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

Given: 
1. The arithmetic mean of scores of 25 students is 50. So, the total score of all students combined is \(25 \times 50 = 1250\). 

2. Five students have the same score (let's call this score \(T\)). So, the combined score of these five students is \(5T\). 

3. The scores of the other 20 students are distinct integers, with the lowest being 30. 
To maximize the score of the toppers, we need to minimize the scores of the other 20 students. 

Let's calculate the total score of the smallest possible distinct integers for 20 students: 

Starting with 30, the smallest 20 distinct integers are: 30, 31, 32, 33 ... up to 49. 

The sum of these scores is: 
\(30 + 31 + 32 + ... + 49\)

This is an arithmetic progression, where: 
\((a_1 = 30), (d = 1), and (n = 20)\). 

The sum of \(n\) terms of an arithmetic progression is given by: 
\(S = \frac{n}{2} [2a_1 + (n-1) d]\)

Using the values provided: 
\(S = \frac{20}{2} [2(30) + (20-1) 1]\)

\(S = 10 [60 + 19] So, S = 10 \times 79 = 790\)

So, the combined score of the 20 students (with the smallest possible distinct integers) is 790. 

Therefore, the combined score of the 5 toppers is: \(1250 - 790 = 460\) 
The score of one topper is: \(T = \frac{460}{5} = 92\)

So, the maximum possible score of each topper is 92.

Answer 2

Let sum of marks of students be \(x\).
\(x = 25\times50 =1250\)
To maximize the marks of the toppers, We will minimize the marks of 20 students.
Then, their scores will be \((30,31,32.....49)\)
Let the score of toppers be y, then
\(5y +\frac {20}{2}(79)=1250\)
\(5y +10 \times 79=1250\)
\(5y +790=1250\)
\(5y=1250-790\)
\(5y=460\)
\(y=\frac {460}{5}\)
\(y=92\)

So, the maximum possible score of the toppers is \(92\).