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(ii) Plan A: (b)
T = a + (n – 1)d Distance (km)
n
= 50 + (n – 1)(10)
= 40 + 10n
2000
Plan B:
T = ar n – 1
n
= 5(2) n – 1
Plan C: 1000
T = n 2 [2]
n
(b) (i) Plan A:
Total: 5(8) + 45(8) = $680
2
0
Plan B: 1300 1500 1700 1900 2100 2300 0100
Total: 5(2 – 1) = $1275 Time (hours) [2]
8
Plan C: 3 (a) 8.45 a.m. [1]
Total: 8(8 + 1)(2 × 8 + 1) = $204 [3] (b) 300 m [1]
6 (c) 11.15 a.m. [1]
(ii) 5(2 – 1) = 4500 (d) 11.15 a.m. – 9.45 a.m. = 1.5 h [2]
n
2 = 901 (e) 1200 m × 2 = 2400 m [1]
n
n = 9.815 (f) Speed = 1200 m
n ∼10th month [2] 1 1 h
= 1 m/s
3
Speed, distance and time 1200 m
Speed =
2
1 (a) (i) Total distance travelled 2 1.5 h
= 18 km + 5 km = 9 m/s
= 23 km 1
Difference = m/s [2]
Total time used v 30 9
18 5 4 (a) =
15
k
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=
+

7.5 4 v = 2k [2]
= 2.4 + 1.25 = 3.65 h 30
Complete = 7.38 a.m. + 3.65 h (b) Deceleration = Time
= 7.38 a.m. + 3 h + (0.65 × 60) min Time = 30
= 11.17 a.m. [3] 1.5
(ii) Average speed = 23 = 20 s
3.65 Time it starts decelerating = 60 – 20
= 6.301 km/h [1] = 40 s [2]
(b) (i) x + x + 7 = 23 (c) Distance = (60 + 25)(30)
x = 8 [1] 2
(ii) Total distance travelled = 23 km = 1275 m
Total time used = 1.275 km [2]

= 8 + 15 = 3.5 h [2] 5 (a) Total distance = [(12 – 5) + 17]v
4 10 2
Average speed 96 = 24v
23 km v = 4 m/s [2]
= (T – 25)(10)
3.5 h (b) = 48
= 46 km/h 2 T – 25 = 9.6
7
(c) Time = 7.38 a.m. + 3.5 h T = 34.6 s [2]
= 7.38 a.m. + 3 h + 30 min (c) (i) Acceleration = 4 m/s 2 [1]
= 11.08 a.m. 5
Susan will complete the journey first. [2] (ii) Acceleration = 10

2 (a) Average speed = 2000 = 363.64 km/h [1] 34.6 – 25
5.5 = 1.04 m/s 2 [2]

Cambridge IGCSE
TM
164 Ace Your Mathematics

Answers.indd 164 15/03/2022 11:08 AM