In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
Question:
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
Options
Answer: 72 & 80
Explanation:
Total students = 250. Let G = number of girls, B = number of boys = 250 − G. Given 44% ≤ %girls ≤ 60% → G between 0.44·250 = 110 and 0.60·250 = 150 (both exact integers). Swimming total = 0.8G + 0.5B. Running total = 0.6G + 0.7B. Everyone chose at least one, so |Swimming ∩ Running| = Swimming + Running − 250. Compute: Swimming + Running = (0.8G + 0.5B) + (0.6G + 0.7B) = 1.4G + 1.2B. Substitute B = 250 − G: 1.4G + 1.2(250 − G) = 1.4G + 300 − 1.2G = 0.2G + 300. So intersection = (0.2G + 300) − 250 = 0.2G + 50 = (G/5) + 50. This is increasing in G, so: Minimum when G = 110: intersection = 110/5 + 50 = 22 + 50 = 72. Maximum when G = 150: intersection = 150/5 + 50 = 30 + 50 = 80. Answer: minimum = 72, maximum = 80.
Explanation:
Total students = 250. Let G = number of girls, B = number of boys = 250 − G. Given 44% ≤ %girls ≤ 60% → G between 0.44·250 = 110 and 0.60·250 = 150 (both exact integers). Swimming total = 0.8G + 0.5B. Running total = 0.6G + 0.7B. Everyone chose at least one, so |Swimming ∩ Running| = Swimming + Running − 250. Compute: Swimming + Running = (0.8G + 0.5B) + (0.6G + 0.7B) = 1.4G + 1.2B. Substitute B = 250 − G: 1.4G + 1.2(250 − G) = 1.4G + 300 − 1.2G = 0.2G + 300. So intersection = (0.2G + 300) − 250 = 0.2G + 50 = (G/5) + 50. This is increasing in G, so: Minimum when G = 110: intersection = 110/5 + 50 = 22 + 50 = 72. Maximum when G = 150: intersection = 150/5 + 50 = 30 + 50 = 80. Answer: minimum = 72, maximum = 80.
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