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Model answers
5 Sequence goes up in 3s so U n
= 3n + k.
Testing e.g. n = 2, 3 × 2 + k = 4
so k = – 2
so U n
= 3n – 2
6 (a) V = 25 × 3·1 × (7·3 2 – 5·9 2 )
= 25 × 3·1 × 18·48 = 1432·2
(b) x = 4y 2 – 3
⇒ x + 3 = 4y 2
⇒
x + 3
–––– = y
4
2
⇒
x
––––
+ 3
= y
4
⇒ y =
x
––––
+ 3
4
7 (a) e = 5d + 3
⇒ e – 3 = 5d
⇒
e
––––
– 3
= d
5
⇒ d =
e
––––
– 3
5
(b) ––––––
3d – 7
= e
4 + 5d
⇒ 3d – 7 = e(4 + 5d)
⇒ 3d – 7 = 4e + 5ed
⇒ 3d = 4e + 5ed + 7
⇒ 3d – 5ed = 4e + 7
⇒ d(3 – 5e) = 4e + 7
⇒ d =
4e
–––––
+ 7
3 – 5e
14·9 × ( – 10·2) – 151·98
8 (a) f = –––––––––––– = –––––––
14·9 + ( – 10·2) 4.7
= – 32·336...
= – 32·3 to 3 s.f.
(b) f = ––––
uv
u + v
⇒ f(u + v) = uv
⇒ fu + fv = uv
⇒ fu = uv – fv
⇒ fu = v (u – f)
⇒ ––––
fu
= v
u – f
⇒ v = ––––
fu
u – f
9 t = 1, n = 2 × 3
t = 2, n = 2 × 3 2
t = 3, n = 2 × 3 3 and so on
n = 2 × 3 t
Direct and inverse proportion
(page 27)
1 y ∝ x 2
so y = kx 2
so 75 = k × 5 2
so k = 3
so y = 3 × 10 2
= 300
1
2 N ∝ ––
d 2
k
so N = ––
d 2
k
so 8000 = –– 2
2
so k = 4 × 8000 = 32 000
32 000
so N = ––––––
4 2
= 2000
3 D ∝ V 2
so D = kV 2
so 12·5 = k × 50 2
12·5
so k = –––– = 0·005
2500
D = 0·005 × 120 2
= 72 m
4 R ∝ V 2
r = kV 2
100 = k × 300 2
100
k = –––––– = 0·001 11
90 000
(a) R = 0·001 11 × 600 2
= 400 N
(b) 200 = 0·001 11 × V 2
V 2 = 180 000
V = 424 m/s
Simultaneous equations and
linear inequalities (page 29)
1 10x + 8y = 26
3x + 8y = 5
Subtract
7x = 21
x = 3
Substitute
3 × 3 + 8y = 5
8y = 5 – 9
y = – 0·5
Solution is x = 3, y = – 0·5
2 4x + 3y = 5
6x + 3y = 3
Subtract
– 2x = 2
x = – 1
Substitute
2 × – 1 + y = 1
y = 1 + 2
y = 3
Solution is x = – 1, y = 3
3 (a) Replace x with 3 and y with 2.
(b) 9p + 11q = 5
(c) 3p + 2q = 5
9p + 11q = 5
9p + 6q = 15
9p + 11q = 5
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
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