Speed, density, pressure and unit pricing

(a) $24$ km/h

1. $$45/60$$
2. $$18/0.75$$
3. 45 minutes is 45/60 = 0.75 hours.
4. Average speed = distance $\div$ time = 18 $\div$ 0.75 = 24 km/h.

• P1 Establishing $45/60$ or an equivalent valid method.
• P1 Establishing $18/0.75$ or an equivalent valid method.
• A1 Correct answer: $24$ km/h

(a) $60$ km/h

1. 2 hours 30 minutes is $2.5$ hours.
2. $$\text{speed} = \frac{150}{2.5} = 60 \text{ km/h}$$

• M1 Writing the time as $2.5$ hours (or 150 minutes) and dividing the distance by it.
• A1 The correct answer, $60$ km/h.

(b) $18$ km

1. 45 minutes is $\frac{45}{60} = 0.75$ hours.
2. $$24 \times 0.75 = 18 \text{ km}$$

• M1 Writing $24 \times 0.75$ (or $24 \times \frac{45}{60}$).
• A1 The correct answer, $18$ km.

(a) $60$ km/h

1. $$54+36$$
2. $$\frac{45+15+30}{60}$$
3. Total distance = 54 + 36 = 90 km.
4. Total time = 45 + 15 + 30 = 90 minutes = 1.5 hours.
5. Average speed = 90 $\div$ 1.5 = 60 km/h.

• P1 Establishing $54+36$ or an equivalent valid method.
• P1 Establishing $\frac{45+15+30}{60}$ or an equivalent valid method.
• A1 Correct answer: $60$ km/h

(a) $6$ N/cm²

1. $$6\times 15$$
2. $$540/90$$
3. Area = 6 $\times$ 15 = 90 cm².
4. Pressure = force $\div$ area = 540 $\div$ 90 = 6 N/cm².

• P1 Establishing $6\times 15$ or an equivalent valid method.
• P1 Establishing $540/90$ or an equivalent valid method.
• A1 Correct answer: $6$ N/cm²

(a) £$728$

1. $$35\times 16$$
2. $$7\times 24$$
3. Basic pay = 35 £$\times$16 = £560.
4. Overtime is 7 hours at £24 per hour, giving £168.
5. Total pay = £560 + £168 = £728.

• P1 Establishing $35\times 16$ or an equivalent valid method.
• P1 Establishing $7\times 24$ or an equivalent valid method.
• A1 Correct answer: £$728$

(a) $2.82$ g/cm³

1. $$80\times 2.4+120\times 3.1$$
2. $$564/200$$
3. Use mass = density $\times$ volume: the masses are 192 g and 372 g.
4. Total mass is 564 g; total volume is 200 cm³.
5. Combined density = 564 $\div$ 200 = 2.82 g/cm³.

• P1 Establishing $80\times 2.4+120\times 3.1$ or an equivalent valid method.
• P1 Establishing $564/200$ or an equivalent valid method.
• A1 Correct answer: $2.82$ g/cm³

(a) $4.8$ km/h

1. $$2/4+2/6$$
2. $$4/(5/6)$$
3. The times are 2/4 = 1/2 hour and 2/6 = 1/3 hour.
4. Total time = 5/6 hour and total distance = 4 km.
5. Average speed = 4 $\div$ (5/6) = 4.8 km/h.

• P1 Establishing $2/4+2/6$ or an equivalent valid method.
• P1 Establishing $4/(5/6)$ or an equivalent valid method.
• A1 Correct answer: $4.8$ km/h

(a) $40$ km/h

1. $$2/(1/v+1/(v+20))=48$$
2. $$v^{2}-28v-480=0$$
3. For each section of length d, the total time is d/v + d/(v + 20).
4. Average speed gives 2d/[d/v + d/(v + 20)] = 48.
5. Cancel d and rearrange: 2v(v + 20) = 48(2v + 20), so $v^{2}$ $-$ 28v $-$ 480 = 0.
6. Factorise (v $-$ 40)(v + 12) = 0. Since v > 0, v = 40 km/h.

• P1 Establishing $2/(1/v+1/(v+20))=48$ or an equivalent valid method.
• P1 Establishing $v^{2}-28v-480=0$ or an equivalent valid method.
• A1 Correct answer: $40$ km/h