Answers and Solutions

1. Arlie's equation describes a unit rate of \($24\) per office. Converting Yannic's rate into a unit rate gives \($28\) per office. Therefore, Arlie's cleaning service offers the best deal.
2. Representing the last two situations using a unit rate we obtain \(30\) km/h for option b) and \(20\) km/h for option c). This shows that option b) represents the greatest speed.
3. Representing each situation using a unit rate of cars per week we get the following.
   1. \(115\) cars/week as given.
   2. From the graph we note the data point \(80\) cars in \(5\) days which is equivalent to \(\dfrac{80}{5} = 16\) cars/day and \(16 \times 7 = 112\) cars/week.
   3. \(30\) cars in \(2\) days which is equivalent to \(\dfrac{30}{2} = 15\) cars/day and \(15 \times 7 = 105\) cars/week.
   From this, we see that the most efficient auto repair shop is represented in option a).
4. We will represent each speed as unit rate, in km/h.
   1. If a snail travels \(1\) km in \(10\) hours then it travels \(0.1\) km/h.
   2. From the graph, the tortoise travels \(5\) m/min. Converting this rate to km/h gives \(0.3\) km/h.
   3. From the table, the sloth travels \(16.5\) cm in \(5\) seconds. As a unit rate, this is \(3.3\) cm/s. Converting to km/h gives approximately \(0.12\) km/h. 
   From this, we see that the snail is the slowest animal, but the sloth comes in as a very close second!
5. Let's display each relationship using an equation. 
   1. \(y = 1.6x\)
   2. \(y = \dfrac{7}{4}x = 1.75 x\)
   3. Notice that there is a multiplicative relationship between the two columns. You can multiply the value of \(x\) by \(1.7\) to obtain the corresponding value of \(y\). Therefore, the equation representing the relationship is \(y = 1.7x\).
   The steepest graph will be seen in option b) where the constant of proportionality is the largest. The least steep graph will be seen in option a) where the constant of proportionality is the smallest. 
6. Writing each flow rate as a unit rate in L/min we get the following:
   1. \(22.7\) L/min
   2. The equation shows a unit rate of \(1580\) L/h which is \(\dfrac{1580}{60} = 26\frac{1}{3}\) L/min.
   3. The graph shows a rate of \(25~000\) mL/min which is \(\dfrac{25~000}{1000} = 25\) L/min.
   The fastest hose is in option b). This hose can fill a \(40~000\) L pool in approximately \(\dfrac{40~000}{26\frac{1}{3}} \approx 1519\) minutes or approximately \(25\) hours.
   The slowest hose is in option a). This hose can fill a \(40~000\) L pool in approximately \(\dfrac{40~000}{22.7} \approx 1762\) minutes or approximately \(29\) hours.
   The hose in option c) can fill a \(40~000\) L pool in \(\dfrac{40~000}{25} =1600\) minutes or approximately \(27\) hours.
7. To compare the two rates, it makes sense to convert the first rate to dollars per square metre. Since one tile has an area of \(10 \times 10 = 100\) cm\(^{2}\), or \(0.1 \times 0.1 = 0.01\) m\(^{2}\), \(8\) tiles have an area of \(0.08\) m\(^{2}\). So the first rate is equivalent to \($2.80\) for \(0.08\) m\(^{2}\) or \(\dfrac{$ 2.80}{0.08} = $35\) per m\(^{2}\). This is the cheapest of the two rates.
   Therefore, the cheapest price to tile a \(25\) m\(^{2}\) floor is \(25 \times $35 = $875\).
8. Let's begin with some trial and error.
   If \(x = 10\), then \(10\%\) of \(45\) is \(4.5\) and \(10\%\) of \(22\) is \(2.2\). Decreasing \(45\) by \(10\%\) would result in a new price of \($45 -$ 4.50 = $40.50\). Increasing \(22\) by \(10\%\) would result in a new price pf \($22 + $2.20 = $24.20\). These changes have not made the prices equal, and it looks like the value of \(x\) will need to be much larger to obtain this. We continue with trial and error and summarize our work in a table:

   Value of \(x\)	\(x\%\) of \(45\)	New Symphony Price	\(x\%\) of \(22\)	New Baseball Price
   \(10\)	\(4.5\)	\($45-$4.50 = $40.50\)	\(2.2\)	\($22 + $2.50 = $24.50\)
   \(20\)	\(9\)	\($45 - $9 = $36\)	\(4.4\)	\($22 + $4.40 = $26.40\)
   \(30\)	\(13.5\)	\($45 - $13.50 = $31.5\)	\(6.6\)	\($22 + $6.60 = $28.6\)
   \(40\)	\(18\)	\($45 - $18 = $27\)	\(8.8\)	\($22 + $8.80 = $30.8\)

   From this table we see that the new prices should be equal for some \(x\) between \(30\) and \(40\). More trial and error can narrow this down further to between \(34\) and \(35\).
   To determine the exact value of \(x\), we solve the following equation.

   \[\begin{align*} 45(100-x) &= 22(100+x)\\ 4500 - 45x &=2200+22x\\ 2300 &= 67x\\ \dfrac{2300}{67} &= x \end{align*}\]

   Thus, \(x = \dfrac{2300}{67} \approx 34.3\).