2. Ratios, Rates, and Proportions (N)
  3. Lesson 8: Proportionality in Tables and Graphs

Answers and Solutions


    1. We have that \(2 \times 3 = 6\) and \(4 \times 3 = 12\), but \(6 \times 3 = 18\), not \(16\). Therefore, the cost is not proportional to the number of tickets.
    2. We have that \(5 \times 4 = 20\), \(25 \times 4 = 100\) and \(62.5 \times 4 = 250\). Therefore, the cost is proportional to the area.
    3. We have that \(0 \times 52 = 0\), \(0.1 \times 52 = 5.2\), \(0.3 \times 52 = 15.6\) and \(0.5 \times 52 = 26\). Therefore, the distance is proportional to the time.
  2. Remember that in a proportional relationship, the data points in the graph will lie on a straight line that passes through the point \((0,0)\), if extended.
    1. Proportional. (Multiply the time by \(2.5\) to obtain the corresponding distance.)
    2. Not proportional. 
    3. Proportional. (Multiply the value of \(x\) by \(0.5\) to obtain the corresponding value of \(y\).)
    1. Looking at the first two rows, we notice that an increase of \(1.5\) in the value of \(x\) produces an increase of \(5.25\) in the value of \(y\). Since the relationship is proportional, this pattern must continue. Since the value of \(x\) is changing from \(4\) to \(6.5\) which is an increase of \(1.5\), the corresponding change in \(y\) must be \(5.25\). Therefore the missing value is \(15.75 + 5.25 = 21\). Alternatively, you could have found this value by determining the multiplicative relationship in the table (we will do this in part b).
    2. Since the relationship is proportional, we can find the multiplicative relationship by looking at the first row. Notice that \(3 \times 3.5 = 10.5\) and so we obtain the \(y\) value by multiplying the \(x\) value by \(3.5\).
      Check that this works for the other rows in the table. Using this relationship, if \(x=100\) then \(y = 100 \times 3.5 = 350\).
    1. From the data points on the graph, we see that the cost can be obtained by multiplying the number of items by \(3\). Therefore, the constant of proportionality is \(3\).
    2. Since \(17 \times 3 = 51\), the cost of \(17\) items is \(51\). 
    3. Using the multiplicative relationship, the cost of \(N\) items would be \(3N\). If the cost is \($204\), then \(N\) must satisfy the equation \(3 N = 204\). Solving this equation for \(N\) gives \(N = 68\).
  5. In the graph in part c), the values of \(y\) are obtained by multiplying the value of \(x\) by \(2.5\) or \(\dfrac{5}{2}\). Therefore, this graph shows a constant of proportionality of \(\dfrac{5}{2}\).
    The graphs in part a) and part b) have constants of proportionality of \(5\) and \(\dfrac{5}{4}\), respectively.

    1. A graph with Time along the horizontal axis and Distance along the vertical axis. Nine points are plotted with a line drawn through seven of them. The points at (45,590) and (120,1400) do not lie on the line representing 12t.
    2. All of the points lie on a single line except for the points corresponding to \(t = 45\) and \(t = 120\). If you ignore these two points then the data in all other rows have the following relationship: Multiply the time by \(12\) to obtain the corresponding distance.
      It appears that the distances for \(t=45\) and \(t=120\) were incorrectly recorded, and the correct values should be \(45\times 12 = 540\) and \(120 \times 12 = 1440\), respectively.
    3. From the table, the car travels \(180\) m every \(15\) seconds which is a unit rate of \(12\) m/s. Converting to kilometres per hour,\[12 \times 60 \times 60 \div 1000 = 43.2\]
      we get an answer of \(43.2\) km/h.
  7. The steepest graph corresponds to the greatest constant of proportionality. Consider the value of \(x\) for which we have all three corresponding values of \(y\). Remember that we obtain the value of \(y\) by multiplying the value of \(x\) by the appropriate constant of proportionality.
    A graph with three lines. The line representing one-half x has 3 points plotted at (1,0.5), (2,1), and (4,2). The line representing 2x has 3 points plotted at (1,2), (2,4), and (5,10). The line representing 4x has 3 points plotted at (2,8), (3,12), and (5,20).
    The largest of the three data points must have a value of \(y\) equal to \(4 x\); the smallest of the three data points must have a value of \(y\) equal to \(\dfrac{1}{2}x\); and the remaining set of data points must have a value of \(y\) equal to \(2x\).
  8. We start by making a table to record the location of each animal at various times. Notice that the cheetah starts at a distance of \(0\), and the hyena starts at a distance of \(435\) m.
    Time (s) Location of Hyena (m) Location of Cheetah (m)
    \(0\) \(435\) \(0\)
    \(1\) \(435+18 \times 1 = 453\) \(33 \times 1 = 33\)
    \(5\) \(435 + 18\times 5 = 525\) \(33 \times 5 = 165\)
    \(10\) \(435 + 18\times 10 = 615\) \(33 \times 10 = 330\)
    \(20\) \(435 + 18 \times 20 = 795\) \(33\times 20 = 660\)
    \(30\) \(435 + 18 \times 30 = 975\) \(33 \times 30 = 990\)
    Notice that after \(30\) seconds, the cheetah has passed the hyena. Can we identify the exact time when the animals cross? 
    Looking at the pattern in the table, we see that the location of the hyena is always given by\[435 + 18\times t\] and the location of the cheetah is always given by\[33 \times t\] where \(t\) is the time. These locations are equal only when \(t = 29\), and so the cheetah overtakes the hyena after \(29\) seconds. Notice that you can solve this problem using algebra by equating the locations to get\[435 + 18t = 33t\] which is equivalent to\[435 = 15t\] since \(33t - 18t = 15t\).
    Dividing both sides of the equation by \(15\) gives an answer of\[t = \dfrac{435}{15} = 29\]