1. ppr maths nbk
SULIT 1449/2
1449/2
NAMA :
Matematik
Kertas 2 TINGKATAN :
Oktober
2007
1
2 jam
2
SEKTOR SEKOLAH BERASRAMA PENUH
BAHAGIAN SEKOLAH
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN DIAGNOSTIK
TINGKATAN EMPAT 2007
Pemeriksa
MATEMATIK
Markah Markah
Bahagian Soalan
Kertas 2 Penuh Diperoleh
Dua jam tiga puluh minit 1 4
2 4
3 5
JANGAN BUKA KERTAS SOALAN INI 4 4
SEHINGGA DIBERITAHU
5 3
A
1. Kertas soalan ini mengandungi dua bahagian : 6 4
Bahagian A dan Bahagian B. Jawab semua
soalan daripada Bahagian A dan empat soalan 7 6
dalam Bahagian B. 8 6
2. Jawapan hendaklah ditulis dengan jelas dalam 9 4
ruang yang disediakan dalam kertas soalan. 10 7
Tunjukkan langkah-langkah penting. Ini boleh
membantu anda untuk mendapatkan markah. 11 5
12 12
3. Rajah yang mengiringi soalan tidak dilukis
mengikut skala kecuali dinyatakan. 13 12
B 14 12
4. Satu senarai rumus disediakan di halaman
2 & 3. 15 12
16 12
5. Anda dibenarkan menggunakan kalkulator
saintifik yang tidak boleh diprogram. Jumlah
Kertas soalan ini mengandungi 24 halaman bercetak.
1449/2 © 2007 Hak Cipta Sektor SBP [Lihat sebelah
SULIT

2. SULIT 1449/2
Untuk
Kegunaan
Pemeriksa
MATHEMATICAL FORMULAE
The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
RELATIONS
1 am x an = a m+ n
2 am ÷ an = a m – n
3 ( am )n = a mn
1 ⎛ d − b⎞
4 A-1 = ⎜
⎜−c a ⎟⎟
ad − bc ⎝ ⎠
n( A)
5 P(A)=
n( S )
6 P ( A′ ) = 1 − P(A)
7 Distance = ( x1 − x2 )2 + ( y1 − y2 ) 2
⎛ x + x 2 y1 + y 2 ⎞
8 Midpoint, ( x, y ) = ⎜ 1 , ⎟
⎝ 2 2 ⎠
9 Average speed = distance travelled
time taken
10 Mean = sum of data
number of data
11 Mean = sum of (class mark × frequency)
sum of frequencies
12 Pythagoras Theorem
c2 = a2 + b2
y 2 − y1
13 m=
x 2 − x1
y -intercept
14 m=
x-intercept
1449/2 2 SULIT

3. SULIT 1449/2
Untuk
Kegunaan
Pemeriksa
SHAPES AND SPACE
1
1 Area of trapezium = × sum of parallel sides × height
2
2 Circumference of circle = πd = 2πr
3 Area of circle = πr2
4 Curved surface area of cylinder = 2πrh
5 Surface area of sphere = 4πr2
6 Volume of right prism = cross sectional area × length
7 Volume of cylinder = πr2h
1 2
8 Volume of cone = πr h
3
4 3
9 Volume of sphere = πr
3
1
10 Volume of right pyramid = × base area× height
3
11 Sum of interior angles of a polygon = ( n – 2) × 180˚
arc length angle subtended at centre
12 =
circumference of circle 360o
area of sector angle subtended at centre
13 =
area of circle 360o
PA'
14 Scale factor , k =
PA
15 Area of image = k 2 × area of object
1449/2 3 [Lihat sebelah
SULIT

4. SULIT 1449/2
Untuk
Kegunaan Section A
Pemeriksa
[52 marks]
Answer all questions in this section.
3 3− x
1 Solve the quadratic equation x=
4 x−3
[4 marks]
Answer :
2 Calculate the value of p and of q that satisfy the following simultaneous linear equations:
2 p − 3q = 6
4 p + 9q = −3
[4 marks]
Answer :
1449/2 4 SULIT

5. SULIT 1449/2
Untuk
3 (a) State whether the following statement is true or false. Kegunaan
Pemeriksa
6 ÷ 2 = 3 and 3 2 = 6
(b) Write down two implications based on the sentence below.
4m > 20 if and only if m > 5
(c) Given the number sequence 5, 18, 39, 68, … which follows the pattern
5 = 1 + 4(12)
18 = 2 + 4(22)
39 = 3 + 4(32)
68 = 4 + 4(42)
……………..
……………..
Form a general conclusion by using the induction method for the numerical sequence
above.
[5 marks]
Answer :
(a)
(b) Implication 1 : _________________________________________________
Implication 2 : _________________________________________________
(c)
1449/2 5 [Lihat sebelah
SULIT

6. SULIT 1449/2
Untuk
Kegunaan 4 Diagram 1 shows a combined solid consists of a cuboid JKLMPQRS and half cylinder
Pemeriksa
JMNTSP.
J 8 cm K
N 7 cm
M L
16 cm
S R
T
P Q
DIAGRAM 1
22
Using π = , calculate the volume, in cm3, of the solid
7
[4 marks]
Answers:
1449/2 6 SULIT

7. SULIT 1449/2
Untuk
5 The Venn diagram in the answer space shows sets P, Q and R such that the universal set, Kegunaan
Pemeriksa
ξ = P ∪ Q ∪ R.
On the diagrams in the answer space, shade
(a) P′ ∩ R′
(b) Q ∩ R ∪ P
[3 marks]
Answer :
(a) Q
R
P
Q
(b) R
P
1449/2 7 [Lihat sebelah
SULIT

8. SULIT 1449/2
Untuk
6 Diagram 2 shows a cuboid with horizontal rectangular base, ABCD.
Kegunaan
Pemeriksa
E H
F G
6 cm
D C
8 cm
A 10 cm B
DIAGRAM 2
Calculate the angle between the plane ABE and the base ABCD.
[4 marks]
Answer :
1449/2 8 SULIT

9. SULIT 1449/2
7 In Diagram 3, O is the origin. KL, PQ and RS are straight lines. PQ is parallel to RS.
Untuk
The equation of the straight line KL is y = −2 x + 6 . Kegunaan
Pemeriksa
y
K R
P
• x
O
S
Q(−3, −3) •
L
DIAGRAM 3
Find the
(a) coordinate of P,
(b) gradient of the straight line PQ,
(c) equation of the straight line RS.
[6 marks]
Answer:
(a)
(b)
(c)
1449/2 9 [Lihat sebelah
SULIT

10. SULIT 1449/2
Untuk
8 (a) Identify the antecedent and consequent in the following implication.
Kegunaan
Pemeriksa ‘ If a triangle has two equal sides, then it is an isosceles triangle.’
(b) State the converse of the following implication.
If x > 5 , then x2 > 25
(c) Complete the premise in the following argument.
Premise 1 : If x is an angle in a semicircle, then x = 90°
Premise 2 : _________________________________________________
Conclusion : x = 90°
(d) Complete the following argument
Premise 1 : _________________________________________________
Premise 2 : M ∩ N ≠ M
Conclusion : M ⊄ N
[6 marks]
Answer:
(a) Antecedent : …………………………………………………………………
Consequent : ………………………………………………………………….
(b) …………………………………………………………………………………
(c) Premise 2 : ……………………………………………………………………
(d) Premise 1 : …………………………………………………………………….
1449/2 10 SULIT

11. SULIT 1449/2
9 Ten cards are placed in an empty box.
Untuk
Kegunaan
Pemeriksa
E X C E L L E N C E
(a) If a card is selected at random from the box, state the probability that the card marked
C is selected.
(b) A number of cards marked C are added into the box. If a card is selected at random
3
from the box, the probability of selecting a card marked C is .
7
Find the number of cards that has been added into the box.
[4 marks]
Answer :
(a)
(b)
1449/2 11 [Lihat sebelah
SULIT

12. SULIT 1449/2
Untuk
10 In Diagram 4, ABOD is a square and PQR is a semicircle with centre O.
Kegunaan
Pemeriksa
A B
P
Q
D 14 cm O
R
DIAGRAM 4
22
OR = 7 cm. Using π = , calculate
7
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.
[7 marks]
Answer :
(a)
(b)
1449/2 12 SULIT

13. SULIT 1449/2
Untuk
11 Diagram 5 shows a right pyramid with rectangular base PQRS . Kegunaan
Pemeriksa
The apex V is 5 cm vertically above the point T.
V
S
R
T
6 cm
P 8 cm Q
DIAGRAM 5
Calculate the angle between the line VP and the base PQRS.
[5 marks]
Answer:
1449/2 13 [Lihat sebelah
SULIT

14. SULIT 1449/2
Untuk
Kegunaan Section B
Pemeriksa
[48 marks]
Answer any four questions in this section.
12 Table 1 in the answer space shows the distribution of the ages of 200 participants in a big
walk event.
(a) Using the data in Table 1, complete the table provided in the answer space.
[4 marks]
(b) By using a scale of 2 cm to 5 years on the x-axis and 2 cm to 20 participants on
the y-axis , draw an ogive for the data.
[5 marks]
(c) Based on your ogive in (b),
(i) find the inter quartile,
(ii) explain briefly the meaning of the third quartile.
[3 marks]
Answer:
(a)
Age (years) Frequency Cumulative Upper
Frequency Boundary
15 − 19 10
20 − 24 20
25 − 29 50
30 − 34 60
35 − 39 36
40 − 44 18
45 − 49 6
TABLE 1
(b) Refer graph on page 15.
(c) (i)
(ii)
1449/2 14 SULIT

15. SULIT 1449/2
Graph for Question 12
Untuk
Kegunaan
Pemeriksa
1449/2 15 [Lihat sebelah
SULIT

16. SULIT 1449/2
Untuk
Kegunaan 13 (a) Diagram 6 shows sectors OTQR, OABC and OCP with the same centre O.
Pemeriksa TAOP and OCR are straight lines.
Q
R
B
C
P
T A O
DIAGRAM 6
21
OR = 2OC , OC = cm and ∠COP = 60°.
2
22
Using π = , calculate
7
(i) the perimeter, in cm, of the whole diagram,
[3 marks]
(ii) the area, in cm2, of the shaded region.
[4 marks]
(b) Diagram 7 shows some number cards.
45 46 47 48 49 50 51 52
53 54 55 56 57 58 59 60
DIAGRAM 7
A card is picked at random. State the probability of choosing a card containing
(i) digit 5,
(ii) a multiple of 5,
(iii) a number such that when divided by 5, the remainder is 1.
[5 marks]
1449/2 16 SULIT

17. SULIT 1449/2
Untuk
Answer: Kegunaan
Pemeriksa
13 (a) (i)
(ii)
(b) (i)
(ii)
(iii)
1449/2 17 [Lihat sebelah
SULIT

18. SULIT 1449/2
Untuk
14 The data in Diagram 8 show the marks of 40 students for the Mathematics monthly test.
Kegunaan
Pemeriksa
46 53 44 60 42 38 31 55
35 37 54 32 46 56 40 60
52 40 34 45 52 35 50 36
47 38 40 48 45 42 53 44
50 44 58 51 36 48 56 32
DIAGRAM 8
(a) Using the data in Diagram 8 and a class interval of 5 marks, complete Table 2
in the answer space.
[4 marks]
(b) Based on Table 2 in (a),
(i) state the modal class,
(ii) calculate the mean mark of the Mathematics monthly test and give your answer
correct to 2 decimal places.
[4 marks]
(c) For this part of the question, use the graph paper provided on page 19.
By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the
vertical axis, draw a histogram for the data.
[4 marks]
1449/2 18 SULIT

19. SULIT 1449/2
Answer:
Untuk
Kegunaan
14 (a) Pemeriksa
Class interval Midpoint Frequency
31 – 35
36 - 40
TABLE 2
(b) (i)
(ii)
(c) Refer to the graph on page 20.
1449/2 19 [Lihat sebelah
SULIT

20. SULIT 1449/2
Graph for Question 14
Untuk
Kegunaan
Pemeriksa
1449/2 20 SULIT

21. SULIT 1449/2
15 (a) Diagram 9 is a Venn diagram showing the number of students in a class who play at least
Untuk
one of the three games. Given the universal set, ξ = F ∪ R ∪ H. Kegunaan
Pemeriksa
F
x
7 5
R 2x H
10 + x 18 – 2x
10
DIAGRAM 9
Given that set F = { students who play football }
set R = { students who play rugby } and
set H = { students who play hockey}.
(i) Find the number of students who play hockey.
(ii) If the number of students who play football is 30, find
(a) the number of students who play rugby,
(b) the number of students who play one game only.
[6 marks]
Answer:
(a) (i)
(ii) (a)
(b)
1449/2 21 [Lihat sebelah
SULIT

22. SULIT 1449/2
Untuk
15 (b) In Diagram 10 , TQRS, NR and OQ are straight lines drawn on a Cartesan plane. O is the
Kegunaan
Pemeriksa origin and OQ is parallel to NR. The gradient of NR = 2.
y
S
•R
• Q(p, 3)
T
• • x
N(-6, 0) 0
DIAGRAM 10
Find
(i) the value of p,
(ii) the coordinate of R,
(iii) the equation of the straight line ST.
[6 marks]
Answer:
(b) (i)
(ii)
(iii)
1449/2 22 SULIT

23. SULIT 1449/2
2w 2 − 3
16 (a) Solve the equation = 1. Untuk
5w Kegunaan
Pemeriksa
[4 marks]
(b) Calculate the values of x and y that satisfy the following simultaneous linear
equations:
2x + y = 3
4x - 3y = 11
[4 marks]
(c) (i) Form a true compound statement by combining the two statements
given below.
(a) 52 = 10
1
(b) = 0.25
4
(ii) Form a general conclusion by induction based on the numerical sequence
below.
2, 9, 16, 23 …
2 = 2 + 7 (0)
9 = 2 + 7 (1)
16 = 2 + 7 (2)
23 = 2 + 7 (3)
……………….
(iii) Complete the following sentence using a suitable quantifier to make it a true
statement.
‘………………. prime numbers are odd numbers.’
[4 marks]
1449/2 23 [Lihat sebelah
SULIT

24. SULIT 1449/2
Untuk
Answer:
Kegunaan
Pemeriksa
16 (a)
(b)
(c) (i)
(ii)
(iii)
1449/2 24 SULIT