Download LiuPoShanMemorialCollege - 廖寶珊紀念書院Liu Po Shan Memorial

LIU PO SHAN MEMORIAL COLLEGE
(2010 – 2011)
SECOND TERM TEST
SECONDARY THREE
MATHEMATICS
MARKING SCHEME
Date of Exam. : 29 – 3 – 2011
Time allowed : 1 hour
Name : ________________ Class: ______
Class no.:_____
Marks:
/ 105
Instructions:
1.
2.
3.
4.
5.
Answer ALL questions in this paper.
Write your name, class and class number in the spaces provided on this cover.
This paper must be answered in English.
For Section A, the answers should be written in the boxes provided.
For Section B and Section C, the solutions should be shown clearly in the spaces provided below
the questions.
6. Calculators in the approved list can be used.
7. This paper consists of 13 pages.
1
Section A: Multiple Choice Questions (30 marks)
Choose the correct answer and write the letter in the boxes provided.
1.
2.
(4) 3 
A.
B.
–64.
64.
C.

D.
1
.
64
C
Which of the following must have x + y as a factor?
I. x2 – y2
II. x2 + y2
III. x(x + y) – x – y
A.
B.
C.
D.
3.
1
.
64
I only
II only
I and III only
II and III only
C
In a purse, there are three $2 coins and x $5 coins. The total value of the coins is not
more than $26. Which of the following inequalities can be used to find the range of
values of x?
4.
A.
6 + 5x < 26
B.
C.
6 + 5x  26
6x + 5 < 26
D.
6x + 5  26
B
A coin is tossed 400 times and the results are as follows:
Outcome
Head
Tail
Frequency
228
172
The coin is tossed once again, find the probability of getting a tail.
A.
B.
C.
D.
0.64
0.43
0.34
0.27
B
2
5.
6.
7.
8.
If the mode of 2, 2, 2, 4, 6 and x is x, find the mean of this set of data.
A.
2
B.
8
3
C.
D.
3
Cannot be found
A sum of $14 000 is deposited at 4% p.a. for 5 years, compounded yearly.
correct to the nearest dollar.
A.
B.
C.
$2 378
$2 800
$3 033
D.
$3 034
Find the interest
C
In the figure, AB = AC, BF = CD and BD = CE. Which of the following must be true?
A.
B.
C.
AFE ~ DEF
BDF  CED
DF // CE
D.
ABC
B
is equilateral
In △ABC, AB  BC and BD  AC. Which of the following must be true?
A.
B.
AB  CD = BC BD
AB  BC = AD BD
C.
BC 2  AB  CD
AB AC

BC BD
D.
9.
C
A
The orthocentre of a triangle is the point of intersection of the three
A.
B.
C.
D.
perpendicular bisectors.
medians.
angle bisectors.
altitudes.
D
3
10. Which net can be folded into the 3-D object below?
A.
B.
C.
D.
A
11. In the figure, ABCD is a parallelogram.
A.
B.
C.
D.
If x1 = x2 , find the value of y.
100°
110°
120°
130°
x1
x2
12. In the figure, ABCD is a square and ABE is an equilateral triangle.
Find∠BEG.
A.
B.
C.
D.
40°
45°
55°
65°
D
DEG is a straight line.
B
13. Referring to the figure, find y.
A.
B.
C.
D.
3
4
6
8
D
4
14. Referring to the front, top and side views of the given object, choose the corresponding
objects on the given information.
A.
B.
C.
D.
A
15. The angle between the planes BCD and ABC is
A.
B.
C.
D.
DMA.
DAM.
ABD.
ACD.
A
5
Section B : Conventional Questions (70 marks)
Solutions should be shown clearly in the spaces provided below the questions.
1.
Factorize the following expressions.
(a) 4h2 – 81k2
(b) 6x2 + 7xy – 5y2
(4 marks)
1M
(a) 4h2 – 81k2 = (2h)2 – (9k)2
1A
= (2h + 9k)(2h – 9k)
(b) 6x2 + 7xy – 5y2 = (3x + 5y)(2x – y)
2.
Simplify
( x 3 y 3 ) 1
( x 3 y 3 ) 1
y
2
y2
=
1A
and express the answer with positive indices.
x 3 y 3
y
1A
1A
2
for (ab)  a b
1A
n
n n
(3 marks)
or
am
a
n
 a m  n or a  n 
= x 3 y 5
=
y5
x
3.
1A
3
Express the following numbers in scientific notation.
(a) 486 000 000
(b) 0.000 265
(2 marks)
4.
(a) 486 000 000 = 4.86  108
1A
(b) 0.000 265 = 2.65  10–4
1A
In a class of 40 students, there are 22 boys. There are 10 boys and 8 girls wearing glasses.
If one student is selected at random, find the probability that
(a) the student is a girl,
(b) the student does not wear glasses.
(4 marks)
(a) P(girl) =
40  22
40

18
9
=
20
40
1A
for numerator
1A
(b) P(does not wear glasses) =
=
40  10  8
40
1A
11
20
1A
6
for numerator
1
an
5.
The cumulative frequency polygon below shows the numbers of pages of the books on a bookcase.
(a) How many books are there on the bookcase?
(b) Find the median of number of pages of the books.
(3 marks)
6.
(a) There are 38 books
1A
(b) median = 165
1M
1A
(a) Solve the inequality 2x + 13  5x  1.
(b) Find the greatest integer that satisfies the inequality in (a).
(3 marks)
(a)
2x + 13  5x  1
2x – 5x  1 – 13
–3x  14
x
(b)
14
3
1M
1A
The greatest integer that satisfies the inequality in (a) is 4.
7
1A
7.
Draw the front, top and right views of the following object.
2A
2A
Front view
8.
(6 marks)
2A
Top view
Right view
Refer to the following triangular prism ABCDEF. ADE is a right-angled triangle.
(a) Name the angle between the planes CDEF and ABCD.
(b) Name the angle between the planes ADE and ABFE.
(c) Find the projection of the line segment DF on the plane ABCD.
(d) Find the projection of the line segment DF on the plane ABFE.
(4 marks)
(a) ADE or BCF
1A
(b) BAD
1A
(c) BD
1A
(d) AF
1A
8
9.
The figure below shows a net which can be folded into a cube. If we can see that the numbers on
the front and the top of the cube are 4 and 3 respectively, what would be the number shown on each
of the following faces?
 right
(a) The right face.
(b) The left face.
(c) The back face.
(d) The bottom face.
(4 marks)
(a) 5
1A
(b) 2
1A
(c) 6
1A
(d) 1
1A
10. The figure below shows the rhombus ABCD. Find the values of x and y.
DA = DC
D
(property of rhombus)
4 cm
4=9–x
y°
A
(property of rhombus)
C
30°
1A
DA = DC
DAC = DCA = 30
y + DAC + DCA = 180
(base s, isos. )
( sum of )
y = 180 – 2  30
y = 120
(9 – x ) cm
1A
x=5
DAC = BAC = 30
(5 marks)
1A
9
1M
1A
B
11. Referring to the figure below, find the values of p and q.
(3 marks)
C
AE = EC
(given)
DE // BC
(given)
 AD = DB
p cm
3 cm
3 cm
(intercept theorem)
1A
q=2
B
1
DE = BC
2
2 cm
D
q cm
1A
1A
=6
12. (a) Prove that PQRS is a parallelogram.
(b) Find the values of x and y.
PS = QR
(3 marks)
(3 marks)
6
6
48o
 PQRS is a parallelogram.
Q
(opp. sides equal)
PSR = PQR = 48
S
x
y
1A
(given)
10
P
1A
(given)
PQ = SR
(b)
1.5
Acm
(mid-point theorem)
p=23
(a)
E
1.5
cm
10
1A
(property of // gram)
1A
x = 48
QPS + PQR = 180 (int. s, PS // QR)
1M
y + 48 = 180
1A
y = 132
10
R
13. In the figure, DC // AB and AD = CD.
(a) Find CAB.
(b) Prove that ABC is isosceles.
(a)
DA = DC
DAC = DCA
(4 marks)
(3 marks)
(given)
(base s, isos. )
DAC + DCA + 130 = 180
DCA =
( sum of )
180  130
= 25
2
1M
1A
CAB = DCA = 25
(alt. s, DC // AB)
CBA = 180 – ACB – CAB 
(b)
1A
1A
( sum of )
1A
= 180 – 25 – 130
= 25
1A
CBA = CAB

CA = CB
(sides opp., equal s)
1A
ABC is isosceles.
14. In BDF as shown, AB = 18 cm, BC = 12 cm, CD = 15 cm, DE = 10 cm, AC = 15 cm and
CE = 12.5 cm. Prove that
(a) ABC ~ CDE,
(b) FB = FD.
(4 marks)
(3 marks)
(a) In ABC and CDE,
AB 18 6


CD 15 5
BC 12 6


DE 10 5
1A
1A
any two
AC 15
6


CE 12.5 5

AB BC AC


CD DE CE

ABC ~ CDE
(3 sides prop.)
(b) 
ABC ~ CDE
(proved)
ABC = CDE
(corr. s, ~s)
1
(sides opp., equal s)
1

FB = FD
1
1
11
1
15. In the figure, the diagonals of the quadrilateral
and BO = OD, prove that
ABCD
intersect at O.
(a) ABO  CDO,
(b) AB // DC,
(c) ABCD is a parallelogram.
(a)
(4 marks)
(2 marks)
(3 marks)
In ABO and CDO,
OB = OD
OAB = OCD = 90
1A
(given)
1A
(given)
AOB = COD
(vert. opp. s)
1A
 ABO  CDO
(AAS)
1
(b) OAB = OCD = 90
AB // DC
(c)


1A
(given)
1
(alt. s equal)
AB // DC
(proved)
ABO  CDO
(proved)
AB = CD
If AB  AC, DC  CA
1A
(corr. s, s)
ABCD is a parallelogram.
1A
(2 sides equal and //)
12
1
Section C : Bonus Questions (5 marks)
Solutions should be shown clearly in the spaces provided below the questions.
1. Convert the decimal number 210 + 25 + 21 to a binary number.
1A
210 + 25 + 21 = 10000100010 2
2. Calculate
(1 mark)
1116  112 and express the answer as a decimal number.
1116  112 = (116 + 1) – (12 + 1)
1A
(2 marks)
any one
= 17 – 3
= 14
1A
3. Factorize x 3  ( x  2)3 .
(2 marks)
x 3  ( x  2)3 = [ x  ( x  2)][ x 2  x( x  2)  ( x  2) 2 ]
1M
= 2( x 2  x 2  2 x  x 2  4 x  4)
= 2(3x 2  6 x  4)
1A
- END OF PAPER -
13