MATHEMATICS – SEC LEVEL
MS10
MARKING SCHEME – PAPER IIA
MAY 2010 SESSION
Question
Number
1(i)
(ii)
Answer
Mark
26.54 ‒ (0.43 × 57) = €2.03
2.03 ‒ 0.43 = 1.60
58 days
1.60 ‒ 0.43 = 1.17
Less than €1 in 60 days
59 days
Notes
A1
€2.03 or 203cents
M1
Repeated subtraction/trial and error [one enough]
A1
60
Total: 3 marks
2
a = 32 (OM = OR radii – ∆OMR isosceles)
b = 32 (Alternate segment theorem)
c = 2 × 32 = 64 (Angle at centre = 2×angle at circumference)
d = 90 ‒ 64 = 26 (ORP = 90o and angles in ∆ORP = 180o)
B1
M1
A1ft
B1
M1
A1ft
32
Using b = a, including reason
32 [ft for incorrect value of a]
64
Using angles in ∆ORP = 180o
26 [ft for incorrect b and/or c]
Total: 6 marks
3
Width: 6.5 ≤ 7 < 7.5
Area: 57.5 ≤ 58 < 58.5
Maximum length = Maximum area ÷ Minimum width
=
= 9cm
M1
B1
B1
A1
Using maximum area ÷ minimum width
Maximum area = 58.5, 58.49, 58.499 or higher accuracy
Minimum width = 6.5
9
No marks for 58 ÷ 7
Page 1 of 10
MATHEMATICS – SEC LEVEL
MS10
Total: 4 marks
4(a)(i)
f(‒ 9) = 2(‒ 9) ÷3 ‒ 5
= ‒ 6 ‒ 5 = ‒11
M1
A1
Up to
‒ 11
(ii)
M1
M1
A1
Up to
or multiplying throughout by 3
Up to x subject
or equivalent [not in terms of y]
(iii)
f‒1(x) =
(b)(i)
(ii)
3(x + 5) + 4x = 16
3x + 15 + 4x = 16
7x = 1
x=
x > ‒2
y > ‒1
M1
M1
A1ft
Eliminating denominator
Expanding brackets
M1
A1
[ft for incorrect inverse in (ii). Award only if the resulting
equation is of comparable difficulty]
B1
B1
B1
For obtaining gradient
or equivalent
x > ‒2
y > ‒1
or equivalent
Page 2 of 10
MATHEMATICS – SEC LEVEL
MS10
Total: 13 marks
5(i)
Speed = distance ÷ time
x=6÷t
M1
Correct substitution up to x = 6 ÷ t
A1
t=
hours
(ii)
x+3=6÷t
t=
M1
M1
A1
Using x + 3
Correct substitution
hours
(iii)
M1
Correct relationship established [in words]
return time = outward ‒ 20 minutes
M1
LCM and correct numerators [only if equation is of comparable
difficulty]
M1
A1
M1
M1
A1
(iv)
(x + 9)(x ‒ 6) = 0
Expanding both brackets
Collecting terms to obtain quadratic equation in the required form
Factorisation or using formula [including substitution of values]
Adding positive solution to 3
9
x = ‒9 not suitable
Page 3 of 10
MATHEMATICS – SEC LEVEL
MS10
x=6
Return journey = 6 + 3 = 9km/h
Total: 12 marks
6(i)
5 × 4 × 4 = 80cm2
(ii)
160 = x2sin 60
x2 = 160 ÷ sin 60 = 184.752
x = 13.5924
= 13.6cm
(iii)
h = 37 × 3 ÷ 20
= 5.55cm
(iv)
B1
80cm2
M1
Using ½ absinC or equivalent method
M1
Substitution and simplification up to x2 subject
A1ft
13.6cm [ft for incorrect (i)]
M1
B1
M1
A1
Formula and substitution
20
h subject
5.55cm
M1
Award for other similar and valid reasons
Volume is only given to 1 d.p. or Inaccurate due to thickness of
paper when forming pyramid
Total: 9 marks
Page 4 of 10
MATHEMATICS – SEC LEVEL
7(i)
MS10
Method A
247o ‒ 180o = 67o
121o alternate angle [from A to B]
121o ‒ 67o = 54o [at B]
M1
M1
M1
A1
For 247o ‒ 180o
Using 121o at B
For at least one reason
54o
M1
M1
M1
A1
Using 121o at B
Method B
121o alternate angle [from A to B]
180o ‒ 121o = 59o [at B]
360o ‒ 247o ‒ 59o = 54o [at B]
M1
Method C
121 ‒ 90 = 31 [at A]
o
(ii)
o
180o ‒ 121o
For at least one reason
54o
o
270o ‒ 247o = 23o [at B]
31o alternate angle [from A to B]
31o + 23o = 54o [at B]
M1
M1
A1
M1
M1
For 121o ‒ 90o or 270o ‒ 247o
Using 31o at B
For at least one reason
54o
For Cosine rule
Page 5 of 10
MATHEMATICS – SEC LEVEL
MS10
AC2 = AB2 + BC2 ‒ 2(AB)(BC)cosB
A1ft
AC2 = 3102 +1932 ‒ 2(310)(193)cos54
(iii)
AC = 96100 + 37249 ‒70334.383 = 63014.617
AC = 251.027 = 251km
2
Substitution of values
251km [ft for incorrect part (i)]
M1
M1
M1
A1ft
For Sine rule and substitution
SinBAC subject of formula [Award even without numerical values]
For adding angle BAC to 121o
159o [ft for incorrect Angle BAC and/or AC]
, so BAC = 38.46o
121 + 38.46
=159.46 = 159o
Total: 11 marks
8(i)
M1
M1
(ii)
k = 18 × 4 = 72
A1
So
M1
A1ft
(iii)
For
or equivalent
Substituting values in equation
[Accept also up to k = 72]
Substitution of d and k in equation
0.72 or equivalent [ft for incorrect value of k only]
M1
A1ft
(iv)
d=½
A1ft
Up to
½ or equivalent [ft for incorrect value of k only]
M1
Page 6 of 10
MATHEMATICS – SEC LEVEL
When d = 2, F = 18 (given)
When d = 1, F = 72 ÷ 12 = 72
Since 18 × 4 = 72
Then F is four times as large
MS10
M1
72 [ft for incorrect value of k only]
Using 18 in reasoning
Conclusion [up to 18 × 4 = 72]
Total: 10 marks
9(i)
(ii)
(iii)
(iv)
sin32 = AB ÷ 16.5
AB = 16.5 × sin32
AB = 8.743
AB = 8.7m
M1
Trigonometric ratio and substitution
A1
8.7m
cos32 = BD ÷ 16.5 [or Pythagoras: BD2 = 16.52 ‒ 8.7432]
BD = 16.5 × cos32
BD = 13.992
BD = 14.0m
M1
Trigonometric ratio and substitution
A1
14.0m or 14m
M1
Trigonometric ratio and substitution
A1ft
35.6o or 35.5o (in case 8.7 is used) only [ft for incorrect AB in (i)]
M1
M1
Pythagoras’ theorem and substitution
CD2 subject
tanACB = 8.743 ÷ 12.2
ACB = 35.626
ACB = 35.6o
CD2 = BD2 ‒ BC2
Page 7 of 10
MATHEMATICS – SEC LEVEL
MS10
CD2 = 13.9922 ‒ 12.22 = 46.936
CD = 6.850
CD = 6.9m
A1ft
6.9m [ft for incorrect BD in (ii)]
Total: 9 marks
10(i)
or
M1
M1
M1
Up to
Product of two brackets
Multiplying by x
All working must be shown
(ii)
P = 48
Q = 96
R = 105
S=0
B1
B1
B1
Any one correct values
Another one correct value
Remaining two correct values
(iii)
[Refer to graph]
M1
B1
B1
A1
Correct scale on both axes
Values from table plotted correctly. Deduct one mark for each two
incorrectly plotted values
Overall accuracy of graph [all values plotted must be correct, even if
a different scale is used]
(iv)
Vmax = 105.027 = 105.0cm3
B1
Accept values from 103 to 107, both values inclusive
Page 8 of 10
MATHEMATICS – SEC LEVEL
(v)
MS10
At x = 5.055 = 5.1cm
B1
Accept values from 4.9 to 5.3, both values inclusive
V = 80 when x = 3.205 = 3.2cm
or x = 6.585 = 6.6cm
B1
Both correct. Accept values from 3.1 to 3.3 and 6.5 to 6.7, all values
inclusive
Total: 13 marks
11(i)
(ii)
Test 1: 40
Test 2: 55
B1
B1
Accept values from 39 to 41, both inclusive
Accept values from 54 to 56, both inclusive
Test 1: 60 ‒ 24 = 36
M1
A1
A1
Using upper ‒ lower quartile
Accept values from 35 to 37, both inclusive
Accept values from 27 to 29, both inclusive
Test 2: 68 ‒ 40 = 28
(iii)
M1
360 × 4 ÷ 9 = 160 passed or
(iv)
A1
360 ‒ 160 = 200 failed
Pass mark = 44
A1ft
Test 1
100
0
24
40
60
Test 2
0
A1ft
For 360 × 4 ÷ 9 = 160 or 360 × 5 ÷ 9 = 200
Accept values from 43 to 45, both inclusive
Box plot for Test 1 [0 and 100 not required and ft for incorrect part
(i) and/or (ii)]
Box plot for Test 2 [0 and 100 not required and ft for incorrect part
(i) and/or (ii)]
100
40
55
68
Page 9 of 10
MATHEMATICS – SEC LEVEL
MS10
M1
Accept other similar reasons
Students preformed better in Test 2 – higher median mark and
higher upper/lower quartiles in Test 2
Total: 10 marks
Page 10 of 10