In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
Question:
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
Options
Answer: 21
Explanation:
Let G be the mark of each girl and B the mark of each boy. Given average of 4 girls and 6 boys is 24, so total = 10 * 24 = 240. Hence 4G + 6B = 240, which simplifies to 2G + 3B = 120. (1) We want possible integer values of the total marks of 2 girls and 6 boys: T = 2G + 6B. From (1) we have 2G = 120 − 3B, so T = (120 − 3B) + 6B = 120 + 3B. (2) Constraints: each girl’s mark is at least any boy’s mark and at most twice any boy’s mark, so B ≤ G ≤ 2B. Express G from (1): G = (120 − 3B)/2. Now apply inequalities: G ≥ B → (120 − 3B)/2 ≥ B → 120 ≥ 5B → B ≤ 24. G ≤ 2B → (120 − 3B)/2 ≤ 2B → 120 ≤ 7B → B ≥ 120/7 ≈ 17.142857. Thus B can range (as a real number) over the closed interval [120/7, 24]. From (2), T = 120 + 3B ranges over [120 + 3*(120/7), 120 + 3*24] = [1200/7, 192] which is approximately [171.42857, 192]. We need the number of integer values T can take in that interval. Integers from 172 up to 192 inclusive: count = 192 − 172 + 1 = 21. So the correct choice is 21
Explanation:
Let G be the mark of each girl and B the mark of each boy. Given average of 4 girls and 6 boys is 24, so total = 10 * 24 = 240. Hence 4G + 6B = 240, which simplifies to 2G + 3B = 120. (1) We want possible integer values of the total marks of 2 girls and 6 boys: T = 2G + 6B. From (1) we have 2G = 120 − 3B, so T = (120 − 3B) + 6B = 120 + 3B. (2) Constraints: each girl’s mark is at least any boy’s mark and at most twice any boy’s mark, so B ≤ G ≤ 2B. Express G from (1): G = (120 − 3B)/2. Now apply inequalities: G ≥ B → (120 − 3B)/2 ≥ B → 120 ≥ 5B → B ≤ 24. G ≤ 2B → (120 − 3B)/2 ≤ 2B → 120 ≤ 7B → B ≥ 120/7 ≈ 17.142857. Thus B can range (as a real number) over the closed interval [120/7, 24]. From (2), T = 120 + 3B ranges over [120 + 3*(120/7), 120 + 3*24] = [1200/7, 192] which is approximately [171.42857, 192]. We need the number of integer values T can take in that interval. Integers from 172 up to 192 inclusive: count = 192 − 172 + 1 = 21. So the correct choice is 21
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