CEMC Euclid, 2022

Suppose that the volume of the jug is $V$ L.

Then $\frac{1}{4}V + 24 = \frac{5}{8}V$.

Multiplying by 8, we obtain $2V + 24 \cdot 8 = 5V$ which gives $3V = 192$
and so $V = 64$.

Therefore, the volume of the jug is 64 L.
Suppose that Stephanie starts with $n$ soccer balls.

Since Stephanie can divide the $n$
balls into fifths and into elevenths, then $n$ is a multiple of both 5 and 11.

Since 5 and 11 are both prime numbers, then $n$ must be a multiple of $5\cdot 11 = 55$.

Thus, $n = 55k$ for some positive
integer $k$.

In this case, $\frac{2}{5}n = \frac{2}{5} \cdot 55k = 22k$ and $\frac{6}{11}n = \frac{6}{11} \cdot 55k = 30k$.

When Stephanie has given these balls away, she is left with $55k - 22k - 30k = 3k$ balls.

Since $3k$ is a multiple of 9, then
$k$ is a multiple of 3.

Therefore, the smallest possible number of balls is obtained when $k = 3$, which means that Stephanie started
with $n = 55\cdot 3 = 165$ soccer
balls.
Suppose that the number of students in the Junior section is
$j$ and the number of students in the
Senior section is $s$.

The number of left-handed Junior students is 60% of $j$, or $0.6j$.

The number of right-handed Junior students is 40% of $j$, or $0.4j$.

The number of left-handed Senior students is 10% of $s$, or $0.1s$.

The number of right-handed Senior students is 90% of $s$, or $0.9s$.

Since the total numbers of left-handed and right-students are equal, we
obtain the equation $0.6j + 0.1s = 0.4j + 0.9s$ which gives $0.2j = 0.8s$ or $j = 4s$.

This means that there are 4 times as many Junior students as Senior
students, which means that $\frac{4}{5}$ of the students are Junior
and $\frac{1}{5}$ are Senior.

Therefore, 80% of the students in the math club are in the Junior
section.