SEC 23 Maths 2020

Marking Scheme SEC MATHEMATICS

Special September Session 2020

MATSEC Examinations Board

Marking Scheme (Special September Session 2020): SEC Mathematics

Page 1 of 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 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 2 of 19

Paper I – Section A Non-Calculator section

Question Suggested Answer Marks

Distribution Marks

1

Accept only

fully correct

answer

1

2 2674 million or 2 674 000 000 or equivalent 1

3

1

4 x = 4 1

5 Any four of the following: 1,2,3,6,9,18

Do not penalise when they give 5 or 6 correct factors

1

6 23 or 29 1

7 OR 1

8 or equivalent… Accept 1.3 1

9 0.88 or equivalent 1

10 or equivalent… do not accept 0.1 1

11 15 1

12 105o 1

13 70.35 1

14 €375 1

15 (a) kite, (d) rhombus 1

16 5 cm 1

17 x = 5 1


Page 3 of 19

18 See diagram

1

19 12% or 12 1

20 0.5 or ½ 1

Total: 20

y = x + 12

x

y


Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”

Page 4 of 19

Paper I - Section B Calculator section

Question Solution Criteria Mark

1 a i 11.2 Or equivalent B1 8

ii = 881383.8 7826.688 or 0.00888

881383.8 or more accurate

Accept 881383

B1

B1

fmnw

b = 1 + 25 + 15625

= 15651

Two of correctly

evaluated

15651

M1

A1

fmnw

c i -36, 49 A1

ii Accept 0.33, 0.11 A1

iii 0.00004, 0.000005 A1

2

a 5a+5b−3a+6b

2a+11b

M1

A1

7

b 3d =5(0) for substitution

d = 0

M1

A1

c i A1

ii

M1

M1

fmnw

3

a Belgium 8.679x108

Peru 6.679x108

8.679 x108, 6.679108…accept

rounding to one or more dp

Both correct

A1 8

b Eight hundred sixty seven million and nine hundred thousand

Accept variations like 867 million and 9 hundred thousand

A1

c Highest: Brazil

Lowest: Peru

A1

A1

d 3.5031x109 Accept answers rounded to 1

or more dp

A1

e 103.7% 4.371 x 109

100% =

4.215x109 (4215043394.4)

4.371x109 103.7%

Sets out the proportion

4.22x109 or more accurate

M1

M1

A1

4 a

On map, HG = 9.2cm

Actual dist HG = 9.2 x 500 = 4600 cm 9.2 0.1 cm

46 0.5m or equivalent

Do not award A1 when there

is inconsistency of units

M1

A1

7



Page 5 of 19

b

i 50 m actual is 5000cm/500=10cm on

the map

Calculates radius of circle to

be 10 cm

The arc drawn

M1

A1

ii Draws perp bisector of EG only

Accurate construction, with arcs shown

M1

A1

iii Labels the point of intersection of the two loci as X

Mark not awarded to candidates obtaining zero marks in both 4bi and 4bii

M1

5

a Exterior ∠ = = 45o

Interior ∠ = 180o −45o = 135o

OR

=135o

OR

Finding sum of interior angles

Dividing by 8

Note:

No marks awarded when candidate

assumes int angle to be 135o

M1

M1

OR

M1

M1

9

b

i The shape enclosed by the octagons:

At each its four vertices:

Int Angle = 360o − 2x135o = 90o

Since octagons are regular and of the

same size, the four sides of the shape

are equal.

Quad is a square, having all sides

equal and all angles of size 90o

Two conditions mentioned:

i.e. quad has angles of 90o

and quad has all sides equal

Justifies the two conditions

OR

M1

M2



Page 6 of 19

OR

Quotes that enclosed shape has

reflective symmetry about the 4 lines

shown

Mentions two lines of

symmetry

A full explanation mentioning

four lines of reflective

symmetry

OR

Award full marks for shape has

rotational symmetry of order 4

Award 1 mark for candidates

mentioning rotational

symmetry only

OR

M1

M2

ii

11.5 cm

M1

A1 fmnw

iii Outer perimeter is made up of 20 octagon sides

Finds (20 1)x side of octagon (as worked out in 5bii)

Accurate answer here is 230 cm

M1

M1

6 a Area =

Area = cm2

= 1125 cm2

M1

A1

fmnw

9

b Let X be the foot of the perp from E

to DF

BX is now the perp from B to AC

DF:AC as EX:BX

25:30 as EX:75

5:6 as EX: 75

EX = = 62.5 cm

OR

Let X be the foot of the perp from E

to DF

BX is now the perp from B to AC

DF:AC as EX:BX

25:30 as EX:75

5:6 as EX: 75

EX = = 62.5cm

Writes or uses the fact:

that DXE is similar to AXB

OR that XFE is similar to

XCB

DF:AC as EX:BX

62.5 cm

OR

Identifies a sufficient condition

for finding EX; e.g.:

25:30 as EX:75 = 5:6

OR

DEF is an enlargement by

5/6 of ABC

OR

EX = 62.5 cm

M1

M1

A1

OR

M2

A1



Page 7 of 19

c Shaded Area =

Area ABC − AreaDEF

= 1125 −

=1125 −781.25 = 343.75 cm2

Area ABC − AreaDEF

Area DEF = 781

Ans 343.75 cm2

M2

M1

A1 ft

7 a 4, 12, 20, 28

Give A1 for one or two correct

values

Give A2 for 3 or 4 correct values

A2 5

b nth square has sides of (2n −1) cm

Perimeter of nth square is (8n − 4) cm

8n−4 or equivalent

Candidates just finding 2n −1 for

the size of the nth square get 2

marks

OR

Diff. bet. successive trms = 8n -4, one mark for 8n and the other for -4

3

M1

A2

8 a i 75 ± 2 B1 10

ii 24 ± 2 B1

b − 40 0 10 38 1 100

−40 2 32 2 50 100 212 2

B4

c Intercept = 32 2

Gradient = 1.8 0.1

Correct eqt , e.g. F= 1.8C + 32; F= 1.7C + 30

M1

M1, A1

A1

9 a 40 m /s = 0.04km x 3600 sec

= 144 km/h

Converts hrs to sec

appropriately

Converts m to km

appropriately

144 km/h

M1

M1

A1

8

b km = 108 km Distance = time x speed...

108 km

M1

A1

ft

c 8:00 – 8:45 108 km

8:45 – 9:15 112 km

8:00 - 9:15 220 km

Total dist =220km, Time 1 ¼ hour

Av speed = = 176 km/h

OR

Total dist =220 000m,

Time 75x60s = 4500s

Av speed = 48.889m/s or 2933.33m/min



Page 8 of 19

Total dist =220km or total time =1 ¼ hour

Divides total dist by total time

176 km/h OR 49m/s or 2933m/min...with the appropriate units stated

M1

M1

A1

10 a i 30 A1 9

ii Highest mark = 9

Lowest mark = 4 Range =5

Correct highest mark or lowest

mark

Range = 5

M1

A1

b

= 6.8

Deduction of Sum of marks

Sum of marks correct...using

fully correct expression

Divides sum of marks by no of

students

M1

A1

M1

c

Anna (in 3A) could have got a higher

mark than Bridget (in 3B) even

though the mean mark for 3A is

lower than for 3B

Not Correct

Considers mean to depend on

the performance of the class

as a whole

Considers the performance of

a low achieving student in a

high performing class or vice-

versa

B1

M1

M1



Page 9 of 19

Paper IIA


1 a i t = – 4 B1 9

ii

2t =

t =

Expresses as

Expresses as

t =

M1

M1

A1

b x + x2 + x3 + x4 + x5 minus

(1 + x + x2 + x3 + x4 )

= x5 − 1

− (1 + x + x2 + x3 + x4 ) =

− 1 − x − x2 − x3 − x4 …allow

one error at most

x5 − 1

M1

A1

c Expression =

Simplified Expression =

A1

A1

M1

2 a AC2 = 82 + 102 − 2 x 8 x10 x cos150o

= 64 + 100 + 160 x 0.86603

= 164 + 138.56 = 302.56

AC = 17.39 cm

AC2 = 82 + 102 − 2 x 8 x 10 x

cos150o

Cos 150o = − 0.8660

Totalling components of AC2

(302.56)

AC= 17cm or more accurate

M1

M1

M1

A1

8

b

ACB = 29o

Subject of the formula

Correct evaluation

M1

M1 M1

A1

3 a Price after one yr

€15000 x 0.84 = €12600

Price after second yr

€12600 x 0.9 = €11340

Price after one yr = 15000 0.84…

Or (15000 −16% of 15000)

Price after second yr

€11340

M1

M1

A1

7

b After 1yr

2yrs

3yrs

4yrs

5yrs

6yrs

After 6yrs

12600

11340

11340x0.9 = 10206

10206x0.9 = 9185.4

9185.4x0.9 = 8266.86

8266.86x0.9 = 7440.17

7440.17 ½ of 15000

Finds price after 3yrs



ANS After 6yrs

M1

M1

M1

A1



Page 10 of 19

4 a i p 1 10 q 0.3 3 5

B1

B1

12

ii x 1 10 6

y 30 3 5

B1

B1

iii H 1 4 5

R 20 S T

HR2 = 400, so 4s2 =400 and s = 10

and 5t2 = 400, t2 = 80 and t =

HR2 = 400 or equivalent

s = ±10 or s = 10

t = ±√80 or √80 or equivalent

M2

M1

A1

b In 1 hr, supply A fills ½ tank

In 1 hr, supply B fills 1/3 tank

Supplies A&B fill ½ + 1/3 = 5/6 tank in 1hr

Full tank takes 6/5 hr to fill,

i.e. 1hr 12mins

Finds amounts for a fixed time for

A and B

Adds amounts for fixed time 6/5 hr or equivalent

For successive approx method,

allow 3 marks for method, 1

mark accuracy, accept 1 hr 12

mins 3mins as accurate

M2

M1

A1

5 a 0.5, 0.56, 0.81 and −1 B2 10

b

Allow 1 mark for 2 or 3 correct

points

Allow 2 marks for 4 or 5 correct

points


points


points

B4

c Graph represents

When − = − 0.5 (eqt 1),

Solutions to eqt 1 are therefore:

−1, −0.6, 1.6

Concludes that the solutions are

to be found at the points of

intersection of the curve with y =

0.5

Accept −1, −0.6 0.2, 1.6 0.2

M1

M1

d _0.45__ < < _1.52_ 0.45 0.1, 1.52 0.1 B2

-2 -1 0 1 2

3

2

1

-1

y

x



Page 11 of 19

6 a Let x, x + 2, be the cost of the

magazine in June & July respectively

Multiplying by x(x + 2) gives:

40(x +2) = 40x + x(x + 2); i.e.

No of magazines in June =

No of magazines in July =

Eliminating the denominator in

the eqt and simplifying:

M1

M1

M1

M1

A1

8

b (x + 10)(x − 8) = 0

x = −10 or x = 8

Only x = 8 is permissible here

(x+10)(x−8)=0

x = 8.

A1

A1

fmnw

c Cost in July is €10 A1

ft

7 a ABCD is a cyclic quad

∠ ABC = 90o (angle between tangent

and radius. Similarly ∠ CDA =90o

∠ ABC and ∠ CDA are opposite angles

of quadrilateral ABCD.

Since the sum of these angles is

180o, ABCD is cyclic

OR

BC = CD (radii of same circle)

BA = DA (tangents to circle &

meeting at A)

So ABCD is a kite

Classifies ABCD correctly;

e.g. says ABCD is cyclic

OR

Says ABCD is a kite

Justification kite:

1 mark for equal radii, 1 mark for

equal tangents -subject to their

claim that shape is kite

Justification cyclic quad:

1 mark for naming B and D as 90º

1 mark for full justification...with

opp angles being supplementary

B1

M2

9

b i ∠ BCD = 130o

Angle at centre is twice angle at circumference

B1

M1

ii CB= CD(radii), so CBD is isosceles...

OR

ABC is rt angled at B and has an angle of 65o at C

CBD =25o

M1

B1

c BAD = 50o

Justification:

e.g. using opposite angles of a cyclic quad

OR

Using that the sum of the angles of a quad is 360o

B1

M1



Page 12 of 19

8 a i 3 B1 9

ii Let P be the foot of the

perpendicular from X to AB. Because

of symmetry AP = ½ AB = 6 cm

andXAP = 30º.

6/R = cos 30º and R =

=6.928 cm

Drops a perpendicular from X to

one of the sides of the triangle

Uses appropriate trig eqt to find

R

7 cm or more accurate

OR

by using sine formula in AXB:

Determines that base angle = 30º

Correct substitution in sine rule

7 cm or more accurate

M1

M1

A1

M1

M1

A1

b i

AD2 = 182 + 6.9282 = 324 + 48 = 372

AD = 19.2873 cm

AD2 = 182 + 6.9282

Award only one mark for

diagram representing AXD or

BXD or CXD as a , with a rt

angle at X, if Pythagoras is not

used.

AD =19 cm or more accurate

Note: Using AX=7cm, gives AD =

19.31

M2

A1

ft

ii Tan ∠ DAX = =2.5981

∠ DAX = 68.95º

Use of correct trig ratio to find ∠

DAX

∠DAX = 69º or more accurate

Note: If AX=7cm, ∠DAX

=68.7º

M1

A1

ft

9 a 2(x +3x) +4y = 700

8x +4y = 700

2x + y = 175

y = 175 − 2x

Perimeter of square = 4y

Perimeter of rectangle

= 2(x +3x) or equivalent

y = 175 - 2x

M1

M1

A1

10

b y2 = (3x)(x) = 3x2

(175 – 2x)2 = 3x2

4x2 – 700x + 30625 = 3x2

x2 – 700x + 30625 = 0

Area of rectangle = 3 x 2

Area of square = (175 - 2x)2

x2 -700x + 30625 = 0 fully justified

M1

M1

A1

c x = =

653.11 or 46.89

Correct substitution in formula

653.11

46.89

M1

A1

A1

fmnw

A P

X

6 cm



Page 13 of 19

d When x = 653.11, y is negative

So only x = 46.89 is possible

Width of the rectangle is 46.89m

47m or more accurate

Mark not given when answer

includes the negative solution

A1

NOT ft

10 a x +2x + 3x + 4x = 1

so x = 0.1

M1 9

b First spin:

p(Odd) = 0.4

Second Spin, Top Branch:

p(Odd) = 0.4, p(Even) =0.6

Second Spin, Bottom Branch:

p(Odd) = 0.4, p(Even) =0.6

B1

B1

B1

c i Let P(Even, Even) denote prob of an

even no on the first spin and an even

no on the second spin; etc...

P(Even, Even) = 0.6 x 0.6

= 0.36

M1

A1

ii P(Even, Odd) = 0.6 x 0.4 = 0.24

P(Odd, Even) = 0.4 x 0.6 = 0.24

Required prob = 0.48

Adds P(Even,Odd)

andP(Odd,Even)

0.48

M1

A1

d Probability of landing in sector 4 is

0.4. When tossed 200 times, spinner

is expected to land 200x0.4 = 80

times in this sector

80 A1

11 a x + y ≤ 11

40x +20y ≥ 280 or 2x + y ≥ 14

One mark for a condition about

no of drivers with condition

including both x+y and 11

One mark for a condition abt

no of students

Award the mark for just one

accurate inequality

M1

M1

A1

9

b

Correct graph of x + y = 11

Correct graph of 2x + y = 14

Correct graphs of x =10 and y =

5

Correct shading for x =10 and y

= 5

Correct shading for x + y = 11

and for 2x + y = 14

A1

A1

A1

A1

A1

y

0

5

5

10

15

10 15

x

The shading is in the direction of the unwanted side of the line

x y + = 11

2 + = 14x y



Page 14 of 19

c Possible (no of buses, no of

minivans)

are shown in grey discs on the figure.

The possibilities are:

(5, 5); (6, 5); (5, 4); (6, 4); (7, 4); (6,

3); (7, 3); (8, 3); (6, 2); (7, 2); (8, 2);

(9, 2); (7, 1); (8, 1); (9, 1); (10, 1); (7,

0); (8, 0); (9, 0); (10, 0)

Any two correct combinations

Accept a correct solution by trial

and error, rather than using linear

programming

B1



Page 15 of 19

Paper IIB


1 4600

2.67

5.08

23.57 m or 2357 cm or 2357 or 23.57

B1

B1

B1

B1

4

2 A = - 1.5

B = 0

C = 3.5

B1

B1

B1

3

3 a Range: -5oC -(-17oC) = 12oC 12oC; accept also 12 and -12 B1 5

b Mean:

−11oC

Adds temperatures

- 11oC, - 11

M1

A1

fmnv

c -17, -13, -13, -10, -8, -5

Median is mean of -13 and -10

Median is -11.5 oC

Puts in order and attempts to find the

middle number

-11.5oC

M1

A1

4 a p = 74o (vert opp angles) Explanation

74o

M1

B1

6

b q = 74o (corr angle with p) …

accept F angles

Explanation

74o

M1

B1

c r=106o (supplementary angle to q) Explanation

106o

M1

B1

5 0.52, 55%, 0.59 , B2 2

6 a Monday

06:30, 6:30 in the morning

Accept 6:30 B1

B1

4

b Tuesday

01:00,1:00, 1 am

B1

B1

7 a Red: Blue: Green as 8: 3: 1

Red paint needed is 8/12×600=400 g

Uses the fraction 8/12 or sets up an

appropriate proportion or finds one

share

400 g

M1

A1

5

b 20 g blue is mixed with:

g of red paint

g of green paint

g of red paint

Mark given for correct method

Mark still given if there is a

computational mistake

g of green paint

M1

M1



Page 16 of 19

Total paint in the mixture:

+ 20 + = 80 g

80 g

Award full marks to candidates using

decimals and answer 79.9

A1

8 a 15% of €65 = €9.75

Sale price = €65 - €9.75 = €55.25

OR

Sale price is 85% of original price

85% of €65 = €55.25

15% of €65

Subtracts reduction from €65

€55.25

OR

Subtracts 15% from 100%

85%

€55.25

M1 M1 A1 OR M1 M1 A1

6

b Discount = €70 - €45.50 = €24.50

% reduction = ×100 = 35%

OR

×100 = 65%

100% - 65% =

35%

Subtracts €45.50 from €70

×100

35%

OR

×100

100% - 65%

35%

M1

M1

A1

OR

M1

M1

A1

9 a 2 feet = 0.6096 m

Wall is 3.95 m long

No of cabinets: = 6.48

6 cabinets fit along the wall

OR

Wall is = 12.96 feet long

No of cabinets: 12.96 ÷ 2 = 6.48

6 cabinets fit along the wall

2 feet = (0.3048 x 2) m

= 6.48

6: (Unrounded ans not accepted)

OR

12.96 ÷ 2

6: (Unrounded ans not accepted)

M1

M1

A1

fmnw

OR

M1

M1

A1

5

b 6 cabinets take 6 x 0.6096 = 3.657m

Length of uncovered wall =

3.95 – 3.66 = 0.29 m = 29 cm

No of cabinets x 0.6096

29 cm

Accept correct answers which have

not been rounded to the nearest cm

M1

A1

10 a ∠RAT = 360/3 = 120o 120o A1 6

b In s ARS, ATS:

AR = AT

AS is common

∠RAS = ∠TAS

Triangles are congruent SAS

Names a suitable condition for

congruency

For 2 correct explanations, award

M1, for 3 correct explanations award

M2

M1

M2

c RAT and RST RAT and RST A1, A1



Page 17 of 19

11

7

a Shape B is a translation of Shape A

B has correct shape and position

M1

A1

b Shape C is a rotation of shape A by 90º clockwise

C has correct shape and position

M1

A1

c Shape D has correct shape and orientation

D has correct shape and position

M1

A1

d y = x A1

12 a 78.53982 cm278.5 cm2

Uses r = 5

Accept 78 -79 cm2

M1

A1

5

b Height of water = =12.7 cm Uses 1 litre = 1000cm3

Uses

Height = 12.7 -13 cm

M1

M1

A1

13 a Thickness of one magazine=16.8/20=0.84 Thickness of one page = 0.84/24cm =0.035cm = 0.35mm OR

No of pages in 20 mag = 20x24 = 480 Thickness of one page = 16.8/480 cm =0.035cm = 0.35mm

Finds thickness of magazine Uses thickness of magazine to find thickness of a page 0.035cm Converts cm to mm appropriately

OR

Finds total no of pages in 20 mag Uses total no of pages to find thickness of a page 0.035cm Converts cm to mm appropriately

M1

M1

A1

M1

OR

M1

M1

A1

M1

5

b 0.035 x 36 cm = 1.26 cm Thickness x 36

1.26 cm

M1

fmnw



Page 18 of 19

14 a AC2 = 252 + 52.32 =

625 + 2735.29 = 3360.29

AC=57.9680 = 58 m (to nearest m)

Correct use of pythagoras

AC = 57.97 m

rounding to the nearest metre

M1

A1

M1

7

b Tan ∠BAC = =0.4780

∠BAC = 25.5o

∠BAD = 25.5o + 75o = 100.5o

Tan ∠BAC = 25/52.3

∠BAC = 25.5o or 26o

Uses ∠BAD = ∠BAC + 75o

∠BAD = 100o - 101o

M1

A1

M1

A1

15 a Orange: purple as 1:2 or equivalent A1 3

b (Orange, purple) beads = (8,16)

There are 24 beads in total

OR

Let x be the no of orange beads

2x is the no of purple beads

2x-x = 8…So x = 8

Total no of beads is 3x = 24

(Orange, purple) = (8,16)

There are 24 beads in total

OR

Uses that there are 8 more purple

beads to derive a suitable eqt

involving orange and purple beads

24

M1 A1

OR

M1

A1 fmnw

16 a If x is the larger no;

x − y = 8

3x + 2y = 59

OR

If y is the larger no;

y − x = 8

3y + 2x = 59

One eqt correct

Second eqt correct and consistent

with the first eqt

A1

A1

5

b x – y = 8 … (i)

3x + 2y = 59 … (ii)

3x - 3y = 24 … (mult (i) by 3)

Subtr last two eqts:

5y = 35

y = 7

x = 15

Attempts to Eliminate one unknown

appropriately

Attempts to Eliminate other

unknown appropriately

15, 7

M1

M1

A1

17 a Yes

Ann spent a bigger portion of the day

working…

Yes

For a valid explanation, the day needs

to be mentioned… here the pie

charts give fractions of the same

whole- the day

B1

M1

4

b Ann spent a smaller portion of her

salary on food. But we do not know

her salary. If she had a much higher

salary than Brian, she could still have

spent more on food

We cannot say

If candidate mentions that it depends

on the salary, give both marks…

otherwise zero

2



Page 19 of 19

18 a ∠BCD = 130o

Angle at the centre is twice angle at circumference

B1

M1

6

b ∠ABC = 90o

The tangent makes an angle of 90o with the radius

B1

M1

c ∠BAD = (360 − 130 − 90 − 90)o = 50o

Angles of a quad add up to 360o

50o

Reason

B1

M1

19 a B1 6

b A straight line drawn and two points on line correctly identified

Correct Line

OR

Identifies gradient as -1/2

Identifies intercept as 1.5

Correct plot

M1,M1

A1

OR

M1

M1

A1

c (3,0)

(-0.5, 1.75)…Accept (-0.5, 1.7) and (-0.5, 1.8)

A1 ft

A1 ft

20 a 13 B1 6

b 31 B1

c 4+3(n - 1) or equivalent A2

d 3n + 1 = 195

3n = 194

194 is not divisible by 3

So it is not possible

Explanation

Trial and error accepted

M2

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SEC 23 Maths 2020