MUNSANG COLLEGE
2016-2017 First Term Examination
F. 3 Mathematics
Marking Scheme
Paper 1
Class : _______
Name : _____________________
Time allowed :
1 hour 15 minutes
Full mark
80
:
Class Number : _____
This question-answer book consists of 12 printed pages.
Instructions to candidates:
1. This paper must be answered in English with a blue / black ball pen.
2. Write your name, class and class number in the space provided on this cover.
3. This paper consists of TWO sections, A and B.
Section A carries 40 marks and Section B carries 40 marks.
4. Answer ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer
Book. Do not write in the margins. Answers written in the margins will not be marked.
5. Graph paper and supplementary answer sheets will be supplied on request. Write your name, class and
class number on each sheet, and fasten them INSIDE this book.
6. Unless otherwise specified, all working must be clearly shown.
7. The diagrams in this paper are not necessarily drawn to scale.
8. Unless otherwise specified, numerical answers must be exact or correct to 3 significant figures.
9. Calculator pad printed with the “HKEA Approved” / “HKEAA Approved” label is allowed.
Remove the calculator cover / jacket.
1
Section A (40 marks)
1. Solve
x x4

 1 and write down the smallest integral solution of x.
3
2
(4 marks)
x x4

1
3
2
 x x4
 
  6  1 6
2 
3
2 x  3x  4   6
1M
1A
 x  6
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 The smallest integral solution of x is 6.
1A
1A
____________________________________________________________________________________
2. The prices of a box of candy and a box of biscuits are $35 and $48 respectively. Mrs Wong wants to
spend not more than $400 to buy a total of 10 boxes of candy and biscuit. In order to buy as many
boxes of biscuits as possible, how many boxes of candy should she buy?
(4 marks)
Let x be the number of boxes of candy she should buy.
35 x  48(10  x)  400
1M+1A
35 x  480  48 x  400
 13 x  80
x  6.1538
She should buy 7 boxes of candy.
1A
1A
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2
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x6
3.
The number of cars in a city increases steadily at a rate of 3% per year. It is known that the number
of cars in the city was 636540 at the beginning of 2013.
(a) What was the number of cars at the beginning of 2011?
(b)
Find the percentage increase in the number of cars over these two years.
(a)
Let x be the number of cars at the beginning of 2011.
2
x1  3%  636540
1M
(4 marks)
x  600000
1A
The number of cars at the beginning of 2011 is 600000.
The required percentage
636540  600000
 100 %
600000
 6.09%
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
1M
1A
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4.
Bank A offers a simple interest rate of r% per annum. If Mary deposits $25 000 into bank A for
3 years, she will get back an amount of $29 875.
(a) Find the value of r.
(b) Bank B offers an interest rate of (r – 0.5)% per annum compounded monthly. If Mary wants to
deposit $25 000 into a bank such that she can receive more interest after 3 years, which bank
should she choose? Explain your answer.
(6 marks)
(a)
25000 1  r %  3  29875
1M
1  r %  3  1.195
r %  0.065
r  6.5
1A
(b) The amount that Mary receive in bank B
 (6.5  0.5)% 
 250001 

12


 29917 .01312
36
1M+1A
1A
 29875
Mary should choose bank B.
1A
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3
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(b)
5.
In Figure 1, ABC is a right-angled triangle and BD is the median of  ABC.
It is given that AB = 10 cm and BC = 24 cm.
Figure 1
(a)
Find BD.
(b)
Suppose X is the centroid of ABC. Find BX.
(6 marks)
DE // CB
(corr.s equal)
AD = DC
(given)
 AE = EB
(intercept theorem)
1
BC
2
ED =
(mid-point theorem) 1M
1
 24
2
=
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1A
= 12
2
1A
2
2
BD = ED + EB
BD2 = 122 + 52
1A
BD = 13 cm
(b) BX : XD = 2 : 1
BX 
2
 13
3
1M
26
cm
1A
3
____________________________________________________________________________________
6. Figure 2 shows a net of a right prism with AI = AJ = EG = FG = 5cm. A student folds the net into
=
a solid.
(a)
Which vertices will coincide with H?
(b)
Determine the number of planes of reflection of the solid.
(c)
(i)
Name and find the angle between planes ABEG and BCDE.
(ii) Name the projection of HB of the solid on plane ABEG.
(5 marks)
(a)
Vertices F and D
(b)
2 planes
(c)
The required angle is IJA = 45
Figure 2
1A(both)
1A
1A+1A
(d) GB
1A
____________________________________________________________________________________
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4
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(a)
7.
In Figure 3, ABCD is a square. BCEF is a parallelogram. CD = DF. BDF is a straight line.
Find (a) ECF.
Figure 3
(b) CFE.
(6 marks)
(a)
BDC = 45
(prop. of square)
1A
DCF = DFC (base s, isos )
 DCF + DFC = BDC
(ext.  of )
1A
2DFC = 45
DFC = 22.5
1A
ECF = DFC (alt. s, BF // CE)
(b)
1A
BCF =  BCD + DCF
= 90 + 22.5
1A
= 112.5
CFE =  BCF
(alt. s, FE // BC)
= 112.5
1A
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5
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= 22.5
8.
Figure 4 is the orthographic view of a solid.
Figure 4
Draw the solid on isometric grid paper.
(1 mark)
(b)
(c)
How many planes of reflection are there in this solid?
How many axes of rotation are there in this solid?
(2 marks)
(2 marks)
(b)
5 planes of reflection
2A
(c)
5 axes of rotation
2A
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(a)
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6
Section B (40 marks)
9.
A hawker buys 150 pears at a cost of $240. Later he finds that 12% of them are rotten and cannot be
sold. What selling price should he set for each of the remaining pears so that the overall percentage
gain is not less than 30%? (Give the answer correct to the nearest dollar.)
Number of pears sold = 150  (1 – 12%)
1M
1A
Let $x be the selling price of each remaining pears
132 x  240
 100 %  30%
240
132 x  240  72
1M+1A
132 x  312
x  2.36 (cor to 3 sig. fig.)
He should set the selling price not less than $3.
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7
1A
1A
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= 132
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(6 marks)
10. In Figure 5, PQR, RST and PUT are straight lines. QU // RT, ∠PRT = 90, ∠URT = a and
UR = UT. Prove that
(a)
QU is an angle bisector of ∠PUR, and
(b)
QU is a perpendicular bisector of PR.
(6 marks)
(a)
QUR = a
(alt. s, BF // CE)
PTR = a
(base s, isos )
1A
Figure 5
 QUR = PTR
1A
 PTR = PUQ (corr.s, QU//RT)
QUR = PUQ
1A
(b)
In PQU and RQU,
QUR = QUP (proved)
QU = QU (common)
 PQU =  QRT=90
(corr. s, QU // RT)
PQU =  QRU = 90
PQU  RQU
 PQ = RQ
(ASA)
1A
(corr.sides,  )
1A
and PR  QU
1A
 QU is a perpendicular bisector of PR.
____________________________________________________________________________________
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8
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 QU is an angle bisector of PUR.
11. The top view of an object X is the square (including the diagonals) in Figure 6A. The front, left and
right views of X are the same as the rhombus (including the horizontal diagonal) in Figure 6B.
Figure 6B
Figure 6A
(a)
Name and sketch (with labels) the object X in the following box.
(a) Object X :
Octahedron
(3 marks)
1A
(b)
(c)
(d)
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1A (shape)
1A(label)
If the mid-points of BC, CD, DE and EB are P, Q, R and S respectively(not shown in the figure),
and that BCDE and ACFE are two planes of reflection of object X, name the other planes of
reflection of object X.
(2 marks)
AF is an axis of rotational symmetry of X and with respect to AF, the order of rotational
symmetry is 4. Name the other axes of rotational symmetry of X and the corresponding order of
rotational symmetry with respect to each of the axes.
(3 marks)
Name the angle between the plane ABE and FBE.
(1 mark)
(b) The planes of reflection of object X are ABFD, ASFQ and ARFP.
(c) The axes and the order of rotational symmetry are
(d) The required angle is ASF.
1A
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9
PR, 2;
1A (any one) + 1A(all)
SQ, 2; CE, 4; BD, 4.
1A + 1A + 1A (all)
12. Figure 7 shows a rectangular block ABFE-HDCG. P, Q, R and S are the mid-points of AD, EH, FG
and BC respectively. It is given that AB = BF = 4 cm and AD = 8 cm.
(a) (i) Name the angle between the line BH and the plane BFGC.
(1 mark)
(ii) Name and find the angle between the plane PGF and the plane EFGH.
(2 marks)
(iii) Name and find the angle between the plane PQGC and the plane PQFB.
(2 marks)
(a) (i) The required angle is HBG.
1A
(ii) The required angle is PRQ = 45
1A+1A
(iii) The required angle is GQF / CPB = 90
1A + 1A
front
____________________________________________________________________________________
(b)
If the block is cut along the plane PQFB, draw the orthographic views of the remaining part
PBCD-HQFG.
(3 marks)
1A+1A+1A
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10
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Figure 8
13. In Figure 8, D and E are mid-points of AB and AC respectively. F and G are on BC such that
(a)
Prove that 2DE = 3FG.
(2 marks)
(b)
Find HG : GE.
(4 marks)
(c)
Prove that GCE  GBH.
(2 marks)
(d)
Hence prove that ABHC is a parallelogram.
(3 marks)
(a)
DE =
1
BC
2
(mid-point theorem)
=
1
(BF + FG + GC)
2
=
1
3  FG
2
2 DE = 3 FG
1M
Figure 8
1
(b) DE // BC (mid-point theorem)
FGH = DEH
(corr. s, FG // DE)
EDF = GFH
(corr. s, FG // DE)
GHF = EHD
(common)
HGF HED
1A
1A
(AAA)
HG GF

HE ED
(corr. sides, s)
1A
HG
2

HG  GE 3
3HG  2( HG  GE)
HG = 2GE
HG
2
GE
HG : GE = 2 : 1
(c)
1
EG 1
(proved)

GH 2
CG 1

GB 2

(given)
EG CG

GH GB
1M
EGC = HGB
GCE GBH
(vert.opp s)
(ratio of 2 sides, inc. )
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11
1
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BF = FG = GC. DF and EG are produced to meet at H.
EC 1

BH 2
BH = 2EC
AC = 2EC
BH = AC
ECG = GBH
 BH // AC
1A
(corr. s, )
(alt. s equal)
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ABHC is a //gram. (opp. sides eq & //)
1A
1A
End of Paper
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12
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(d)