# Problem of the Week Problem E and Solution Painting is Fund (Raising)

## Problem

The POTW High School Painting Club has $$30$$ student members, some from Grade $$11$$ and the remainder from Grade $$12$$. Over the year, they want to fundraise money for the school art department. Each possible pairing of students from the club will create exactly one painting together. When two Grade $$11$$ students paint together, they will sell the painting for $$\20$$. When a Grade $$11$$ and a Grade $$12$$ student paint together, they will sell the painting for $$\30$$. When two Grade $$12$$ students paint together, they will sell the painting for $$\40$$.

When all the paintings are sold, the students will have raised $$\13\,920$$. How many of the $$30$$ members of the club are Grade $$11$$ students?

## Solution

If $$3$$ students, Student $$A$$, Student $$B$$, and Student $$C$$, are in the same grade, then there will be $$(3\times2)\div2=3$$ pairings, namely $$AB$$, $$AC$$, and $$BC$$. (If we look at this using a counting argument, there would be $$3$$ choices for the first student and for each of these choices, there would be 2 choices for the second student, a total of $$3\times 2=6$$ pairings, namely $$AB$$, $$AC$$, $$BA$$, $$BC$$, $$CA$$, and $$CB$$. However, notice that each pairing appears twice. Since order is not important we must divide by $$2$$, getting us $$3$$ possible pairings.)

If $$3$$ students, Student $$A$$, Student $$B$$, and Student $$C$$, are in Grade $$11$$ and $$2$$ students, Student $$D$$ and Student $$E$$, are in Grade $$12$$, then there will be $$3\times2=6$$ pairings of students in different grades, namely $$AD$$, $$AE$$, $$BD$$, $$BE$$, $$CD$$, and $$CE$$. (There are $$3$$ choices for the Grade $$11$$ student in the pairing and for each of these choices, there are $$2$$ possibilities for the Grade $$12$$ student. This gives a total of $$3\times 2=6$$ pairings.)

Similar arguments will now be applied to our problem.

Let $$a$$ represent the number of Grade $$11$$ students in the club and $$(30-a)$$ represent the number of Grade $$12$$ students in the club.

Since there are $$a$$ students in Grade $$11$$ and each must paint with every other student in Grade $$11$$, there will be $$a\times (a-1)\div2$$ paintings from pairs where both students are in Grade $$11$$. Thus, the amount raised by paintings from pairs where both students are in Grade $$11$$ would be $20\times \left(\dfrac{a(a-1)}{2}\right)$ Similarly, since there are $$(30-a)$$ students in Grade $$12$$ and each must paint with every other student in Grade $$12$$, there will be $$(30-a)\times (30-a-1)\div2=(30-a)\times (29-a)\div2$$ paintings from pairs where both students are in Grade $$12$$. Thus, the amount raised by paintings from pairs where both students are in Grade $$12$$ would be $40\times \left(\dfrac{(30-a) (29-a)}{2}\right)$ Since every Grade $$11$$ student must paint with every Grade $$12$$ student, there will be $$a\times (30-a)$$ paintings from pairs with one student from each grade. Thus, the amount raised by paintings from pairs with one student from each grade would be $30\times(a (30-a))$ Therefore, \begin{aligned} 13\,920&=20\left(\frac{a(a-1)}{2}\right)+40\left(\frac{(30-a) (29-a)}{2}\right)+30(a(30-a))\\[1mm] 13\,920&=10(a^2-a)+20(870-59a+a^2)+30(30a-a^2)\\ 1392&=(a^2-a)+2(870-59a+a^2)+3(30a-a^2)\\ 1392&=a^2-a+1740-118a+2a^2+90a-3a^2\\ 1392& = -29a+1740\\ 29a&=348\\ a&=12 \end{aligned} Therefore, $$12$$ of the students in the club are in Grade $$11$$.