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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

Entered answer:

✅ Correct Answer: 92

We have 25 students with an arithmetic mean of 50.

Since arithmetic mean = total sum ÷ number of students, we can find:

Total sum of all scores = 25 × 50 = 1250

We also know:

5 students are toppers with the same score (let's call this score x)

20 other students have distinct integer scores

The lowest score among the 20 students is 30

To find the maximum possible score of the toppers, we need to think strategically.

Since the total sum is fixed at 1250, if we want to maximize the toppers' scores, we must minimize the scores of the other 20 students.

Think of it like this: If you have a fixed amount of money to distribute among people, to give the maximum to some people, you must give the minimum to others.

Let the score of each topper = x

Sum of 5 toppers = 5x

Sum of 20 other students = 1250 - 5x

To maximize x, we need to minimize the sum of the 20 other students.

The 20 students have:

Distinct integer scores (no two students can have the same score)

Lowest score = 30

To minimize their total, we give them the smallest possible distinct integers starting from 30:

30, 31, 32, 33, ..., 49 (these are 20 consecutive integers)

We need to find: 30 + 31 + 32 + ... + 49

This is an arithmetic progression with:

First term (a) = 30

Number of terms (n) = 20

Last term (l) = 49

Formula for sum of arithmetic progression:

Sum = $\tfrac{n}{2} \times$ (first term + last term)

Sum = $\tfrac{20}{2} \times (30 + 49) = 10 \times 79 = 790$

Now we can find the maximum possible sum of the 5 toppers:

Sum of 5 toppers = Total sum - Sum of 20 other students

5x = 1250 - 790 = 460

Therefore: x = 460 ÷ 5 = 92

Therefore, the maximum possible score of the toppers is 92.

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The arithmetic mean of scores of 252525 students in an examination is 505050. 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 303030, then the maximum possible score of the toppers is

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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