Day 1 - Solutions

Puzzle 1 - Music Festival

We can solve this problem by “simulating” the process.

Let’s consider: how many full rows will need to get off before I do? First the cellists/bassists have to go, then all the violinists also need to go since they all sit in front of me, and then the \(2\) violist rows will go before it’s my row’s turn to get off. That gives us a total of \(12 + 13 + 2 = 27\) full rows. Since each row has \(2\) benches of \(2\) players, there are \(4\) players in each row, so \(27 \times 4 = 108\) players will go first.

After the \(108\) players leave, it’ll be my row’s turn. However, I cannot leave as I am sitting next to the window! So at the least I need to wait for my bench-mate to get off first. Therefore, in the best case, I will be the \(110\mathrm{th}\) person to leave.

The \(1\mathrm{st}\) person leaves at \(0\) seconds past 3 o’clock. The \(2\mathrm{nd}\) leaves at \(10\) seconds past three. The \(3\mathrm{rd}\) leaves at \(20\) seconds past three. The \(n\)-th person seems to leave at \(10(n-1)\) seconds past three! As the \(110\mathrm{th}\) person, I will leave at \(10(110-1) = 10 \times 109 = 1090\) seconds after three. \(1090\) seconds is equivalent to \(18\) minutes and \(10\) seconds. Thus, I’ll leave at \(\boxed{03:18:10}\).

Problem Credit: Wenda Mao

Puzzle 2 - Average Marks

Let’s review the definition of average:

$$ \mathrm{average} = \frac {\mathrm{sum\ of\ all\ numbers}} {\mathrm{how\ many\ numbers\ there\ are}} $$

Let’s assume that \(A\) is the sum of the marks of everyone taking MCR3U. Since there are \(6\) classes and each class has \(30\) students, we know the total number of students taking MCR3U is \(30 \times 6 = 180\). Using our definition of average, we know that \(82 = \frac{A}{180}\), which simplifies to \(A = 180 \times 82 = 14760\).

Similarly, let’s assume that \(B\) is the sum of the marks of students in our class. Then we get \(77 = \frac{B}{30}\), which simplifies to \(B = 30 \times 77 = 2310\).

When we exclude our class from the six classes, there are \((6-1) \times 30 = 150\) students left, and the sum of marks of all students except the thirty students in my class is just \(A - B = 12450\). Using the definition of average again, we get the course average excluding our class: \(\frac{12450}{150} = \boxed{83}\).

Extension: You might have noticed that the number of students in each class (\(30\)) does not matter! As long as every class has the same number of students (making sure that each class is “weighted equally”), the answer will be the same.

Problem Credit: Wenda Mao

Puzzle 3 - A Very Algebra-ish Problem

Notice that the left hand side and the right hand side of the equation look similar, except that the numerator and the denominator switch places, which forms reciprocals. This suggests us to introduce two new letters (let’s call them \(p\) and \(q\)):

$$ \begin{cases} p = \frac{x - A}{B} \\ q = \frac{x - B}{A} \end{cases} $$

Now the equation can be rewritten as:

$$ p + q = \frac{1}{p} + \frac{1}{q} $$

It’s easier to work with now. Let’s eliminate the denominators by multiplying both sides by \(pq\):

$$ \begin{align} pq(p + q) & = \frac{pq}{p} + \frac{pq}{q} \\ & = q + p \end{align} $$

Now we may subtract \(p + q\) on both sides, and factor it out:

$$ \begin{align} pq(p + q) - (q + p) & = 0 \\ (pq - 1)(p + q) & = 0 \end{align} $$

Two things are multiplied together and give you a product of zero. When can this happen? Well, one of the things have to be zero, thus making the whole expression zero! In this case, either \(pq - 1 = 0\) or \(p + q = 0\) (or both). Time to do some case work.

• Case 1: \(p + q = 0\)

  Let’s plug our definition of \(p\) and \(q\) back in.

  $$ \begin{align} \frac{x - A}{B} + \frac{x - B}{A} & = 0 \\ (x - A)A + (x - B)B & = 0 \\ Ax - A^{2} + Bx - B^{2} & = 0 \\ (A + B)x & = A^{2} + B^{2} \\ x & = \frac{A^{2} + B^{2}}{A + B} \end{align} $$

  We have obtained one solution for \(x\) in this case.

• Case 2: \(pq - 1 = 0\)

  This is just equivalent to \(pq = 1\). Again, we plug the definition of \(p\) and \(q\) back in:

  $$ \begin{align} \frac{x - A}{B} \times \frac{x - B}{A} & = 1 \\ \frac{(x - A)(x - B)}{AB} & = 1 \\ (x - A)(x - B) & = AB \\ x^{2} - Ax - Bx + AB & = AB \\ x^{2} - Ax - Bx & = 0 \\ x(x - A - B) & = 0 \end{align} $$

  We now run into a similar situation: the product of two things is zero. That means either \(x = 0\) or \(x - A - B = 0\). We have obtained two more solutions: \(x = 0\) and \(x = A + B\).

In total, we have three possible values for \(x\): \(0\), \(A + B\) and \(\frac{A^{2} + B^{2}}{A + B}\). The problem requires us to multiply the nonzero solutions:

$$ \frac{A^{2} + B^{2}}{A + B} \times (A + B) = A^{2} + B^{2} = 20^{2} + 21^{2} = \boxed{841} $$

Notice that the \(A\) and \(B\) can be arbitrary numbers in this problem, and we have solved the problem for this generic case. Do be careful that since \(x - A\) and \(x - B\) are denominators in the original problem and denominators can’t be zero, \(x \neq A\) and \(x \neq B\). For some values of \(A\) and \(B\), this constraint will eliminate some of the three solutions.

Note: Introducing \(p\) and \(q\) isn’t necessary, but that does simplifies the problem a lot! There are other ways to solve the problem.

Problem Credit: The Art of Problem Solving, Intermediate Algebra, Exercise 4.61; adapted by Wenda Mao