UNIVERSITY OF MALTA
SECONDARY EDUCATION CERTIFICATE
SEC
MATHEMATICS
May 2011
MARKING SCHEME
MATRICULATION AND SECONDARY EDUCATION
CERTIFICATE EXAMINATIONS BOARD
SEC Mathematics May 2011
Extracts from General Marking Procedures
1.
Abbreviations used in Marking Schemes:
M denote method marks. Such marks are awarded for using a suitably correct
method to answer the relevant question.
A denote accuracy marks following from a correct method. A marks are only
awarded if there is explicit or implied evidence that the candidate is using an
appropriate method.
B denote accuracy marks, given independent of method used. In awarding B
marks, no evidence is sought as to the method employed to answer the question.
Aft denotes follow through accuracy marks. In some questions, the use of a
previous answer is necessary in order to work out the final answer. When an the
accuracy mark is deemed as ft, candidate will get the mark for an inaccurate answer
as long as the given answer follows from correct accurate working using previous
incorrect answer.
FNW: In some questions/ part questions, although the total mark is broken down
into M and A marks, the total mark is designated as FNW (full marks no
working).
In
such cases, candidates still get full marks even if they do not show the method used
provided that candidates give a FULLY CORRECT ANSWER.
2.
Trial & Error Methods: Unless the marking scheme specifies to the contrary,
candidates obtaining correct answers using trial & error methods are NOT penalised.
3.
Do not follow the marking scheme blindly. If a candidate uses a valid method
that is not mentioned in the marking scheme, use your discretion to give credit to the
candidate’s work within the parameters of the marks allotted to the question. When
allotting marks in this manner, specify this by writing OS (Outside Scheme) next to
the marks awarded for that question.
4.
Misreading of questions: Sometimes questions are misread, e.g. a number is
copied incorrectly from the exam paper. If the question has been simplified, do not
award marks. Otherwise, mark initially so as not to penalise the misreading. Add the
component marks and subtract one mark from the total mark obtained. Indicate this
by writing MR next to the mark awarded for that question.
2
MATHEMATICS
SEC MAY 2011 SESSION
Mental Paper Marking Scheme
ALL QUESTIONS CARRY ONE B-MARK *IGNORE ALL UNITS*
1
1
5
2
131
3
132°
4
5
7
150°
3
5
5

2
10
15cm
11
28
2
13
12
Accept 132
(b) The diagonals of a rhombus are perpendicular.
20
9
Accept 0.2
0.76
6
8
COMMENTS
13
½
14
55°
15
21
16
5%
17
4
18
04:30 or half past four
19
12.11
20
€3
Accept 0.6 or
6
10
Accept – 2.5 or – 2 ½
Units can be left out
Accept
4
8
or
26
52
Accept 0.5 or
2
4
Accept 5
Accept 4.30 or 4:30
Accept 3.00 or 3 or 3.0
3
MAY 2011 SESSION (Core Paper)
1
(i) 350,407 or 350407
(ii) 23:53 or 23.53 or 23 hr 53 min
(iii) 1
.
8
23
or
or 1.53
15
15
Do not accept 1.5
(iv) 96 or
960
10
(v) 9.9
1 B mark each = 5 marks
______________________________________________________________________________
2
(a) (i) Butter : Icing.Sugar = 3: 7
Butter = 103  250 g = 75 gms
for 103  250
1M mark
Full marks even for no working
for 75 (gms)
1A mark
For mult. by 14
1M mark
(ii) Icing sugar = 14 x 750 = 10,500 gms = 10.5 kg
Icing sugar : Butter-cream = 7 : 10
10
10
for 10.5 kg
1M mark
Hence butter-cream =
10.5 kg = 15 kg
7
7
Do not accept 15000 gms
for 15(kg)
1 A mark
______________________________________________________________________________
4 x  3 5 x  12 3(4 x  3) - (5 x  12) 12 x  9  5 x  12 7 x  21 x  3
(b) (i)





.
35
105
105
105
105
15
up to
12 x  9  5 x  12
or equivalent
105
x3
for
.
15
1 M mark
1 B mark
4 x  3 5 x  12
x3
=7 

 7  x  3  105  x  102
35
105
15
for cross-multiplying 1 M mark
for 102 1 A mark
______________________________________________________________________________
3
(i) The perimeter = 2 circles = 2  2   12.5  50 = 157.0796m
To 1 decimal place, answer = 157.1m
for 2  2    r 1M mark
for r = 12.5
1B mark
(ii)
4
for 157.1(m) only
1 A mark
1M mark
(ii) Area of lawn = 2 circles + square
2
2
Area of Circle =  r and area of square = 25
1 M mark
Area of lawn = 981.748 + 625 = 1606.748m2
To 4 significant figures, answer = 1607(m2). for rounded answer 1A mark
(b) Fencing cost 8 rolls @ € 70 = € 560
Turf cost 1607m2 @ € 25 = € 40,175
1M mark
(or 1606.748m2 @ € 25 = € 40,169)
1M mark
Labour expenses cost € 7,600
Total cost to the nearest Euro, is € 48,335 (or €48329) or in between.
For addition 1 M mark
For correct answer 1A mark
_____________________________________________________________________________
4
(i) 3 packets (or lbs) weigh 1.364 kg
1.364  22
kg = 10.002 kg
1 M mark
22 packets weigh
3
FMNW
To the nearest kg, answer is 10 kg
1A mark
(ii) 1.364kg = 3 packets
3 7
7 kg =
1M mark
1.364
= 15.396 packets
1A mark
Number of packets = 16
FMNW
1B mark
______________________________________________________________________________
5
(i) 18% of €480 = € 86.40
OR 118% of € 480 = € 566.40
1M mark
Total cost = 480 + 86.40 = € 566.40(N0 marks for 18% only) FMNW
1A mark
OR 90% of € 480 = € 432
1M mark
(ii) 10% of € 480 = € 48
Showroom price = € (480 – 48) = € 432(NO marks for 10% only)
1A mark
FMNW
VAT = 21% of € 432 = € 90.72
1M mark
Final cost = € (432 + 90.72) = € 522.72
OR 121% of € 432 = € 522.72
1A mark
If final 522.72 only is given and no working shown throughout give 2 marks only.
(iii) Savings = € (566.40 – 522.72) = € 43.68
1M mark
43.68
Percentage savings =
1M mark
100 % = 7.71%
566.40
To 1 dec. place answer = 7.7(%)
1A mark
__________________________________________________________________________
6
(i) Constructing an equilateral triangle, ABC of side 6 cms. ARCS seen
(ii) Bisecting angle BAC.
ARCS seen
(iii) Constructing circle, centre O, passing through A, B and D.
5
2A marks
2A marks
2A marks
(iv) Marking E or D correctly.
Measuring the length of EB = 3.45 cm ± 0.2cm
Measuring the length of OC = 5.2cm ± 0.2cm
1A mark
1A mark
1A mark
C
D
E
A
B
_______________________________________________________________________
7
2x + 4
S
(i) 4y = 2x + 4
y = ½ x + 1 or equivalent
R
1B mark
x
x
P
4y
Q
1B mark
(ii) perimeter = 2x + 2(2x + 4) = 50
or
2x + 2(4y) = 50
2x + 4x + 8 = 50
6x + 8 = 50
6x = 42
x = 7 (cm),
y = 4.5 (cm)
1Mmark
1M mark
1A mark
1A mark
(iii) SP = 7cm and SR = 18cm
1Aft mark
Area = 7 x 18 = 126(cm2 )
1Aft mark
______________________________________________________________________________
6
Instrument
guitar
piano
drums
violin
none
Total
Frequency
8x
5x
4x
2x
x
20x
No. of students
200
125
100
50
25
500
(i) Every two correct entries get 1 mark
(ii) Guitar =
200
500
of 3600  1440 ; Piano =
Violin =
50
500
of 3600  36 0 ; None =
3A marks
of 3600  90 0 ; Drums =
125
500
25
500
100
500
of 3600  72 0
of 3600  180
For at least 1 correct
1M mark
guitar
piano
drums
violin
none
For drawing a Pie chart with labels in degrees or names
1M mark
For all correct angles drawn
2A marks
[For every two wrong angles subtract 1A mark]
25
x
18
(iii) Probability =
or
or
1M mark
500
20 x
360
1
=
or equivalent but not in terms of x
1A mark
20
________________________________________________________________________
7
9
(a) (i) A: 2A marks
(ii) B: 2A ft marks (iii) C: 2A ft marks
C
A
B
1B mark
(b) Reflection in the y-axis or equivalent.
_______________________________________________________________________________________
1A mark
10
(i) Marking both angles of elevation correctly on the given diagram.
1B mark
(ii) Taking 12.5m not 13.5m
T
tan 36° =
B
26°
36
G
1M mark
GF .tan 36° = 12.5
12.5m
GF =
J
12.5
GF
°
12.5
tan360
1M mark
= 17.205or 17.2m or more accurate
1A mark
F
(iii) tan 26° =
TX TX
TX


BX GF 17.205
1M mark
TX = 17.2(05) tan 26° = 8.39(14)
1M mark
BG = TF – TX
1M mark
BG = 13.5 – 8.39(14) = 5.1(086)m
1A FTmark
8
MAY 2011 SESSION
Paper 2A
1
 x > 1.5
1.5  x  3
or in words correctly
(a)
2x > 3
1B mark
1A mark
(b) Let x be no. of hrs worked.
for both expressions correct 1A mark
Either he will be paid 9x Euro or 175 + 5x Euro
9x > 175 + 5x
for inequality
1M mark
4x > 175
x > 43.75 hrs
1M mark
Least number of hours is 44. Accept also any number between 43.75 and 44
1A mark
__________________________________________________________________________________
(i) f(18) =
2
1
4
(18)  12
g(– 3) = 3(– 3)2
1M mark
1 A mark
1M mark
1A mark
=5
= 27
(ii) y =
1
4
x  12

1
4
x  y  12
or multiplying by LCM
x  4y  2
f
–1
(x) = 4x – 2 . Do not accept final answer in terms of y
1M mark
1A mark
1A mark
FMNW
g(x) + 4f – 1(x) = 4.
for substituting f- -1(x)and g(x)
1M mark
3x2 + 4(4x – 2) = 4
2
3x + 16x – 8 = 4  3x2 + 16x – 12 = 0
1M mark
(3x – 2)(x + 6) = 0
or using formula
1M mark
1M mark
3x = 2 or x = – 6 or correct values in formula
x = 23 or x = – 6 (both correct)
1A mark
__________________________________________________________________________________
(iii)
3
(i)
volume of jewel box = 1/3 area of base x height =
1
 15 x 2
3
For substituting correct values
= 5x2 (cm3)
(ii)
(iii)
volume of lid =
1
x2 5x2
 5 =
3
9
27
5x2 –
5x2
= 495
27
1M mark
1A mark
forming equation
130 x 2
= 495 for multiplying by correct LCM
27
9
1M mark
1A mark
1M mark
1M mark
x2 = 102.808
1A mark
x = 10.139
finding root
1M mark
x = 10.14(cm) to 2 d.pl. only
1A mark
________________________________________________________________________________
4
(i) A = P(1 + r/100)n or for correct long method
1M mark
1M mark
A = 20,000(1 + 4.2/100)5 for correct substitution in either method
A = 24567.93 or 24568 €
1A mark
(ii) Interest = 24,568 – 20,000 = 4,568 € Accept more accurate answer
(iii) Tax = 20% of 4568€ = 913.60 € (or 913.59)
New amount left is 24,568€ – 913.60 € – 12,000 €
= 11,654(.40) €
1A mark
1A mark
1M mark
1A mark
2P = P(1 + 5/100)n gives 2 = 1.05n
If LHS is twice Principal or correct substitution in long method or formula
Now 1.0512 = 1.796 Or amount = €20929(.53) which is less than double.
1M mark
(iv)
1A mark
Hence 12 years are not enough
1B mark
__________________________________________________________________________________
5
15
25
35
45
55
65
75
85
Age
40
140
95
80
60
65
35
5
No.
40
180
275
355
415
480
515
520
Total
For finding totals
1M mark
(i) Drawing of cumulative frequency curve.
Correct Scale and axes
1A mark
4 correct values and points plotted
1A mark
All plotted points correct
1A mark
For general correct shape 1A mark
600
Number
500
400
300
200
100
0
0
20
40
60
80
100
Age
(ii) the median age is reading on 260, which is about 33 years, (31 – 35) 1A mark
(iii) the percentage of people under 18 years that attend this gym is 70 (60 – 80)
1A mark
This amounts to
70
100 %
520
= 13.5%, (11.5% – 15.5%)
10
1M mark
(iv) the number of people under 60 years is around 450 (440 – 460)
1A mark
So over 60 years no. is 520 – 450 = 70 (60 – 80) (subtracting from 520)
1M mark
__________________________________________________________________________________6
(a)
(i)
(x – 1)
1B mark
1B mark
(x + 1)
(ii) Show that the sum of any three consecutive numbers is a multiple of 3.
For adding correct terms
(x – 1) + x +(x + 1) = 3x
If sum is shown as multiple of 3
M1 mark
A1 ft mark
Do not accept proof with values instead of x.
(iii) Show that the sum of any four consecutive numbers is always even.
(x – 1) + x + (x + 1) + (x + 2) = 4x + 2 = 2(2x + 1) For adding 4 consecutive terms M1 mark
For showing 2 is a factor A1 mark
(b) (i) p =
k

x

10 = k/( 201 )
10 = 20k
 k=½
1M mark
FMNW
1
when x = 2, p = ¼ .
p=

2x
p = cy  10 = c  15 
 c = 50 so that p = 50y
(ii)
y=
1
p
and x =
50
2p

xy =
1A mark
1 p

or
2 p 50
1
20
.
1
=
100
 15 
1M mark
1A mark
Ignore explanation
(iii)
When p = 15, x =
1
30
and y =
15
50
 103
1M mark
23
60x + y = 60( 301 ) + 103 = 10
or 2.3
1A mark
__________________________________________________________________________________
7
(i) Pts of intersection are (2.3, 2.7), ( – 0.8, – 3.7) or near enough (∓ 0.2).
11
1B, 1B marks
(ii)
4
 2x  2
x
x  4  2 x2  2 x
y  1
2 x  3x  4  0
2
x
3  9  32
4
1A mark
1M mark
1A
mark
(Even if = 0 left out)
or correct substitution in own equation
1M mark
1A, 1A marks
x = – 0.851 or 2.351
(iii)
Must have 3 d.pl. but correct if two d.pl are correct
(– 0.851, – 3.702) and (2.351, 2.702)
1A ,1A marks
1M mark
(iv) Area of triangle OAC = ½ base x height
= ½ x1 x 2.702
1M mark
= 1.35(1) Ignore rounding and units
1A mark
__________________________________________________________________________________
8
(i)
Angle OAB = x (OA = OB)
w = 90° – x; (PQ tangent, OA radius)
for reason either OA = OB or tgt perp to radius
1M mark
1A mark
y = 90° – x; (180° – 90° – x)
for reason involving triangle ODA 1M mark
1A mark
(ii)
Δs ODA, ODB are congruent (RHS or other correct reasons)
for reason of congruency 1M mark
1A mark
for RHS or SAS proof
Hence angle BOA = 2y
But angle at circumference = ½ angle at centre
1M mark
Therefore z = y
1A mark
°
But y = w (both = 90 – x)
1A mark
So w = z
If alternate segment theorem is used without proof give 1A mark
_______________________________________________________________________________
12
9
A
5.26m
(i) angle ABE = 45° (90° – 45°)
B
1B mark
Angle BEC = 90° – ( 45 + 21)° = 24°
1B
C
mark
Angle ECD = 180° – ( 100 + 21)° = 59°
1M mark
E
(ii)
D
7.38m
BE2 = AB2 + AE2 = 5.262 + 5.262 = 55.3352
For
correct substitution in Pythagoras Th.
1M mark
1A mark
BE = 7.4388 so that to 3 s.f , BE = 7.44m
For using Sine rule to find CE
1M mark
giving CE =
7.38 sin1000
sin590
For subst. and cross-mult.
CE= 8.479m or 8.48(m)
1M mark
1A mark
1M mark
(iii) Area using ½ bc sinA
1M mark
Area = ½ (7.4388)(8.479)(sin 240) for substitution
2
= 12.8m or more accurate
1A mark
__________________________________________________________________________________
10
H
triangle HCB rt-angled at C 1A mark
1A mark
Labelling x and x + 0.5
HB + x + x + 0.5 = 6
x
HB = 5.5 – 2x or equivalent 1A mark
C
x + 0.5
B
30.25 – 22x + 4x2 = x2 + x2 + x + 0.25
Pythagoras Theorem with correct substitutions
(5.5 – 2x)2 = x2 + (x + 0.5)2
1M mark
For correct expansion of at least 1 bracket
2x2 – 23x + 30 (= 0)
(2x – 3)(x – 10) = 0 or using formula
x = 10 or 1.5 km
x = 1.5 km
(x has to be less than 6)
x + 0.5 = 2km only
Ignore extra answer with x = 10
13
1M mark
1A mark
1M mark
1A mark
MAY 2011 SESSION
Paper 2B
1
Write down:
(i)
a prime number between 20 and 28,
(i)
23
1B mark
(ii) 7 , 13 ,1 or 91
1B mark
(ii) a factor of 91,
(iii) 25, 50 or 75 or 100 … 1B mark
(iii) a multiple of 25.
__________________________________________________________________________________
2
FMNW
1M mark
2(3x) = 486 so that 3x = 243
x=5
1A mark
__________________________________________________________________________________
3
91 litres =
91
gallons = 20 gallons
4.55
FMNW
1M mark
1A mark
__________________________________________________________________________________
4
820  13.12 
63
= 2190.92
5
FMNW
= 2191 to 4 significant figures.
1M mark
1A mark
1Aft mark
In standard form, answer = 2.19(1)x103.
If only final answer in standard form given, without previous answers, give 2 marks in all.
__________________________________________________________________________________
5
Jan
Feb
Mar Apr May Jun
Jul
Aug Sep
Oct
Nov Dec
31
20
21
25
28
26
30
32
26
22
24
37
(i) 20, 21, 22, 24, 25, 26, 26, 28, 30, 31, 32, 37 in order
1B mark
The median is 26
(ii) Total no. = 31 + 20 + 21 + 25 + + 28 + 26 + 30 + 32 + 26 + 22 + 24 + 37 = 322
The mean =
322
= 26.833
12
FMNW
1M mark
To the nearest integer mean = 27
1A mark
___________________________________________________________________________
6
7
PTR
100 I
P
100
TR
100  3645
P
6  4 .5
I=
1M mark
1M mark
P = € 13,500
1B mark
__________________________________________________________________________________
2
(i)
1B mark
5 of 80 = 32 are green
(ii) What fraction of the crayons are red or yellow?
1
4 of 80 = 20 are red and24 are yellow
For 20 red or total = 44
1B mark
44 are red or yellow
14
Fraction =
44 11

80 20
1A mark
1B mark
(iii) Blue = 80 – ( 32 + 20 + 24) = 4
__________________________________________________________________________________
8
(i)
(ii)
Pattern
Tiles
1
1
2
4
3
9
For correct pattern
1B mark
For every 3 correct entries
1B mark
For all correct
1B mark
4
5
6
16
25
36
n2
1B mark
(iii)
__________________________________________________________________________________
If f(x) = x2 and g(x) = 2x,
9
(i) f(6) – g(–6) = 62 – 2(–6) or 6 2 – 2.– 6 or 36 + 12
1M mark
= 48
FMNW
1A mark
(ii) f(x) + g(x) = x2 + 2x
1M mark
= x(x + 2)
1A mark
__________________________________________________________________________________
A = 2 r 2  2 rh
10
A – 2r2 = 2rh
1M mark
h=
A  2r 2
2r
or equivalent
FMNW
1A mark
__________________________________________________________________________________
11
(i)
1
2
3
4
5
6
A (A, 1)
(A, 2)
(A, 3)
(A, 4)
(A, 5)
(A, 6)
E (E, 1)
(E, 2)
(E, 3)
(E, 4)
(E, 5)
(E, 6)
I
(I, 1)
(I, 2)
(I, 3)
(I, 4)
(I, 5)
(I, 6)
O (O, 1)
(O, 2)
(O, 3)
(O, 4)
(O, 5)
(O, 6)
U (U, 1)
(U, 2)
(U, 3)
(U, 4)
(U, 5)
(U, 6)
10 correct entries 1A mark
Another 10 entries 1A mark
(ii)
15 1

30 2
or equivalent
1B mark
(iii) For counting 9 possibilities
1B mark
9
3
or equivalent

30 10
1A mark
__________________________________________________________________________________
(i) 35% on € 450 = € 157.50
or 135% on € 450
1M mark
12
15
Total insurance = 450 + 157.50 = € 607.5(0)
FMNW
Method 1:New insurance = 112% of € 450 = € 504 or 12% of 450 + 450
1A mark
1M
mark
No-claim bonus = 25% of € 504 = € 126
or 25% of 450 or previous answer 1M mark
Amount to be paid = €(504 – 126) = € 378
1A mark
Method 2: 75% of € 450 = €337.50
112% of €337.50 or 12% of 337.50 + 337.50
= €378
M1, M1, A1
__________________________________________________________________________________
(ii)
13
(a) (i) Time = [(3x24) + 3] +
= 75
49
60
= 75.8167 hrs
(ii) Speed =
49
60
1M mark
To 3 places of decimal time = 75.817 hrs
distance
380000
=
km/hr
time
75.8167
Answer = 5012 km/hr
(b)
1A mark
1M mark
FMNW
1Aft mark
380000
Time =
44272
1M mark
1A mark
= 8.58(33) hrs
= 8 hrs 35 mins
1A mark
__________________________________________________________________________________
14
G
B
x
25°
y
z
A
D
E
F
C
EITHER
Triangle BCE is isosceles (or radii)
Hence angle x = 25°.
1M mark
1A mark
Angle y = 25° + 25° = 50° (exterior angle of triangle or angle at centre)
1M mark
1A mark
1M mark
Angle ABE = 90° + 25° = 115°
For using 90°
°
°
°
In Δ ABE, angle z = 180 – (115 + 25) or z = 180 – (90 + 50) °
For using either angle sum of triangle or angle between radius and tangent
1M mark
angle z = 40°
1A mark
__________________________________________________________________________________
16
15
(ii)
BC2 = AC2 – AB2 (Pythagoras’ Theorem)
= 132 – 122 = 169 – 144 = 25
BC = 5(cm)
Area of Δ ABC = ½ (5x12) (including correct substitution)
= 30cm2
1B mark
1M mark
1A mark
Rotational symmetry of 90° or Rotational symmetry of order 4
1A mark
(i)
(iii) the area of the tile
= 24 x 24
= 576cm2
1M mark
1A mark
FMNW
(iv) Horizontal line
1A mark
Vertical line
1A mark
Subtract 1 mark if any extra lines of symmetry are added
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16
For scale drawing ,
for using the correct given scale
1A mark
for placing letters correctly
1M mark
for correct diagram
1A mark
(i) BS = 2.5 ± 0.1cm representing 250m ± 10m or just for 2.5cm or 4.3cm
BD = 4.3 ± 0.1cm representing 430m ±10m .
or either 250m or 430m
FMNW
1M mark
1A mark
.
(ii) Bearing of B from Y is 263° ± 5°
For measuring just inner angle between 4o
and 10o
1A mark
For giving final answer
FMNW
17
1A mark
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17
(i) the gradient of the line l1 =
6  0
 3 .
0  2
1B mark
l1
(ii) the equation of the given line l1 is y = – 3 x + 6. (gradient or intercept)
1M mark
Equation exact (follow through from (i))
1Aft mark
1M mark
(iii) the equation of a line l2 is y = – 3 x – 3.
follow through from (i)
1Aft mark
(iv) Drawing: For each correct intercept on axes
2A marks
If the line is drawn correctly from wrong equation in (iii), then subtract 1A mark
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18
(i) Let a and b be the number of hours assigned to machines A and B respectively.
a + b = 24 ...........(i)
1A mark
50a + 40b = 1000 ...(ii)
1A mark
(ii)
(iii)
Eqtn (i)  40 :
Subtracting:
40a + 40b = 960
= 40 or reduce to 1 equation
10a
a = 4 hrs or b = 20 hrs
(iv) Cost is €(30x4) + €(20x20)
= €120 + €400 = €520
(v) Machine A takes
FMNW
1000
= 20 hrs.
50
1M mark
1A mark
1A mark
Cost = €(20x30) = € 600
Extra cost = €(600 – 520) = € 80
Percentage increase =
1M mark
1A mark
80 100
% = 15.384%
520
To 1 decimal place answer = 15.4(%)
18
For both steps
1M mark
1A mark
19
(i) Vol =
92
   4 2  22 = 1017.373 cm3
100
For correct volume of cylinder
. For 92% of volume
1M mark
1M mark
Answer = 1017cm3 or more accurate
(ii) No. of lamps =
For correct answer
1A mark
1017.37
125
1M mark
= 8.14 but correct answer = 8 lamps
(iii) Let x be number of lamps
24x + 125x = 1017(.37)
149 x = 1017(.37)
1Aft mark
For 149 or 149x
1M mark
For vol/149
1M mark
x = 6.83
1Aft mark
giving number of lamps = 6
1A mark
Accept trial and error method
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20
(i) To show that the triangles, BCD and ABE are similar:
In Δs BCD, ABE
A
angle BDC = angle BEA = 30° (given)
B
x
1A mark
no reason needed here
x
angle BCD = angle ABE (alternate s)
or
30°
D
C
Remaining angles CBD = BAE
Reason must be given
1A mark
Hence
Δs
are
equiangular,
so
similar
or
30°
Δ BCD and ΔABE are similar in that order
Or the remaining angles are equal.
1A mark
E
(ii) To calculate the length of CD:
BE – CE = BC
1M mark
BC = 10 – 6 = 4cm
BC CD BD
=
(=
)
AB BE AE
so that CD =
1M mark
4 10
= 8cm
5
FMNW
19
1A mark