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Algebra Difficulty 2.1 Junior Prove it Canada

IMG0 A large water jug is $\frac{1}{4}$ full of water. After 24 litres of water are added, the jug is $\frac{5}{8}$ full. What is the volume of the jug, in litres? Stephanie starts with a large number of soccer balls. She gives $\frac{2}{5}$ of them to Alphonso and $\frac{6}{11}$ of them
to Christine. The number of balls that she is left with is a multiple of
9. What is the smallest number of soccer balls with which Stephanie
could have started?
Each student in a math club is in either the Junior
section or the Senior section.

No student is in both sections.

Of the Junior students, 60% are left-handed and 40% are
right-handed.

Of the Senior students, 10% are left-handed and 90% are
right-handed.

No student in the math club is both left-handed and right-handed.

The total number of left-handed students is equal to the total number of
right-handed students in the math club.

Determine the percentage of math club members that are in the Junior
section.

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Solution

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

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Figure for this problem n soccer balls. Since Stephanie can divide the

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Figure for this problem n balls into fifths and into elevenths, then

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Figure for this problem n is a multiple of both 5 and 11. Since 5 and 11 are both prime numbers, then

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Figure for this problem n $must be a multiple of$ $5\cdot$ 11 = 55 $. Thus,$ n = 55k for some positive integer

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Figure for this problem 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

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Figure for this problem $\frac{4}{5}$ of the students are Junior and

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Figure for this problem $\frac{1}{5}\$$ are Senior.

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

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.