Proportional graphs and rates of change

(a) $30$

1. $y$ is always $3$ times $x$, because $12 \div 4 = 3$.
2. When $x = 10$, $y = 3 \times 10 = 30$.

• M1 Finding the multiplier 3, or scaling by $\frac{10}{4}$.
• A1 The correct answer, $30$.

(b) $3 : 1$

1. $12 : 4 = 3 : 1$, and this ratio is the same for every pair of values.

• B1 The correct answer, $3 : 1$.

(a) $60$ m/s

1. $$\frac{840-480}{17-11}$$
2. The final segment lasts 17 $-$ 11 = 6 seconds and covers 840 $-$ 480 = 360 metres.
3. For a straight distance-time segment, speed is the gradient: 360 $\div$ 6 = 60 m/s.

• P1 Establishing $\frac{840-480}{17-11}$ or an equivalent valid method.
• A1 Correct answer: $60$ m/s

(a) $4.5$ m/s

1. $$\frac{38-11}{10-4}$$
2. Speed is the gradient of a distance-time graph.
3. Gradient = (38 $-$ 11)/(10 $-$ 4) = 27/6 = 4.5 m/s.

• P1 Establishing $\frac{38-11}{10-4}$ or an equivalent valid method.
• A1 Correct answer: $4.5$ m/s

(a) $5$ litres/min

1. Use change in volume divided by change in time.
   $$\frac{54-14}{8}$$
2. Therefore $5$ litres/min.

• M1 Use change in volume divided by change in time.
• A1 Correct answer: $5$ litres/min

(b) No. The tank already contains water at time zero, so the straight line does not pass through the origin.

1. No. The tank already contains water at time zero, so the straight line does not pass through the origin.

• C1 Correct conclusion with supporting reasoning: No. The tank already contains water at time zero, so the straight line does not pass through the origin.

(a) $n=6t$

1. Find the number of labels printed each second.
   $$42/7$$
2. Therefore $n=6t$.

• M1 Find the number of labels printed each second.
• A1 Correct answer: $n=6t$

(b) $19$ seconds

1. Divide the required output by the constant rate.
   $$114/6$$
2. Therefore $19$ seconds.

• P1 Divide the required output by the constant rate.
• A1 Correct answer: $19$ seconds

(a) £$7$

1. Find the change in fare for four more kilometres.
   $$\frac{46-18}{6-2}$$
2. Therefore £$7$.

• M1 Find the change in fare for four more kilometres.
• A1 Correct answer: £$7$

(b) £$4$

1. Subtract the distance charge from either plotted fare.
   $$18-2\times 7$$
2. Therefore £$4$.

• P1 Subtract the distance charge from either plotted fare.
• A1 Correct answer: £$4$

(a) Company B: its graph is a straight line through the origin.

1. Line B is a straight line through $(0, 0)$, so B's fare is directly proportional to distance.
2. Line A crosses the fare axis at £3, so A's fare is not proportional: 2 km costs £7 but 4 km costs £11, not £14.

• C1 Company B, because its line passes through the origin (or because A has a fixed charge of £3).

(b) 2: the fare rises by £2 for each extra km.

1. Line A goes from $(0, 3)$ to $(10, 23)$.
2. $$\text{gradient} = \frac{23 - 3}{10 - 0} = 2$$
3. The fare increases by £2 for every kilometre travelled.

• M1 Using rise over run from two points on line A.
• A1 The correct answer, $2$.
• C1 Saying the fare rises by £2 per km (the cost per kilometre).

(c) $6$ km

1. Set the fares equal.
   $$2d + 3 = 2.5d \Rightarrow 0.5d = 3 \Rightarrow d = 6$$
2. Both charge £15 at 6 km, where the lines cross.

• P1 Reading the crossing point from the graph, or forming $2d + 3 = 2.5d$.
• A1 The correct answer, $6$ km.

(a) 18 litres per hour

1. 300 cm³ is 0.3 litres.
2. Multiply the per-minute rate by 60: 0.3 $\times$ 60 = 18 litres per hour.

• B1 Correct answer: 18 litres per hour