Marking Scheme - um.edu.mt€¦ · 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

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.


Page 3 of 29

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


Page 4 of 29

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


Page 5 of 29

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


Page 6 of 29

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


Page 7 of 29

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


Page 8 of 29

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


Page 9 of 29

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


Page 10 of 29

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


Page 11 of 29

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


Page 12 of 29

Paper 2A


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


Page 13 of 29

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


Page 14 of 29

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


Page 15 of 29

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


Page 16 of 29

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


Page 17 of 29

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


19.6% 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


19.6% 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


Page 18 of 29

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


Page 19 of 29

Ħaġar Qim

Ta’ Kandja

Tal-Providenza

(a)

(b)

(c)

(d)


Page 20 of 29


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


Page 21 of 29

Paper 2B


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


Page 22 of 29

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


Page 23 of 29

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


Page 24 of 29

(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


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°


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°


FBE = 𝑥 = 62°


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


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


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


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


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

Documents

Marking Scheme - um.edu.mt€¦ · 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