MATHEMATICS – SEC LEVEL
MS10
MARKING SCHEME – PAPER IIB
MAY 2010 SESSION
Question
Number
1(a)
(b)
Answer
Mark
630 = 63 × 10
= 3 × 21 × 5 × 2
=2×3×3×5×7
27 = 3 × 3 × 3
36 = 2 × 2 × 3 × 3
99 = 3 × 3 × 11
LCM = 2 × 2 × 3 × 3 × 3 × 11 = 1188
Notes
M1
A1
Factorisation process [at least two steps or 2 factors]
Accept also in index form and in any order
M1
For two correct steps
A1
1188
Total: 4 marks
2(i)
M1
,
,
,
,
Comparing size of fractions and/or changing all fractions to
decimal form
A1
(ii)
(iii)
Fraction nearest to
or 0.6 [or any equivalent value]
is
M1
A1
0.8 ‒ 0.2
= 0.6
Subtracting smallest from largest fraction
0.6
M1
Using decimal form or LCM method to obtain ‘boundaries’
In decimal form:
and
A1
Any value written as a fraction or decimal
Examples:
LCM method:
and
Page 1 of 9
MATHEMATICS – SEC LEVEL
MS10
Examples:
Total: 6 marks
3(i)
M1
A1
Multiplying both sides by 100
(ii)
M1
Substitution in formula obtained in (i) or substitution in original
formula
4%
= 4%
A1
Total: 4 marks
4(i)
(ii)
4km in 1 hour
1km in ¼ hour
2.6km in ¼ hour × 2.6
= (13 ÷ 20) × 60 minutes = 39 minutes
8.15am ‒ 39min
= 7.15am + 60min ‒ 39min
= 7.15am + 21min
= 7.36am
M1
M1
A1
Direct proportion method up to 1 ÷ 4 × 2.6 [or equivalent in
minutes]
× 60 to convert from hours to minutes
39 minutes
M1
A1ft
Correct subtraction of minutes from minutes
7.36am [ft for incorrect part (i)]
Total: 5 marks
Page 2 of 9
MATHEMATICS – SEC LEVEL
MS10
5(i)
Side of square = 15 ÷ 5 = 3cm
x = 7 × 3 = 21cm
M1
A1
For 15 ÷ 5
21cm
(ii)
Perimeter = 15 + 15 + 21 + 21 = 72cm
M1
A1
Concept of Perimeter [Adding outer sides of figure]
72cm
Total: 4 marks
6
M1
M1
M1
Substituting given information to form an equation in either r or h
Multiplying to remove denominator
subject
A1
7.7cm
r = 3.8551
So h = 2 × 3.8551 = 7.7102 = 7.7cm
Total: 4 marks
7(i)
(ii)
x+2
x+4
B1
B1
x+2
x+4
M1
Adding x to the sum of both answers in part (i) to form equation
M1
A1
Collecting like terms [up to 3x = 408 ‒ 6 or 3x ‒ 408 = 0]
134
Total: 5 marks
Page 3 of 9
MATHEMATICS – SEC LEVEL
8(i)
MS10
P(vowel) =
B1
B1
6 in numerator
13 in denominator
P(A or I) =
B1
B1
4 in numerator
13 in denominator
(ii)
Total: 4 marks
9
x = 40 [corresponding angles]
A1
M1
40
Any equivalent reason
M1
M1
A1
M1
For angle BFC or EBF
EFC = 40o [alternate angles]
BFC = 40 ‒ 15 = 25o
In ∆BFC: y = 180 ‒ (90 + 25) [angles in ∆ sum = 180]
y = 65
For y = 180 ‒ (90 + BFC) or y = 90 ‒ EBF
65
Award if all reasons for obtaining y are given
Total: 6 marks
10(i)
(ii)
(iii)
A1
Ascending or descending order and selecting 10 and 15 or 5.5th
term or between 5th and 6th term
12.5 or equivalent
Mean = sum of ages ÷ total number of clients
= (5 + 6 + … +81) ÷ 10
= 240 ÷ 10 = 24
M1
For knowing how to use formula for determining mean
A1
24
Mode = 7
B1
7
5, 6, 7, 7, 10, 15, 24, 40, 45, 81
Median = (10 + 15) ÷ 2 = 12.5
M1
Page 4 of 9
MATHEMATICS – SEC LEVEL
(iv)
Mode is not a good ‘average’ since sample is too small
[or spread of data is too large]
MS10
M1
Award for any reasonable explanation
Total: 6 marks
11(a)(i)
(ii)
20 and 21 [or 1 and 2]
B1
Both correct and in this order
2‒3 + 2‒2 + 2‒1 + 20 + 21
M1
A1
Conversion to fractions/decimals and addition of all five terms
3.875 or equivalent
B1
B1
B1
2 in numerator
in numerator
in numerator or
(b)
in denominator
Total: 6 marks
12
Longest distance = 90 + 70 + (90 ‒ 40) = 210m
B1
210m
By Pythagoras’: AB2 = 402 + 702
AB2 = 6500
AB = 80.623m
M1
Pythagoras’ theorem and substitution
A1
80.623m
M1
Subtracting shortest from longest distance
A1
130m
Shortest distance = 80.623m
Difference = 210 ‒ 80.623
Page 5 of 9
MATHEMATICS – SEC LEVEL
MS10
= 129.377 = 130m
Total: 5 marks
13(i)
(ii)
Since D is mid-point of AC, then
∆ABD and ∆ACD are congruent by SSS or RHS
AB = BC [given]
AD = CD [D bisects AC]
BD common side [or BD = BD]
Or BDA = BDC = 90o [property of isosceles ∆]
AC ÷ 2 = 6.7cm
In ∆ABD:
sinABD = 6.7 ÷ 8.2
ABD = 54.793o
Hence ABC = 2 × ABD
= 109.586 = 110o
M1
M1
M1
For correct proof [that is; 3 statements, even without reasons]
For at least one reason
For stating SSS, SAS or RHS in conclusion
M1
M1
Using ½ AC
Up to sinABD = AD ÷ AB
M1
A1
Multiplying ABD by 2
110o
Total: 7 marks
14(i)
a = 106
[corresponding angles]
b = 106
[ext. angle = int. opp. angle of cyclic quad.]
c = 180 ‒ 106 = 74
A1
M1
A1
M1
A1
106
Reason
106
Reason
74
Page 6 of 9
MATHEMATICS – SEC LEVEL
(ii)
MS10
[opp. angles of cyclic quad. are supp.]
M1
Reason
PR is not a diameter since PQR is not a right angle
M1
Or similar
Total: 7 marks
15
£0.876 = €1
£1 = €1 ÷ 0.876
£270 = €1 ÷ 0.876 × 270
= €308.219
= €308
M1
M1
A1
B1
Dividing by 0.876
Multiplying by 270
308.2 or higher accuracy
€308
Total: 4 marks
16
(9 × 3) ‒ (4 × 2) ‒ (2 × 1)
= 27 ‒ 8 ‒ 2
=17 points
M1
M1
A1
Multiplying any one score with corresponding points
+ and ‒ assigned correctly
17
Total: 3 marks
17
10.4cm × 250 000
= 2 600 000cm
= 26 000m = 26km
M1
M1
A1
For 10.4 × 250 000
÷ 100 to convert to metres and ÷ 1000 to convert to kilometers
26km
Page 7 of 9
MATHEMATICS – SEC LEVEL
MS10
Total: 3 marks
18
Area large circle = 22 × 12 × 12 ÷ 7 = 452.571
Area of small circle = 22 × 5 × 5 ÷ 7 = 78.571
Shaded area = ¾ (452.571 ‒ 78.571)
= 280.5cm2
M1
Obtaining area of one of the circles
M1
M1
A1
Area of largest ‒ area of smallest circle
Obtaining ¼ or ¾ of difference in areas
280.5cm2
Total: 4 marks
y
19(i)
M1
A1
A1
M1
A1
(ii)
A1ft
A1ft
x
Substituting any 2 values of x in equation to obtain the
corresponding values of y
Coordinates of one point on line
Coordinates of another point on line
Plotting the obtained two points correctly
Accuracy of line drawn
x = 1.6 [accept values from 1.5 to 1.7, both values inclusive]
y = ‒1.8 [accept values from ‒1.9 to ‒1.7, both values inclusive]
[ft only if coordinates of point of intersection are read off correctly
from graph]
Page 8 of 9
MATHEMATICS – SEC LEVEL
MS10
Total: 7 marks
20(i)
(ii)
Area of trapezium = ½ (9.2 + 14.8) h
= ½ × 24h = 12h cm2
M1
M1
A1
Formula for area of trapezium
Substitution of all values
12h cm2
12h × 21.8 = 1831.2
h = 1831.2 ÷ 21.8 = 7cm
M1
M1
A1
For formula and substitution
h subject
7cm
Total: 6 marks
Page 9 of 9