Download Answers to pages 39

Model answers
Linear programming (page 41)
1 (a) Type A takes 1 hour of machine time and type B takes 2 hours. There are 28 hours of
machine time in a day.
x + 2y 28
Type A takes 3 hours of workers’ time and type B takes 1 hour. There are 24 hours of workers’
time in a day.
3x + y 24
y
(b)
30
24
20
14
10
x + 2y = 28
3x + y = 24
8
O
10
28
20
x
30
(c) (i) The greatest profit will result from either the greatest number of type A, the greatest number of type
B or the greatest total number of items so test the points (8, 0), (0, 14) and (4, 12).
Profit at (8, 0) = $160.
Profit at (0, 14) = $140.
Profit at (4, 12) = $80 + $120 = $200.
So the greatest profit that can be made is $200.
(ii) 4 type A and 12 type B.
2 (a) (i) There is $1000 available; paperback books cost $10 and hardback books cost $25.
So 10x + 25y 1000.
Dividing by 5 gives 2x + 5y 200.
(ii) y > x and x + y 50
(b)
y
y=x
50
40
20
O
x
50
x + y = 50
100
2x + 5y = 200
(c) 33 hardback books
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
9
Model answers
3 (a) (i) x small vehicles will carry 5x people and y large vehicles will carrry 8y people. 60 people must
fit into these vehicles so 5x + 8y 60.
(ii) 600x + 300y 4500 or 2x + y 15
x8
y7
y
(b)
x=8
15
10
7.5
7
5
y=7
2x + y = 15
O
5x + 8y = 60
7 8
5
x
12
10
15
(c) (i) The points (4, 7), (5, 7), (6, 7), (7, 7), (8, 7), (5, 6), (6, 6), (7, 6), (8, 6), (5, 5), (6, 5), (7, 5),
(8, 5), (6, 4), (7, 4), (8, 4), and (8, 3) marked.
(ii) 10
(iii) 5 of each type or 6 small and 4 large vehicles.
4 (a) x > y
10x + 30y 300 or x + 3y 30
(b)
y
y=x
20
10
(9, 7)
x + 3y = 30
O
10
20
28
x
30
(c) (i) The shortest amount of time spent must result from either the highest or lowest integer
point on the line x + y = 30 so test the points (9, 7) and (30, 0).
Time at (9, 7) is 9 + 14 = 23 hours.
Time at (30, 0) is 30 hours.
So the shortest time is 23 hours.
(ii) 9 individual photos and 7 sets.
10
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
Model answers
Interpreting graphs (page 43)
(
)
1
1 (a) 6 km/h means that in 10 minutes 6– th of an hour Steve walks 1 km.
32 km/h for 8 km.
1
Time = 8 ÷ 32 = 4– hour = 15 mins.
10
9
8
Distance in km
7
6
5
4
3
2
1
0800
0820
0810
0830
Time
2
(b) 33– km or 3·7 km
2 e.g. A car accelerates and then travels at a steady speed and then stops suddenly (crashes, etc.).
3 (a)
(b)
d
t
d
(c)
t
d
t
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
11
Model answers
Quadratic and other functions (page 47)
1 (a) –5 0 [3] 4 3 [0] –5
y
4
3
2
1
–
O
–
2
1
1
2
3
4
x
5
–
1
–
2
–
3
–
4
–
5
(b) (i) 2
(ii) 0·6 and 3·4
2 (a) T
(b) Q
3 (a)
(c) U
x
1
2
3
4
5
6
V
7
32
81
160
275
432
v
500
450
400
350
300
250
200
150
100
50
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
(b) 4·4 cm, 4·4 cm, 10·4 cm
12
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
6
x
Model answers
4 (a)
y
5
4
3
2
1
–5
–4
–3
–2
O
–1
1
2
3
4
x
5
–1
–2
–3
–4
–5
(b) 2·5
5 (a) (–2, –1) in table
y
10
–2
–3
–1
O
1
2
3
x
–10
(b) –2 to –1·7, –0·4 to –0·2, 2·0 to 2·2
(c) y = 2x + 2 drawn,
–
2·3 to –2·0, –0·6 to –0·3, 2·5 to 2·8
6 Manipulate x 3 – 2x – 1 = 0 to give x 3 = 2x + 1.
The line to draw is y = 2x + 1.
y
8
4
–
2
–
1
O
4
y = 2x + 1
1
2
x
–
y = x3
–
8
The solution is where the line cuts the curve.
x = –1, –0·6 or 1·6
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
13
Model answers
2x 3
7x 2
+
– 6 = 0 gives
7 Manipulating
6 – 2x 3 = 7x 2.
Dividing by 3x 2 gives
2 2x 7
––
– –– = – .
x2
3
3
Manipulating this gives
–1
2
1
––
– x = –– x + 2 – .
3
x2
3
y
2
x2
y=
–x
4
3
y = – 1x + 2 1
2
3
3
1
–
4
–
3
–
2
–
1 O
–
1
1
2
y=
3
2
x2
x
4
–x
The solution is x = –3·2, –1·1 or 0·8.
Graphs – gradient and area (pages 51–52)
4
1 (a) ––– = 1.6 ms–2
2·5
1
(b) Area under graph = 4 × 3.5 + – × 2.5 × 4 = 14 + 5 = 19 m.
2
2 (a)
Time (t hours)
0
1
2
3
4
Number of bacteria (n)
20
60
180
540
1620
5
4860
(b) Each hour the number of bacteria multiplies by 3 so n = 20 × 3t.
(c)
n
5000
4000
3000
2000
1000
0
1
2
3
4
5
t
2600 – 450
(d) (i) Gradient = –––––––––– = 1075
5–3
(ii) The rate of increase in bacteria per hour.
48
3 t = 0, d = 12 + –– – 20 = 29; p = 29
1
48
t = 5, d = 62 + –– – 20 = 24; q = 24
6
48
t = 7, d = 82 + –– – 20 = 50; r = 50
8
14
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
Model answers
(b)
Approximate distance = 19.85 km or 20 km
d (m)
(b) The acceleration is greatest when the graph is
steepest. This is approximately at 9.45 a.m.
50
40
Functions (page 55)
1 (a)
30
20
F
f(x) = 2x – 1
y = 2x – 1
2x = y + 1
y+1
x = –––––
2
10
0
1
2
3
4
5
6
7 t (s)
(c) F on graph (d must be increasing because the fish
is swimming away from Dimitra.)
t = 3·6 seconds
(d) 3·35 – 0·8 = 2.55 s
(e) Tangent at t = 2·5 drawn on graph.
18·5 – 0
Gradient = speed = ––––––– = 3·7 m/s.
6–1
4 (a) Tangent at t = 30.
33·5 – 12
Gradient = acceleration = –––––––– = 1·08 m/s2
35 – 15
(b) Taking strips of width = 10 s
1
– × 5 × 10
2
1
– (5 + 14) × 10
2
1
– (14 + 28) × 10
2
1
– (28 + 32) × 10
2
10 × 32
10 × 32
1
– (32 + 25) × 10
2
1
– × 25 × 10
2
Total
x+1
f –1(x) = –––––
2
(b) gf(x) = g[f(x)]
= g[2x – 1]
= (2x – 1)2 – 1
= 4x 2 – 4x
2 (a) (i) g(4) = 2 × 42 – 5
= 27
1
–
(ii) fg(4) = 27 3
=3
(b) (i) gf(x) = g[f(x)]
1
–
= g[x 3]
1
–
2
–
= 2x 3 – 5
=
25
=
95
=
210
3 (a)
=
300
(b)
=
=
320
320
=
285
=
125
(ii)
(c)
= 1680
(d)
Approximate distance = 1680 m
1
– hour
5 (a) Using strips of width 2
1 1
– × – ×2
= 0·5
2 2
1
– (2 +
2
1
– (8 +
2
1
– (9.1
2
1
– (8.7
2
1
– (8.2
2
Total
1
–
8) × 2
1
–
9.1) × 2
1
–
+ 8.7) × 2
1
–
+ 8.2) × 2
1
–
+ 7.4) × 2
=
2.5
=
4.275
=
4.45
=
4.225
=
3.9
= 19.85
1
–
= 2 × x3 × x3 – 5
4 (a)
1
–
x3 = y
x = y3
–1
f (x) = x3
–
f( 1) = 3 × –1 – 5
= –8
3x – 5 = y
y+5
x = –––––
3
x+5
f –1(x) = –––––
3
fg(x) = f[x + 1]
= 3(x + 1) – 5
= 3x – 2
3f(x) = 5g(x)
3(3x – 5) = 5(x + 1)
9x – 15 = 5x + 5
4x = 20
x=5
1
– (2 × –4 + 5)
f(–4) = 3
= –1
1
– (2x + 5) = y
(b) 3
2x + 5 = 3y
2x = 3y – 5
3y – 5
x = ––––––
2
3x
–5
f –1(x) = ––––––
2
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
15
Model answers
5 (a) 3x + 1 = y
3
S
3x = y – 1
y–1
x = –––––
3
x–1
f –1(x) = –––––
3
(b) fg(2) = f[g(2)]
= f[2 × 22]
=3×8+1
= 25
(c) gf(x) = g[3x + 1]
= 2(3x + 1)2
= 2(9x 2 + 6x + 1)
= 18x 2 + 12x + 2
(d) ff(x) = f[3x + 1]
= 3(3x + 1) + 1
= 9x + 4
5
6
9
10
Total = 5 + 6 + 9 + 10 = 30
4
J
K
L
(b) S' ∪ T
(c) Let y students study both.
6 (a) f(–1) = 2 × –1 + 1
= –1
(b) 2x + 1 = y
2x = y – 1
y–1
x = –––––
2
x
–
1
f –1(x) = –––––
2
x
––5 =y
(c)
3
x
– =y+5
3
x = 3y + 15
–
1
g (x) = 3x + 15
0
(d) fg(0) = f – –5
3
= f[–5]
= 2 × –5 + 1
= –9
x
(e)
fg(x) = f – – 5
3
x
=2 ––5 +1
3
2x
= –– – 9
3
C
F
12 – y
y
n() = 32
12 – y + 18 + 9 = 32
39 – y = 32
y=7
12 – y = 5
so 5 students study chemistry but not French
Matrices (page 61)
3×5 + 4×1
2×5 + 4×1
(
=( )
1 (a) AB =
[ ]
[ ]
)
19
14
(b)
1
C
B
|A| = (3 × 4) – (2 × 4)
=4
–1
Sets and Venn diagrams (page 58)
A
=
4
–
4
–4
–2
3
–
4
( )
–
4
–
4
=
1
–1
–1
3
–
4
( )
–
2
2 (a) The elements in the corresponding positions in A
and M add up to zero.
–
(b)
3
1
2
2
( )
( ) is the identity matrix so N is the inverse of A.
M=
–
–
1 0
0 1
|A| = (3 × 2) – (1 × –2)
=8
2 (a) {6. 12}
(b) {1, 5, 7, 11}
A–1 = N
1
–
4
1
–
4
–1
3
–
8
( )
–
8
16
18 – y
9
[ ]
A
W
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
Model answers
3 (a)
1
x
(
0
0
)+ (
–
3
2
–
–
2
3
4
6
)=(
–
–
2
3
)
–
So x + 2 = 6
x = –4
(b) |R| = 4 × 3 – (–6 × –2)
=0
The determinant of R is zero.
(b) |Q| = 3 × 3 – (–2 × –2)
=5
Q–1
=
3
–
5
2
–
5
2
–
5
3
–
5
( )
4 (a) XY = Z
So 3p – 8 = q
1
2
–
and 4p + 4 = 16
2
From –
4p = 12
–
p = 3
Substituting in 1
3 × –3 – 8 = q
–
q = 17
(b) |X| = 3 × –1 – (–4 × 2)
=5
X–1 =
–1
–2
4
–
5
3
–
5
( )
–
5
–
5
5 (a) The number of columns
in A is not the same as the number of rows in B.
3×3 + –4×–2 3×–4 + –4×5
(b) (i) B2 = –2×3 + 5×–2 –2×–4 + 5×5
(
=
)
17
–
16
(
–
32
33
)
(ii) |B| = 3 × 5 – (–2 × –4)
=7
B–1
=
5
–
7
4
–
7
2
–
7
3
–
7
( )
Properties of triangles and other shapes (page 64)
1 (a) Five equal sides so centre will split in to five equal angles.
360°
x = ––––
5 = 72°
(b) As the pentagon is regular, all the lines from the vertices to the centre are equal.
So triangle ABC is isosceles.
2 x = 53° (corresponding angles are equal)
y = 43° + 53° = 96° (exterior angle = sum of opposite interior angles)
3 Hexagon: exterior angle =
360º
–––
6
= 60° (sum of exterior angles = 360°)
interior angle = 180° – 60° = 120° (exterior + interior = 180°)
n-sided polygon: interior angle = 120° + 48° = 168° (hexagon + 48°)
exterior angle = 180° – 168° = 12º (exterior + interior = 180°)
number of sides, n =
360
–––
12
= 30 (sum of exterior angles = 360°)
4 Interior angles of triangle are 64°, 64° and 52°. (isosceles triangle)
So x = 180° – 64° = 116°, y = 180° – 52° = 128°.
5 x = 96° – 27° = 69° (exterior angle = sum of interior angles)
Both base angles are y (isosceles triangle)
2y = 27° (exterior angle = sum of interior angles)
y = 13·5°
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
17
Model answers
6 In triangle ABC and triangle ECD
BC = CD (given)
angle ABC = angle EDC (alternate angles)
angle ACB = angle DCE (vertically opposite)
So triangle ABC is congruent to triangle EDC
7 (a) Angle ADE = angle ABC (corresponding angles, DE parallel to BC)
Angle AED = angle ACB (corresponding angles, DE parallel to BC)
Angle A is common
Triangles are similar as corresponding angles are equal.
AD 1
(b) ––– = –, BC = 4 × DE = 12 cm
AB 4
Pythagoras and trigonometry (page 67)
1 (a) BC2 = BD2 + DC2 – 2 × BD × BD × cos160°
= 102 + 82 – 2 × 10 × 8 × cos160°
= 314·35
BC = 17·7 m
(b) PA2 = PD2 + DA2
= 4 2 + 82
= 16 + 64
= 80
PA = 8·94 m
2 (a) AB2 = AC2 + BC2 – 2AC × BC × cos15°
= 82 + 32 – 2 × 8 × 3 × cos15°
= 64 + 9 – 48 cos15°
= 26·64
AB = 5·16 cm or 5.2 cm
1
(b) Area = 2– × BC × AC × sinC
1
= 2– × 8 × 3 × sin15°
= 3·11 cm2 or 3.1 cm2
2
3 (a) BC = 152 + 232 – 2 × 15 × 23 × cos64°
= 451.52
BC = 21.2 km
1
(b) Area = 2– × 15 × 23 × sin64°
= 155 km2
Properties of circles (page 70)
1 x=
y=
z=
2 (a)
(b)
(c)
36° (angles in the same segment)
2 × 36° = 72° (angle at centre = twice angle at circumference)
90° – 28° = 62° (radius at right angles to tangent)
Angle ACO = 34° (isosceles triangle) ⇒ angle OCB = 90° – 34° = 56° (angle in a semi-circle)
Angle CBO = 56° (isosceles triangle) ⇒ angle CBT = 180° – 56° = 124°
Angle BCT = 90° – 56° = 34° (tangent and radius) ⇒ angle CTA = 180° – 34° – 124° = 22°
or angle COB = 2 × 34° = 68° (angle at centre = twice angle at circumference) ⇒ angle
CTA = 180° – 90° – 68° = 22°
3 a = 90° – 63° = 27°
b = 63° (isosceles triangle)
c = 180° – 63° – 63° = 54°
d = 180° – 90° – 54° = 36°
4 (a) Angle ADE = x° (angles in the same segment)
(b) Angle DAE = 180° – 90° – x° = 90° – x° (angle in a semi-circle)
(c) Angle EAB = 90° – (90° – x°) = 90° – 90° + x° = x° (tangent and radius)
(d) Angle AOE = 2 × x° = 2x° (angle at centre = twice angle at circumference)
5 (a) Angle ACB = 90° (angle in a semi-circle)
(b) Angle AOB = 2x° (angle at centre = twice angle at circumference)
Angle ATB = 360° – 90° – 90° – 2x° (since angle TAO = angle TBO = 90°)
= 180° – 2x°
18
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational