Module 1 Topic Test

Linear Relationships

Maths Standard Year 11 · All 5 lessons · MC checkpoint plus separate short-answer practice

L1, Coordinates, Tables & Linear Patterns L2, Gradient as Rate of Change L3, Intercepts & Linear Equations L4, Direct Variation
20 MC 6 SA ~50 min
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Part A, Multiple Choice (1 mark each, 25 marks total)
1 A table has inputs $0, 1, 2, 3$ and outputs $6, 9, 12, 15$. What is the ordered pair for the row where the input is $2$? L1
A $(12, 2)$
B $(2, 12)$
C $(2, 9)$
D $(9, 2)$
B, $(2, 12)$. An ordered pair is written $(x, y)$ with the input first. When the input is $2$, the matching output is $12$, so the pair is $(2, 12)$.
2 A table shows a linear pattern when equal steps in the input produce which of the following? L1
A Outputs that double each step
B A constant ratio in the output
C Outputs that always decrease
D A constant difference in the output
D, a constant difference in the output. A relationship is linear when equal input steps give equal output changes, that is, a constant first difference. A constant ratio or doubling indicates a non-linear pattern.
3 Which of these tables (input then output) is a linear relationship? L1
A $1{\to}3,\ 2{\to}6,\ 3{\to}12,\ 4{\to}24$
B $1{\to}1,\ 2{\to}4,\ 3{\to}9,\ 4{\to}16$
C $1{\to}5,\ 2{\to}8,\ 3{\to}11,\ 4{\to}14$
D $1{\to}2,\ 2{\to}5,\ 3{\to}10,\ 4{\to}17$
C. In C the outputs go $5, 8, 11, 14$ with a constant first difference of $+3$ each step, so it is linear. A doubles, B are squares, and D has differences $3, 5, 7$, so all three are non-linear.
4 On the Cartesian plane, in which quadrant does the point $(-3, 2)$ lie? L1
A Quadrant 1
B Quadrant 2
C Quadrant 3
D Quadrant 4
B, Quadrant 2. The $x$-value is negative and the $y$-value is positive. A negative $x$ with positive $y$ places the point in the upper-left region, Quadrant 2.
5 A taxi fare starts at $\$6$ and increases by $\$3$ per kilometre. Using kilometres as the input, what ordered pair represents a $4\text{ km}$ trip? L1
A $(4, 18)$
B $(4, 12)$
C $(18, 4)$
D $(4, 24)$
A, $(4, 18)$. Fare $= 6 + 3 \times 4 = 6 + 12 = 18$. The input (distance) comes first, so the pair is $(4, 18)$.
6 A savings balance changes from $\$300$ at week $0$ to $\$450$ at week $5$. What is the gradient with correct units? L2
A $\$5$ per week
B $\$150$ per week
C $\$30$ per week
D $\$75$ per week
C, $\$30$ per week. $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{450 - 300}{5 - 0} = \dfrac{150}{5} = 30$ dollars per week.
7 What is the gradient of the line joining the points $(2, 7)$ and $(6, 19)$? L2
A $3$
B $\dfrac{1}{3}$
C $4$
D $12$
A, $3$. $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{19 - 7}{6 - 2} = \dfrac{12}{4} = 3$.
8 Water drains from a tank so its volume falls from $80\text{ L}$ to $56\text{ L}$ over $6$ minutes. Which statement about the gradient is correct? L2
A $+4\text{ L/min}$, the volume is increasing
B $-4\text{ L/min}$, the volume is decreasing
C $0\text{ L/min}$, the volume is constant
D $-24\text{ L/min}$, the volume is decreasing
B, $-4\text{ L/min}$, the volume is decreasing. $m = \dfrac{56 - 80}{6 - 0} = \dfrac{-24}{6} = -4\text{ L/min}$. The negative sign shows the volume is falling by $4$ litres each minute.
9 A straight line is horizontal (it neither rises nor falls as the input increases). What is its gradient? L2
A Positive
B Negative
C Undefined
D Zero
D, zero. A horizontal line has no vertical change, so the rise is $0$ and $m = \dfrac{0}{\text{run}} = 0$. A zero gradient means the output stays constant.
10 A car travels so its distance rises from $40\text{ km}$ to $190\text{ km}$ between hour $1$ and hour $4$. What is the gradient, interpreted as a speed? L2
A $50\text{ km/h}$
B $150\text{ km/h}$
C $30\text{ km/h}$
D $47.5\text{ km/h}$
A, $50\text{ km/h}$. $m = \dfrac{190 - 40}{4 - 1} = \dfrac{150}{3} = 50\text{ km/h}$.
11 A bike hire is modelled by $C = 5h + 12$, where $C$ is the cost in dollars and $h$ is the number of hours. What does the $12$ represent? L3
A The cost per hour
B The fixed cost paid before any riding
C The total cost for the hire
D The number of hours ridden
B, the fixed cost paid before any riding. In $y = mx + b$, the constant $b$ is the vertical intercept, the output when $x = 0$. Here $C = 12$ when $h = 0$, so $\$12$ is the fixed starting cost. The $5$ is the rate per hour.
12 In the equation $y = mx + b$, what is the value of $y$ when $x = 0$? L3
A $m$
B $0$
C $b$
D $m + b$
C, $b$. Substituting $x = 0$ gives $y = m(0) + b = b$. The value of $b$ is the vertical intercept, the output where the line crosses the $y$-axis.
13 A phone plan costs $\$25$ per month plus $\$0.10$ per text message. Which equation models the monthly cost $C$ for $t$ text messages? L3
A $C = 25t + 0.10$
B $C = 25.10t$
C $C = 0.10t - 25$
D $C = 0.10t + 25$
D, $C = 0.10t + 25$. The rate ($\$0.10$ per text) is the gradient and multiplies the input $t$; the fixed monthly charge ($\$25$) is the intercept added on. So $C = 0.10t + 25$.
14 A gym membership is modelled by $C = 8v + 40$, where $v$ is the number of visits. What is the cost of $15$ visits? L3
A $\$120$
B $\$48$
C $\$160$
D $\$720$
C, $\$160$. Substitute $v = 15$: $C = 8(15) + 40 = 120 + 40 = 160$ dollars.
15 A straight-line graph crosses the $y$-axis at $(0, 6)$ and has a gradient of $-2$. What is its equation? L3
A $y = 6x - 2$
B $y = 2x + 6$
C $y = -2x + 6$
D $y = -2x - 6$
C, $y = -2x + 6$. The gradient $m = -2$ and the vertical intercept $b = 6$, so $y = mx + b = -2x + 6$.
16 Which equation represents a direct variation relationship? L4
A $y = 5x + 12$
B $y = 4x$
C $y = x^2$
D $y = 3x - 1$
B, $y = 4x$. Direct variation has the form $y = kx$ with no added constant, so the graph passes through the origin. A and D have non-zero intercepts, and C is not linear.
17 If $y$ varies directly with $x$ and $y = 20$ when $x = 5$, what is the constant of variation $k$? L4
A $100$
B $0.25$
C $15$
D $4$
D, $4$. For direct variation, $k = \dfrac{y}{x} = \dfrac{20}{5} = 4$. The relationship is therefore $y = 4x$.
18 The cost of apples varies directly with the mass bought, with $y = 4x$ (dollars for $x$ kilograms). If the mass is doubled, the cost is: L4
A Unchanged
B Increased by $\$4$
C Halved
D Doubled
D, doubled. Direct variation is perfectly proportional with no fixed fee, so doubling the input doubles the output. For example, $x = 3$ gives $\$12$ and $x = 6$ gives $\$24$.
19 Which table (input then output) shows direct variation? L4
A $2{\to}9,\ 4{\to}18,\ 6{\to}27$
B $2{\to}10,\ 4{\to}16,\ 6{\to}22$
C $2{\to}8,\ 4{\to}12,\ 6{\to}16$
D $2{\to}5,\ 4{\to}20,\ 6{\to}45$
A. Direct variation needs a constant ratio $y/x = k$. In A: $\tfrac{9}{2} = \tfrac{18}{4} = \tfrac{27}{6} = 4.5$, so $k = 4.5$. In B and C the ratio changes (they have a non-zero intercept), and in D it is not constant.
20 Which statement about direct variation is correct? L4
A A direct variation graph passes through the origin $(0, 0)$
B A direct variation graph never passes through the origin
C Every straight line is a direct variation
D Direct variation must have a fixed starting fee
A, a direct variation graph passes through the origin $(0, 0)$. Because $y = kx$ has no added constant, an input of $0$ gives an output of $0$. Not every straight line is direct variation, for example $y = 5x + 12$ is linear but has a non-zero intercept.
Part B, Short Answer (show all working)
1 L1
A market stall records the cost of buying different numbers of oranges. The table is: $1$ orange $\to \$2$, $2 \to \$4$, $3 \to \$6$, $4 \to \$8$.
(a) Write the ordered pair for buying $3$ oranges.
(b) Show that the table is linear by finding the first difference.
(c) Predict the cost of buying $7$ oranges.
(a) The input comes first, then the output: $(3, 6)$.
(b) The outputs are $2, 4, 6, 8$. Each step increases by $8 - 6 = 6 - 4 = 4 - 2 = 2$. The first difference is a constant $\$2$, so the table is linear.
(c) Continuing the constant difference (or $\$2$ per orange): $7 \times 2 = \$14$.
2 L2
A water tank is being filled. At $2$ minutes it holds $50\text{ L}$ and at $8$ minutes it holds $170\text{ L}$.
(a) Calculate the gradient of the volume-time relationship, with units.
(b) Interpret what this gradient means in the context of the tank.
(a) $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{170 - 50}{8 - 2} = \dfrac{120}{6} = 20\text{ L/min}$.
(b) The tank is filling at a rate of $20$ litres per minute. The positive gradient shows the volume is increasing steadily.
3 L3
A plumber charges a $\$60$ call-out fee plus $\$45$ per hour of work. Let $C$ be the total cost in dollars for $h$ hours.
(a) Write a linear equation for $C$ in terms of $h$.
(b) State the gradient and the vertical intercept, and explain what each represents.
(c) Find the total cost of a job that takes $3$ hours.
(a) $C = 45h + 60$.
(b) Gradient $= 45$, the rate of $\$45$ per hour. Vertical intercept $= 60$, the fixed $\$60$ call-out fee paid when $h = 0$.
(c) $C = 45(3) + 60 = 135 + 60 = \$195$.
4 L4
The distance $d$ (in kilometres) a car travels varies directly with the fuel used $f$ (in litres). A car travels $84\text{ km}$ on $6\text{ L}$.
(a) Find the constant of variation $k$.
(b) Write the equation linking $d$ and $f$.
(c) How far can the car travel on $10\text{ L}$ of fuel?
(a) $k = \dfrac{d}{f} = \dfrac{84}{6} = 14$ (kilometres per litre).
(b) $d = 14f$ (direct variation, graph through the origin).
(c) $d = 14 \times 10 = 140\text{ km}$.
5 L1 & L2
A candle burns at a steady rate. Its height is recorded as: $0\text{ h} \to 24\text{ cm}$, $1\text{ h} \to 21\text{ cm}$, $2\text{ h} \to 18\text{ cm}$, $3\text{ h} \to 15\text{ cm}$.
(a) Explain how the table shows the relationship is linear.
(b) Find the gradient, with units, and explain its sign.
(c) After how many hours will the candle burn out (height $0\text{ cm}$)?
(a) The heights fall by the same amount each hour: $24 \to 21 \to 18 \to 15$ is a constant first difference of $-3\text{ cm}$, so the relationship is linear.
(b) $m = \dfrac{21 - 24}{1 - 0} = \dfrac{-3}{1} = -3\text{ cm/h}$. The negative sign shows the height is decreasing (the candle shortens by $3\text{ cm}$ each hour).
(c) Height model: $H = 24 - 3t$. Set $H = 0$: $24 - 3t = 0 \Rightarrow 3t = 24 \Rightarrow t = 8$ hours.
6 L3 & L4
Two mobile data offers are being compared. Offer 1: $y = 2x$ (dollars for $x$ gigabytes, no fixed fee). Offer 2: $y = 1.5x + 6$ (dollars for $x$ gigabytes).
(a) Which offer is a direct variation, and how can you tell?
(b) State the vertical intercept of Offer 2 and explain what it represents.
(a) Offer 1 is direct variation. It has the form $y = kx$ with no added constant, so its graph passes through the origin. Offer 2 has a non-zero intercept of $6$, so it is linear but not direct variation.
(b) The vertical intercept of Offer 2 is $6$. It represents the fixed $\$6$ charge paid when $x = 0$ (before any data is used).
Module 1 Complete

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