MARKING SCHEME - University of Malta...MARKING SCHEME: SEC MATHEMATICS (MAIN SESSION 2018) 3 PAPER 1...

MARKING SCHEME

SEC MATHEMATICS

MAIN SESSION 2018

MARKING SCHEME: SEC MATHEMATICS (MAIN SESSION 2018)

2

Contents PAPER 1 .................................................................................................................................................................3

Section A – Non Calculator Section ..................................................................................................................3

Section B - Calculator Section ...........................................................................................................................4

PAPER 2A ........................................................................................................................................................... 11

PAPER 2B ........................................................................................................................................................... 19

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 verification/moderation processes to ensure consistent and accurate application of the marking scheme.

In the case of marking schemes which include expected 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.


3

PAPER 1 Section A – Non Calculator Section

QN ANSWER

1. €28.04 or 28.04

2. 23

12 or 1

11

12 or 1.9166…

3. 11.15 p.m. or 23:15 or quarter past eleven or 11.15

4. 130° or 130

5. 2

12 or

1

6 or

4

24

6. €24 or 24

7. 10

8.

14

24 or

7

12

Possible answers 14:24 or 7:12 14 of 24 or 7 of 12 58% or more accurate

9. 60° or 60

10. 1 (bottom) and 3 (top) both correct

11. 16

12. 125

13. B or valid reference to statement B

14. 1

15. 50 or 50

1

16. 320° or 320

17. 245° or 245

18. 11 or 22

19. 51 cm2 or 51 or (√512

)2

20. 22 years or 22


4

Section B - Calculator Section

QN Solution Criteria Marks

1(a) 20 5 = 4 B1

6

1(b) 2.75 × 60 M1

= 165 minutes A1

fmnw*

OR OR

120 + 45 M1

= 165 minutes A1

Note: 45 mins seen , award M1 A0

1(c) (50 × 1000) (60 × 60) M1 for converting km to m M1 for converting hours to seconds

M1 M1

= 13.9 m/s A1

2(a) 8x 8 6x 10 M1

10

2x 18 Accept 2(𝑥 9) A1

2(b) 3y2(5 + y) Award M1 for one correct factor Note: Factorizing followed by solving Award M1 A0

M1 A1

2(c) (3𝑥 − 1) − 3(𝑥 − 2)

6 Correct numerator matching denominator M1

3𝑥 − 1 − 3𝑥 + 6

6 Correct expansion M1

= 5

6 A1

2(d) 2x = 8 + 8 M1

2x = 16

2x = 24 M1

x = 4 A1

3(a) Asia Accept also 3.71 × 109 B1

7

3(b) Africa, America, Asia, Europe and Oceania 1 e.e.o.o.** B2

3(c) (1.23 × 109) (8.14 × 108) × 100 M1

151.1%

increase of 51.1% accept also 51% or more accurate A1

OR

(1.23 × 109) (8.14 × 108) = 4.16 × 108

(4.16 × 108) (8.14 × 108) × 100 M1

increase of 51.1% accept also 51% or more accurate A1


5

3(d) Adds values for population in 2016 seen or implied M1

7469101110 = 7.5 × 109 Award A1 for answer in standard form only Accept 7.5 × 109 or more accurate

A1 fmnw

4(a) XZY = 90° Identifies Z as 90° M1

9

ZYX = a and ZXY = 2a

3a + 90 = 180 or 3a = 90 seen or implied M1

a = 30° A1

OR OR

XZY = 90° M1

30 + 60 = 90° or 30 + 60 + 90 = 180° M1 A1

4(b) Sin 30° = 𝑋𝑍

40 OR using Sine rule

40

𝑆𝑖𝑛 90=

𝑋𝑍

𝑆𝑖𝑛 30 M1

XZ = 40 × Sin 30° M1

XZ = 20 cm A1

OR OR

Cos 60° = 𝑋𝑍

40 M1

XZ = 40 × Cos 60° M1

XZ = 20 cm A1

OR OR

Note: Accept responses that identify XZ = 20 cm (since XZ forms part of an isosceles triangle with sides as radii of length 20 cm)

M1 M1 A1

4(c) Cos 30° =

𝑌𝑍

40

M1

YZ = 40 × Cos 30° M1

YZ = 34.6410… cm = 34.64 cm Accept 34.6 or more accurate A1

OR OR

Sin 60° = 𝑌𝑍

40 OR using Sine rule

40

𝑆𝑖𝑛 90=

𝑌𝑍

𝑆𝑖𝑛 60 M1

YZ = 40 × Sin 60° M1

YZ = 34.6410… cm = 34.64 Accept 34.6 or more accurate A1

OR OR

Tan 60° = 𝑌𝑍

20 M1

YZ = 20 × Tan 60° M1

YZ = 34.6410… cm = 34.64 A1

OR OR

YZ2 = XY2 XZ2

YZ2 = 402 – 202 = 1200 M1


6

YZ = √1200 M1

= 34.6410… cm = 34.64 Accept 35 or more accurate f.t. from part (b)

A1 ft

5

Mark is not correct Mark is partially correct Mark is not always correct

Accept any of these statements or equivalent Mark is correct - Award B0

B1

3

Counter example:

Base angle = 32°

Vertex angle = 180 (32 × 2) M1

Vertex angle = 116° ≠ 74° A1

OR OR

If 32° is one of the base angles, then one of the other angles (base angle) must also be 32°.

2M

6(a) 12500 – 9000 = 3500 M1

7

15% of 3500 M1

= €525 A1

6(b) first €9000 0 tax

next €5500 15% tax

15% of 5500 = €825 M1

935 – 825 = 110

25% = €110

100% = €440 M1

Income = 9000 + 5500 + 440 M1

Income = €14 940 A1

7(a) 60° drawn at A ( 2°) Award even if uses protractor A1

8

All arcs drawn M1

Line AC = 10 cm ( 0.2 cm) M1

7(b) Arc/s centre B radius 9.5 cm ( 0.1 cm) M1

DE measured according to diagram M1

7(c) Angle bisector of any angle at B All arcs drawn

This M1 is awarded for any correct bisection of angle at B

M1


7

BF measured according to diagram Award only for F lying on line segment DE This M1 is awarded ONLY for F resulting from bisection of angle DBE

M1

EBF measured according to diagram

Award only for F lying on line segment DE This M1 is awarded ONLY for F resulting

from bisection of angle DBE

M1

8(a) Equation 1:

8

A+ C = 180° x + y + 18 + y + 2 = 180

Award M1 for evidence that opposite angles of cyclic quad add up to 180°

M1

x + 2y = 160 Simplifying equation M1

Equation 2: Option (i) for Equation 2

A + B + C+ D= 360° (x + y + 18) + (2x + 10) + (y + 2) + (3x + y) = 360

Award M1 for evidence that angles in a quadrilateral add up to 360°

M1

2x + y = 110 Simplifying equation M1

Option (ii) for Equation 2 OR

B + D= 180°

2x + 10 + 3x + y = 180

5x + y = 170

5x + y + x + 2y = 170 + 160

Award for adding both equations i.e. A + B + C + D= 360°

M1

2x + y = 110 Simplifying equation A1

8(b) Eliminating x or y M1

x = 20° or y = 70° A1

Substituting to obtain second unknown M1

y = 70° or x = 20° A1

OR OR

Subtracting equations: (A + B + C + D= 360°) (B + D= 180°)

M1

x = 20° A1

Substituting to obtain y M1

y = 70° A1

9(a) (i) 07:20 or 7.20 a.m. or 7.20 twenty past seven B1

8 (ii) 8:15 7:20 seen or implied M1

= 55 min A1

fmnw


8

9(b) refer to graph below Represents ferry journey B1

9(c) 26×30

60 = 13 km seen or implied M1

refer to graph below Represents waiting time on ferry M1

refer to graph below Represents journey from Mġarr to Wied il- Mielaħ

M1

9(d) 54 km B1


9

10(a) Refer to sample answer below. (Accept any other correct responses) Blue lines indicate line of reflective symmetry.

A perfectly symmetrical shape Line of symmetry seen or implied

M2

4 10(b) Third rectangle in place M1

Two other sectors in place M1

10a

10b


10

11(a) P(not from Mosta) =

60

100 =

3

5

Accept 0.6 or 60% or 3:2 o.e.*** M1 A1 fmnw

10

OR OR

P(from Mosta) = 40

100 =

2

5

M1

P(not from Mosta) = 1 - 2

5 =

3

5

Accept 0.6 or 60% or equivalent A1 fmnw

11(b) Balzan = 20% M1

Lija = x %

Attard = (x + 20)% M1

40 + 20 + x + x + 20 = 100 or 20 + x + x + 20 = 60 or x + x + 20 = 40

M1

x = 10 A1

fmnw

11(c) Mosta = 40% of 360 = 144° For any one correct angle (seen or implied)

M1

Balzan = 20% of 360 = 72° For any one correct angle

(seen or implied) M1

Lija = 10% of 360 = 36°

Attard = 30% of 360 = 108°

Correct construction of any two sectors

(1°) Correct construction of remaining

sectors (1°)

A1

A1

* full marks no working ** each error or omission *** or equivalent


11

PAPER 2A


1(a) 1000 Accept 1000 B1

9

1(b) 32𝑛 −1 = 34 M1

2𝑛 – 1 = 4

𝑛 = 2.5 o.e. A1

1(c) (i) A = 10 000 × (1.024)5 Substituting in formula or using multiplying factor

M1

A = 11 258.99907 seen or implied M1

A = €11 259 Award A1 for rounded answer only A1

Note: Ignore extra working

OR OR

after 1st year, A = (1.024 × 10 000) = 10240 Amount for 1st year M1

after 2nd year, A = (1.024 × 10 240) A = 10485.76

Using A = 10240 to work out amount for 2nd year

M1

after 3rd year, A = (1.024 × 10485.76) A = 10737.418...

after 4th year, A = (1.024 × 10737.418..) A = 10995.116...

after 5th year, A = (1.024 × 10995.116...) A = 11258.999...

A = €11 259 Correct to nearest euro A1

(ii) Interest = 11258.999... 10 000 = €1258.999…

M1

P = 100 × 1258.999…

2.4 × 5 M1

P = 10 491.6589 = €10 500

P = €10 500 Correct to nearest 100 euro A1 ft

Note: Award M0 M1 A0 for

P = 100 × 11258.999…

2.4 × 5 =

€93 824.99..


12

2

8

2(a) Reflection of Shape A in any vertical line M1

Accurate size and position A1

2(b) Reflection of Shape B Shape in correct orientation in 1st quadrant

A1

Accurate size and position A1

2(c) Rotation by 90° about O by 90° anticlockwise about O

by 270° about O by 270° clockwise about O

Award M1 if only ‘Rotation’ stated (anticlockwise understood) (clockwise understood)

M1 M1

2(d) Shape D is 1

2 the size of Shape A M1

Accurate drawing and positioning of Shape D A1

3(a) 5𝑥 − 1 ≥ 14 M1

13

5𝑥 ≥ 15

𝑥 ≥ 3 A1

fmnw

Note:

Ans only 𝑥 = 3 M0 A0

Ans only 𝑥 > 3 M1 A0

Working using ‘=’ throughout

Ans 𝑥 = 3 M1 A0


13

3(b) 𝑦(𝑥 − 1) = 4 − 6𝑥 Award also for 𝑥 − 1 (𝑦) = 4 − 6𝑥 (𝑦)𝑥 − 1 = 4 − 6𝑥

M1

𝑦𝑥 − 𝑦 = 4 − 6𝑥

𝑦𝑥 + 6𝑥 = 𝑦 + 4 Collect like terms 𝑥 on one side and correct signs

M1

𝑥(𝑦 + 6) = 𝑦 + 4

𝑥 =𝑦 + 4

𝑦 + 6 A1

3(c) 2(2𝑎 − 3) + 14

(𝑎 + 2)(2𝑎 − 3) OR

2(𝑎+2)(2𝑎−3)+14(𝑎+2)

(𝑎+2)2(2𝑎−3) M1

= 4𝑎 + 8

(𝑎 + 2)(2𝑎 − 3) OR

4𝑎2+16𝑎+16

(𝑎+2)2(2𝑎−3) M1

=

4 (𝑎 + 2)

(𝑎 + 2)(2𝑎 − 3)

Factorising completely the numerator

OR 4(𝑎+2)2

(𝑎+2)2(2𝑎−3)

M1

= 4

(2𝑎 − 3) A1

3(d) 𝑦 ∝ 𝑥3 Seen or implied 5 23

40 ?

M1

𝑦 = 𝑘𝑥3

𝑘 = 5

8 Obtaining the value of 𝑘 𝑥3 =

40 × 8

5 M1

40 = 5

8𝑥3 Substituting values in equation 𝑥 = √64

3 M1

𝑥3 = 64

𝑥 = 4 𝑥 = 4 A1

Award A0 for 𝑥 = ±4

4(a) Time (Upper bound) = 45.5 min OR 45 min 30 sec B1

7

Time (Lower bound) = 44.5 min OR 44 min 30 sec B1

OR Award B2 for

44.5 time 45.5

Award B2 for 44.5 < time < 45.5

4(b) Volume (Upper bound) = 255 litres B1

Volume (Lower bound) = 245 litres B1

4(c) Rate (Lower Bound) = 𝑉𝑜𝑙

𝑇𝑖𝑚𝑒 Seen or implied M1


14

Rate (Lower Bound) =

𝐿𝐵 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑉𝑜𝑙

𝑈𝐵 𝑣𝑎𝑙𝑢𝑒𝑓𝑜𝑟 𝑇𝑖𝑚𝑒 =

245

45.5

M1

= 5.3846… litres/min Accept 5.38 litres/min or more accurate

A1 ft

Note:

When both workings for the UB and the LB values for Rate are given and answer for LB is not identified award M1 M1 A0

5(a) RPQ = QRT Alternate segment theorem

OR working through the alternate segment theorem derivation

B1

8

5(b) PQR = QRT = 70° Award even if no reason supplied M1

PRQ = 180 (70 + 70) Award even if no reason supplied M1

Alternate angles (for PQR = 70°)

Angles in a triangle (for PRQ)

Interior angles (for PRQ = 40°)

Angles on a straight line (for PRQ = 40°)

Mentioning at least one correct reason Straight line (TR) extended on diagram

M1

PRQ = 40° A1

5(c) 𝑆𝑖𝑛 20° = 𝑥

8 (𝑥 =

1

2PQ) M1

𝑥 = 8 × 𝑆𝑖𝑛 20° M1

𝑥 = 2.73616…

PQ = 2𝑥

PQ = 2 × 2.73616 = 5.472… Accept 5.5 cm or more accurate

ft from incorrect PRQ in 5(b) A1 ft

OR OR

𝐶𝑜𝑠 70° = 𝑥

8 (𝑥 =

1

2PQ) M1

𝑥 = 8 × 𝐶𝑜𝑠 70° M1

𝑥 = 2.73616…

PQ = 2𝑥

PQ = 2 × 2.73616 = 5.472… Accept 5.5 cm or more accurate

ft from incorrect RPQ or PQR in 5(a) or 5(b)

A1 ft

Accept other valid methods

6(a) S.A. of sphere = 4 (𝑥 + 3)2 o.e. B1

9

6(b) slant height = 5𝑥 M1

area of curved surface = (3𝑥)(5𝑥) = 15𝑥2 ft from incorrect slant height M1

area of base = (3𝑥)2 = 9𝑥2 M1

Total S.A of cone = 15𝑥2 + 9𝑥2 = 24𝑥2 Adding areas M1


15

6(c) 4 (𝑥 + 3)2 = 24𝑥2 Equating the two expressions OR expression (a) = expression (b)

M1

for (𝑥 + 3)2 = 𝑥2 6𝑥 9 M1

5𝑥2 6𝑥 9 = 0 o.e. A1

𝑥 = 6 ± √216

10

𝑥 = 2.06969… or 𝑥 = 0.86969…

𝑥 = 2.07 Accept 2.1 or more accurate

Award even if 0.87 is not crossed out

A1

OR OR

4 (𝑥 + 3)2 = 24𝑥2 Equating the two expressions

OR expression (a) = expression (b) M1

(𝑥 + 3)2 = 6𝑥2

𝑥 + 3 = (2.449…) 𝑥 A1

𝑥(1 2.449…) = or 𝑥(1 2.449…) = M1

𝑥 = 2.06969… or 𝑥 = 0.86969…

𝑥 = 2.07 Accept 2.1 or more accurate

Award even if 0.87 is not crossed out

A1

7(a) (i) 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10

Both 6 and 10 correct

B1

11

1 + 2 + 3 + 4 + 5 = 15 For 15 B1

(ii) 5th term = 1

25(5 + 1) M1

5th term = 15 A1

(iii) 𝑛 th term =

1

2𝑛(𝑛 + 1) = 120

M1

𝑛2 + 𝑛 240 = 0

(𝑛 15)( 𝑛 + 16) = 0 OR 𝑛 = −1 ± √961

2 M1

𝑛 = 15 or 𝑛 = 16 Award even if 16 is not crossed out

A1

Note: For answer only (no working), i.e. 𝑛 = 15 M0 M0 A0

7(b) 13 + 23 = 9 13 + 23 + 33 = 36 13 + 23 + 33 + 43 = 100

All correct B1

13 + 23 + 33 + 43 + 53 = 225 For 225 B1


16

7(c) terms of sequence B = the square of the terms of sequence A

Seen or implied M1

𝑛th term of Sequence B = [1

2𝑛(𝑛 + 1)]

2 OR

1

4𝑛2(𝑛 + 1)2, o.e. A1

8(a) 𝑥 8 6 4 3 2 1 −

1

2

1

2 1 2 3 4 6 8

𝑦 1 1.5 3 12 12 6 2 0.75

Award B1 for any 4 correct entries.

B2

12

8(b) Correct scale and labelling of 𝑥 and 𝑦 axis Plotting of points and drawing curve

B1 for 7 points correctly plotted B1 drawing of curve

B1 B2

8(c)

𝑦 becomes very large for +ve 𝑥 OR 𝑦 becomes very small for –ve 𝑥 Other responses: 𝑦 increases/decreases 𝑦 is inversely proportional to 𝑥

it changes from ve to +ve

Accept valid answer that refer to this response

M1

8(d) Evidence of one of the following:

obtaining two points on the line

obtaining equation 𝑦 = 1

2𝑥 − 1

M1

Correct drawing of line passing through

(0, 1) and (2, 0)

B1

8(e) At points of intersection

6

𝑥=

𝑥 − 2

2 M1

12 = 𝑥(𝑥 − 2) Any correct intermediate step M1

𝑥2 − 2𝑥 − 12 = 0

8(f) 𝑥 = 2.6 0.3 (Accept 2.9 𝑥 2.3) Accurate value 𝑥 = 2. 60555… B1

𝑥 = .6 0.3 (Accept 4.3 𝑥 4.9) Accurate value 𝑥 = 4.60555… B1

Note: Accept also coordinates

Answers derived algebraically are not accepted

9 For 9(a) and 9(b)

If only one statement + reason given RHS stated OR If two statements without reasons given RHS stated OR If three statements + two reasons given RHS stated

M1 M0 M1 M1 M0 M1 M1 M1 M1


17

9(a) In Δs YDM and XBM

9

YM = XM (given)

MYD = MXB = 90° (given) Accept angles on a straight line

MD = MB (diagonals bisect each other) M2

ΔYDM ΔXBM (RHS) RHS seen M1

9(b) In Δs AYM and ΔAXM

YM = XM (given)

MYA = MXA = 90° (given) Accept angles on a straight line

AM common M2

ΔAYM ΔAXM (RHS) RHS seen M1

9(c) Using sides of congruent triangles

DY = BX (ΔYDM ΔXBM from (a)) M1

YA = XA (ΔAYM ΔAXM from (b)) M1

DY + YA = BX + XA

DA = BA

ABCD is a parallelogram with equal sides Identifies property of rhombus A1

ABCD is a rhombus

OR OR

Using angles of congruent triangles

DMY = BMX (ΔYDM ΔXBM from (a)) M1

AMY = AMX (ΔAYM ΔAXM from (b)) M1

DMY + AMY = BMX + AMX

DMA + BMA = 180 2 = 90°

ABCD is a parallelogram with diagonals intersecting at 90°

Identifies property of rhombus A1

ABCD is a rhombus

10(a) Area small : Area large = R2 : (2R)2 o.e. M1

8 Area small : Area large = 1 : 4 A1

OR OR

Area small : Area large = R2 : (2R)2 o.e. M1

Area small : Area large = 1 : 4 A1

OR

R2 : 4R2 OR R2 : 4R2 M1 A1

For 1 : 3 Award M1 A0

10(b) area of inner circle = R2

area of one inner part = 𝜋𝑅2

5 M1


18

area of ring = 4R2 R2 = 3R2

For subtracting OR uses area of inner cake: area of ring = 1 : 3

M1

area of one outer part = 3𝜋𝑅2

15 =

𝜋𝑅2

5 For dividing to find one part M1

area of inner part = area of outer part

all parts have the same area A1

Note:

If assuming a particular value for R Award M3 A0

10(c) Area of large cake = (2R)2

= × 162 = 804.2477…

Area of one part = 804.2477… 20 M1

Area of one part = 40.2123… cm2 Accept 40.2 cm2 or more accurate A1

OR OR

area of one part = 𝜋𝑅2

5=

𝜋 × 82

5 Substituting in equation M1

Area of one part = 40.2123… cm2 Accept 40.2 cm2 or more accurate A1

11(a) P(one colour appears three times) = 1

4×

1

4×

1

4 Accept also 0.015625, 1.5625% M1

6

P(any colour appears three times) = 1

64 × 4 =

1

16 Accept also 0.0625, 6.25% A1

fmnw

11(b) Income = 640 × €0.50 = €320 M1

No. of winners = 1

16 × 640 = 40 players

Expenditure = 40 × €5 = €200 M1

Expected money to be raised = €320 €200

= €120 ft from answer in (a) A1 ft

11(c) Increase the number of sectors (colours)

Increase the number of wheels

Raise the fee

Reduce the winning prize Accept any valid reason B1


19

PAPER 2B


1(a) Rectangle Accept Parallelogram B1

4 1(b) Trapezium B1

1(c) Parallelogram B1

1(d) Kite B1

2(a) 7800 m B1

4 2(b) 3.8 litres B1

2(c) 1.085 kg B1

2(d) 31

2 hours or 3.5 hours only B1

3 39.304 − 6.9282 …

5.1=

32.37579 …

5.1= 6.34819 … M1 2

fmnw = 6.35 A1

4(a) 15

40,

30

40,

8

40,

28

40 OR 0.375, 0.75, 0.2, 0.7 M1

6

1

5,

3

8,

7

10,

3

4 All correct A1

Answer in descending order Award M1 A0

4(b)

Fraction Percentage Decimal

9

20 45% 0.45

6

5 120% 1.2

Accept 45

100

Accept 6

5 𝑜𝑟 1

1

5

B1 B1 B1 B1

5(a) 24 × 30 M1 4

= 720 litres A1

fmnw

5(b) 4200 24 = 175 min M1

175 min = 2 hours 55 min Do not award A1 for 175 min or 2.91… hours

A1 fmnw

6(a) 400 ml B1

6

6(b) (30 4) × 6 M1

= 45 g

A1

fmnw

6(c) (750 240) × 4 = 12.5 servings M1

= 12.5 servings A1

12 servings A1

fmnw


20

7(a) 150 : 30 M1

5

5 : 1 A1

Accept 5 cm : 1 cm

Accept 5

1

7(b) 2 + 3 + 4 = 9

270 9 = 30 Dividing 270 by 9 M1

Anna 2 × 30 = €60

Brenda 3 × 30 = €90 Multiplying by 30 M1

Carla 4 × 30 = €120 All correct A1

OR OR

Anna 2

9× 270 = €60 dividing by 9 M1

Brenda 3

9× 270 = €90

multiplying by 2, 3 or 4

M1

Carla 4

9× 270 = €120

€60, €90 and €120 All correct A1

8(a)

8(b)

8(c)

8(d)

Reflection of Shape A in y-axis Accurate size and position Reflection of Shape B in y-axis Accurate size and position ft from answer (a) Rotation (or turn) Rotation of 180° about the origin. Rotation of Shape A Accurate size and position

M1

A1

M1

A1 ft

M1 A1

M1 A1

8

9 Sum of interior angles = 540° M1

4

Sum of four angles = 424° M1

fifth angle = 540 424 ° M1

fifth angle = 116° A1

OR OR

Sum of exterior angles = 360° M1

Sum of 4 exterior angles = 55 + 108 + 85 + 48 = 296° M1

Fifth exterior angle = 360 296 = 64° M1

Fifth interior angle = 180 64 = 116° A1


21

10(a) 1800 × 1.05 M1

4

= €1890 A1

OR OR

5% of 1800 = 90 M1

1800 + 90

= €1890 A1

10(b) (1998 1800) × 100 = 111 M1

111 100 = 11

11% A1

OR OR

1998 1800 = 198 M1

(198 1800) × 100 = 11

= 11% A1

11(a) PQS = 180 (90 + 47) = 43° M1

5

PRS = PQS angles in the same segment

Award only for reason given

M1

PRS = 43° A1

OR OR

QSR = QPR = 47° angles in the same segment

Award only for reason given

M1

PRS = 180 (90 + 47) M1

PRS = 43° A1

11(b) angle at centre = twice angle at circumference Award only for reason given M1

POS = 86° A1 ft

Note: If uses at centre = 2 × at circumference

to state POS = 180°

Award M0 A0

12(a) 7th term = 44 B1

4 12(b) 𝑛th term = 7𝑛 5

or

𝑛th term = 2 + 7(𝑛 1)

Award M1 for identifying 7𝑛

M1 A1

12(c) 100th term = 695 B1

13(a) 12 in numerator M1

6

12

36 𝑜𝑟

1

3 or 0.3 or 33

1

3% A1

13(b) 8 in numerator M1

8

36 𝑜𝑟

2

9 or 0.2 or 22

2

9% A1


22

13(c) 22 in numerator M1

P(not almond) = 22

36 𝑜𝑟

11

18 0.61 or 61

1

9% or equivalent A1

OR OR

P(almond) = 14

36 𝑜𝑟

7

18 0.38 or 38

8

9% or equivalent M1

P(not almond) = 1

14

36 =

22

36

OR

P(not almond) = 1 7

18 =

11

18 A1

14(a) 34 = 81 B1

3

14(b) 504 (8 × 7) seen or implied M1

3𝑥 = 9

𝑥 = 2 A1

fmnw

Accept also: 32 M1 A1 𝑥 = 32 M1 A1 23 × 32 × 7 M1 A1

15(a) No B1

8

15(b) line 𝑦 = 2 − 𝑥 drawn on graph line passing through (2, 0) and (0, 2)

B2

15(c) gradient = 1 B1

15(d) 𝑥-intercept (2, 0) B1

𝑦-intercept (0, 2) B1

15(e) 1.3 (0.1)

2.3 (0.1)

Accurate answers

1.30277…, 2.30277… B1 B1

16 2(𝑥 + 5) or 2(3𝑥) or 4(𝑥) Identifying any one M1

4

77 25 = 52 Seen or implied M1

2(𝑥 + 5) + 2(3𝑥) + 4(𝑥) + 25 = 77 Forming equation (o.e.) A1

2𝑥 + 10 + 6𝑥 + 4𝑥 + 25 = 77

12𝑥 = 42

𝑥 = 3.5 or 31

2 A1

17(a) 𝐶 = 2𝜋𝑟 = 2 × 𝜋 × 2.8 M1

7

𝐶 = 17.59291… cm A1

17(b) (i) 105 = 1

3 × 𝜋 × 2.82 × ℎ Substituting values M1

ℎ = 3 × 105

𝜋 × 7.84 Seen or implied M1

ℎ = 12.78923… A1


23

(ii) 𝑟2 = 3𝑉

𝜋ℎ or for any one correct step M1

𝑟 = √3𝑉

𝜋ℎ or 𝑟 = √

𝑉

1.05ℎ or 𝑟 = √

0.95𝑉

ℎ A1

18 10 × 60 × 10 = 6000 cm3 In this question, accept any M1

5

20 × 12 × 10 = 2400 cm3 valid combination of parts M1

(3 X 2400) = 7200 M1

sum of parts = 6000 + 7200 M1

= 13 200 cm3 A1

OR OR

30 × 12 × 10 = 3600 cm3 M1

10 × 12 × 10 = 1200 cm3 M1

3600 × 3 = 10800 1200 × 2 = 2400

any one correct multiplication

M1

sum of parts = 10800 + 2400 M1

= 13 200 cm3 A1

Accept other valid methods

19(a) In ΔPAB and ΔPCB

6

AB = CB (given) and PA = PC (given) at least one ‘given’ seen M1

BP common M1

ΔPAB ΔPCB (SSS) SSS seen or 3 sides given M1

19(b) In ΔPBC

BPC = 58 2 = 29°

PCB = PAB = 36° (ΔPAB ΔPCB) any angle from BPC or

PCB identified M1

PBC = 180 (29 + 36) M1

PBC = 115° A1

20(a) True B1

5

20(b) False B1

20(c) False B1

20(d) True B1

20(e) True B1

MARKING SCHEME - University of Malta...MARKING SCHEME: SEC MATHEMATICS (MAIN SESSION 2018) 3 PAPER 1...

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Transcript of MARKING SCHEME - University of Malta...MARKING SCHEME: SEC MATHEMATICS (MAIN SESSION 2018) 3 PAPER 1...