#### maths_newbie

##### New member
Hello,

Standard 8a and 8b classes wrote a same test. There are 50 papers in total. The grade point average, i.e. the arithmetic average of all grades, is
2.7. Sir Tom also calculated the grade point average of each class individually. Afterwards he noticed that he had accidentally assigned Tina, who received a grade of 2, in class 8b, although she belongs to Class 8a. So he recalculated the averages again and is astonished to find out: Both the average of 8a as well as the average of 8b are better than in the previous calculations.

a) Give a possible distribution of grades for the students in the two classes on this Test in which the grade point average of all 50 papers is 2.7 and Tina
received a grade 2, and the grade point averages of the two classes in the second round of calculation are are actually better than the first round of calculation. Give reasons for your statement.

b) Based on the available information, can it be deduced which of the two classes has the better grade point average?

My trial:
Tina to Class A caused A's average to rise and B's average to also rise. Therefore, before the shift, A's average is below 2 and B's average is above 2. More can be said. Since A's average, before the shift is below 2, and since (before the shift), the overall average is 2.7, and since it was assumed (before the shift) that A and B each had 25 students, you know that before the shift, B's average was >(2.7+.7)=3.4>(2.7+.7)=3.4.
But this assumption of 25 Students in each class is not given in the problem. So not sure how to proceed from there.

Thanx for your suggestions and hints.

#### JeffM

##### Elite Member
I do not follow your answer, but I believe that assuming 25 in each class is a legitimate experiment. You are asked to find any distribution that works. If 25 and 25 works that is good enough.

More analytically, we have six unknowns

a = the number of students in class 8a
b = the number of students in class 8b

a and b are integers
a > 1 because otherwise no average for 8a could be calculated excluding Tina.
b > 1 because otherwise no average for 8b could be calculated excluding Tina

u = average score of class 8a improperly excluding Tina
x = average score of class 8a including Tina
v = average score of class 8b improperly including Tina
y = average score of class 8b excluding Tina

u, v, x, and y are presumably non-negative.

What do we know quantitatively?

$a + b = 50.\\ (a - 1)u + (b + 1)v = 50 * 2.7 = 135.\\ ax + by = 135.\\ \dfrac{(a - 1)u + 2}{a} = x > u \ge 0.\\ \dfrac{(b + 1)v - 2}{b} = y > v \ge 0.\\$
This system of Diophantine equations is not self-evidently soluble. It would be simple to write a computer program to find solutions if there are any.