Top One (Maths F1) Penerbitan Pelangi Sdn Bhd

Top One (MATHS F1) Penerbitan Pelangi Sdn Bhd

Mathematics Form 1 Answers

5.

Obtuse-angled All angles in the
triangle triangles are acute

angle.

Acute-angled One of the angle
triangle in the triangles is

obtuse angle.

Right-angled One of the angle
triangle in the triangles is

right angle.

6. (a) 1 axis of symmetry 8. (a) ∠EBD = 180° – 2(72°)
(b) No axis of symmetry = 180° – 144°
= 36°
7. (a) x + 44° = 62°
x = 62° − 44° 102° + 36° + y = 180°
= 18° 138° + y = 180°
y = 180° − 138°
(b) x + 90° + 58° = 180° = 42°
x + 148° = 180°
x = 180° − 148° (b) ∠ACB = 180° – 65° – 40°
= 32° = 75°

(c) x + 35° = 105° y + 75° + 35° = 180°
x = 105° − 35° y + 110° = 180°
= 70° y = 180° − 110°
= 70°
(d) x + 43° + 75° = 180°
x + 118° = 180° 9. (a) ∠PRQ = 180° – 2(63°)
x = 180° − 118° = 180° – 126°
= 62° = 54°

(e) x + 35° + 35° = 180° ∠SRT = ∠PRQ
x + 70° = 180° = 54°
x = 180° − 70°
= 110° y + 54° + 54° = 180°
y + 108° = 180°
(f) x + 54° + 37° = 180° y = 180° – 108°
x + 91° = 180° = 72°
x = 180° − 91°
= 89° RTS is an isosceles triangle because it
has two same angle.

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A2

Mathematics Form 1 Answers (b)

(b) ∠POQ = 180° – 100° 2
= 80°
11. (a) Rectangle ; 2
80° – ∠TOU = 40° (b) Kite ; 1
∠TOU = 40° (c) Parallelogram ; 0
(d) Rhombus ; 2
y = (180° – 40°) ÷ 2 (e) Square ; 4
= 70°
Parallelogram • All sides are of equal length.
10. (a) Rectangle • Pair of opposite sides are

4 parallel.
• All its interior angles are
12
• Its opposite sides are of 90°.
equal length and parallel. • The diagonals are of equal
• All of its interior angles
are 90°. length and perpendicularly
• The diagonals are of bisectors of each other.
equal length and are
bisectors each other.

• Two pairs of adjacent Square • Its opposite sides are of
sides are of equal length. Rhombus equal length and parallel.
Trapezium
• Has a pair of opposite • Opposite angles are equal.
angles of the same value. • The diagonals are bisectors

• One of the diagonals of each other.
is the perpendicular
bisector of the other. • All sides are of equal
length.
• One of the diagonals is
the angle bisector of the • Its opposite sides are
angles at the vertices. parallel.

• Only one pair of Kite • Opposite angles are equal.
opposite sites are parallel 40 • The diagonals

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each other.
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13. (a) x + 117° = 180° Mathematics Form 1 Answers
x = 180° − 117°
= 63° PT3 Standard Practice 9

(b) x + 123° + 48° + 66° = 360° Section A
x + 237° = 360° 1. Answer: D
x = 360° − 237°
= 123° 2. ∠FBA = 180o – 145o = 35o
∠FAD = 180o – 68o – 22o – 35o = 55o
(c)
Answer: B

78° 360° − 135° 3. ∠FHG = 180o – 128o = 52o
= 225° x = 180o – 90o – 52o = 38o

15° x Answer: B
135°

x + 225° + 15° + 78° = 360° 4. g = 90o
x + 318° = 360° e + f + g + h + i = 180o + 90o = 270o
x = 360° − 318°
= 42° Answer: B

14. (a) x = 2 × 30° 5. ∠SQR = 180o – 74o = 106o
= 60°
m = 180o – 106o = 37
∠PRS = 180° − 60° 2
= 120°
Answer: B
∠PRQ =180° − 60°
= 120° 6. 3(3 – 3) = 0
2
y = 60° ÷ 2
= 30° Answer: A

(b) x + 60° = 180° Section B
x = 180° − 60°
x = 120° 1. (a) Parallelogram; 0
(b) Square; 4
120° + 60° + 60° + y + 36° = 360°
y + 276° = 360° 2. 9
y = 360° − 276° Hexagon
y = 84°

15. (a) (i) The number of triangle = 5 Heptagon 8
The number of quadrilateral = 6

(ii) BCDF is a kite because it has two Octogon 7
pairs of adjacent sides that are
equal in length. A kite has a pair of Nonagon 6
opposite angles of the same value.
Therefore, ∠CBF and ∠CDF are
same.

65° + 105° + x + x = 360°
170° + 2x = 360°
2x = 360° − 170°
= 190°
x = 190° ÷ 2
= 95°

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A2

Mathematics Form 1 Answers (c) (i) p = 180o – 78o = 51o
2
Section C
1. (a) q = 180o – 65o – 65o = 50o

(ii) x + 115o = 70o + 4x

115o – 70o = 4x – x

3x = 45o

x = 45o
3
Heptagon
x = 15o
(b) ∠PRQ = 75o
m = 180o – 35o – 75o HOTS Challenge
= 180o – 110o = 70o
n = 34o + 75o = 109o C E
m + n = 70o + 109o = 179o
BF
(c) (i) x = 180o – 57o – 57o = 66o 60° 102°

(ii) y = 57o
z = 180o – 57o = 123o
z – y = 123o – 57o = 66o

2. (a) Number of sides = 7 18°

Number of vertices = 7 7(7 – 3)
2
Number of diagonals = = 14 A

(b) ∠QSR = 90o ÷ 2 = 45o ∠BAF = 180o – 60o – 102o

∠TSR = 180o – 45o = 135o = 18o

180o – 135o ∠CAE = 60o – 18o = 21o
2 2
∠STR = = 22.5o

∠RTU = 180o – 22.5o = 157.5o

CHAPTER Perimeter and Area

10 Perimeter dan Luas

1. (a) Perimeter
= AB + BC + CD + DE + EF + FG + GH + HA

(b) Perimeter
= AB + BC + CD + DE + EF + FA

2. (a) Perimeter
= 3 + 3 + 3 + 3 + 3 + 10 + 3 + 3 + 3 + 3 + 3 + 10 = 50 cm

(b) Perimeter
= 12 + 12 + 6 + 6 + 5 + 5 + 6 + 12 = 64 cm

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(c) Perimeter 5.2 cm Mathematics Form 1 Answers
= 2 + 2 + 3.15 = 7.15 cm
2 cm
(d) Perimeter 2 cm
= 1.5 + 2.8 + 1.5 + 2.8 = 8.6 cm
2 cm
3. 2 cm

1 cm
1 cm

1 cm 12 cm
1 cm 11 cm
5 cm

4. (a) Total length of fence = 4 × 15 = 60 m (b) Area
Perimeter of ABC = 16 + 23 + 25
= 64 m = 1 × (5 + 12) × 7
As 60 , 64, 2
4 rolls of fence are not enough to fence
the area. = 1 × 17 × 7
2
(b) Perimeter = 1 × 18
= 18 units = 59.5 cm2

(c) Area

= 1 × (8 + 16) × (6 + 6)
2

= 1 × 24 × 12
2

5. (a) 11 unit2 = 1 × 288
(b) 10 unit2 2
(c) 18 unit2
= 144 cm2

(d) Area

6. (a) Area = 1 × b × d = 1 × 8 × 6
2 2

(b) Area = 1 × (b + d) ×a = 24 cm2
2
(e) Area
(c) Area = b × c = 16 × 10
= 160 cm2
7. (a) Area
= 6 × 7 = 42 cm2

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Mathematics Form 1 Answers

(f) Area 10. (a) Area of side of the sofa
1
= ( 1 × 20 × 8) + (20 × 10) = 50 × 60 – 2 × (30 + 40) × 30
2
1
= 80 + 200 = 280 cm2 = 3 000 – 2 × 70 × 30

(g) Area = 3 000 – 1 050

= ( 1 × 6 × 6) + 1 (6 + 10) ×6 = 1 950 cm2
2 2

= 18 + 48 = 66 cm2 (b) Area of both walls

(h) Area =2× 1 × (3 + 1.5) × 4.5
2
= 1 (11 + 13) 11 + (11 15)
2 × × = 2 × 1 × 4.5 × 4.5
2
= 132 + 165 = 297 cm2

1 = 20.25 m2
2
8. (a) × (8 + 12) × h = 80 The number of cans of paint needed
= 20.25 ÷ 4.5 = 4.5 ≈ 5
h= 80 × 2 = 8 cm
20
Total cost = 5 × 25.50
(b) 0.9 × h = 8.1 = RM127.50

h = 8.1 =9 m The allocation is enough because the
0.9 total cost of the paint is RM127.50
which is less than the allocated money.
1
(c) 2 × (11 + 14) × h = 237.5 11. (a) .
= (b) ,
1 × 25 × h = 237.5
2 237.5 ×
2
h 25
12. (a) ,
= 19 cm (b) .

(d) 1 ×6×h = 57 13. (a) (i) Let the height of the trapezium is
2 h x m.
57 × 2
= 6 = 19 cm x + 5 + 2 + 10 + 13 + 10 + 2 = 54
x + 42 = 54
(e) 1 × (5 + 9) × h = 70 x = 54 – 42
2 h = = 12
1 h = 70
2 × 14 × 70 ×

14 2 Area of flower crop

= 10 cm = 1 × 10 × 12 = 60 m2
2

9. (a) Area of shaded region Area of pond

= 1 × (8 + 12) × 8 – 6 × 4 = 1 × 5× 12 = 30 m2
2 2

= 1 × 20 × 8 – 24 (ii) Area of shaded region
2 = 2 × 12
= 24 m2
= 80 – 24 = 56 cm2
Numbers of bag of pebbles needed
(b) Area of shaded region = 24 ÷ 3
= 8 bags
1 2= 1
10 × 15 – 2 × 15 × 10 ÷2 Total cost = 8 × 40 = RM320
The cost to buy the pebbles is
= (150 – 75) ÷ 2 RM320.

= 37.5 cm2

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(b) The possible rectangles Mathematics Form 1 Answers

Length Perimeter Area 5. Let ML = x
and width 1×9=9 x + x + 6 + 6 + 15 + 15 = 60
2 × 8 = 16 2x + 42 = 60
1 and 9 1 + 9 + 1 + 9 3 × 7 = 21 2x = 60 – 42
= 20 4 × 6 = 24 2x = 18
x=9
2 and 8 2 + 8 + 2 + 8
= 20 Answer: C

3 and 7 3 + 7 + 3 + 7 6. 1 × (6 + 12) ×p = 180
= 20 2 9p = 180

4 and 6 4 + 6 + 4 + 6 p = 20
= 20
Answer: C

Area of Amir’s paper = 10 × 10 Section B
= 100 cm2 1.

100 ÷ 9 = 11.11 ≈ 11 (the most) 3 cm • • 18 cm
100 ÷ 16 = 6.25 ≈ 6
100 ÷ 21 = 4.76 ≈ 4 6 cm • 16 cm
100 ÷ 24 = 4.16 ≈ 4
•3 cm
Therefore, the length and width of the
rectangle is 1 cm and 9 cm respectively
and the most number of the pieces is
11.

PT3 Standard Practice 10 2 cm • • 12 cm
3 cm • 10 cm
Section A 3 cm

1. Area of triangle PQR •

= 1 × 14 × 12 = 84 cm2
2

Answer: B

2. Side of the square

= 96 ÷ 4 = 24 1
2
Area of the whole diagram 2. A: × 5 × 5 = 12.5 cm2
1
= 24 × 24 + 2 × 24 × 24 = 864 cm2 B: 4 × 5 = 20 cm2

Answer: A C: 1 × (5 + 3) × (3 + 3) = 1 × 8 ×6
2 2

3. 12 units → 36 cm 1 = 24 cm2
1 unit → 3 cm 2
3 units → 9 cm D: × (3 + 5) × 4 = 16 cm2

Answer: C A→D→B→C

4. Perimeter of the whole diagram Section C
= 10 + 10 + 12 × 3 = 56 cm
1. (a) (i) Length of side
Answer: C = 28 ÷ 4 = 7 cm
Area of the square
= 7 × 7 = 49 cm2

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A2

Mathematics Form 1 Answers (b) (i) A
(ii) C
(ii) (iii) The greater the difference between
the lengths, the greater the
11 + 10 + 6 = 27 cm perimeter of the rectangle.

10 cm (c)

11 cm 12 cm

12 + 8 + 6 = 26 cm 18 cm C 27 cm
8 cm 9 cm

6 cm 9 cm B

Perimeter 24 cm A
= 27 + 26 + 27 + 26 = 106 cm 15 cm
(b)
Length of a side of square B
14 cm 14 cm = 36 ÷ 4 = 9 cm
Length of a side of square A
14 cm = 9 + 15 = 24 cm
Length of a side of square C
28 cm = 18 + 9 = 27 cm

Width of the rectangle = 28 ÷ 2 = 14 cm Area of square A = 24 × 24 = 576 cm2
Area of each identical rectangle Area of square C = 27 × 27 = 729 cm2
= 28 × 14 = 392 cm2 Area of square B = 9 × 9 = 81 cm2

(c) D Cʹ D Area of the whole diagram
= 576 + 729 – 81 = 1 224 cm2
C
HOTS Challenge
AB A

HG H U T
D 1m C

F E Fʹ E 2m 2m 2m

Perimeter of square C’DEF’ = 60 cm Garden F E

Length of each side of square C’DEF’ 3m

= 60 ÷ 4 = 15 cm A 1m GB
Length of each small square R S

= 15 ÷ 2 = 7.5 cm Length of rectangle RSTU = 7 × 24 = 42 m
Area of each small square AB = 42 – 2 – 2 = 38 m 4

= 7.5 × 7.5 = 56.25 cm2
The area of the whole diagram

= 3 × 56.25 = 168.75 cm2

1 BC = 24 – 1 – 1 = 22 m
2
2. (a) Total area = × 2r × (r + s) + r × s Area of rectangle ABCD = 38 × 22 = 836 m2

= r(r + s) + rs Area of rectangle BEFG = 2 × 3 = 6 m2

= r2 + rs + rs = r2 + 2rs Area of the garden = 836 – 6 = 830 m2

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Mathematics Form 1 Answers

CHAPTER Introduction to Set

11 Pengenalan Set

1. (a) Colour

Red Blue Green Yellow Orange
(b) Number

Four Five Seven Nine Ten
(c)
Polygon
Triangle
Trapezium Rhombus Pentagon Hexagon

2. (a) P is a set of prime numbers which is (d) 
less than 20. (e) 

(b) P is a set of the colours of a rainbow. 7. (a) n(Q) = 5
(b) n(P) = 7
3. (a) Q = {1, 2, 3, 4, 6, 8, 12, 24} (c) T = {O}
(b) Q = {M, L, Y, S} n(T) = 1
(d) U = {1, 4, 9, 16, 25, 36, 49, 64, 81}
4. (a) S = {x : x is an odd number and x , 10} n(U) = 9
(b) S = {March, May}
8. (a) ≠
5. (a) 3 (b) =
(b) 7 (c) ≠
(c) 3 (d) =
(d) 3
9. (a) x = 6
6. (a)  (b) A = {4, 9, 16}
(b)  x=9
(c) 

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A2

Mathematics Form 1 Answers

10.

A = {2, 4, 8, 12, 16} • • j = {x : x is a factor of 30}

B = {G, A, P} • • j = {x : x is an integer and x , 13}

C = {7, 9, 11} • • j = {Alphabets}

D = {A, B, E, G, H} • • j = {x : x is an even number and x , 20}

F = {x : x is a factor of 10} • • j = {P, E, L, A, N, G, I}

11. (a) A = {1, 2, 3, …, 10} (b) P = {j, p, s, y}
(b) A = {1, 2, 4, 8} P’ = {c, v, z}
(c) A’ = {1, 2, 4, 5, 7, 8, 10}
14. (a) 
12. (a) ξ •S (b) 
•I (c) 
Q (d) 
•H
15. (a) { };
•G •E {5};
•N {6};
{7};
•L {5, 6};
{5, 7};
(b) ξ •15 {6, 7};
•16 {5, 6, 7}.
•12 R
•11 •18 (b) { };
•13 {p};
{q};
•17 {r};
•14 {p, q};
{p, r};
(c) ξ {q, r};
{p, q, r}.
S •25
•15 •17

•19 •23

•21 •27

(d) ξ 16. (a)

T •1 U S T
•0 •2 •5
•4 •8
•3
•6 •7

13. (a) P = {2, 4, 6}
P’ = {1, 7}

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Mathematics Form 1 Answers

(b) X (ii) Set A
Z (iii) N  G  A
Y
(b) (i) T
17. (a) H •Rani
•3 B •Emma
•5 F •Malik •Sofia
•Anas •Ravi •Mei Mei
•Muthu •Yoke Lin

•Chow

K •7 (ii) Set T
•2 •6 •9 (iii) F  B  T
•4
•8

(b) N PT3 Standard Practice 11

•36 M •2 Section A
•1
1. Factors of 40 = 1, 2, 4, 5, 8, 10, 20, 40
•18 •3 •4 •6 Set P = {1, 2, 4, 5, 8, 10}
Answer: B
•9 •12

(c) P 2. Answer: D

•P Q 3. Vowels in set M: u, i
•A Set N = {u, i}

•R Answer: A
•I

•L

18. (a) ξ Q 4. 25 = 32
P Answer: D

T

5. Set M = {6, 14}
Answer: C

(b) ξ F 6. The set that has the same elements as set A
D is set R.
E
Answer: C

19. (a) (i) Z9  j Section B N
(ii) Y  Z  X 1. (a) (i) Set M

(b) (i) B  A ξ
(ii) C  B9 M

20. (a) (i) A
G •Fiza
•Ali
•Nisa N •Ammar
•Mikael
•Lela •Ani •Anaqi
•Umar •Linda •Danny

•Ros

•Chan •Ella

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Mathematics Form 1 Answers N 2. (a) (i) n() = 6, n(G) = 2
n(G’) = 6 – 2 = 4
Set N’
(ii) n() = 14, n(H) = 2
ξ n(H’) = 14 – 2 = 12
M

(ii) (a)  (b) S = {4, x + 7, 13, 16, 19}
(b)  P = {y, 10, 4, z – 4 , 16}
If y = 13, then x + 7 = 10
2. (a) 
(b)  So, x = 3, z – 4 =19 → z = 23
(c)  OR
(d) 
If y = 19, then x + 7 = 10
Section C
1. (a) (i) A = {16, 18, 20, 22, 24} So, x = 3, z – 4 = 13 → z = 17
(ii) B = {–5, –4, –3, –2, –1} OR
(iii) C = {8, 10, 12, 14}
If y = 13, then x + 7 = 19

So, x = 8, z – 4 = 10 → z = 14
OR

If y = 19, then x + 7, = 13

So, x = 6, z – 4 =10 → z = 14

(c) (i) ξ A
B

(b) (i) E’ = {3, 4, 5, 7, 10} (ii) (a) M  L
(ii) F = {3, 5, 7, 9} (b) M  N
(iii) G’ = {1, 2, 3, 4, 6, 8, 9, 10}
HOTS Challenge
(c) (i) ξ P •b
•d Q B = {roti canai, nasi lemak}
•g •a {roti canai, fried chicken}
•e •h •c {roti canai, noodles soup}
{roti canai, burger}
•f {nasi lemak, fried chicken}
{nasi lemak, noodles soup}
(ii) ξ N {nasi lemak, burger}
{fried chicken, noodles soup}
M {fried chicken, burger}
{noodles soup, burger}

Number of choices = 10

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Mathematics Form 1 Answers

CHAPTER Data Handling

12 Pengendalian Data

1.
Methods of collecting data

Experiment (a) (b) (c)
Interview Survey Observation

2. The hair colour of the
residents in a village.
The favourite food of a Numerical
group of teachers. data The shoe sizes of
the children in a
The marks of Categorical kindergarten.
Mathematics test obtained data
by form 1 students. The type of vehicles
owned by the teachers.
The heights of the
workers in a factory.

3. (a) (b)
The sizes
Number of Tally Frequency of shirt Tally Frequency
members
| 1 S ||| 3
1 |||| | 6 |||| |||| 9
2 5 M |||| ||| 8
3 |||| 10 4
4 |||| |||| 3 L ||||
5
||| XL

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Mathematics Form 1 Answers Data representations

4.

Bar chart (b) (d) (f)
Line graph Stem-and-leaf plot Histogram

(a) (c) (e)
Pie chart Dot plot Frequency polygon

5. (a) Sales of A Car Company 7. (a) Colour Angle of sector
Blue
Number of cars (thousand) Jualan Sebuah Syarikat Kereta 75 × 360° = 125°
Black 216
Bilangan kereta (ribu) 200
Red
150 Total 60
216
100 × 360° = 100°

50 81 × 360° = 135°
216

0 Year 360°
2012 2013 2014 2015 Tahun
Sales of Pens
Horizontal bar chart is acceptable.

(b) Tourists in Four Islands

Redang Blue

Islands Rawa 125° Black
100°

135°

Red

Pangkor

Langkawi (b) Flavour Angle of sector

0 200 Original 130 × 360° = 65°
400 720
600
800
1000
1200

Number of tourists 300
720
Vertical bar chart is acceptable. Chocolate × 360° = 150°

6. (a) Participants in A Sport Tournament Strawberry 220 × 360° = 110°
720

20 Coffee 70 × 360° = 35°
18 720
Number of participant 16
14 Total 360°
12
Student’s Fovourite Milk
10
8 Coffee Original Chocolate
6 65° 150°
4
2 35° 110°
0
Red Yellow Blue Green Strawberry
Sport house

Male

Female

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8. (a) Form One Students of A School Mathematics Form 1 Answers

300 (c) Mass of A Group of Kindergarten Students

Number of students 200 14 15 16 17 18 19 20 21 22 23 24
Mass (kg)

100

10. (a) Mass of Form 1 Students

0 2012 2013 2014 2015 Stem Leaf
2011 Year
2 9
(b) Sales of A Food Stall 3 4253873
4 07128
600 5 0

500

Sales (RM) 400 Mass of Form 1 Students

Stem Leaf

300 29
3 2334578
200 4 01278
50

0 Fri. Sat. Key: 2 | 9 means 29 kg
Mon. Tue. Wed. Thur.
Day

9. (a) (b) Height of A Group of Teenagers

Time Taken to Solve A Mathematics Question Stem Leaf

45 50 55 60 65 70 75 80 13 6638
Time (second) 14 0
15 85
(b) 16 9281
17 890
Grade of A Group of Students
Height of A Group of Teenagers

ABCDE Stem Leaf
Grade
13 3668
14 0
15 58
16 1289
17 089

Key: 13 | 3 means 133 cm

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Mathematics Form 1 Answers

11. (a) Quiz Marks of A Group of Monthly saving
Students = 3 037.50 – 2 700 = RM337.50

Stem Leaf (c) (i) 2014

2 044488 (ii) 2 000 + 1 500 + 2 500 + 2 500 +
3 22266 3 000 = 11 500 visitors
4 044
(iii) Difference in 2014 and 2015
Key: 2 | 0 means 20 marks = 2 500 – 1 500
= 1 000 visitors

Stem-and-leaf plot is suitable to show Percentage of increasing
the marks of each students.
= 1 000 × 100% = 66.67%
1 500
(b) Sales of A Car Company

Jualan Sebuah Syarikat Kereta (d) (i) 8 durians

Number of cars (thousand) 200 (ii) 5 – 1 = 4 participants

Bilangan kereta (ribu) 150 (iii) 5 + 7 + 2 = 14
The total number of participants
100 who ate 9 or 8 or 7 durians is 14.
So the least number of durians
50 eaten by a participant is 7.

0 Year (e) (i) 6.6 m
2012 2013 2014 2015 Tahun
(ii) 6.6 – 4.3 = 2.3 m
Line graph is suitable to show the
changes in the sales for the four (iii) Distance of leap qualified to
consecutive years. semifinal, are 6.6 m, 6.2 m, 5.7
m, 5.5 m, 5.4 m and 5.4 m. Zamir
12. (a) (i) 25 + 10 = 35 patients is the sixth athlete qualified to
semifinal. Therefore, his leap
(ii) Percentage of female patients distance is 5.4 m.

= 25 + 15 + 10 × 100% (f) (i) 7 students
20 + 25 + 30 + 15
(ii) Percentage of students who write
+ 25 + 10
less than 25 minutes
50
= 125 × 100% = 40% = 2 × 100%
2+6+8+7+2

(iii) 125 – 50 = 75 patients = 2 × 100% = 8%
25

(b) (i) Food (iii) Most of the students take about 25
minutes to 40 minutes to write an
(ii) Angle of food sector essay.

= 360° – 40° – 70° – 90° = 160°

Percentage of food expenses (g) (i) Number of papaya
=0+4+5+8+7+0
= 160° × 100% = 24 papayas
360°

= 44.44% (ii) Percentage of papaya weigh more

(iii) 2 700 = 360° – 40° × Salary than 1.4 kg
360°
= 8+7 × 100%
2 700 × 360° 24
Salary = 320
= 15 100% = 62.5%
= RM3 037.50 24 ×

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(iii) The mass of most of the papayas Mathematics Form 1 Answers
in the orchard is between 1.4 kg to
1.8 kg. 3. Total marks for Mathematics quiz
= 25 + 18 + 46 + 48 = 137
(h) (i) Total marks for Science quiz
= 32 + 35 + 37 + 43 = 147
Year Number of babies Total Total marks difference
Boy Girl = 147 – 137 = 10
2014 12 18 30
2015 15 21 36 Answer: A
2016 18 24 42
4. x + 2x + 60o + 120o = 360o
Birth of National Baby 3x = 180o
x = 60o

60o = 30 students
120o = 60 students

Answer: D

42 5. Total profit
= 4 000 + 2 000 + 8 000 + 10 000 + 6 000
38 = 30 000
The fraction of the profit in May over the
Number of babies total profit in five months

34 6 000 = 1
30 000 5

Answer: C

30 6. Answer: D

0 2015 Year Section B
2014 2016
1. (a) (i) Survey
(ii) The number of national babies (ii) Experiment (iii) Observation
born at the government hospital (iv) Interview
is increasing for every year.
2. (a) Marks Tally Frequency
(iii) The pattern of births is an increase 10 8
of 6. 20
Prediction of the number of 30 7
national babies in 2017 40
= 42 + 6 50 3
= 48
Prize money = 48 × 1 000 12
= RM48 000
10

PT3 Standard Practice 12 Total 40

Section A Section C
1. Answer: C
1. (a) Largest angle = 7 × 360o = 100.8o
25

2. 26 × 100% = 65% Smallest angle = 3 × 360o = 43.2o
40 25

Answer: D 100.8o – 43.2o = 57.6o

55 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 1 Answers

(b) (i) 360 ÷ 24 = 15 2. (a) (i) 200 – 60 – 46 – 34 = 60
Number of cars of type C April:
March:
= 15 × 8 = 120 4 units → 60
1 unit → 60 ÷ 4 = 15 (March)
(ii) Type A: 90o × 360o = 90o 3 units → 3 × 15 = 45 (April)
360o
(ii) January = 60
Type B: 150o × 360o = 150o February and May = 46 + 34 = 80
360o 60 : 80 = 3 : 4

Type C: 120o × 360o = 120o
360o

Type C (b) (i) Stem Leaf

Jenis C 1 011
2 144457789
Type B 120° 3 2667
4 001
Jenis B 90°
Type A
150°
Jenis A
Key: 1 | 0 means 10 days

(ii) 12 × 100 = 66.67%
18
(c) Time taken to arrive to the office
(c) (i) Range = 20 – 10 = 10

(ii) Most of the teachers bought 13

until 15 pens

21 22 23 24 25 HOTS Challenge
Time (minutes)

The interpretation of the bar chart is inaccurate.
The vertical axis of the bar chart does not start
with zero. The value of the vertical axis decreased
by 2 only. Thus, the difference between the year
2018 and 2019 is small.

CHAPTER The Pythagoras Theorem

13 Teorem Pythagoras

1. (a) QR 3 2. (a) HJ2 + HK2 = JK2
(b) q 3 (b) a2 + b2 = c 2
(c) x 3 (c) m2 + p2 = n2

3. (a) x = √12.52 – 122
= √156.25 – 144
= √12.25 = 3.5

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Mathematics Form 1 Answers

(b) x = √92 + 122 QR = √202 – 162
= √81 + 144 = √400 – 256
= √225 = √144
= 15 = 12 cm

(c) x = √142 – 10.52 PQ = PR – QR
= √196 – 110.25 = 30 – 12
= √85.75 = 18 cm
= 9.26
Area of PQS
(d) x = √72 – 32
= √49 – 9 = 1 × 18 × 16
= √40 2
= 6.32
= 144 cm2
(e) x = √52 + 132
= √25 + 169 (b) Length of diagonal
= √194 = √0.52 + 1.22
= 13.93 = √0.25 + 1.44
= √1.69
4. (a) For ∆RST, = 1.3 m
x = √102 – 82
= √100 – 64 Total length of steel needed
= √36 = 3(1.2) + 2(1.0) + 2(1.3)
=6 = 3.6 + 2.0 + 2.6
= 8.2 m
For ∆PQR
PR = √122 + 52 Total cost = 8.2 × 30
= RM246
= √144 + 25
= √169 6. (a) 3
= 13 Longest side = 29 cm
292 = 841
y = 13 – 6
=7 202 + 212 = 400 + 441
= 841
(b) For ∆RST,
x = TR = √132 – 52 292 = 202 + 212
= √169 – 25 Thus, the triangle is a right-angled
= √144 triangle.
= 12
(b) 7
For ∆PQU Longest side = 18 cm
PQ = √152 – 122 182 = 324

= √225 – 144 102 + 152 = 100 + 225
= √81 = 325
=9
y = 9 + 8 + 5 = 22 182 ≠ 102 + 152

5. (a) PR = √342 – 162 7. (a) 9 cm, 40 cm, 41 cm
= √1 156 – 256
= √900 Longest side = 41 cm
= 30 cm
412 = 1 681

92 + 402 = 81 + 1 600
= 1 681

412 = 92 + 402

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A3

Mathematics Form 1 Answers 5. PU = √152 + 202 = 25 cm
Area of square STUV = 169 cm2
(b) 15 cm, 20 cm, 26 cm
SV = √169 = 13 cm
Longest side = 26 cm PV = PU – VU
262 = 676
152 + 202 = 225 + 400 = 25 – 13 = 12 cm
Area of square PQRV
= 625 = 12 × 12 = 144 cm2
262 . 152 + 202
Answer: B

8. (a) 2.52 = 6.25 6. 82 = 64
52 + 62 = 25 + 36 = 61 cm
1.52 + 2.02 = 2.25 + 4 82 > 52 + 62
= 6.25
Answer: B
2.52 = 1.52 + 2.02
So, ∠XYZ = 90° Section B
Therefore, the wall is vertically on the
ground. 1. (a) (i) PR
(ii) a
(b) 162 = 256
(b) (i) 3
7.22 + 13.52 = 51.84 + 182.25 (ii) 3
= 234.09
2. (a) (i) Right-angled triangle
162 ≠ 7.22 + 13.52 (ii) Obtuse-angled triangle
So, the triangle is not a right-angled
triangle. (b) 9, 40, 41; 15, 20, 25

√234.09 = 15.3 cm Section C
16 – 15.3 = 0.7 cm 1. (a) (i) Z = √122 + 162 = 20 cm
(ii) Z = √(18)2 – 42 = √2 cm
To make a right-angled triangle, Esther
has to cut out 0.7 cm from the longest (b) Height of the trapezium
straw.

PT3 Standard Practice 13 = √102 – 62 = 8 cm
Area of the trapezium
Section A
1. Answer: C = 1 × (9 + 15) × 8 = 96 cm2
2

2. x2 = (√45)2 – 62 (c) PR = √82 + 152 = 17 cm
= 45 – 36 = 9
RT = √162 + 302 = 34 cm
x = √9 = 3 cm
PT = √172 + 342 = √1 445
Answer: B = 38.01 cm

3. VR = √102 – 62 = 8 cm 2. (a) (i) Let the side be x cm,
Area of triangle VUR x2 + x2 = 242
2x2 = 576
= 1 × 8 × 6 = 24 cm2 x2 = 288
2 x = √288
x = 16.97 cm
Answer: C
(ii) Area of square
4. √182 + 802 = √6 724 = 82 cm = 16.97 × 16.97
Answer: C = 287.98 cm2

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Mathematics Form 1 Answers

(b) R 8QR = 5RS
RS = 58QR
S 8
60 m = 5 × 15

74 m = 24 cm
25 m
QS = 15 + 24
P TQ
= 39 cm
PQ = √742 – 252
= √4 851 PS = √17.442 + 392
= 69.65 m
= 42.72 cm
PT = √602 – 252
= √2 975 HOTS Challenge
= 54.54 m
14 cm 48 cm 26 cm
RS = 69.65 – 54.54 48 cm
= 15.11 m 26 cm

(c) PQ = √232 – 152 AB = √142 + 482
= √304 = 50 m
= 17.44 cm

Year-End Assessment

Section A 3. A: (22 – 5) ÷ 5 = –0.2
B: (32 – 5) ÷ 5 = 0.8
1. A: 3 = 0.60 C: (42 – 5) ÷ 5 = 2.2
5 D: (52 – 5) ÷ 5 = 4

B: 59= ÷0.56 Answer: D

C: 8 = 0.89 4. 8 – (2 + 3) = 3 units
9 3 units → RM90
1 unit → RM30
D: 12 = 0.71 2 units → RM60
17
Answer: C
Answer: B

2. 31, 37, 41, 43, 47 5. – 35a2b = –35 × a × a × b
m = 31, n = 47 14ab2 14 × a × b × b
m + n = 31 + 47
= 78 = – 5a
2b
Answer: C
Answer: B

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A3

Mathematics Form 1 Answers

6. p + q = 1 -------- ➀ 18. R 3 cm S
p – q = –11 ------ ➁
➀ + ➁ 2p = –10 9 cm 11 cm
Q
Answer: A

7. – 6g > 12
5
5 7 cm
g < 12 × – 6
g < –10
1 1 P
Answer: A 2 × 7 × 9 + 2 × × 3
11

8. 2x + 3x = 180o = 48 cm

5x = 180o Answer: C

x = 180o 19. Answer: A
5

x = 36o 20. ∠FIH = 180o – 70o = 110o
∠FHI = 90o – 38o = 52o
Answer: A ∠HFI = 180o – 110o – 52o = 18o

9. Answer: A Answer: A

10. Answer: C Section B

11. 5 + 5 + 5 + 5 + 7 + 7 + 2 + 2 = 38 cm 1. (a) (i) Corresponding angle
Answer: C (ii) Interior angle

12. j = {2, 3, 4, 5, 6, 7, 8, 9, 10} (b) 2
P = {2, 4, 6, 8, 10}
Q = {2, 3, 5, 7} 1
42
Answer: B

13. Answer: D 21 42 3

14. Answer: D

15. x = √162 – 132 = 9.33 cm 14 6
Answer: B 7

16. Answer: C 2. (a) 74, –1.5
(b) 9 < 15, –9 > –15
17. X: 1×3 = 3
10 × 3 30

Z: 1 × 5 = 5 3. (a) Polygon Heptagon
6 × 5 30 Number of sides 7
Number of diagonals 14
Y: 4 = 2
30 15

Answer: C

(b) 3 : 7; 6 : 14
(Accept any reasonable answers)

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4. –2 < x < 1 Mathematics Form 1 Answers
–2 < x  1
–2  x < 1 Number of boys = 15
–2  x  1 Number of girls = 21

5. (a) False (c) True 3. (a) (i) 10  x
(b) False (d) False (ii) (a)

Section C –8 –7 –6 –5 –4 –3

1. (a) (i) a = 4, c = 8 (b)

(ii) b = 7.6, d = 8.4 12 13 14 15 16 17

( ) ( )(b)31 + 2 1 2= 1 + 11 2 (b) (i) y = 180o – (130o – 90o)
125 5 5 5 = 180o – 40o = 140o

= 1 + 121 = 126 = 5215 (ii) z = 360o – 25o – 90o – 30o
5 25 25 = 215o

(c) Total number of buns sold last week (c) ∠RSP = 180o – 74o = 106o
∠TSP = 180o – 35o = 145o
= 2n × 5 + (n + 10) + (3n – 2) ∠RST = 360o – 106o – 145o = 109o
= 10n + n + 10 + 3n – 2 = 14n + 8
4. (a) (x + 1) + (x + 1) + (5x –1) + (5x – 1)

2. (a) Multiples of 6: = 1 × 12 × 6
6, 12, 18, 24, 30, 36, 42, 48, 54 2
Multiples of 9:
9, 18, 27, 36, 45, 54 12x = 36
The common multiples of 6 and 9 are
18, 36 and 54. x = 36
12

x=3

(b) (i) 4 800 : 6 000 (b) (i) (a) j = {1, 2, 3, 4, 5, 6, 7, 8, 9}
4:5 K = {1, 3, 5, 7, 9}
K’ = {2, 4, 6, 8}
(ii) 4 800 + 1 200 = 4
6 000 + x 5 (b) j = {20, 24, 28, 32, 36, 40, 44}
L = {24, 32, 40}
6 000 = 4 L’ = {20, 28, 36, 44}
6 000 + 5
x (c) Stem

2 400 + 4x = 30 000 Leaf
4x = 30 000 – 24 000
4x = 6 000 4 81698
x = 1 500 5 7192
6 68
The increase in Bing Yau’s 7 370
investment = RM1 500 84

(c) Let boys = b and girls = g

g – b = 6 --------- ➀ Duration of badminton match

g + b = 36 ------- ➁

➀ + ➁: 2g = 42 Stem Leaf

g = 42 4 16889
2 5 1279
6 68
g = 21 7 037
84
Substitute g = 21 into ➁

21 + b = 36

b = 36 – 21

b = 15 Key: 4 | 1 means 41 minutes

61 © Penerbitan Pelangi Sdn. Bhd.

A3

Mathematics Form 1 Answers

5. (a) Q (c) (i) (6mn)2

= 62m2n2

5 cm = 36m2n2

PM R (ii) –2abc × 5
4a2
5bc
= – 2a

6. (a) (i) –3p + 5(5) = –17
2
52 cm

–3p – 25 = –17 × 2

–3p – 25 = –34

S –3p = –34 + 25

PM = 6 ÷ 2 = 3 cm –3p = –9

QM = √52 – 32 = 4 cm p = –9 = 3
–3
MS = (√52)2 – 42 = 6 cm
(ii) 4 + 2d = –8
QS = 4 + 6 = 10 cm
2d = –8 – 4

(b) (i) Sales obtained 2d = –12

= 2 750 + (–565 × 2) + 1 337 d = –12
2
= +2 957
d = –6
Ahmad obtains a profit of
(b) –2x < –2 – 2 4x  11 + 9
RM2 957. –2x < –4 4x  20
x>2
( ) ( )(ii) x5
–213 × 5 – 261 + 5 2<x5
18
( )=
– 7 × 5 – 13 + 5
3 6 18

= – 7 × 17 + 5
3 6 18

= – 119 + 5 = – 114 25
18 18 18

= – 19 = –613
3

(c)

E

30° 6 cm
C
D
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