Answers and Solutions


    1. The rate is \(6\) metres in \(3\) seconds which is a unit rate of \(2\) m/s.
    2. The rate is \($6\) for \(4\) items which is a unit rate of \($1.50\) per item.
  2. An equation is \(C = 15 m\). 
  3. The rate of \(175\) shoes per week is equivalent to producing \(175\) shoes in \(7\) days. As a unit rate, this is \(25\) shoes per day. Therefore an equation is \(P = 25 d\). 
  4. The graph in option c) represents the snow fall of \(1\) mm every \(6\) minutes. In this graph, a change in \(6\) minutes results in an extra \(1\) mm of snow accumulation.
    The graphs in option a) and b) represent rates of \(1\) mm per minute and \(6\) mm per minute, respectively.
  5. Biker 1 travels \(60\) km in \(3\) hours which, as a unit rate, is \(20\) km/h. Biker 2 travels \(45\) km in \(3\) hours which, as a unit rate, is \(15\) km/h.
    1. Leg 1: \(0\) - \(15\) seconds
      Leg 2: \(15\) - \(50\) seconds
      Leg 3: \(50\) - \(75\) seconds
      Leg 4: \(75\) - \(90\) seconds
    2. Each leg of the race is represented by the following graph:  
      A graph that shows that Leg 1 and Leg 4 have the same steepness. Leg 3 is less steep than both Leg 1 and Leg 4. Leg 2 is less steep than Leg 3.
      The steepness of each line segment represents the rate or speed of each runner. The steeper the line, the more distance the runner is covering in a certain amount of time. Therefore, a steeper line indicates a faster runner. It appears that the fastest runners ran Leg 1 and Leg 4, and the slowest runner ran Leg 2.
    3. Runner 1 ran \(100\) m in \(15\) seconds or at a speed of \(6\dfrac{2}{3}\) m/s.
      Runner 2 ran \(100\) m in \(35\) seconds or at a speed of \(2\dfrac{6}{7}\) m/s.
      Runner 3 ran \(100\) m in \(25\) seconds or at a speed of \(4\) m/s.
      Runner 4 ran \(100\) m in \(15\) seconds or at a speed of \(6\dfrac{2}{3}\) m/s.
    4. One way to represent this race using a single equation would be to obtain a single unit rate by averaging the four rates from part c):\[\frac{6\frac{2}{3} + 2 \frac{6}{7} + 4 + 6 \frac{2}{3}}{4} = \frac{106}{21} = 5\frac{1}{21} \]This gives the equation \(d = \dfrac{106}{21}t\) where \(d\) is the distance, in metres, and \(t\) is the time, in seconds.
      Another way to represent this race using a single equation would be to obtain a single unit rate by taking what is called the average speed over the race:\[\frac{\text{total distance}}{\text{total time}} = \frac{400 \text{ m}}{90 \text{ s}} = \frac{40}{9} \text{m/s}=4 \frac{4}{9} \text{m/s}\]This gives the equation \(d = \dfrac{40}{9}t\).
      Plotting these two relationships on our original graph gives the following:
      The two equations, d equals 106 over 21 t and d equals 40 over 9 t are plotted on a graph.
      Is it clear which of these two linear relationships better represents the race?
    1. Number of Guests Cost (\($\))
      \(0\) \(20\)
      \(10\) \(60\)
      \(20\) \(100\)
      \(30\) \(140\)
      \(40\) \(180\)
      \(50\) \(220\)

      We have that \(10 \times 6 = 60\) in the second row of the table, but \(20 \times 6 = 120\) not \(100\) in the third row, so there is no multiplicative relationship between the two quantities. Therefore this is not a proportional situation.
    2. Looking at the rows of the table, we can see that adding \(10\) more guests always results in \($40\) more dollars. For example, changing from \(10\) to \(20\) guests results in \($100 - $60 = $40\) more. We can represent this relationship as a rate:\[\frac{$40}{10 \text{ guests}} \text{ or } \frac{$4}{1\text{ guest}}\] This is a unit rate of \($4\) per guest.
    3. The rental cost cannot be accurately represented using only the unit rate \($4\) per guest. This would mean that an event with \(10\) guests would cost \($4 \times 10 = $40\) which is not true. We need to take into account that there is a "base fee" to the rental. Following the pattern that \(10\) more guests should result in \($40\) more, or \(10\) less guests should result in \($40\) less, we would expect the cost of \(0\) guests to be \($60 - $40 = $ 20\). This is the base fee.
      So the rental cost can be described using the base fee of \($20\) and a unit rate of \($4\) per person after that.
    1. Let \(P\) represent the amount of the pool filled in \(t\) hours.
      Hose A fills at a rate \(\dfrac{1 \text{ pool}}{a \text{ hours}}\) giving the equation \(P = \dfrac{1}{a} t\).
      Hose B fills \(\dfrac{1 \text{ pool}}{b \text{ hours}}\) giving the equation \(P=\dfrac{1}{b}t\).
    2. \(P =\left( \dfrac{1}{a}+\dfrac{1}{b}\right) t\).
    3. Using the equation from part b), we know that when \(t = 6\), we must have\[P = \left(\frac{1}{a} + \frac{1}{b}\right)(6) = \frac{6}{a} + \frac{6}{b}\]

      Since we are told that the two hoses fill a total of \(1\) pool in \(6\) hours, we must have

      \[\frac{6}{a}+\frac{6}{b} = 1\]Knowing that \(a\) and \(b\) must both be positive integers, what are the possible values of \(a\) that can make this equation true? The nine possible values for \(a\) can be found by systematic trial and error and are shown below:\[a = 7,~8,~9,~10,~12,~15,~18,~24,~42\]