In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is:
Question:
In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is:
In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is:
Options
Answer: 63
Explanation:
Let the possible overall average marks be a, so 32 < a < 60. If the overall average is a, then the ratio of the number of students in section A to section B is: Number in A : Number in B = (60 - a) : (a - 32) If the sections have equal number of students, then a = (32 + 60)/2 = 46. Since section B has 10 more students than section A, the overall average a > 46. The extreme values of the number of students depend on extreme values of a: For a = 47: Ratio A:B = 60-47 : 47-32 = 13 : 15 Let the multiplier = x → 15x - 13x = 10 → 2x = 10 → x = 5 Number of students in A = 13 × 5 = 65 For a = 56: Ratio A:B = 60-56 : 56-32 = 4 : 24 = 1 : 6 Let multiplier = x → 6x - x = 10 → 5x = 10 → x = 2 Number of students in A = 1 × 2 = 2 Difference between maximum and minimum possible students in A: 65 - 2 = 63
Explanation:
Let the possible overall average marks be a, so 32 < a < 60. If the overall average is a, then the ratio of the number of students in section A to section B is: Number in A : Number in B = (60 - a) : (a - 32) If the sections have equal number of students, then a = (32 + 60)/2 = 46. Since section B has 10 more students than section A, the overall average a > 46. The extreme values of the number of students depend on extreme values of a: For a = 47: Ratio A:B = 60-47 : 47-32 = 13 : 15 Let the multiplier = x → 15x - 13x = 10 → 2x = 10 → x = 5 Number of students in A = 13 × 5 = 65 For a = 56: Ratio A:B = 60-56 : 56-32 = 4 : 24 = 1 : 6 Let multiplier = x → 6x - x = 10 → 5x = 10 → x = 2 Number of students in A = 1 × 2 = 2 Difference between maximum and minimum possible students in A: 65 - 2 = 63
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