MATHEMATICS – SEC LEVEL
MS10
MARKING SCHEME – PAPER I [CORE]
MAY 2010 SESSION
Question
Number
Answer
1
Year 1: 160 000 × 1.08 = €172 800
Year 2: 172 800 × 1.08
= €186 624
Mark
M1
M1
M1
A1
Notes
8% of 160 000
Adding increase to 160 000
8% of year 1 ‘new’ price
€186 624
Total: 4 marks
2
25km × 6 = 150km
150km ÷ 18km =
× 1.15 = €9.583
= €10
litres
M1
M1
M1
Multiplying by 6
Dividing by 18
Multiplying by 1.15
A1
€10
Total: 4 marks
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MATHEMATICS – SEC LEVEL
MS10
3(a)
0.001 =
(b)(i)
M1
A1
=
Convert to fraction or standard form
10‒3 or ‒3
‒3
= 10
M1
A1
Pattern: +85
8 nights: 450 + (3 × 85)
= €705
For identifying pattern +85
€705
(ii)
Night:
1
110
2
3
110+85 110+(2×85)
nth term = 110 + 85(n ‒1)
= 25 + 85n
(iii)
1215 = 25 + 85n
85n = 1190
n = 14 nights
4
…
110+(3×85)
M1
M1
A1
M1
M1
A1ft
Writing at least 3 consecutive values in terms of multiples of 85
Ability to generalize pattern [up to 85n seen]
25 + 85n [or equivalent]
Equating nth term in part (ii) with 1215 [Award as long as n is included in equation]
Making n subject
14 nights [ft for incorrect part (ii). Award only if the final value of n is given as a
positive whole number, without any rounding]
Total: 10 marks
4(a)
(b)
Red : Blue : Yellow = 1 : 3 : 4
(×1.5) = 1.5 : 4.5 : 6
Tins: 2 + 5 + 6 = 13 tins
Banana : apples and
apples : oranges
=2:3
=5:6
(×5) = 10 : 15
(×3) = 15 : 18
So 10 bananas ≡ 18 oranges
(÷2) 5 bananas ≡ 9 oranges
B1
B1
M1
A1
4.5 litres of blue paint
6 litres of yellow paint
Rounding from litres to number of whole tins of paint [One case enough]
13 tins
M1
M1
Multiply or divide ratios to obtain same number of apples in each ratio
Obtain equivalence between weight of bananas and oranges
A1
5 bananas
Total: 7 marks
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MATHEMATICS – SEC LEVEL
5(i)
MS10
M1
For circle and chord drawn accurately [± 2mm]
[Ignore labeling and arc not required for BC]
(ii)
M1
A1
Arcs drawn such that ABC is an equilateral triangle
Labelling points B, C and D [as long as ABC is equilateral, of side 8cm and with two
vertices A and B on circumference]
(iii)
M1
A1
Arcs drawn to obtain perpendicular bisector of AB
Labelling point P [even without arcs as long as accurate]
(iv)
M1
A1
Arcs drawn to obtain bisector of angle BCD
Labelling point Q [even without arcs as long as accurate]
A1
Accept PQ in the range from 4.4cm to 4.8cm, both values inclusive
(v)
[Refer to Construction]
PQ = 4.6cm (approx.)
Total: 8 marks
6(a)(i)
(ii)
BPC = xo (alternate angles)
PBC = 90o (interior angles)
x + y + 90 = 180 (sum of angles in triangle PBC = 180o)
x + y = 90
M1
M1
M1
M1
AQB = PBC (alternate angles or equivalent reason)
ABP = BPC (alternate angles)
Since ∆AQB and ∆CBP are equiangular then they are
similar.
M1
M1
M1
Including reason
Including reason
For proper conclusion [Or award the third M1 if 3rd set of equal angles is given,
including reason]
M1
B1
Dimensions of shape B increased by a factor of 3
Shape, position and labeling [labeling of either B or C enough]
B1
Shape and position [labeling of either B or C enough]
If using triangle AQB, second M1 is automatically awarded
Award only if at least two reasons are given for previous statements
10
(b)(i)
C
5
B
(ii)
A
0
5
10
15
20
Page 3 of 7
MATHEMATICS – SEC LEVEL
MS10
Total: 10 marks
7(a)(i)
2x + 2y = 37
2y = 37 ‒ 2x
M1
Relationship between x and y established
A1
[or equivalent, provided y subject]
(ii)
(iii)
A1
x = 2y + 5 [or equivalent]
x = 2y + 5
x = 37 ‒ 2x +5
3x = 42
x = 14
M1
M1
Substituting to obtain an equation in one unknown
Substituting value of x (or y) obtained to find remaining unknown
A1
Values of x and y both correct
y = (37 ‒ 2×14) ÷ 2
(b)
= (37 ‒ 28) ÷ 2
= 4.5
B1
M1
M1
Up to
A1
Substituting coordinates of point in line to obtain equation in terms of c. [Award even
if value of gradient is incorrect]
Gradient =
‒6=
or 0.2
+c
‒6=1+c
[or equivalent]
c = ‒7
Total: 10 marks
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MATHEMATICS – SEC LEVEL
8(i)
MS10
[Possibility Space]
B3
Deduct one mark for each two incorrect entries
P(sum = 8) =
B1
B1
5 in numerator
36 in denominator [or award B2 for
B1
B1
Correct numerator
Correct denominator
(ii)
(iii)
P(difference is 1) =
(iv)
[or award B2 for
B1
,
or
%]
%]
0 or ‘impossible’ [or equivalent]
P(sum = 8 and difference is 1) = 0
Total: 8 marks
9(a)(i)
M1
Eliminating denominators
M1
Squaring both sides [up to
M1
A1
(ii)
c = ± 0.57735
c = ± 0.58
enough]
Collecting like terms and c2 subject
± 0.58 only
M1
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MATHEMATICS – SEC LEVEL
(b)
for any value of c
So p is always smaller than 1
(×30)
MS10
M1
M1
M1
A1
15x ‒ 10(x ‒ 3) = 6(7)
LCM and correct numerators [even with LCM not crossed out]
Expanding brackets
Collecting like terms
2.4 [or equivalent]
15x ‒ 10x + 30 = 42
5x = 12
x = 12 ÷ 5 = 2.4
Total: 9 marks
10(i)
(ii)
Tan14 = 20÷ AD
AD = 20 ÷ tan14
= 80.2156
= 80.2m
M1
Trigonometric ratio and substitution
A1
80.2m
Cos(28 + 14) = (AD + DC) ÷ AB
Cos42 = (16 + 80.2156) ÷ AB
AB = 96.2156 ÷ cos42
AB = 129.4708
M1
Trigonometric ratio and substitution
M1
A1ft
AB subject
Accept values in range 129.4m to 129.5m [ft for incorrect AD in (i)]
M1
Trigonometric ratio or Pythagoras Theorem, including substitution
A1ft
86.6m [ft for incorrect AD and/or AB in (i) and/or (ii)]
M1
M1
A1ft
Subtracting 20m from BC, provided BC > 20m
Trigonometric ratio and substitution
76.5o [ft for incorrect BC in (iii). Award only if first M1 is awarded]
AB = 129.5m [Range: 129.4 ‒ 129.5]
(iii)
Sin42 = BC ÷ 129.4708
[or by Pythagoras: BC2 = 129.4702 ‒ 96.21562]
BC = 86.6328
(iv)
BC = 86.6m [Range: 86.5 ‒ 86.7]
tanBEF = (86.6328 ‒ 20) ÷ 16
BEF = 76.497
= 76.5o
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MATHEMATICS – SEC LEVEL
MS10
Total: 10 marks
Page 7 of 7