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The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
Correct answer is '92'. Can you explain this answer?
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The arithmetic mean of scores of 25 students in an examination is 50....
To find the maximum possible score of the top students, we need to first determine the total sum of the scores of all the students.

Total Sum of Scores:
Given that the arithmetic mean of scores of 25 students is 50, we can calculate the total sum of scores using the formula:
Total Sum = Mean × Number of Students
Total Sum = 50 × 25 = 1250

Scores of the Top 5 Students:
Since five students top the examination with the same score, let's assume their score as 'x'. Therefore, the sum of their scores would be 5x.

Scores of the Other Students:
The remaining 20 students have distinct integer scores, with the lowest being 30. These scores must be greater than or equal to 30. Let's assume the scores of the other students as 'a1', 'a2', 'a3', ..., 'a20'.

Total Sum Equation:
We can write an equation for the total sum of scores using the scores of the top students and the other students:
Total Sum = 5x + a1 + a2 + a3 + ... + a20

Maximum Possible Score of the Top Students:
To find the maximum possible score of the top students, we need to minimize the scores of the other students. Since the lowest score is 30, we assign the lowest possible score to each of the remaining students:
Total Sum = 5x + 30 + 31 + 32 + ... + (30 + 19) (Sum of the first 20 positive integers starting from 30)
Total Sum = 5x + (30 + 31 + 32 + ... + 49)

Sum of First 20 Positive Integers:
Using the formula for the sum of the first n positive integers, we can simplify the equation:
Sum of First 20 Positive Integers = n(n+1)/2
Sum of First 20 Positive Integers = 20(20+1)/2 = 20(21)/2 = 210

Simplifying the Total Sum Equation:
Total Sum = 5x + 210

Since the total sum needs to be 1250, we can equate the equation:
5x + 210 = 1250

Solving the Equation:
5x = 1250 - 210
5x = 1040
x = 1040/5
x = 208

Therefore, the maximum possible score of the top students is 208. However, the answer provided in the question is 92, which contradicts the given information and calculations.
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Community Answer
The arithmetic mean of scores of 25 students in an examination is 50....
Let sum of marks of students be x
Now therefore x = 25 * 50 = 1250
Now to maximize the marks of the toppers
We will minimize the marks of 20 students
so their scores will be (30, 31, 32.....49)
let score of toppers be y
so we get 5y + 20/2 (79) = 1250
we get 5y + 790 =1250
5y = 460 y = 92
So scores of toppers = 92
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The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers isCorrect answer is '92'. Can you explain this answer?
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