallest [tex]\(m\)[/tex] making [tex]\(w\)[/tex] an integer:
2. Increasing [tex]\(m\)[/tex] step by step:
3. If we set [tex]\(m = 301\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 301 = \frac{3010}{9} \approx 334.44 \quad \text{(not an integer)} \][/tex]
4. If we set [tex]\(m = 302\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 302 = \frac{3020}{9} \approx 335.56 \quad \text{(not an integer)} \][/tex]
5. Continuing this process:
6. We find that when [tex]\(m = 306\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 306 = \frac{3060}{9} = 340 \quad \text{(an integer)} \][/tex]
Thus, the least possible number of adult men in the province, ensuring that the number of adult women is an integer, is:
[tex]\[ \boxed{306} \][/tex]
1. According to the problem, [tex]\(\frac{2}{3}\)[/tex] of the adult men are married to [tex]\(\frac{3}{5}\)[/tex] of the adult women.
2. Let [tex]\(m\)[/tex] represent the number of adult men and [tex]\(w\)[/tex] represent the number of adult women.
3. The relationship translates to:
[tex]\[ \frac{2}{3} \cdot m = \frac{3}{5} \cdot w \][/tex]
4. Simplifying this equation, we get:
[tex]\[ 2m = \frac{9}{5}w \][/tex]
Multiplying both sides by 5 to clear the fraction:
[tex]\[ 10m = 9w \][/tex]
5. Solving for [tex]\(w\)[/tex] in terms of [tex]\(m\)[/tex], we have:
[tex]\[ w = \frac{10}{9}m \][/tex]
We need the number of women, [tex]\(w\)[/tex], to be an integer. Starting with at least 300 men, we check if there is an integer value for [tex]\(w\)[/tex] corresponding to an integer [tex]\(m \geq 300\)[/tex]:
1. If we set [tex]\(m = 300\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 300 = \frac{3000}{9} \approx 333.33 \quad \text{(not an integer)} \][/tex]
Since [tex]\(333.33\)[/tex] is not an integer, [tex]\(m = 300\)[/tex] is not a solution. We increment [tex]\(m\)[/tex] until we find the smallest [tex]\(m\)[/tex] making [tex]\(w\)[/tex] an integer:
2. Increasing [tex]\(m\)[/tex] step by step:
3. If we set [tex]\(m = 301\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 301 = \frac{3010}{9} \approx 334.44 \quad \text{(not an integer)} \][/tex]
4. If we set [tex]\(m = 302\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 302 = \frac{3020}{9} \approx 335.56 \quad \text{(not an integer)} \][/tex]
5. Continuing this process:
6. We find that when [tex]\(m = 306\)[/tex]:
[tex]\[ w = \frac{10}{9} \cdot 306 = \frac{3060}{9} = 340 \quad \text{(an integer)} \][/tex]
Thus, the least possible number of adult men in the province, ensuring that the number of adult women is an integer, is:
[tex]\[ \boxed{306} \][/tex]
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