Since Joseph is the only person who gets average marks. His marks cannot be the most frequent marks. With 4 students, median is the arithmetic mean of the second and third numbers. Joseph cannot be the median too as no one else has got same marks as Joseph for the arithmetic mean of Joseph and that guy to be same as Joseph's marks. So, given our conditions, Joseph's average is the arithmetic mean.

Tyler's average score cannot be mode as then no one would get mode marks, which is not possible since mode is the highest frequency number.

So, Tyler's average score is median and Sadia's average score is mode.

So, the three statements are:
No one scores higher than mode
No one scores the median
Joseph is the only person with marks as arithmetic mean

Marks can be of the form 2x - (50-x) = 3x-50

Let the marks be h1>=h2>=h3>=h4

Max frequency is highest marks. So, h1=h2

Now, median is (h3+h1)/2 not equal to h1
So, h3 h4 and is < h1
So, Joseph gets h3 and since arithmetic mean=h3

3*h3=2*h1+h4
So, The marks are
h1, h1, h3, 3*h3-2*h1

So,
Joseph's average is the arithmetic mean
Tyler's average score is median
Sadia's average score is mode

Joseph's score is x
Highest score is say y
Lowest score is 3x-2y

## Solution - 4 Students

Since Joseph is the only person who gets average marks. His marks cannot be the most frequent marks. With 4 students, median is the arithmetic mean of the second and third numbers. Joseph cannot be the median too as no one else has got same marks as Joseph for the arithmetic mean of Joseph and that guy to be same as Joseph's marks. So, given our conditions, Joseph's average is the arithmetic mean.

Tyler's average score cannot be mode as then no one would get mode marks, which is not possible since mode is the highest frequency number.

So, Tyler's average score is median and Sadia's average score is mode.

So, the three statements are:

No one scores higher than mode

No one scores the median

Joseph is the only person with marks as arithmetic mean

Marks can be of the form 2x - (50-x) = 3x-50

Let the marks be h1>=h2>=h3>=h4

Max frequency is highest marks. So, h1=h2

Now, median is (h3+h1)/2 not equal to h1

So, h3 h4 and is < h1

So, Joseph gets h3 and since arithmetic mean=h3

3*h3=2*h1+h4

So, The marks are

h1, h1, h3, 3*h3-2*h1

So,

Joseph's average is the arithmetic mean

Tyler's average score is median

Sadia's average score is mode

Joseph's score is x

Highest score is say y

Lowest score is 3x-2y

So, the scores are of the form y,y,x,3x-2y (x>y)

Cheers!

Pratik

http://www.pratikpoddarcse.blogspot.com