SEC 23 Maths 2020
Transcript of 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.
Marking Scheme (Special September Session 2020): SEC Mathematics
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
Marking Scheme (Special September Session 2020): SEC Mathematics
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
Marking Scheme (Special September Session 2020): SEC Mathematics
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
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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
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7
b 3d =5(0) for substitution
d = 0
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c i A1
ii
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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
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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
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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
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A1
7
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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ii Draws perp bisector of EG only
Accurate construction, with arcs shown
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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
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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
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OR
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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
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M2
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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M2
ii
11.5 cm
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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
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6 a Area =
Area = cm2
= 1125 cm2
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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
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A1
OR
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Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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8
b km = 108 km Distance = time x speed...
108 km
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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10 a i 30 A1 9
ii Highest mark = 9
Lowest mark = 4 Range =5
Correct highest mark or lowest
mark
Range = 5
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b
= 6.8
Deduction of Sum of marks
Sum of marks correct...using
fully correct expression
Divides sum of marks by no of
students
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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
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M1
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
Page 9 of 19
Paper IIA
Question Solution Criteria Mark
1 a i t = – 4 B1 9
ii
2t =
t =
Expresses as
Expresses as
t =
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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
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c Expression =
Simplified Expression =
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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
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A1
8
b
ACB = 29o
Subject of the formula
Correct evaluation
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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
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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
Finds price after 4yrs
Finds price after 5yrs
ANS After 6yrs
M1
M1
M1
A1
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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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
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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
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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
Allow 3 marks for 6 or 7 correct
points
Allow 4 marks for 8 or 9 correct
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
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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:
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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.
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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
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9
b i ∠ BCD = 130o
Angle at centre is twice angle at circumference
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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
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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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
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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
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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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
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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
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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
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
Page 15 of 19
Paper IIB
Question Solution Criteria Mark
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
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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
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4 a p = 74o (vert opp angles) Explanation
74o
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6
b q = 74o (corr angle with p) …
accept F angles
Explanation
74o
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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
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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A1
OR
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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
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A1
fmnw
OR
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
Page 17 of 19
11
7
a Shape B is a translation of Shape A
B has correct shape and position
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A1
b Shape C is a rotation of shape A by 90º clockwise
C has correct shape and position
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A1
c Shape D has correct shape and orientation
D has correct shape and position
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d y = x A1
12 a 78.53982 cm278.5 cm2
Uses r = 5
Accept 78 -79 cm2
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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
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A1
M1
OR
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5
b 0.035 x 36 cm = 1.26 cm Thickness x 36
1.26 cm
M1
fmnw
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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
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A1
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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
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A1
M1
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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
Marking Scheme (Special September Session 2020): SEC Mathematics
Fmnw stands for “full marks, no working” Ft stands for “follow through mistakes in previous part questions”
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