Question

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

Answer 1

Correct Answer: 92

Given: 

• The arithmetic mean of scores of 25 students is 50.
• Total score of all students: 
  $25 \times 50 = 1250$
• Five students have the same score. Let that score be $T$. 
  Combined score of 5 students: $5T$
• The scores of the remaining 20 students are distinct integers with the lowest being 30.

Objective:

To maximize the score of the toppers (the 5 students), we need to minimize the scores of the other 20 students.

Step 1: Choose the 20 smallest distinct integers starting from 30:

These are: 
30, 31, 32, ..., 49

Step 2: Sum of the arithmetic progression:

First term: $a_1 = 30$ 
Common difference: $d = 1$ 
Number of terms: $n = 20$

Sum formula: 
$S = \dfrac{n}{2} \left[2a_1 + (n - 1)d\right]$

Substituting the values: 
$S = \dfrac{20}{2} \left[2 \times 30 + (20 - 1) \times 1\right]$ 
$S = 10 \left[60 + 19\right] = 10 \times 79 = 790$

Step 3: Calculate the combined score of 5 toppers:

Total score of all students: 1250 
Total score of 20 students: 790 
So, score of 5 toppers: $1250 - 790 = 460$

Step 4: Score of each topper:

$T = \dfrac{460}{5} = 92$

Conclusion:

The maximum possible score of each topper is 92.