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Marking Scheme SEC Mathematics
Main Session 2019
MATSEC Examinations Board
Marking Scheme (Main Session 2019): SEC Mathematics
Page 2 of 29
Marking schemes published by the MATSEC Examination Board are not intended to be standalone documents. They are an essential resource for markers who are subsequently monitored through a verification process to ensure consistent and accurate application of the marking scheme.
In the case of marking schemes that include model solutions or answers, it should be noted that these are not intended to be exhaustive. Variations and alternatives may also be acceptable. Examiners must consider all answers on their merits, and will have consulted with the MATSEC Examinations Board when in doubt.
Marking Scheme (Main Session 2019): SEC Mathematics
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Paper I Section A – Non Calculator Section
QN ANSWER
1. 52 034
2. 44
3. 1 hour 35 min or 95 mins
4. C or 9 and 11
5.
𝑦 = 𝑥+ 3
2 or 𝑦 =
1
2(𝑥 + 3) or 𝑦 = (𝑥 + 3) 2 or 2𝑦 = 𝑥 + 3
or 𝑦 =𝑥
2+ 1.5
Do not accept: 𝑦 = 𝑥 + 3 ÷ 2
6. 5
3 or 1 2
3 or 1
70
105 or 1
175
105 or 1
14
21 or 1. 6 or more accurate
7. 7.32054 × 105
8. 116° or 116
9. 52
10. 5
4
11. €480 or 480
12. 37
13. 9
14. 𝑦 = 2𝑥 + 3
15. 6𝑥3 − 15𝑥
16. 𝑎
𝑏2 or 𝑎𝑏−2 or (𝑎 × 𝑏−2) or (𝑎 × 1
𝑏2) or equivalent
17. €200 or 200
18.
𝑥 = 9𝑦 − 5
2 or 𝑥 =
1
2(9𝑦 − 5) or 𝑥 = (9𝑦 − 5) ÷ 2
9 × 𝑦 − 5
2 or
9𝑦 − 5
2 or
𝑦9 − 5
2 or equivalent
Do not accept: 9y – 5 ÷ 2
19. 21° or 21
20. C or AAS
Marking Scheme (Main Session 2019): SEC Mathematics
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Paper I Section B – Non Calculator Section
QN Solution Criteria Marks
1(a) 35 ml Accept answers in the range 34 – 35
ml B1
5
50 ml Accept answers in the range 49 – 50
ml B1
Cylinder Q …….. Cylinder P B1
1(b) −3.9, −0.37, 0.378, 0.38 Award M1 for any two values in
correct position OR M2
2(a) 9 × 4
0.2 + 0.4
Award M1 for 9 or 9.2, 3.9 or 4
Award M1 for (0.2 + 0.4) or 0.6 M2
5
60 Accept answers in the range 58.5 –
61.4 A1
2(b) 60.723559… M1
60.723559… − 60 = 0.723559… Accept 0.72 or more accurate
OR 60 − 60.723559… = −
0.723559….
Accept ans in the range 0.67 to 2.22
or −2.22 to −0.67 A1
3(a) 2
5 × 125 OR 40% of 125 M1
7
50 ml
A1
Marking Scheme (Main Session 2019): SEC Mathematics
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3(b) (i) 120 3 = 40 ml (Method 1) M1
40 ml × 5 M1
= 200 ml A1
OR OR
120 3 = 40 ml (Method 2) M1
40 ml × 2 = 80 ml (vinegar)
120 ml + 80 ml adding oil and vinegar M1
= 200 ml A1
OR OR
Working on vinegar only
(1000 × 3) 2 = 1500 ml of oil M1
M0
A0
(ii) Vinegar used = 200 – 120 = 80 (following Method 1 in (i))
1000 − 80 M1
920 ml of vinegar A1
OR OR
1000 − 80 (following Method 2 in (i)) M1
920 ml of vinegar A1
4 correct method for eliminating x or eliminating 𝑦 M1
4 𝑦 = −1 or x = 3.5 o.e. A1
correct method for eliminating 𝑦 or eliminating x M1
x = 3.5 o.e. or 𝑦 = −1 A1
5(a) 3(2𝑎 + 1) − 4(3𝑎 − 2)
12 for correct numerator M1
10 6𝑎 + 3 − 12𝑎 + 8
12
for correct expansion of both
brackets
(including −4 × −2 = 8)
M1
11− 6𝑎
12 or
−6𝑎+ 11
12 for collecting like terms A1
Marking Scheme (Main Session 2019): SEC Mathematics
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5(b) (i) 9 − 3 + 5 M1
= 11 A1
(ii) f(2) = 4 − 2 + 5 = 7 Working must be shown to show M1
f(−1) = 1 + 1 + 5 = 7 f(2) = f(−1) M1
f(2) = f(−1) = 7
(iii) 𝑥2 – 𝑥 + 5 = 37 − 𝑥 − 𝑥2 For trial and error method: M1
2𝑥2 = 32 OR 2𝑥2 − 32 = 0 Award M1 M1 A0 if only 𝑥 = 4
obtained M1
𝑥2 = 16 Award 3 marks if 𝑥 = 4 obtained
𝑥 = 4 seen A1
6(a) Volume of prism = 1
2 (30 × 40) × 80 M1
8
= 48 000 cm3 Ignore units A1
6(b) 1
2 (30 × 40) × 2 = 1200 cm2
Award for correct area of two
triangles
Accept also (30 × 40)
M1
50 × 80 = 4000 cm2 Hypotenuse = 50 cm seen or implied M1
30 × 80 = 2400 cm2
40 × 80 = 3200 cm2 Award for both 2400 cm2 and 3200
cm2 M1
Total SA = 1200 + 2400 + 3200 +
4000
Total surface area = 10 800 cm2 A1
6(c) 10 800 × 1.3 Award for ans in (b) in cm2 × 1.3 M1
= 14 040 g
= 14 kg or 14 000 g
This ft mark is only awarded for
correct working which is also
followed correct rounding to the
nearest kg.
A1ft
Marking Scheme (Main Session 2019): SEC Mathematics
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7(a) 49 = 40 (basic pay) + 9 (overtime)
7
40 × 9.50 = €380 (basic pay) Multiplies 40 by 9.50 M1
Overtime pay per hr = 5
3 × 9.50 =
€15.83
M1
9 × €15.83 9 × overtime pay per hour obtained
previously M1
= €142.50 (overtime)
Total = 380 + 142.50 = €522.50 A1
7(b) 506.66 – 380 = 126.66 Subtracting 380 from 506.66 M1
Overtime pay per hr =
5
3 × 9.50 =
€15.83
126.66 15.83
dividing overtime payment by
overtime rate per hour …
mark is still given if the overtime rate
per hour as calculated is incorrect
M1
= 7.9995…or 8.00126
8 hours
40 + 8 = 48 hours
48 hrs. Accept 47.8 hours
Accept 47 h 48 mins or more
accurate
A1
OR
€522.50 − €15.83 = €506.67
49 − 1 = 48 hours
3
Marking Scheme (Main Session 2019): SEC Mathematics
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8(a)
M1 for three correct coordinates
M2 for all correct M2
10
Graph – correct labelling of axes M1
Graph – correct plotting of points and
curve
M1 for three correct points
plotted M2
Note:
In awarding the M2 marks for plotting… plotting is considered correct if
candidates plot correctly the coordinates they obtained… even if these are
incorrect.
8(b) 𝑦 = 9 OR (0, 9) Award no marks for (9, 0) B1
8(c) Any two coordinates on straight line 𝑦 =
6 − 𝑥 M1
Plotting line 𝑦 = 6 − 𝑥 A1
Marking Scheme (Main Session 2019): SEC Mathematics
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8(d) 𝑥 = −1.3 0.2 , 𝑦 = 7.3 0.2 x = −1.3027…. (by calculation) B1
𝑥 = 2.3 0.2 , 𝑦 = 3.7 0.2 x = 2.3027…. (by calculation) B1
Marking Scheme (Main Session 2019): SEC Mathematics
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9(a) Sum of interior angles = (8 − ) ×
180° or (16 −4) × 90° M1
8
= 1080°
one interior angle = 1080° 8 M1
= 135° A1
OR OR
One exterior angle = 360° 8 M1
= 45°
One interior angle = 180° − 45° M1
= 135° A1
9(b) (i) 𝑏 = 360° – (75° + 135°) M1
𝑏 = 150° ft for b less than 180° A1 ft
(ii) one exterior angle = 180° − 150°
= 30° M1
360° 30° M1
= 12 sides A1
OR OR
180 − 135 = 45
75 − 45 = 30° M1
360° 30° M1
= 12 sides A1
OR OR
180𝑛 − 360 = 150𝑛 M1
180𝑛 − 150𝑛 = 360
30𝑛 = 360 M1
𝑛 = 12 sides A1
10(a) (i) 90° + 27° = 117° B1
10 (ii) 90° − 27° = 63° Accept any valid method M1
360° − 63° = 297° A1
Marking Scheme (Main Session 2019): SEC Mathematics
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10(b) Sin B = 4.16
6.45 M1
Sin B = 0.64496…
ABC = 40.16… = 40.2° Accept 40° or more accurate A1
OR OR
Cos C = 4.16
6.45 = 0.64496…
ACB = 49.83…
ABC = 90 − 49.83… M1
ABC = 40.16… = 40.2° Accept 40° or more accurate A1
10(c) Cos 27° = AD
4.16 M1
AD = 4.16 × Cos 27° M1
AD = 3.70658… = 3.7 km Accept 3.7 or more accurate A1
10(d) 270° − 63° M1
= 207° A1
OR OR OR
180° + 27° 360° − (90°
+ 63°) M1
= 207° = 207° A1
11(a) (i) September B1
6
(ii) 950 animals Accept answers in the range 940 −
960 B1
11(b) Total animals = 950 + 750 + 580 +
600 (950 10), (750 10), (580 10) M1
Total animals = 2880 Answer may occur in range 2850 −
2910
Dogs represented by 158° 2° seen or implied M1
No of dogs = 158
360 × 2880
Award for angle measured
360 × total
animals M1
No of dogs = 1264 Accept answers in the range 1235 to
1294 A1
Marking Scheme (Main Session 2019): SEC Mathematics
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Paper 2A
QN Solution Criteria Marks
1(a) 𝑥 = 3 ± √9 − (4 × 2 × −7)
4 for substitution M1
13
𝑥 = 3 ± √65
4 seen or implied M1
𝑥 = −1.27 or 𝑥 = 2.77
Award only for answer correct to 2
d.p.
Exact values -1.26556…. or
2.76556….
A1
1(b) (i) (3𝑎 − 2)(2𝑎 − 1) (3𝑎 2)(2𝑎 1) M1
Give full marks if given in terms of x A1
Award M1 A0 if values for a are given
after factorization
Award M1 A1 for 6 (𝑎 − 2
3) (𝑎 −
1
2)
No marks for (𝑎 − 2
3) (𝑎 −
1
2)
(ii) 2(𝑥6 − 4𝑦4) Correct factorisation by 2 M1
2(𝑥3 − 2𝑦2)( 𝑥3 + 2𝑦2)
Award M1 for any of the following:
(𝑥3 − 2𝑦2)( 𝑥3 + 2𝑦2)
(2𝑥3 − 4𝑦2)( 𝑥3 + 2𝑦2)
(𝑥3 − 2𝑦2)(2𝑥3 + 4𝑦2)
M1
1(c) (i) 𝑐 = 4𝑛 OR 4𝑛 OR 𝑛 =𝑐
4 OR
𝑐
4
forming equation ① seen or
implied M1
(𝑐 − 8) = 10(𝑛 − 8) forming equation ② or equivalent M1
4𝑛 − 8 = 10𝑛 − 80 substituting eqn ① in eqn ② M1
𝑛 = 12 years
𝑐 = 4 × 12 = 48 years A1
No marks awarded for incorrect
method leading to correct answer
(ii) 48 + 𝑡 = 3(12 + 𝑡)
M1
2𝑡 = 12
𝑡 = 6 years No marks awarded for answer only
and no working. A1
Marking Scheme (Main Session 2019): SEC Mathematics
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2(a)
correct enlargement of shape and
distance from O, by scale factor 2.5
correct orientation
M1
accurate shape and position A1
2(b) (i) correct orientation M1
accurate diagram A1
(ii) correct translation M1
accurate diagram A1
(iii) Rotation of 180° OR Enlargement
by −1 B1
about the point (12, 9) or reference to any point X at (12,
9) B1
3(a) 35
100 × 13600 M1
9
= €4760 A1
3(b) 13600 − 4760 = €8840 M1
€8840 36 = €245.56 or 24556c accurate answer only A1 ft
3(c) 13 600 × 0.852 M1
(13 600 × 0.852) × 0.93 M1
= €7163.15 €7163 or more accurate A1
Marking Scheme (Main Session 2019): SEC Mathematics
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3(d) Loss = 13 600 − 7500 = €6100
6100
13600 × 100 M1
= 44.9% Accept 45% or more accurate A1
OR OR
7500
13600 × 100 = 55.15…%
100 − 55.15 = 44.85… M1
= 44.9% Accept 45% or more accurate A1
4(a) 242 = 322 + 212 − (2 × 32 × 21)Cos P M1
7
Cos P = 322 + 212 − 242
2 × 32 × 21 M1
Cos P = 0.66145…
P = 48.5888… = 48.6°
Accept ans in the range 48.6° to
48.7° and 49°
A1
4(b) 58
𝑆𝑖𝑛 48.588 …=
𝑃𝑅
𝑆𝑖𝑛 53 for substitution M1
PR = 58×𝑆𝑖𝑛 53
𝑆𝑖𝑛 48.588… = 61.762… for PR subject of the formula M1
TR = PR − PT = 61.762… − 21 for subtracting PT from PR M1
TR = 40.762…
Accept answers in the range
40.66 cm to 40.8 cm and
40.4 cm or more accurate
A1
OR OR
Using similar triangles (INCORRECT)
Correct subject of the formula for PR
TR = PR − 21
Answer
M0
M1
M1
A0
5(a) (i) 17 , 28 obtaining 5th and 6th terms M1
10
Sum = 68 A1
(ii) 4 × (5th term) = 4 × 17 = 68 M1
S6 = 68 A1
OR OR OR
68 4 = 68 17 M1
5th term = 17 = 4 A1
Marking Scheme (Main Session 2019): SEC Mathematics
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5(b) (i) 5th term = (𝑎 + 𝑏) +(𝑎 + 2𝑏) seen or implied M1
5th term = 2𝑎 + 3𝑏 A1
6th term = (𝑎 + 2𝑏) +(2𝑎 + 3𝑏) = 3𝑎 +
5𝑏 A1
(ii) S6 = 𝑎 + 𝑏
+(𝑎+𝑏)+(𝑎+2𝑏)+(2𝑎+3𝑏)+(3𝑎+5𝑏)
seen or implied Sum of 6 terms
ONLY M1
S6 = 8𝑎 + 12𝑏 Correct sum for any algebraic sum
given M1
S6 = 4(2𝑎 + 3𝑏) LHS = RHS M1
S6 = 4 × (5th term)
OR OR
For answers of the form 8𝑎+12𝑏
2𝑎+3𝑏 Award M2 for 8𝑎 + 12𝑏 M2
Award M1 for (2𝑎 + 3𝑏) = 4 M1
6(a) Tony is not correct B1
7
since HA : HB = √13
: √83
OR Reference to linear ratios M1
HA : HB = 1 ∶ 2
6(b) AA : AB = 12 : 22 = 1 : 4 seen or implied M1
AA = 96 4 Award for 4 only M1
AA = 24 cm2 A1
6(c) WA : WB = VA : VB = 1 : 8
No marks awarded for working
involving length or area
WB = 15 × 8 M1
WB = 120 kg A1
Marking Scheme (Main Session 2019): SEC Mathematics
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7(a) (i) Cos MOP = 16
32=
1
2 M1
12
MOP = 60°
POQ = 2 × 60° = 120° Accurate answer A1
(ii) Area = 1
2 × 32 × 32 × Sin 120° M1
Area = 443.405… cm2 = 443.4 cm2 Accept 443 cm2 or more accurate A1
OR OR
MP = √322 − 162 = 27.7128…
Area = 1
2 × (27.7128… × 2) × 16 Award M1 A0 for using MP as base M1
Area = 443.405… cm2 = 443.4 cm2 Accept 443 cm2 or more accurate A1
7(b) (i) Cross-sectional area of tank
Award M1 for area of sector with
ANY angle and correct radius
substituted
M1
= area of sector + area ΔPOQ
= (240
360× 𝜋 × 322) + 443.405…
Award M1 for addition of sector +
triangle OR circle – (sector –
triangle)
M1
= 2588.0655… cm2
Volume = 2588.0655… × 120 Any area × 120 M1
Volume = 310 567.87… cm3
No. of litres = 310 567.87… 1000 Volume 1000 M1
No. of litres = 310.56787… = 311
litres
7(b) (ii) Cross-sectional area = 𝜋 × 322 Total Vol = 𝜋 × 322 × 120 M1
= 3216.99… = 38 6038.90… cm3
Empty area = 3216.99… − 2588.0655… Empty Vol = 386 038.90… − 310
567.87… M1
Empty area = 628.925… Empty Vol = 75 471.03… cm3
628.925…
3216.99… X 100
75471.03…
386038.90… X 100 M1
= 19.55% = 19.6% or more accurate Accept ans in the range 19.4% −
19.6% A1
OR Accept also 19% or 20% OR
Marking Scheme (Main Session 2019): SEC Mathematics
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OR OR
2588.0655…
3216.99… X 100 = 80.449…%
310 567.87…
386 038.90… X 100 = 80.449…% M2
100 – 80.449 100 – 80.44 M1
= 19.55% = 19.6% or more accurate Accept ans in the range 19.4% −
19.6% A1
OR Accept also 19% or 20% OR
Area of sector = (120
360× 𝜋 × 322) M1
Area of segment = 1072.33… −
443.405… i.e empty area M1
= 628.925…
628.925…
3216.99… X 100 Accept also 19% or 20% M1
= 19.55% = 19.6% or more accurate Accept ans in the range 19.4% −
19.6% A1
OR OR
311 litres
386 litres X 100 = 80.569…% M2
100 – 80.569… Accept also 19% or 20% M1
= 19.43…% Accept ans in the range 19.4% −
19.6% A1
8(a) 10 = 𝑝 𝑞2−2 = 𝑝 𝑞0 OR 10 = 10 𝑞2−2 M1
9
10 = 𝑝 × 𝑞0 OR 10 = 𝑝 × 1 OR 10 = 10 𝑞0 OR 10 = 10 ×
1 M1
𝑝 = 10 A1
8(b) 0.01 = 10 𝑞5−2 for substituting M1
0.01
10 = 𝑞3 M1
1
1000 = 𝑞3 OR 0.001
𝑞 = √1
1000
3 OR 𝑞 = √0.001
3 Award for cubic root M1
𝑞 = 1
10 OR 0.1 Award A0 for 0.1 A1
Marking Scheme (Main Session 2019): SEC Mathematics
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8(c) 𝑦 = 10 × (1
10)
7 − 2 for substituting M1
𝑦 = 10 ×
1
105
𝑦 =
1
10 000 OR 10−4 OR 1 × 10−4
OR 1
104 OR 0.0001
ft from (b)
A1 ft
9 Refer to diagram on page 8
8
9(a) Drawing perpendicular bisector M1
Accurate construction (including arcs) A1
9(b) Circle, centre Tal-Providenza M1
Radius = 2 cm 0.2 cm A1
9(c) Circle, centre Hagar Qim, radius 3 cm
0.2 cm A1
9(d)
M3
Marking Scheme (Main Session 2019): SEC Mathematics
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Ħaġar Qim
Ta’ Kandja
Tal-Providenza
(a)
(b)
(c)
(d)
Marking Scheme (Main Session 2019): SEC Mathematics
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QN Solution Criteria Marks
10(a) graph Q B1
4 10(b) graph S B1
10(c) graph P B1
10(d) graph R B1
11(a) (i) €1100 B1
13
(ii) €2600 B1
(iii) €3200 − 1900 For UQ – LQ with substitution M1
= €1300 A1
(iv) No
Awarded only if reason indicates
Mario cannot reach this conclusion
A1
because the boxplot shows
information
on the monthly salary not on the
number of employees.
M1
11(b)
(i) First set:
2
5,
1
5,
2
5
Second set: 3
5, 0,
2
5 Accept 0/5
Third set: 3
5,
1
5,
1
5
Award B1 for 5 in denominator B1
Any one completely correct set B1
All correct B1
(ii) 3
6×
2
5 M1
6
30 =
1
5 Accept also
6
30 A1 ft
(iii) 1 − (6
30+
2
30)
OR Add any relevant probabilities
M1
1 − ( 8
30)
22
30 =
11
15
Accept also 22
30
OR 0.73 OR 73% or more accurate
A1
Marking Scheme (Main Session 2019): SEC Mathematics
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Paper 2B
QN Solution Criteria Marks
1(a) 19, 17 both correct B1
3 1(b) 128, 256 both correct B1
1(c) 2.75, 3.25 both correct B1
2 6.15 - 3.55 = 2 h 20 min Award for evidence that 1 hr = 60
min M1
3
2 h 20 min = 120 + 20 = 140 min
140 25 = 5.6 episodes Award M1 M1 A0 for answers
such as M1
5 episodes 5 episodes and …..mins of
another A1
OR OR
3:55 + 25 min 4.20 correct carrying M1
4.20 + 25 min 4.45 + 25 min 5:10
5:10 + 25 min 5:35 + 25 min 6.00 repeated addition M1
5 episodes A1
3(a) 0.000 012 B0 for 000 012 (i.e. no decimal pt) B1
3 3(b) Cell type R OR 7.5 × 10-6 B1
3(c) 1.5 × 10−4
5.0 × 10−5 = 3 B1
Marking Scheme (Main Session 2019): SEC Mathematics
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4(a)
(i)
Award M1 for completing the
reflection on one side, i.e.
(a and b) or (c)
M1
M1
5
(ii)
Completing reflection in one line
of symmetry, i.e.
(a and d) or (b and c)
Completing reflection in second
line of symmetry
M1
M1
4(b) 4 B1
5(a) 3
100 × 6000 Award M1 for
103
100 × 6000 M1
6
= €180 Award A0 for 6180 A1
5(b) 15
100 × 180 Award M1 A0 for
15
100 × 6180 M1
= €27
ft awarded only for working out
tax on interest gained
A1 ft
Award M0 A0 for the following:
85% of 180 or 15% of 6000
5(c) 6000 + 180 − 27 6000 + (a) - (b) M1
= €6153 ft awarded only if ans (c) > €6000 A1 ft
a
b
c
a
b
d
c
Marking Scheme (Main Session 2019): SEC Mathematics
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6
Award full marks for arrows in correct position.
Correct position of 0.4 arrow – anywhere in first half of the box
but not on the 6/16 position.
B1
B1
2
7(a)
Walk along Sage Street, take the second turning on the left (or anticlockwise).
Walk along Mint Street, take the first turning on the left (or anticlockwise).
Walk along Fennel Street, take the second turning on the right (or clockwise).
Walk along Laurel Avenue, until you reach E.
Award B1 for left
Award B1 for one complete statement i.e first turning on the left OR second
turning on the right
Award B1 for all correct
B1
B1
B1
6
7(b) 6.4 + 3.5 + 6.6 + 2.1 = 18.6 cm 0.4 cm M1
18.6 × 50 M1
930 m 20 m Accept ans in the range 910 m −
955 m
A1
FMNW
OR OR
6.4 cm × 50 = 325 m (5 m) Award for at least 3 measurements
in cm M1
3.5 cm × 50 = 175 m (5 m)
6.6 cm × 50 = 330 m (5 m)
2.1 cm × 50 = 105 m (5 m) Award for × 50 M1
Total = 935 m (20 m) Accept ans in the range 910m −
955 m A1
FMNW
OR OR
0 1
0.4 5
8
Marking Scheme (Main Session 2019): SEC Mathematics
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(6 × 50) + (3 × 50) + (6 × 50) + (2 × 50) + (1.5 ×
50) = 925
Award if candidate works with
approximate lengths for the blocks
and streets.
M1
M1
A1
8(a) 120 × 4 M1
5
= 480 km A1
8(b) 480 90 M1
= 5.3 or 51
3
1
3 of 60 mins = 20 mins M1
5 h 20 min Do not accept 5.20 A1
9(a) 945 = 3 × 3 × 3 × 5 × 7 Award correct method for finding
prime factors.
M1
7
945 = 33 × 5 × 7 A1
9(b)
(i) multiples of 3 = 3, 6, 12, 15, 18, 21, 24, 27,
30,
33, 36.
OR any other correct method M1
multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32,
36, …
Award M1 A0 for a correct
multiple, even if not least
multiple.
multiples of 9 = 9, 18, 27, 36, … e.g. 72, 108, 144 ...
multiples of 18 = 18, 36, …
LCM = 36 A1
(ii)
3
4=
27
36 ,
13
18=
26
36,
7
9=
28
36 ,
2
3=
24
36
Award M1 for each two correct
equivalent fractions M2
2
3,
13
18,
3
4,
7
9 All correct and in ascending order
Award A1 for equivalent fractions
in correct ascending order.
A1
10(a) 10𝑥 + 2𝑦 + 5𝑥 − 3𝑦 expanding brackets (10𝑥 + 2𝑦) M1 4
15𝑥 − 𝑦 both terms correct A1
Marking Scheme (Main Session 2019): SEC Mathematics
Page 25 of 29
10(b) 5𝑎4
𝑏 Award B1 for correct numerator B1
Award B1 for correct
denominator B1
OR OR
5 𝑎4 𝑏−1 Award B1 for correct 5 𝑎4 B1
Award B1 for correct 𝑏−1 B1
11 𝑥 = 62° B1
5
alternate angles
correct reason
Do not accept Z angles
M1
FBE = 𝑥 = 62°
(base angles of isosceles triangle)
ft for incorrect value of 𝑥.
M1
𝑦 = BFE + FBE = 62° + 62°
(exterior angle of triangle)
M1
𝑦 = 124° A1 ft
OR OR
FBE = 𝑥 = 62°
(base angles of isosceles triangle)
M1
BEF = 180° − (62° + 62°) = 56°
(angles in a triangle)
𝑦 = 180° − 56°
(angles on a straight line)
M1
𝑦 = 124° A1 ft
OR OR OR
FBE = 𝑥 = 62°
(base angles of isosceles triangle)
FBE = 𝑥 = 62°
(base angles of isosceles triangle)
M1
CBE = 180° − (62° + 62°) = 56°
(angles on a straight line)
ABE = 62° + 62°
𝑦 = 180° − 56°
(interior angles)
ABE = BED = y
(alternate angles)
M1
𝑦 = 124° 𝑦 = 124° A1 ft
Marking Scheme (Main Session 2019): SEC Mathematics
Page 26 of 29
12(a) 𝑝 + 11𝑝 = −2 × − 3
4
12𝑝 = 6 Award M1 for 𝑝 + 11𝑝 = 6 OR −2 × −3
12
M1
𝑝 = 1
2 or 0.5 A1
12(b) 𝑝(1 + 𝑞) = 𝑠𝑡 M1
𝑝 = 𝑠𝑡
1 + 𝑞 A1
13
4
Award B1 for correct properties of a square B1
Award B1 for correct properties of a parallelogram B1
Award B1 for correct properties of a kite B1
Award B1 for correct properties of a trapezium B1
14(a)
6 Award B1 for any 3 outcomes
Accept answers presented as a tree
diagram B3
14(b) 1
9 o.e. B1
14(c) Numerator = 4 M1
OR 4 possible outcomes identified (20 + 10), (20 + 5), (20 + 0), (5 +
10)
4
9 o.e. A1
15(a) gradient = 8 −3
4 + 6 o.e. M1
4 =
1
2 A1
Win €20, Win €10 Win €5, Win €10 No Win, Win €10
Win €20, Win €5 Win €5, Win €5 No Win, Win €5
Win €20, No Win Win €5, No Win No Win, No Win
Marking Scheme (Main Session 2019): SEC Mathematics
Page 27 of 29
15(b) 𝑦 = 1
2𝑥 + 6 OR 𝑦 = 0.5 𝑥 + 6
for using (positive) gradient obtained
in (a) M1
OR for correct equation including A1 ft
2𝑦 = 𝑥 + 12 𝑦-intercept
16 a = 90° − 50° = 40° A1
6
angle between tangent and radius = 90° M1
b = 25° A1
angles in the same segment M1
c = 90° − 25° = 65° A1
angle in semicircle = 90° M1
17 Construction below
8
17(a)
Award for AB = 10 cm 0.1 cm
B1
17(b) Construction of 60° angle showing arcs M1
Accurate construction ( 1°) at A A1
AD = 8 cm 0.1 cm A1
17(c) Construction of 90° angle showing arcs M1
Accurate construction ( 1°) at B A1
BC = 9 cm 0.1 cm A1
17(d) CD = 6.4 cm 0.2 cm CD = 6.392 cm (by calculation) A1
18(a) Paved space = 1
2× 𝜋 × 62 M1
8 = 56.5486… m2 Accept 56.5 m2 or more accurate A1
Marking Scheme (Main Session 2019): SEC Mathematics
Page 28 of 29
18(b) height of trapezium = 6 M1
Area whole garden = 1
2(12 + 24) × 6 M1
= 108 m2 Award M0 M1 A0 for 18ℎ A1
18(c) Stephen is not correct A1
Area of flowerbeds = 108 − 56.5 = 51.5 m2 (b) − (a)
Half area of garden = 108 2 = 54 m2 (b) 2 Award for both
steps M1
Area flowerbeds (51.5m2) < Half Area garden (54m2) Conclusion comparing both areas M1
OR OR
Stephen is not correct A1
Area of flowerbeds = 108 − 56.5 = 51.5 m2 (b − a)
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑙𝑜𝑤𝑒𝑟𝑏𝑒𝑑𝑠
𝑎𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑙𝑓𝑔𝑎𝑟𝑑𝑒𝑛=
51.5…
108 = 0.47
(b − a) (b) Award for both
steps M1
0.47 < 0.5 Conclusion comparing both areas M1
Area flowerbeds < Half Area garden
OR OR
Stephen is not correct A1
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑣𝑒𝑚𝑒𝑛𝑡
𝑎𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑙𝑓𝑔𝑎𝑟𝑑𝑒𝑛=
56.5…
108 = 0.52 (a) (b) M1
area of pavement > Half Area garden Conclusion comparing both areas M1
Area flowerbeds < Half Area garden
19(a) In ΔPQR and ΔRST
Do not award marks for stating
PQR = RST (corresponding
angles) if given instead of one of
first two statements unless
proved.
5
QPR = SRT (corresponding
angles) M1
PRQ = RTS (corresponding
angles) M1
PQR = RST (3rd angle
equal)
ΔPQR is similar to ΔRST
A1
Marking Scheme (Main Session 2019): SEC Mathematics
Page 29 of 29
19(b) 33
44 =
24
RT M1
RT =
24 × 44
33
RT = 32 cm A1
20(a) constructs perpendicular bisector M1
6
accurate construction 90° 1° and lying at 0.1 cm from
mid-pt A1
20(b) radius 6 cm seen or implied M1
Constructs arc centre tal-Providenza, 𝑟 =
6cm A1
20(c) P to Ta’ Kandja = 3.6 cm 0.2 cm Accept ans in the range 3.4 − 3.8
cm M1
3.6 × 500
= 1800 m 100 m
OR 1.8 km 0.1 km
Accept answer in the range
1700 to 1900 m OR 1.7 km to 1.9
km
A1