Problem E and Solution

Painting is Fund (Raising)

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?

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 \(11\).