1. finance
2. personal finance
3. lending
4. credit cards

L.17/20

Pre-Leaving Certificate Examination, 2015
Mathematics
Higher Level
Marking Scheme
Paper 1
Pg. 2
Paper 2
Pg. 36
Page 1 of 68
DEB
exams
Pre-Leaving Certificate Examination, 2015
Mathematics
Higher Level – Paper 1
Marking Scheme (300 marks)
Structure of the Marking Scheme
Students’ responses are marked according to different scales, depending on the types of response anticipated.
Scales labelled A divide students’ responses into two categories (correct and incorrect).
Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on.
These scales and the marks that they generate are summarised in the following table:
Scale label
A
B
C
D
No of categories
2
3
4
5
5 mark scale
10 mark scale
15 mark scale
0, 5
0, 2, 5
0, 5, 10
0, 2, 4, 5
0, 4, 7, 10
0, 6, 11, 15
0, 2, 3, 4, 5
0, 4, 6, 8, 10
0, 6, 10, 13, 15
A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting
the scales in the context of each question are given in the scheme, where necessary.
Marking scales – level descriptors
DEB 2014 LC-H
Scale label
A-scales (two categories)

incorrect response (no credit)

correct response (full credit)
No of categories
B-scales (three categories)

response of no substantial merit (no credit)

partially correct response (partial credit)

correct response (full credit)
5 mark scale
10 mark scale
15 mark scale
20 mark scale
A
B
C
D
2
3
4
5
0, 2, 5
0, 2, , 5
0, 2, 3, 4, 5
0, 5, 10 0, 3, 7, 10 0, 2, 5, 8, 10
0, 7, 15 0, 5, 10,15 0, 4, 7, 11, 15
C-scales (four categories)

response of no substantial merit (no credit)

response with some merit (low partial credit)

almost correct response (high partial credit)

correct response (full credit)
D-scales (five categories)

response of no substantial merit (no credit)

response with some merit (low partial credit)

response about half-right (middle partial credit)

almost correct response (high partial credit)

correct response (full credit)
In certain cases, typically involving  incorrect rounding,  omission of units,  a misreading that does not
oversimplify the work or  an arithmetical error that does not oversimplify the work, a mark that is one mark
below the full-credit mark may also be awarded. Such cases are flagged with an asterisk.
Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct.

The * to be applied once only per question.

The * penalty is not applied to currency solutions.
Unless otherwise specified, accept correct answer with or without work shown.
Accept a student’s work in one part of a question for use in subsequent parts of the question, unless this
oversimplifies the work involved.
2015.1 L.17/20_MS 27/68
Page 2 of 67
DEB
exams
Summary of Marks – LC Maths (Higher Level, Paper 1)
Q.1
(a)
(i)
(ii)
(iii)
(b)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
10C (0, 4, 7, 10)
Q.7
(a)
(b)
(i)
(ii)
(iii)
(i)
25
Q.2
(ii)
(iii)
(iv)
10C (0, 4, 7, 10)
10C (0, 4, 7, 10)
5C (0, 2, 4, 5)
(a)
(b)
(c)
(c)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5B (0, 2, 5)
5C* (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C* (0, 2, 4, 5)
5C (0, 2, 4, 5)
10C* (0, 4, 7, 10)
50
25
Q.8
Q.3
(a)
(b)
(i)
(ii)
(i)
(ii)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
10D (0, 4, 6, 8, 10)
5C (0, 2, 4, 5)
(a)
(b)
25
Q.4
5C (0, 2, 4, 5)
10D (0, 4, 6, 8, 10)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
10C (0, 4, 7, 10)
5C (0, 2, 4, 5)
5B* (0, 2, 5)
5B (0, 2, 5)
10C (0, 4, 7, 10)
10C (0, 4, 7, 10)
(a)
(b)
(c)
50
25
Q.5
(i)
(ii)
(iii)
(i)
(ii)
(iii)
(iv)
(v)
Q.9
5C (0, 2, 4, 5)
10C (0, 4, 7, 10)
10D (0, 4, 6, 8, 10)
(a)
(b)
(c)
(a)
(b)
(i)
(ii)
(iii)
(i)
(ii)
(iii)
5C (0, 2, 4, 5)
10C* (0, 4, 7, 10)
10C* (0, 4, 7, 10)
10C* (0, 4, 7, 10)
10D (0, 4, 6, 8, 10)
5B (0, 2, 5)
25
Q.6
(a)
(b)
(i)
(ii)
(iii)
50
5B (0, 2, 5)
10D (0, 4, 6, 8, 10)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
25
Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the
underlying assessment principles remain the same, the exact details of the marking of a particular type of question may
vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question
to the overall examination in the current year. In setting these marking schemes, we have strived to determine how
best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment
from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these
examinations are subject to change from past SEC marking schemes and from one year to the next without notice.
General Instructions
There are two sections in this examination paper.
Section A
Section B
Concepts and Skills
Contexts and Applications
150 marks
150 marks
6 questions
3 questions
Answer all questions.
Marks will be lost if all necessary work is not clearly shown.
Answers should include the appropriate units of measurement, where relevant.
Answers should be given in simplest form, where relevant.
2015.1 L.17/20_MS 3/68
Page 3 of 67
DEB
exams
DEB
exams
Pre-Leaving Certificate Examination, 2015
Mathematics
Higher Level – Paper 1
Marking Scheme (300 marks)
Section A
Concepts and Skills
150 marks
Answer all six questions from this section.
Question 1
1(a)
(25 marks)
The number of users that log on to a new social networking website doubles every week.
In week 1, there are 24 users and in week 2, there are 48 users.
(i)
How many users access the website in week 12?


Scale 5C (0, 2, 4, 5)
(5C)
Number of users = geometric progression
=
ar n – 1
Tn
a
=
24
r
=
2
T12
=
=
=
=
24(2)12 – 1
24(2)11
24(2,048)
49,152
Low partial credit: (2 marks)
–
–
–
High partial credit: (4 marks)
2015.1 L.17/20_MS 4/68
Page 4 of 67
–
Any relevant first step, e.g. states that
example of geometric sequence, writes
down relevant correct formula.
Finds a = 24 or r = 2.
Some correct substitution into relevant
formula.
Substitution into relevant formula fully
correct, but fails to evaluate or evaluates
incorrectly.
DEB
exams
1(a)
(ii)
In which week does the number of users first exceed 10 million?
Tn









a
r
24(2)n – 1
=
=
=
(5C)
ar n – 1
24
2
>
10,000,000
10,000,000
>
(2)
24
>
416,666⋅666666...
>
log2 416,666⋅666666...
log2 (2) n – 1
(n – 1)log2 (2)
>
18⋅668534...
n–1
>
18⋅668534...
n
>
19⋅668534...
n
=
20
number of users first exceeds 10 million in the 20th week
n–1
or











Scale 5C (0, 2, 4, 5)
24(2)n – 1
>
10,000,000
10,000,000
>
(2)
24
>
416,666⋅666666...
ln (2) n – 1
>
ln 416,666⋅666666...
(n – 1)ln(2)
>
12⋅940041...
(n – 1)(0⋅693147...) >
12⋅940041...
12 ⋅ 940041...
n–1
>
0 ⋅ 693147...
n–1
>
18⋅668534...
n
>
19⋅668534...
n
=
20
number of users first exceeds 10 million in the 20th week
n–1
Low partial credit: (2 marks)
–
Any relevant first step, e.g. states that
Tn = 24(2)n – 1 > 10,000,000 or equivalent
and stops.
High partial credit: (4 marks)
–
Finds log2 (2) n – 1 > log2 416,666⋅666666...
or ln(2) n – 1 > ln 416,666⋅666666..., but
fails to finish or finishes incorrectly.
Final answer not rounded up to 20.
–
2015.1 L.17/20_MS 5/68
Page 5 of 67
DEB
exams
1(a)
(iii)
Users access the website on average 10 times per week. The website earns 0⋅4 cent
per user visit. How much revenue does the website generate in the first 12 weeks?


Number of hits = sum of a geometric progression
a(1 − r n )
Sn
=
1− r
a
=
24 × 10
=
240
r
=
2
S12
=
Scale 5C (0, 2, 4, 5)
Revenue
240(1 − 212 )
1− 2
=
=
=
=
240(1 − 212 )
−1
240(212 – 1)
240(4,096 – 1)
240(4,095)
982,800
=
=
=
982,800 × 0⋅4 cent
393,120 cent
€3,931⋅20
=

Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
–
–
*
2015.1 L.17/20_MS 6/68
(5C)
Any relevant first step, e.g. writes down
relevant correct formula.
Some correct substitution into relevant
formula.
Substitution into relevant formula fully
correct, but fails to evaluate or evaluates
incorrectly.
Finds S12 = 982,800 or 98,280, but fails
to find or finds revenue incorrectly.
No deduction applied for the omission of or incorrect units involving currency.
Page 6 of 67
DEB
exams
1(b)
The sum to infinity of the series 5 + 10x + 20 x2 + 40 x3 + … is 100.
Find the value of x.
=
a
1− r
5
10 x
5
2x
5
1 − 2x
100
=
100
5
5
200x
=
=
=
=
x
=
100(1 – 2x)
100 – 200x
100 – 5
95
95
200
0⋅475
=
S∞
a
=
r
=
=

S∞

5
1 − 2x




=
=
Scale 10C (0, 4, 7, 10)
Low partial credit: (4 marks)
–
–
–
High partial credit: (7 marks)
2015.1 L.17/20_MS 7/68
Page 7 of 67
–
(10C)
Any relevant first step, e.g. writes down
relevant correct formula.
Finds a = 5 or r = 2x.
Some correct substitution into relevant
formula.
5
= 100, but fails to finish
1 − 2x
or finishes incorrectly.
Finds
DEB
exams
Question 2
(25 marks)
The function f is defined as
f : ℝ → ℝ : x ‫→׀‬
2(a)
x−2
, where x ∈ ℝ \ {4}.
x−4
Show that the curve y = f (x) has no local maximum or local minimum point.
f (x)
=
d u
 
dx  v 

=
f ′(x)
=
=
=


Scale 10C (0, 4, 7, 10)
2(b)
( x − 4)(1) − ( x − 2)(1)
( x − 4) 2
x−4−x+2
( x − 4) 2
−2
( x − 4) 2
f ′(x) ≠ 0 as −2 ≠ 0
f ′(x) has no turning points, i.e. no local maximum or local minimum point(s)
Low partial credit: (4 marks)
–
Any relevant first step, e.g. some correct
differentiation or some correct
substitution into quotient rule.
High partial credit: (7 marks)
–
f ′(x) fully correct, but no conclusion
or incorrect conclusion.
Find the equations of the asymptotes and draw a sketch of the curve on the axes below.

(10C)
Vertical asymptote


x–4
x
=
=
0
4
Horizontal asymptote
y
=
=
=
=


y
=
Scale 10C (0, 4, 7, 10)
x−2
x→∞ x − 4
2
1−
x
lim
4
x→∞
1−
x
1− 0
1− 0
1
1
1
lim
Graph
Appropriate graph of f (x) =
2015.1 L.17/20_MS 8/68
(10C)
x−2
x−4
du
dv
v
−v
dx
dx
v2
x−2
x−4
Low partial credit: (4 marks)
–
One asymptote correctly found.
High partial credit: (7 marks)
–
Find both asymptotes correctly, but fails
to draw a sketch or draws an incorrect
sketch of the curve.
Page 8 of 67
DEB
exams
2(c)
Show that no two tangents to the curve are perpendicular to each other.
f ′(x)
=
<
Let
as
and


∴
Scale 5C (0, 2, 4, 5)
−2
( x − 4) 2
0
for all x ∈ ℝ \ {4}
m1 = f ′(a) and m2 = f ′(b)
m1 = f ′(a) < 0 for all x ∈ ℝ \ {4}
m2 = f ′(b) < 0 for all x ∈ ℝ \ {4}
m1 × m2
>
0
m1 × m2
≠
–1
no two tangents are perpendicular
Low partial credit: (2 marks)
High partial credit: (4 marks)
2015.1 L.17/20_MS 9/68
(5C)
Page 9 of 67
–
Any relevant first step, e.g. slope = f ′(x)
or m1 × m2 = −1.
−2
Writes f ′(x) =
< 0 for all x ∈ ℝ \ {4}
( x − 4) 2
and stops.
–
Writes f ′(x) =
–
−2
< 0 for all x ∈ ℝ \ {4}
( x − 4) 2
with m1 = f ′(a) < 0 and m2 = f ′(b) < 0
for all x ∈ ℝ \ {4}, but fails to finish or
finishes incorrectly.
DEB
exams
Question 3
3(a)
(25 marks)
Four complex numbers, z1 , z2 , z3 and z4 , are shown on
the Argand diagram. The same scale is used on both axes.
The number z1 has a modulus less than 1.
1
Im(z)
z4
z3
z2
q
(i)
(ii)
Re(z)
(5C)

z1
z12
=
=
r (cos  + i sin )
r2(cos 2 + i sin 2)

as r
r2
<
<
<
1
r
1
∴
z2 best represents z12
... modulus r, argument 
... modulus r2, argument 2
Low partial credit: (2 marks)
–
Correct answer, but no justification given.
High partial credit: (4 marks)
–
Correct answer with partial but incomplete
justification, e.g. takes a complex number
with modulus < 1 and squares it, e.g.
1 1
( + i)2 or z12 = r2(cos 2 + i sin 2)
2 2
and stops.
Plot the relative positions of iz1 and z1 on the Argand diagram above,
where z1 is the complex conjugate of z1 , and label each point.


z1
1
Which of the numbers, z2 , z3 or z4 , best represents z12 ?
Justify your answer.
Scale 5C (0, 2, 4, 5)
q
=
=
iz1
z1
1
Im(z)
(5C)
90° anti-clockwise rotation
reflection in the real axis
z4
iz1
z3
z2
z1
Re(z)
1
z1
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 10/68
Low partial credit: (2 marks)
–
Any relevant first step, e.g. z1 = a + ib,
 z1 = a – ib or iz1 = –b + ia.
High partial credit: (4 marks)
–
One complex number correctly plotted.
Page 10 of 67
DEB
exams
11
3(b)
π
π

+ i sin 
 cos
18
18 
Let w = 
, where i2 = –1.
4
π
π 

+ i sin 
 cos
36
36


(i)
Express w in the form a + bi, where a, b ∈ ℝ.
(10D)
11

w
=
=
=
π
π

+ i sin 
 cos
18
18


4
π
π 

+ i sin 
 cos
36
36 

11
π
π

cos
+
i
sin


18
18 

2
2π
2π 

+ i sin 
 cos
36
36 

11
π
π

+ i sin 
 cos
18
18 

2
=
π
π

+ i sin 
 cos
18
18 

9
π
π

+ i sin 
 cos
18
18 

9π
9π
+ i sin
cos
18
18
π
π
cos + i sin
2
2
0+i
=
π
π

+ i sin 
 cos
18
18 

=
=
=
or
11

w
=
=
=
=
=
Scale 10D (0, 4, 6, 8, 10)
2015.1 L.17/20_MS 11/68
4
π
π 

+ i sin 
 cos
36
36 

11π
11π
cos
+ i sin
18
18
π
π
cos + i sin
9
9
11π − 2π
11π − 2π
+ i sin
cos
18
18
9π
9π
+ i sin
cos
18
18
π
π
cos + i sin
2
2
0+i
Low partial credit: (4 marks)
–
Any relevant first step, e.g. correct use
of De Moivre’s theorem - top or bottom.
Middle partial credit: (6 marks)
–
De Moivre’s theorem applied correctly
both top and bottom, but fails to reach
π
π
cos + i sin .
2
2
High partial credit: (8 marks)
–
Leaves final answer as cos
Page 11 of 67
π
π
+ i sin .
2
2
DEB
exams
3(b)
(ii)
Find the two complex numbers x + yi for which (x + yi) 2 = w, where x, y ∈ ℝ.



=
=
=
=
w
0+i
0+i
0+i



x2 – y2
x2
x
=
=
=
0
y2
±y

For x = y
2xy
2y2
=
=
1
1
1
2


y2
=

y
=
±
=
=
=
1
1
1
1
2






Scale 5C (0, 2, 4, 5)
(x + yi) 2
(x + yi) 2
x2 + 2xyi + y2 i2
(x2 – y2) + 2xyi
For x = –y
2xy
2(–y)y
–2y2
–y 2
=


=
=
i
1
1
2
y has no solution
complex numbers:
1
2
+
i
2
and –
1
2
–
i
2
Low partial credit: (2 marks)
–
Any relevant first step, e.g. finds
(x + yi) 2 correctly.
High partial credit: (4 marks)
–
Equates x2 – y2 = 0 and 2xy = 1, but fails
to finish or finishes incorrectly.
Finds only one correct solution.
–
2015.1 L.17/20_MS 12/68
2xyi
2xy
(5C)
Page 12 of 67
DEB
exams
Question 4
(25 marks)
The diagrams show the graph of the function g (x) = 4 x − 3 .
y
g (x)
x
4(a)
State whether or not g (x) is injective. Give a reason for your answer.

Answer
–
g(x) is injective

Reason
–
the graph of the function only crosses any horizontal
line at most once
Scale 5B (0, 2, 5)
4(b)
(5B)
Partial credit: (2 marks)
–
Correct answer, but no reason or
incorrect reason given.
Under what criteria is g(x) bijective? For what domain and codomain does g(x) meet
this criteria?
(10C)

Criteria
–
g(x) is bijective if it is both injective and surjective

Domain
–
for g(x) to be surjective, the domain must be defined
g(x) is defined when g(x) ≥ 0
4x − 3
g(x)
≥


4x − 3
≥
0
4x – 3
4x
≥
≥
0
3
3
4



Codomain
–
2015.1 L.17/20_MS 13/68
≥
3
domain of g(x) = [ , ∞)
4
for g(x) to be injective, the codomain must be defined


Scale 10C (0, 4, 7, 10)
x
codomain of g(x) is defined for ℝ+, i.e. [0, ∞)
codomain of g(x) = [0, ∞)
Low partial credit: (4 marks)
–
Any relevant first step, e.g. states that
g(x) is bijective if it is both injective
and surjective or writes down 4x – 3 ≥ 0
and stops.
High partial credit: (7 marks)
–
States criteria for which g(x) is bijective
and finds either correct domain or
codomain, i.e. missing one part of answer.
Page 13 of 67
DEB
exams
4(c)
Find the inverse function g –1(x) and sketch the graph of the function on the diagram above.
Hence, explain how g(x) and g –1(x) relate geometrically to each other.

(10C)
Inverse function

g(x)
=
4x − 3
y
=
4x − 3


2
y
4x
=
=

x
=
–1
g :y→x
g –1(x)

=
Graph of function
y
4x – 3
y2 + 3
y2 + 3
4
x2 + 3
4
g –1(x)
y=x
g(x)
(0,
3
)
4
x
3
( , 0)
4

Scale 10C (0, 4, 7, 10)
2015.1 L.17/20_MS 14/68
Explanation
–
the graph of g –1(x) is determined by reflecting
the graph of g(x) in the line y = x
Low partial credit: (4 marks)
–
Any relevant first step, e.g. y2 = 4x – 3
and stops or draws the line y = x on graph.
High partial credit: (7 marks)
–
Finds g –1(x) correctly and draws sketch of
function, but relationship (geometric)
omitted or incorrect.
Page 14 of 67
DEB
exams
Question 5
(25 marks)
The graphs of the functions f : x ‫ | →׀‬2x – 3 | and g : x ‫ →׀‬3 + 3x – x2 are shown in the diagram.
y
5
f (x)
4
3
2
g (x)
1
x
-1
5(a)
1
4
Use your graph to solve the inequality 3 + 3x – x2 < | 2x – 3 |.
Answer
Scale 5C (0, 2, 4, 5)
5(b)
3
2
–
(5C)
x < 0 and x > 3
Low partial credit: (2 marks)
–
Any relevant first step, e.g. writes down
x = 0 and/or x = 3 and stops.
High partial credit: (4 marks)
–
One inequality correct, i.e. x < 0 or x > 3.
Use algebra to solve the inequality 3 + 3x – x2 < | 2x – 3 |.

Case: 2x – 3 > 0
3 + 3x – x2
6 + x – x2
x2 – x – 6
(x + 2)(x – 3)
x < –2 or x > 3
as 2x – 3 > 0

x>3






Case: 2x – 3 < 0

3 + 3x – x2

3 + 3x – x2

3 – 3 + 3x + 2x – x2

5x – x2

x2 – 5x

x(x – 5)

x < 0 or x > 5
as 2x – 3 < 0

x<0
Scale 10C (0, 4, 7, 10)
2015.1 L.17/20_MS 15/68
(10C)
<
<
>
>
2x – 3
0
0
0
<
<
<
<
>
>
–(2x – 3)
–2x + 3
0
0
0
0
Low partial credit: (4 marks)
–
Any relevant first step, e.g. sets up one
case correctly, 3 + 3x – x2 < 2x – 3 or
3 + 3x – x2 < –(2x – 3).
High partial credit: (7 marks)
–
Finds one equality correctly.
Page 15 of 67
DEB
exams
5(c)
Find the value of m for which | 2x – 3 | = mx + 1 has only one solution.



| 2x – 3 |
(2x – 3)2
4x2 – 12x + 9
x2(4 – m2) – x(12 + 2m) + 8
=
=
=
=








Only one solution
b2 – 4ac
(12 + 2m)2 – 4(4 – m2)(8)
144 + 48m + 4m2 – 128 + 32m2
36m2 + 48m + 16
9m2 + 12m + 4
(3m + 2)(3m + 2)
3m + 2
3m
=
=
=
=
=
=
=
=

Scale 10D (0, 4, 6, 8, 10)
2015.1 L.17/20_MS 16/68
m
=
(10D)
mx + 1
(mx + 1)2
m2x2 + 2mx + 1
0
0
0
0
0
0
0
0
–2
2
–
3
Low partial credit: (4 marks)
–
Any relevant first step, e.g. squares both
sides, (2x –3)2 = (mx + 1)2.
Middle partial credit: (6 marks)
–
Finds x2(4 – m2) – x(12 + 2m) + 8 = 0, but
fails to finish or finishes incorrectly.
High partial credit: (8 marks)
–
Substitutes correct values of a, b and c
into b2 – 4ac = 0, but fails to finish or
finishes incorrectly.
Page 16 of 67
DEB
exams
Question 6
6(a)
(25 marks)
The graphs of four polynomial functions are shown below.
Graph A
Graph B
y
y
x
1
x
3
Graph C
Graph D
y
y
1
3
1
3
x
1
x
3
State which of the graphs above is that of the function f : x ‫ →׀‬x(1 – x) 3(3 – x) 2, where x ∈ ℝ,
and justify your answer.

Answer
–
Graph A

Reason
–
Consider x < 0

1–x

(1 – x)3
also (3 – x)2

f (x)
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 17/68
Partial credit: (2 marks)
Page 17 of 67
–
>
>
>
<
(5B)
0
0
0
0
Correct answer, but no reason or
incorrect reason given.
DEB
exams
6(b)
A cubic function g is defined for x ∈ ℝ as
g : x ‫ →׀‬ax3 + bx2 + cx + d,
where a, b, c, d are constants.
The graph of g(x) has a local maximum point at x = p, a local minimum point at x = q
and a point of inflection at x = t.
(i)
Show that p + q = 2t.




(10D)
Local maximum and minimum points at x = p and x = q respectively
g(x)
=
ax3 + bx2 + cx + d
g′(x)
=
3ax2 + 2bx + c
=
0
3ax2 + 2bx + c
=
0
x
=
=
=
=
=
− 2b ±
(2b) 2 − 4(3a)c
2(3a)
− 2b ±
4b 2 − 12ac
6a
− 2b ±
4(b 2 − 3ac )
6a
−b ±
b 2 − 3ac
3a

p
=
−b − b 2 − 3ac
3a
and
q
=
−b + b 2 − 3ac
3a

p+q
=
=





Point of inflection at x = t
g(x)
=
g′(x)
=
g′′(x)
=
=
6ax + 2b
6ax
=
=

x
=

t
=

2t
=
=
Scale 10D (0, 4, 6, 8, 10)
2015.1 L.17/20_MS 18/68
−b ± b 2 − 4ac
2a
−b − b 2 − 3ac
−b + b 2 − 3ac
+
3a
3a
2b
–
3a
ax3 + bx2 + cx + d
3ax2 + 2bx + c
6ax + 2b
0
0
–2b
2b
–
6a
b
–
3a
2b
–
3a
p+q
Low partial credit: (4 marks)
–
Any relevant first step, e.g. some correct
differentiation.
Middle partial credit: (6 marks)
–
Finds both g′(x) and g′′(x) correctly, but
fails to find x.
High partial credit: (8 marks)
–
Solves both g′(x) = 0 and g′′(x) = 0
correctly for x, but fails to finish or
finishes incorrectly.
Page 18 of 67
DEB
exams
6(b)
(ii)
Given that g(x) has only one turning point, express a in terms of b and c.


If only one turning point, then
p
=
q
p–q
=
0
p
=
−b − b 2 − 3ac
3a
q
=
−b + b 2 − 3ac
3a

−b − b 2 − 3ac −b + b 2 − 3ac
–
3a
3a

–

–

(5C)
b 2 − 3ac
3a
b 2 − 3ac
–
2
b 2 − 3ac
3a
b 2 − 3ac
3a
2
=
0
=
0
=
0
=
0
0
3ac
b2
3c


b – 3ac
b2
=
=

a
=
or



3ax2 + 2bx + c
=
=
=
=

Only 1 solution:
b2 – 4ac
=
0
=
=
=
=
0
0
12ac
3ac
b2
3c





2
(2b) – 4(3a)(c)
4b2 – 12ac
4b2
b2
a
**
Scale 5C (0, 2, 4, 5)
ax3 + bx2 + cx + d
3ax2 + 2bx + c
0
0
g(x)
g′(x)
=
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
–
–
2015.1 L.17/20_MS 19/68
... turning point
Page 19 of 67
Any relevant first step, e.g. states p = q or
writes down b2 – 4ac = 0 and stops.
Some correct substitution into b2 – 4ac
formula without stating b2 – 4ac = 0.
2
b 2 − 3ac = 0 or b 2 − 3ac = 0,
3a
but fails to finish or finishes incorrectly.
Finds (2b)2 – 4(3a)(c) = 0 or equivalent,
but fails to finish or finishes incorrectly.
Finds –
DEB
exams
(iii)
Hence, find the co-ordinates of this turning point in terms of b, c and d.

g(x)
g′(x)
=
=
ax3 + bx2 + cx + d
3ax2 + 2bx + c

x
=
−b ± b 2 − 4ac
2a









Only 1 solution
b2 – 4ac
x
=
0
−b
2a
Substituting values into formula:
−2b
x
=
2(3a)
−b
x
=
3a
b2
a
=
3c
−b
x
=
 b2 
3 
 3c 
c
=
–
b
y
... answer from part (ii)
g(x)
ax3 + bx2 + cx + d
c
c
c
y
=
a(– )3 + b(– )2 + c(– ) + d
b
b
b
ac 3 c 2 c 2
=
– 3 +
–
+d
b
b
b
ac 3
=
– 3 +d
b
b2
... answer from part (ii)
a
=
3c
 b2  c3
y
=
–  3 + d
 3c  b
c2
=
– +d
3b
c c2
co-ordinates of local turning point (– , – + d)
b 3b
**
Scale 5C (0, 2, 4, 5)
=
=
=
Accept student’s answer(s) from previous part(s) if not oversimplified.
−b
and stops.
3a
Low partial credit: (2 marks)
–
Finds x co-ordinate =
High partial credit: (4 marks)
–
Finds x co-ordinate = –
–
2015.1 L.17/20_MS 20/68
(5C)
Page 20 of 67
c
, but fails to
b
find or finds y co-ordinate incorrectly.
Final answer for y co-ordinate not in
terms of b, c and d.
DEB
exams
Section B
Contexts and Applications
150 marks
Answer all three questions from this section.
Question 7
(50 marks)
A quantity is subject to exponential growth or decay if it increases or decreases at a rate proportional
to its current value. Many physical phenomena such as the growth of human population, the spread
of viruses, the rates of radioactive decay, etc. are modelled on exponential functions.
7(a)
dN
= kN, which states that the growth rate
dt
of the quantity N at time t is proportional to the value N (t).
The equation that describes exponential growth is
(i)
Show that N (t) = N0e kt satisfies the equation

N (t)
dN
dt
dN
= kN, where N0 is the quantity at t = 0.
dt
=
N0e kt
=
(N0e kt)k
=
=
Nk
kN
Low partial credit: (2 marks)
Scale 5C (0, 2, 4, 5)
High partial credit: (4 marks)
(ii)
–
Any relevant first step, e.g. some correct
differentiation.
–
Finds
dN
correctly, but not in terms of N.
dt
Deduce that ln N = kt + c, where c is a constant.

(5C)
=
=
=
=
=
=
=
N0e kt
ln N0e kt
ln N0 + ln e kt
ln N0 + kt
c
c + kt
kt + c
=
kN
1
dN
N
=
k
ln N
=
kt + c

N (t)
ln N


ln N0
ln N
ln N
(5C)
... constant
or
dN
dt



Scale 5C (0, 2, 4, 5)


dt
Low partial credit: (2 marks)
–
Any relevant first step, e.g. writes down
dN
ln N = ln N0e kt or
= kN and stops.
dt
High partial credit: (4 marks)
–
Applies ln to both sides and implements
rules of logs correctly without finishing
to required form.
1
dN = k dt, but fails to
Finds
N
–


integrate or integrates incorrectly.
2015.1 L.17/20_MS 21/68
Page 21 of 67
DEB
exams
7(a)
(iii)
Suggest one limitation of using exponential functions to model physical phenomenon
and give a possible reason for your answer.

Limitation

Reason
–
exponential models of physical phenomena only apply
within limited regions / limited period of time, whereas
exponential functions have limitless growth or decay
–
Any 1:
unbounded growth is not physically realistic - while the
initial growth or decay may be exponential, physical
phenomena will eventually enter a region in which
previously ignored factors become significant //
example: growth of human population will be limited
as food production becomes more expensive //
example: the spread of viruses is limited as the host
population diminishes //
example: radioactive decay has a limited lifespan // etc.
**
Accept other appropriate material and examples.
–
–
–
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
7(b)
(5B)
–
Limitation correct, but no reason or
incorrect reason given.
Exponential decay therefore occurs in the same way when the growth rate is negative. The
rates of certain types of chemical reactions, whose rates depend only on the concentration of
one or another reacting substance, consequently follow exponential decay. The concentration
of the reacting substance is represented by the equation ln A = –kt + ln A0, where A0 is the
initial concentration and A is the concentration after time t.
In an experiment to observe the rate of reaction in a chemical process, the concentration
of the reacting substance was measured at various times t from when the experiment began.
The data is shown in the table below.
Time (hours)
–1
Concentration (moles litre )
(i)
0
1
2
3
4
5
0⋅140
0⋅104
0⋅075
0⋅055
0⋅041
0⋅030
Find, correct to two decimal places, the values of ln A for each of the concentration
measurements in the data set shown and plot ln A against t on the axes below.

Values of ln A
Time (hours)
(5C*)
0
1
2
3
4
5
Concentration (moles litre–1)
0⋅140
0⋅104
0⋅075
0⋅055
0⋅041
0⋅030
ln A
–1⋅97
–2⋅26
–2⋅59
–2⋅90
–3⋅19
–3⋅51
Scale 5C* (0, 2, 4, 5)
Low partial credit: (2 marks)
–
One or two values correct.
High partial credit: (4 marks)
–
Three or four values correct.
*
2015.1 L.17/20_MS 22/68
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
Page 22 of 67
DEB
exams
7(b)
(i)
(cont’d.)

Graph
(5C)
1
3
2
5
4
6
t
-1
-2
-3
-4
ln A
Scale 5C (0, 2, 4, 5)
(ii)
Low partial credit: (2 marks)
–
One or two points correctly plotted.
High partial credit: (4 marks)
–
Three or four points correctly plotted.
Using your graph, or otherwise, find the value of k, correct to one decimal place.
ln A


@ t = 0:
ln 0⋅140
=
–kt + ln A0
–k(0) + ln A0
ln A0
0⋅140
–k(1) + ln 0⋅140
ln 0⋅140 – ln 0⋅104
0 ⋅140
ln
0 ⋅104
ln 1⋅346153...
0⋅297251...
0⋅3

A0
=
=
=


@ t = 1:
ln 0⋅104
k
=
=
=

Scale 5C* (0, 2, 4, 5)
=
=
≅
k
Low partial credit: (2 marks)
–
–
–
–
High partial credit: (4 marks)
–
–
*
2015.1 L.17/20_MS 23/68
(5C*)
ln 0⋅140 − ln A
t
Any relevant first step, e.g. writes down
slope = –k and stops.
Finds correct value for A0, but fails to
continue or continues incorrectly.
Finds ln A = –kt + ln 0⋅140, but fails to
continue or continues incorrectly.
Some correct substitution into slope
formula, but fails to continue or continues
incorrectly.
ln 0⋅140 − ln A
, but fails to
t
finish or finishes incorrectly.
Fully correct substitution into slope
formula, but fails to finish or finishes
incorrectly.
Finds k =
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
Page 23 of 67
DEB
exams
7(b)
(iii)
The concentration of the reacting substance when the chemical process is complete
was determined to be 0⋅015 moles litre–1.
Using your value for k, find the time it took for the experiment to be fully completed,
correct to the nearest minute.


ln A
k
ln 0⋅015
0⋅3t
–kt + ln A0
0⋅3
–0⋅3t + ln 0⋅140
ln 0⋅140 – ln 0⋅015
0⋅140
ln
0⋅015
ln 9⋅333333...
2⋅233592...
2 ⋅ 233592...
0 ⋅3
7⋅445307... hours
7 hours, 26⋅718444... minutes
7 hours, 27 minutes
=
=
=
=
=
=
=

t
=
=
=
≅
Scale 5C* (0, 2, 4, 5)
Low partial credit: (2 marks)
–
Any relevant first step, e.g. writes down
ln 0⋅015 = –kt + ln A0 and stops.
High partial credit: (4 marks)
–
Fully correct substitution into formula,
but fails to find t or finds incorrect t.
*
(iv)
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
Show that, according to the exponential model used, the rate of change of A (t)
is always increasing over time.









or

A

dA
dt
=
=
A0e–kt
0⋅140e–0⋅3t
=
(0⋅140e–0⋅3t )(–0⋅3)
=
–0⋅42e–0⋅3t
=
(–0⋅42e–0⋅3t )(–0⋅3)
2
d A
dt 2
=
–0⋅3t


2015.1 L.17/20_MS 24/68
(5C)
ln A
=
–kt + ln A0
ln A – ln A0
=
–kt
A
ln
=
–kt
A0
A
=
e–kt
A0
A
=
A0e–kt
dA
=
–kA0e–kt
dt
d2A
=
k 2A0e–kt
2
dt
as k 2 > 0, A0 > 0 and e–kt > 0
d2A
>
0, for all t
dt 2
dA
is increasing
dt


(5C*)
as 0⋅0126 > 0 and e
d2A
>
dt 2
dA
is increasing
dt
0⋅0126e–0⋅3t
>0
Page 24 of 67
0, for all t
DEB
exams
7(b)
(iv)
(cont’d.)
Low partial credit: (2 marks)
Scale 5C (0, 2, 4, 5)
–
–
–
High partial credit: (4 marks)
7(c)
–
Any relevant first step, e.g. writes down
d2A
> 0 and stops.
dt 2
Finds A = A0e–kt.
Some correct differentiation (first or
second derivative).
d2A
correctly, but no conclusion
dt 2
or wrong conclusion given.
Finds
For any quantity which follows exponential decay, the term ‘half-life’ can be used to describe
the time required for the decaying quantity to fall to one half of its initial value.
Under different experimental conditions, it took 36 minutes for 3 g of the reacting substance
to decay to 0⋅375 g.
Using the formula ln A = –kt + ln A0, or otherwise, find, t1/2 , the half-life of this reaction.

ln A








=
ln A
ln 1⋅5
0⋅057762t
=
=
=
=
t
=
=
≅
Scale 10C* (0, 4, 7, 10)
–kt + ln A0
–0⋅057762t + ln 3
ln 3 – ln 1⋅5
ln 3
1⋅ 5
ln 2
ln 2
0 ⋅ 057762
12⋅000055...
12 minutes
Low partial credit: (4 marks)
–
Any relevant first step, e.g. some correct
substitution into ln A = –kt + ln A0
formula.
High partial credit: (7 marks)
–
Finds correct value for k, but fails to
continue or continues incorrectly.
*
2015.1 L.17/20_MS 25/68
–kt + ln A0
A = 0⋅375 g, A0 = 3, t = 36
ln 0⋅375
=
–36k + ln 3
36k
=
ln 3 – ln 0⋅375
3
36k
=
ln
0⋅375
=
ln 8
36k
=
2⋅079441...
2 ⋅ 079441...
k
=
36
=
0⋅057762...
≅
0⋅057762
=

(10C*)
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
for the omission of or incorrect units - apply only once throughout the question.
Page 25 of 67
DEB
exams
Question 8
8(a)
(50 marks)
The diagram shows a right circular cone with a fixed slant height
of l cm. The slant height makes an angle  with the vertical,
π
where 0 <  < .
2
(i)
l
Express the radius and height of the cone in terms of l and 
πl 3
and, hence, show that the volume of the cone is
(sin  sin 2).
6

r and h in terms of l and 



sin 
=
r
=
cos 
=
h
=
Volume of the cone =
=
=
=
=
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 26/68
q
r
l
l sin 
(5C)
h
h
l
l cos 

l
r
1 2
πr h
3
1
π(l sin )2(l cos )
3
1 3 2
πl sin  cos 
3
1 3 1
πl × sin  × 2sin  cos 
3
2
3
πl
(sin  sin 2)
6
Low partial credit: (2 marks)
–
Any relevant first step, e.g. finds radius
or height in terms of l and .
High partial credit: (4 marks)
–
Finds both radius and height in terms of
l and  with some substitution into
relevant volume formula, but fails to
finish or finishes incorrectly.
Page 26 of 67
DEB
exams
8(a)
(ii)
Find the value of  for which the volume of the cone is a maximum. Give your answer
in the form tan–1 a , where a ∈ ℕ.

V
=
dV
dθ
=
Max. volume when




πl 3
(sin  sin 2)
6
πl 3
[sin (2cos 2) + sin 2(cos )]
6
dV
=0
dθ
πl 3
[sin (2cos 2) + sin 2(cos )]
6
sin (2cos 2) + sin 2(cos )
sin (2cos 2)
sinθ
2
cosθ
=
=
0
–sin 2(cos )
sin 2θ
–
cos 2θ
=
=

2tan 
=








2(1 – tan2 )tan 
2(1 – tan2 )tan  + 2tan 
2(1 – tan2  + 1)tan 
2(2 – tan2 )tan 
2 – tan2 
=
tan2 
=
tan 
=

=
=
=
=
=
0
2
± 2
tan–1 2
Low partial credit: (4 marks)
High partial credit: (8 marks)
(iii)
0
2tan 
Middle partial credit: (6 marks)
8(a)
=

Scale 10D (0, 4, 6, 8, 10)
(10D)
–tan2
2 tan θ
–
1 − tan 2 θ
–2tan 
0
0
0
tan 

=
0
=
tan–1 0
... n/a
–
Any relevant first step, e.g. some correct
differentiation.
–
Finds
–
Finds 2tan  = –tan2 or equivalent, but
but fails to finish or finishes incorrectly.
dV
correctly, but fails to continue
dθ
or continues incorrectly.
Find the maximum volume of the cone. Give your answer in surd form.
(5C)
3
V
=
=
=

Vmax
=
=
=
2015.1 L.17/20_MS 27/68
Page 27 of 67
πl
(sin  sin 2)
6
πl 3
(sin  2sin  cos )
6
πl 3
(2sin2  cos )
6
2
 2
πl 3
 × 1
×2×
 3
6
3


3
πl
2
1
×2× ×
3
6
3
2 πl 3
2 3 πl 3
or
27
9 3
1

3
2
DEB
exams
8(a)
(iii)
(cont’d.)
**
Scale 5C (0, 2, 4, 5)
(b)
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (2 marks)
–
Any relevant first step, e.g. writes down
2 Opp
or equivalent.
tan  =
=
1
Adj
High partial credit: (4 marks)
–
Finds sin  and cos  correctly in surd
form, but fails to find or finds incorrect
maximum volume.
A water tower is an elevated structure which supports a water tank
constructed at sufficient height to pressurise a water supply system
for the distribution of drinking water. It also serves as a reservoir
to help with water needs during peak usage times. The water level
in a tower typically falls during daytime usage, and then a pump
fills it back up during the night.
The reservoir tank in a water tower is in the shape of an inverted
right cone, as shown in the diagram. The slant height of the cone
is l and it makes an angle of 45° with the vertical.
x
During daytime usage, it is observed that the rate at which the
water level drops, in metres per hour, is given by
dx
=
dt
l
45
O
2
where x is the fall in the slant height, in metres, from the maximum capacity (volume) of the tank and t is the time,
in hours, from the instant that the water level begins to fall from this point.
(i)
Find the maximum capacity of the reservoir.

r and h

sin 45°
=
r
=
=

cos 45°
=
h
=
=

Volume of the cone =
=
=
=
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 28/68
(5C)
r
r
l
l sin 45°
l
h
45
O
l
2
h
l
l cos 45°
l
2
1 2
πr h
3
1
l
l
π( )2( )
3
2
2
1 3
1
πl ×
3
2 2
πl 3
6 2
or
2 πl 3
12
Low partial credit: (2 marks)
–
Any relevant first step, e.g. finds radius
or height correctly.
High partial credit: (4 marks)
–
Finds both radius and height correctly
with some substitution into relevant
volume formula, but fails to finish or
finishes incorrectly.
Page 28 of 67
DEB
exams
8(b)
(ii)
Hence, show that the volume of water in the reservoir tank at time t is

r and h

sin 45°
=
r
=
=

cos 45°
=
h
=
=

Volume of the cone =
=
=
=
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 29/68
π
6 2
3
(l − x) .
(5C)
r
r
l−x
h
(l – x)sin 45°
l−x
45
O
l–x
2
h
l−x
(l – x)cos 45°
l−x
2
1 2
πr h
3
1 l−x 2 l−x
π(
)(
)
3
2
2
1
(l − x ) 3
π×
3
2 2
π(l − x) 3
6 2
or
2 π(l − x) 3
12
Low partial credit: (2 marks)
–
Any relevant first step, e.g. finds radius
or height in terms of l and x.
High partial credit: (4 marks)
–
Finds both radius and height in terms of
l and x with some substitution into
relevant volume formula, but fails to
finish or finishes incorrectly.
Page 29 of 67
DEB
exams
8(b)
(iii)
Find, in terms of l and t, the rate at which the volume of water in the reservoir tank
is decreasing with respect to time.


dV
dt
=
V
=
dV
dx
=
=

dx
dt
dV
dt
=








dx
dt
dx
6 2
π
6 2
−π
2 2
(l – x)3
× 3(l – x)2(–1)
(l – x)2
2
... rate of drop in water level
−π
(l – x)2 × 2
2 2
π
– (l – x)2
2
2
=
2 dt
=
x
=
=
=
=
=
Scale 10C (0, 4, 7, 10)
π
=
dx
@ t = 0, x = 0
c
x
dV
dt
dx
dV
×
dx
dt
=
=

(iv)
2t + c
0
2t
π
– (l – x)2
2
π
– (l – 2 t)2
2
Low partial credit: (4 marks)
2015.1 L.17/20_MS 30/68
... in terms of t
–
Any relevant first step, e.g. some correct
differentiation.
–
Finds
dV
in terms of x, but fails to finish
dt
or finishes incorrectly.
Find the rate at which the volume of water in the reservoir tank is decreasing
1
with respect to time when of its capacity is remaining.
8
π

V
=
(l – x)3
6 2
@x=0
π
V
=
(l – 0)3
6 2
πl 3
=
6 2
1
1 πl 3

×V
=
×
8
8 6 2
=
Page 30 of 67
... in terms of x
2 dt
High partial credit: (7 marks)
8(b)
(10C)
(5C)
πl 3
48 2
DEB
exams
8(b)
(iv)
(cont’d.)



π
(l – x)3
=

(l – x)3
=





8(l − x)3
23(l − x)3
2(l − x)
2l − 2x
–2x
=
=
=
=
=

x
=

6 2
dV
dt
=
=
=
=
**
πl 3
48 2
1 3
l
8
l3
l3
l
l
–l
l
2
π
– (l – x)2
2
l
π
– (l – )2
2
2
π l 2
– ( )
2 2
πl 2
–
8
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (2 marks)
Scale 5C (0, 2, 4, 5)
High partial credit: (4 marks)
8(b)
(v)
–
Any relevant first step, e.g. finds volume
1
of tank when of capacity remaining.
8
–
Finds x =
l
, but fails to finish or finishes
2
incorrectly.
Given that a particular example of this type of water tower has a slant height of 30 m,
how long would it take, correct to the nearest minute, to empty the reservoir tank
if no water is replenished?




dx
dt
=
dt
=
dt
=
t
=
=
=
=
Scale 5B* (0, 2, 5)
Partial credit: (2 marks)
*
(5B*)
2
1
2

dx
30
1
2
0
t

2015.1 L.17/20_MS 31/68
... from part (iii)
x
2
30
dx
30
0
2
21⋅213203... hours
21 hours, 13 minutes
–
Any relevant first step, e.g. attempts to
dx
dV
= 2 or with
work with
.
dt
dt
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
Page 31 of 67
DEB
exams
Question 9
(50 marks)
Mary and John took out a mortgage of €400,000 to buy a new house.
They agreed to repay this loan, plus interest, by a series of equal monthly
payments, starting one month after they received the loan and continuing
for 20 years. The effective annual rate of interest charged is 6%.
9(a)
(i)
Show that the rate of interest, compounded monthly, that corresponds to an effective annual
rate of 6% is 0⋅487%, correct to three decimal places.
Scale 5C (0, 2, 4, 5)

F
1⋅06
=
=
P(1 + i)t
(1 + i)12

1+i

i

r
=
=
=
=
=
=
≅
1⋅ 06 12
1⋅004867551...
1⋅004867551... – 1
0⋅004867551...
0⋅004867551... × 100
0⋅4867551...
0⋅487%
1
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
2015.1 L.17/20_MS 32/68
Page 32 of 67
(5C)
–
Any relevant first step, e.g. correct
statement with no work.
Some substitution into correct relevant
formula, but fails to continue or continues
incorrectly.
Finds i correctly, but fails to convert to r
or converts incorrectly.
DEB
exams
9(a)
(ii)
Show how to use the sum of a geometric series to calculate the monthly repayments
that Mary and John have to make on their mortgage, correct to the nearest cent.


F
=
P
=
A
=
=


geometric progression
a
=
r
=
Sn





a (1 − r n )
1− r
P
1


1 −
240 
1 ⋅ 00487 
1 ⋅ 00487 
1
1−
1 ⋅ 00487
1


P1 −
240 
1⋅ 00487 

1⋅ 00487 − 1
1


P1 −
240 
1⋅ 00487 

0 ⋅ 00487
1
P(1 –
)
1⋅ 00487240
(1 + i ) t
Month 1 + Month 2 + ... + Month 240
P
P
P
+
+ ... +
1⋅00487 (1⋅00487 ) 2
(1⋅00487 ) 240
P
1⋅00487
1
1⋅00487
=
=
400,000
=
400,000
=
400,000
=
400,000
=
0.00487(400,000)
1,948
1,948
1948
0 ⋅ 688377...
2,829⋅842100...
€2,829⋅84

P(0⋅688377...)
=
=

P
=
=
≅
Scale 10C* (0, 4, 7, 10)
P(1 + i)t
F
a (1 − r n )
1− r
400,000
=
Low partial credit: (4 marks)
–
–
High partial credit: (7 marks)
*
*
2015.1 L.17/20_MS 33/68
(10C*)
–
Any relevant first step, e.g. correct
statement with no work, reference
to 1⋅00487.
Finds a or r correctly, but fails to continue
or continues incorrectly.
Correct substitution of all values into
formula for Sn , but fails to finish or
finishes incorrectly.
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
No deduction applied for the omission of or incorrect units involving currency.
Page 33 of 67
DEB
exams
9(a)
(iii)
Using amortisation, or otherwise, verify your answer to part (ii) above.
A
=
=
=
=
≅
Scale 10C* (0, 4, 7, 10)
Pi(1 + i )t
(1 + i) t − 1
400,000(0 ⋅ 00487)(1 + 0 ⋅ 00487) 240
(1 + 0 ⋅ 00487) 240 − 1
1,948(1⋅ 00487) 240
(1⋅ 00487) 240 − 1
2,829⋅842101...
€2,829⋅84
Low partial credit: (4 marks)
–
Any relevant first step, e.g. writes down
correct relevant formula.
High partial credit: (7 marks)
–
Correct substitution of all values into
amortisation formula, but fails to finish
or finishes incorrectly.
*
*
9(b)
(10C*)
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
No deduction applied for the omission of or incorrect units involving currency.
After two years of paying their monthly repayments on their mortgage in full, Mary and John’s
financial situation worsened and they were unable to make any further repayments.
(i)
Using amortisation, or otherwise, find the amount outstanding immediately after
Mary and John made their last repayment. Give your answer correct to the nearest cent.


2 years
Outstanding
=
=
=
A
=
2,829⋅84
=
=
=
=

P
=
=
≅
**
Scale 10C* (0, 4, 7, 10)
24 months
240 − 24
216 payments
Pi(1 + i )t
(1 + i) t − 1
P (0 ⋅ 00487)(1 + 0 ⋅ 00487) 216
(1 + 0 ⋅ 00487) 216 − 1
P(0 ⋅ 00487)(2 ⋅855842399...)
2 ⋅855842399... − 1
P(0 ⋅ 013907952...)
1⋅855842399...
P(0⋅007494145...)
2,829⋅84
0 ⋅ 007494145...
€377,606⋅772891595...
€377,606⋅77
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (4 marks)
–
Any relevant first step, e.g. states 216
in the correct context.
High partial credit: (7 marks)
–
Correct substitution of all values into
amortisation formula, but fails to finish
or finishes incorrectly.
*
*
2015.1 L.17/20_MS 34/68
(10C*)
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
No deduction applied for the omission of or incorrect units involving currency.
Page 34 of 67
DEB
exams
9(b)
(ii)
Mary and John’s lender charges an additional rate of interest of 1⋅5% per annum on the
outstanding amounts of loans in arrears, compounded monthly.
How long will it take for the outstanding amount that Mary and John owe to exceed
the original sum of money borrowed?
New annual rate
=
=
6 + 1⋅5
7⋅5%
1⋅075
=
(1 + i)12

1+i

i

r
=
=
=
=
=
=
≅
1⋅ 07512
1⋅006044919...
1⋅006044919... – 1
0⋅006044919...
0⋅006044919... × 100
0⋅6044919...
0⋅604%








1
P(1 + i)t
377,606⋅77(1 + 0⋅00604)t
377,606⋅77(1⋅00604)t
400,000
(1⋅00604)t
=
377,606 ⋅ 77
(1⋅00604)t
=
1⋅059303042...
(1⋅00604)t
=
1⋅059303042...
log1⋅00604 (1⋅00604)t =
log1⋅00604 1⋅059303042...
t
=
9⋅567052...
≅
10 months
i.e. after 10 months the amount outstanding will be more than €400,000
F
400,000
**
Scale 10D (0, 4, 6, 8, 10)
=
=
=
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (4 marks)
–
Any relevant first step, e.g. writes down
new rate = 7⋅5% per annum.
Middle partial credit: (6 marks)
–
Finds new monthly rate, but fails to
continue or continues incorrectly.
Correct substitution of all values into
F = P(1 + i)t formula or equivalent.
–
High partial credit: (8 marks)
–
–
9(b)
(iii)
Justification
–
–
**
2015.1 L.17/20_MS 35/68
Correct substitution of all values into
F = P(1 + i)t formula or equivalent, but
fails to finish or finishes incorrectly.
Final answer not rounded up to 10.
Give one justification why Mary and John’s lender would charge a higher rate of interest on
loans in arrears.
Scale 5B (0, 2, 5)
(10D)
Partial credit: (2 marks)
Page 35 of 67
(5B)
Any 1:
more work created / time for the bank to deal with
individuals in arrears //
needs to be handled in a different way to ‘normal’
mortgages not in arrears // etc.
Accept other appropriate answers.
–
Partial answer given - incomplete or
unsatisfactory, but with some merit.
DEB
exams
DEB
.
exams
Pre-Leaving Certificate Examination, 2015
Mathematics
Higher Level – Paper 2
Marking Scheme (300 marks)
Structure of the Marking Scheme
Students’ responses are marked according to different scales, depending on the types of response anticipated.
Scales labelled A divide students’ responses into two categories (correct and incorrect).
Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on.
These scales and the marks that they generate are summarised in the following table:
Scale label
A
B
C
D
No of categories
2
3
4
5
5 mark scale
10 mark scale
15 mark scale
0, 5
0, 2, 5
0, 5, 10
0, 2, 4, 5
0, 4, 7, 10
0, 6, 11, 15
0, 2, 3, 4, 5
0, 4, 6, 8, 10
0, 6, 10, 13, 15
A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting
the scales in the context of each question are given in the scheme, where necessary.
Marking scales – level descriptors
DEB 2014 LC-H
Scale label
A-scales (two categories)

incorrect response (no credit)

correct response (full credit)
No of categories
B-scales (three categories)

response of no substantial merit (no credit)

partially correct response (partial credit)

correct response (full credit)
5 mark scale
10 mark scale
15 mark scale
20 mark scale
A
B
C
D
2
3
4
5
0, 2, 5
0, 2, , 5
0, 2, 3, 4, 5
0, 5, 10 0, 3, 7, 10 0, 2, 5, 8, 10
0, 7, 15 0, 5, 10,15 0, 4, 7, 11, 15
C-scales (four categories)

response of no substantial merit (no credit)

response with some merit (low partial credit)

almost correct response (high partial credit)

correct response (full credit)
D-scales (five categories)

response of no substantial merit (no credit)

response with some merit (low partial credit)

response about half-right (middle partial credit)

almost correct response (high partial credit)

correct response (full credit)
In certain cases, typically involving  incorrect rounding,  omission of units,  a misreading that does not
oversimplify the work or  an arithmetical error that does not oversimplify the work, a mark that is one mark
below the full-credit mark may also be awarded. Such cases are flagged with an asterisk.
Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct.

The * to be applied once only per question.

The * penalty is not applied to currency solutions.
Unless otherwise specified, accept correct answer with or without work shown.
Accept a student’s work in one part of a question for use in subsequent parts of the question, unless this
oversimplifies the work involved.
2015.1 L.17/20_MS 36/68
Page 36 of 67
DEB
exams
Summary of Marks – LC Maths (Higher Level, Paper 2)
Q.1
(a)
(i)
(ii)
(iii)
(b)
5C (0, 2, 4, 5)
5B (0, 2, 5)
5C (0, 2, 4, 5)
10D (0, 4, 6, 8, 10)
Q.6
(a)
(b)
5B (0, 2, 5) × 3
5B (0, 2, 5)
5C (0, 2, 4, 5)
(i)
(ii)
25
25
Q.7
Q.2
(a)
(b)
(i)
(ii)
(i)
(ii)
(iii)
5C (0, 2, 4, 5)
5C* (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
(a)
(b)
(c)
25
Q.3
(a)
(b)
(i)
(ii)
(i)
(ii)
(iii)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
Q.8
(a)
(b)
10D (0, 4, 6, 8, 10)
5C (0, 2, 4, 5)
10D (0, 4, 6, 8, 10)
(a)
(b)
(c)
10C (0, 4, 7, 10)
5B (0, 2, 5)
5B (0, 2, 5)
5B (0, 2, 5)
5C (0, 2, 4, 5)
5C (0, 2, 4, 5)
5B (0, 2, 5)
5A (0, 5)
5B (0, 2, 5)
50
25
Q.4
(i)
(ii)
(iii)
(i)
(ii)
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv)
(i)
(ii)
10C (0, 4, 7, 10)
5B (0, 2, 5)
15D (0, 6, 10, 13, 15)
5B (0, 2, 5)
10D (0, 4, 6, 8, 10)
5C (0, 2, 4, 5)
50
25
Q.5
(a)
(b)
15D (0, 6, 10, 13, 15)
(i)
(ii)
(iii)
(iv)
(b)
5D (0, 2, 3, 4, 5)
(ii)
(iii)
(i)
(ii)
(iii)
10D (0, 4, 6, 8, 10)
15D (0, 6, 10, 13, 15)
5B (0, 2, 5)
5B (0, 2, 5)
10C (0, 4, 7, 10)
50
5C (0, 2, 4, 5)
25
Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the
underlying assessment principles remain the same, the exact details of the marking of a particular type of question may
vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question
to the overall examination in the current year. In setting these marking schemes, we have strived to determine how
best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment
from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these
examinations are subject to change from past SEC marking schemes and from one year to the next without notice.
General Instructions
There are two sections in this examination paper.
Section A
Section B
Concepts and Skills
Contexts and Applications
150 marks
150 marks
6 questions
3 questions
Answer all questions.
Marks will be lost if all necessary work is not clearly shown.
Answers should include the appropriate units of measurement, where relevant.
Answers should be given in simplest form, where relevant.
2015.1 L.17/20_MS 37/68
Page 37 of 67
DEB
exams
DEB
exams
Pre-Leaving Certificate Examination, 2015
Mathematics
Higher Level – Paper 2
Marking Scheme (300 marks)
Section A
Concepts and Skills
150 marks
Answer all six questions from this section.
Question 1
(25 marks)
OPQR is a parallelogram where O is the origin (0, 0).
The co-ordinates of the point P are (k, 0) and the co-ordinates of the point Q are (4, 2).
The point R is on the line y = 2x.
1(a)
(i)
Find the value of k.
(5C)

OPQR is a parallelogram
OR || PQ

Line OR:
y
y
mOR
=
=
=
mPQ
=
=
=





Scale 5C (0, 2, 4, 5)
mOR
2
4−k
2
2
2k
k
2x
mx + c
2
y2 − y1
x2 − x1
2−0
4−k
2
4−k
=
mPQ
=
2
=
=
=
=
=
2(4 – k)
8 – 2k
8–2
6
3
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
–
–
–
2015.1 L.17/20_MS 38/68
Page 38 of 67
Any relevant first step, e.g. draws sketch
of parallelogram on co-ordinate diagram,
writed down relevant statement, e.g.
parallel lines have equal slopes.
Finds slope of OR.
Finds slope of PQ, but fails to finish
or finishes incorrectly.
Finds equation of PQ, but fails to finish
or finishes incorrectly.
Finds co-ordinates of R, but fails to finish
or finishes incorrectly.
DEB
exams
1(a)
(ii)
Find the co-ordinates of R.

(5B)
OPQR is a parallelogram
→
OR
=
co-ordinates of P

Scale 5B (0, 2, 5)
→
PQ : (3, 0)
x: 3
y:
0
→
OR: (0, 0)
→
PQ
=
=
(k, 0)
(3, 0)
→
→
→
(4, 2)
4
2
→
=
(0 + 1, 0 + 2)
(1, 2)
Partial credit: (2 marks)
↑1
↑2
–
–
–
(iii)
Any relevant first step, e.g. writes down
that R has co-ordinates (x, 2x) and stops.
Finds y-co-ordinate of R only by
observation and stops.
Finds x or y co-ordinate of R by translation,
but fails to find or finds incorrectly other
co-ordinate of R.
Hence, find the area of the parallelogram OPQR.

Area of OPQR
(5C)
=
=
2 × area of Δ OPQ
1
2 × | (3)(2) − (4)(0) |
2
|6 – 0|
6 units2
=
=
=
base × ⊥ height
3×2
6 units2
=
=
or

Scale 5C (0, 2, 4, 5)
Area of OPQR
Low partial credit: (2 marks)
–
Any relevant first step, e.g. writes down
relevant area formula, perpendicular
height indicated on diagram.
High partial credit: (4 marks)
–
Finds base and perpendicular height, but
fails to find or finds incorrectly area of
parallelogram.
Finds correctly area of Δ OPQ or area of
Δ ORQ only.
–
2015.1 L.17/20_MS 39/68
Page 39 of 67
DEB
exams
1(b)
The line OS has positive slope and intersects the line PQ at S such that | ∠ OSP | = 45°.
Find the co-ordinates of S.
=
OS ∩ PQ
=
=
mOR
2
Let mOS
=
m
tan 
=

tan 45°
=

1
=

1 + 2m
=
±1(m – 2)



1 + 2m
2m – m
m
=
=
=
m–2
–2 – 1
–3
Co-ordinates of S

Slope of OS
mPQ
m1 − m2
1 + m1m2
m − ( 2)
±
1 + ( m)( 2)
m−2
±
1 + 2m
±
1 + 2m
or




Equation of OS
Point (0, 0), m =
y – y1
y–0

y


3y
x – 3y



Equation of PQ
y – y1
=
Point (3, 0), slope mPQ = 2
y–0
=
y
=
2x – y
=
m(x – x1)
=
=
0 (×–2)
6
–2x + 6y
2x – y
5y
=
=
=

y
=

x – 3y
x
=
=
0
6
6
6
or 1⋅2
5
or
0
3y
6
3( )
5
18
or 3⋅6
5
=
=
2015.1 L.17/20_MS 40/68
l∩k
=
=
Point (4, 2), mPQ = 2
y–2
=
2(x – 4)
y–2
=
2x – 8
2x – y
=
6

2x – y
2x
=
=
=
=


m
–(m – 2)
–m + 2
2–1
1
1
3



or
2(x – 3)
2x – 6
6
OS ∩ PQ
OS:
x – 3y
PQ: 2x – y

2m + m
3m
=
=
=
=
1
as OS has a positive slope
3
=
m(x – x1)
1
=
(x – 0)
3
1
=
x
3
=
x
=
0


(10D)
x
=
6
6+y
6
6+
5
36
5
18
or 3⋅6
5
18 6
S ( , ) or S (3⋅6, 1⋅2)
5 5
Page 40 of 67
DEB
exams
1(b)
(cont’d.)
Scale 10D (0, 4, 6, 8, 10)
Low partial credit: (4 marks)
–
–
Middle partial credit: (6 marks)
–
Fully correct substitution into tan 
formula, but fails to continue.
High partial credit: (8 marks)
–
Uses negative slope and finishes
correctly.
Finds correct slope and correct equation
of PS and stops.
–
2015.1 L.17/20_MS 41/68
Any relevant first step, e.g. writes down
relevant formula for tan  or draws
relevant diagram with ∠ OSP indicated.
Finds slope of PQ correctly.
Page 41 of 67
DEB
exams
Question 2
2(a)
(25 marks)
In a triangle ABC, | AB | = 6 cm and | BC | = 8 cm.
The area of the triangle ABC is 12 cm2.
(i)
Find the two possible values of | ∠ ABC |.



(5C)
Area Δ ABC
=
Area Δ ABC
=
1
(6)(8) sin | ∠ ABC | =
2
24 sin | ∠ ABC |
=

sin | ∠ ABC |
=

| ∠ ABC |
=
=
1
ab sin C
2
1
(6)(8) sin | ∠ ABC |
2
12
12
1
2
1
2
30° or 150°
sin–1
Low partial credit: (2 marks)
Scale 5C (0, 2, 4, 5)
–
–
High partial credit: (4 marks)
(ii)
–
1
and/or
2
| ∠ ABC | = 30° only and stops.
Finds sin | ∠ ABC | =
In the case that ∠ ABC is obtuse, find | AC |, correct to one decimal place.
| ∠ ABC |
2

a
| AC |2

| AC |
Scale 5C* (0, 2, 4, 5)
=
150°
=
=
=
=
=
=
=
≅
b2 + c2 – 2bc cos A
62 + 82 – 2(6)(8) cos 150°
36 + 64 – 96(–0⋅866025...)
100 + 83⋅138438...
183·138438...
183·138438...
13⋅532865...
13⋅5 cm
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
–
–
*
2015.1 L.17/20_MS 42/68
Any relevant first step, e.g. writes down
correct relevant formula for the area of
a triangle.
Fully correct substitution into relevant
area formula, but fails to continue or
continues incorrectly.
(5C*)
Any relevant first step, e.g. writes down
correct relevant formula for cosine rule.
Some correct substitution into formula
for cosine rule, but fails to continue or
continues incorrectly.
Fully correct substitution into formula
for cosine rule, but fails to finish or
finishes incorrectly.
Finds | AC |2 correctly, but fails to find
or finds incorrect | AC |.
Deduct 1 mark off correct answer only if not rounded or incorrectly rounded
- this deduction should be applied only once throughout the question.
Page 42 of 67
DEB
exams
2(b)
(i)
Prove that tan (A + B) =
tan A + tan B
.
1 − tan A tan B
tan (A + B)
(5C)
=
=
=
=
=
Scale 5C (0, 2, 4, 5)
2(b)
(ii)
Any relevant first step, e.g. writes down
sin A
sin B
, tan B =
or
tan A =
cos A
cos B
sin ( A + B )
tan (A + B) =
and stops.
cos ( A + B )
High partial credit: (4 marks)
–
Expands expression correctly,
sin A cos B + cos A sin B
, but fails
e.g.
cos A cos B − sin A sin B
to finish or finishes incorrectly.
π
= 2 – 1.
8
tan
π
=
4
=
1
=

1 – tan2

tan2

tan
π
8
=
π
π
+ 2 tan – 1 =
8
8
π
8
=
=
2015.1 L.17/20_MS 43/68
(5C)
tan (
π π
+ )
8 8
1
π
π
+ tan
8
8
π
π
1 − tan tan
8
8
π
2 tan
8
2 π
1 − tan
8
π
2 tan
8
tan
=

tan A + tan B
1 − tan A tan B
–
Consider

sin A cos B + cos A sin B
cos A cos B − sin A sin B
sin A cos B
cos A sin B
+
cos A cos B cos A cos B
cos A cos B
sin A sin B
−
cos A cos B cos A cos B
sin A sin B
+
cos A cos B
sin A sin B
1−
.
cos A cos B
Low partial credit: (2 marks)
Hence, or otherwise, show that tan

sin ( A + B)
cos ( A + B)
Page 43 of 67
0
−b ±
b 2 − 4ac
2a
−2 ± 2 2 − 4(1)(−1)
2(1)
DEB
exams
2(b)
(ii)
(cont’d.)

tan
π
8
=
=
>
0
=
–1 + 2
=
2 –1
=
as

Scale 5C (0, 2, 4, 5)
2
−2 ± 8
2
−2 ± 2 2
2
–1 ± 2
=
π
tan
8
π
tan
8
−2 ± 4 + 4
Low partial credit: (2 marks)
–
π
for A or B or both correctly
8
tan A + tan B
formula and stops.
in
1 − tan A tan B
Substitutes
High partial credit: (4 marks)
2 tan
–
(iii)
Finds 1 =
π
8
or equivalent,
π
8
but fails to finish or finishes incorrectly.
1 − tan 2
Given that a line segment of length one unit, show clearly how to construct another line
segment of length 2 – 1 units, using only a compass and straight edge.
(5C)
2
1
Scale 5C (0, 2, 4, 5)
Low partial credit: (2 marks)
High partial credit: (4 marks)
2015.1 L.17/20_MS 44/68
2 -1
Page 44 of 67
–
–
Finds perpendicular bisector of line
segment correctly.
Constructs line segment of length 2 units
correctly, but fails to finish or finishes
incorrectly.
DEB
exams
Question 3
(25 marks)
A ‘call centre’ is a centralised office used for receiving or transmitting
a large volume of requests by telephone. A large mobile phone company
employs an external service provider to operate call centre services to its
customers. The provider manages an inbound centre which administers
information enquiries and product support and an outbound centre which
focuses on telemarketing and market research.
3(a)
The company wishes to change its call centre services provider. Of the 48 providers being
considered, 31 specialise in providing inbound services. 8 of these providers also specialise in
providing outbound services. 11 providers neither specialise in inbound or outbound services.
(i)
Find the probability that a provider selected at random from all those being considered,
specialises in outbound services only.
[ 48 ]
In
(5C)
Out
[ 31 - 8 ]
= [ 23 ]
[8]
[ x]
[ 11 ]



Let x = no. of call centres specialising in outbound services
23 + 8 + x + 11
=
48
42 + x
=
48
x
=
6

P(Out only)
=
=
6
48
1
8
Low partial credit: (2 marks)
Scale 5C (0, 2, 4, 5)
–
–
High partial credit: (4 marks)
(ii)
–
Finds x = 6, but fails to find probability
or finds incorrect probability.
The company wishes to short-list three providers for further consideration. Find the
probability that the three providers selected at random each specialise in providing both
inbound and outbound services.

P(all 3, In & Out)
=
=
Scale 5C (0, 2, 4, 5)
Low partial credit: (2 marks)
High partial credit: (4 marks)
–
–
–
Page 45 of 67
(5C)
8
7
6
×
×
48 47 46
7
336
or
103,776
2,162
–
2015.1 L.17/20_MS 45/68
Any relevant first step, e.g. writes down
#E
or equilavent and stops.
#S
Draws Venn diagram with three or more
entries.
Any correct step, e.g. correct numerator
for each probability, i.e. 8 × 7 × 6.
Finds one correct probability.
8
7
6
×
×
.
48 48 48
Correct probabilities chosen, but
operator incorrect.
Finds
DEB
exams
3(b)
Agents in the outbound centre attempt to contact customers up to a maximum of three times.
After three failed attempts, they are directed to discontinue contact with the customer. All
calls are monitored to measure the effectiveness of agents. On each attempt, the probability
that an agent gets through to a customer is 3 .
10
(i)
Find the probability that an agent gets through to a customer on the second attempt.
3
P(call answered)
=
10
3

P(call not answered) =
1–
10
7
=
10

Scale 5C (0, 2, 4, 5)
P(call answered on 2nd attempt)
7
3
×
=
10 10
21
=
or 0⋅21
100
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
(ii)
–
P(call answered)
=
=
=
=
=
Correct probabilities chosen and operator
correct, but not finished correctly,
7
3
×
and stops.
e.g.
10 10
(5C)
P(ans, 1st att.) + P(ans, 2nd att.) + P(ans, 3rd att.)
3 7
7
3
 3 7
× 
 + × + ×
10
10
10
10
10
10

  
 
147
21
3
+
+
10 100 1,000
300 + 210 + 147
1,000
657
or 0⋅657
1,000
Low partial credit: (2 marks)
–
Any correct relevant step, e.g.
P(call answered) = P(1st att) + P(2nd att)
3
+ P(3rd att) or
+ ans. to part(i).
10
High partial credit: (4 marks)
–
P(ans, 3rd att) correctly calculated, but
fails to add to the other two probabilities.
Correct probabilities chosen and operator
correct, but not finished correctly,
147
21
3
e.g.
+
+
and stops.
10 100 1,000
–
2015.1 L.17/20_MS 46/68
Multiplication of values indicated, but
3
3
× .
one value incorrect, e.g.
10 10
Correct probabilities chosen, but operator
incorrect.
Find the probability that an agent gets through to a customer.
Scale 5C (0, 2, 4, 5)
(5C)
Page 46 of 67
DEB
exams
3(b)
(iii)
An agent has a list of six customers to contact. Find the probability that the agent will
get through to exactly two customers on the first attempt.
P(2 ans on 1st att.)
p=
=
n
  × pr × qn – r
r
7
3
, q = , n = 6, r = 2
10
10
2

P(2 ans on 1st att.)
=
=
=
=
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 47/68
(5C)
 6  3 
 7 
  ×   ×  
 10 
 2   10 
9
2,401
×
15 ×
100 10,000
324,135
1,000,000
0⋅324135
4
Low partial credit: (2 marks)
–
Any correct relevant step, e.g. recognition
3
,
of Bernoulli (binomial) trial, i.e. p =
10
7
q = , n = 6, r = 2 or writes down
10
n
  × pr × qn – r , but no substitution.
r
High partial credit: (4 marks)
–
Finds probability, but with one error in
4
2
 6  3 
 7 
components, e.g.   ×   ×   .
 10 
 2   10 
Page 47 of 67
DEB
exams
Question 4
(25 marks)
P (–1, 11), Q (2, 2) and R (x, y) are points such that | PR | = 2| QR |.
4(a)
Show that the set of all possible points of R lie on a circle.
| PR |








Scale 10D (0, 4, 6, 8, 10)
2
( x − (−1)) + ( y − 11)
2
(10D)
=
2| QR |
=
2 ( x − 2) 2 + ( y − 2) 2
(x + 1)2 + (y – 11)2
=
4[(x – 2)2 + (y – 2)2]
2
2
x + 2x + 1 + y – 22y + 121 =
4[x2 – 4x + 4 + y2 – 4y + 4]
2
2
x + 2x + 1 + y – 22y + 121 =
4x2 – 16x + 16 + 4y2 – 16y + 16
2
2
x + 2x + y – 22y + 122
=
4x2 – 16x + 4y2 – 16y + 32
3x2 + 3y2 – 18x + 6y – 90 =
0
x2 + y2 – 6x + 2y – 30
=
0
circle: 2nd degree equation, coefficients of x and y equal and no xy term
Low partial credit: (4 marks)
–
Correct expression for either | PR | or | QR |.
Middle partial credit: (6 marks)
–
Correct expressions for both | PR |
and | QR |.
High partial credit: (8 marks)
–
Correct equation found, but no reason
given as to why it represents a circle.
Substantially correct work with one error,
but reasons as to why it represents a
circle included.
–
4(b)
Write down the centre and radius-length of the circle.
x2 + y2 – 6x + 2y – 30

centre
=
=

radius-length
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 48/68
=
0
−6 2
(
,
)
−2 −2
(3, –1)
=
g2 + f 2 − c
=
(−3) 2 + (1) 2 − (−30)
=
9 + 1 + 30
=
=
**
(5C)
40
2 10
Accept student’s answer(s) from previous part(s) if not oversimplified.
Low partial credit: (2 marks)
–
Effort at relating one or more
coefficients of equation from part (a)
with general equation of a circle.
High partial credit: (4 marks)
–
Finds correct centre or radius-length.
Page 48 of 67
DEB
exams
4(c)
Find the slopes of the two tangents to the circle through the point (2, –8) outside the circle.




Equations of tangent
Point (2, –8), slope m
y – y1
=
m(x – x1)
y – (–8)
=
m(x – 2)
y+8
=
mx – 2m
mx – y – (2m + 8) =
0
⊥ distance from centre of circle to tangent
Circle
centre
=
(3, –1)
radius
=
2 10

⊥ distance
=
=
=
ax1 + by1 + c
a 2 + b2
m(3) − 1(−1) − (2m + 8)
(m) 2 + (−1) 2
3m + 1 − 2m − 8
m2 + 1
m−7
=

m−7
m2 + 1
m2 + 1
=
2 10
=
2 10

(m – 7)2
=
(2 10 )2( m 2 + 1 )2






(m – 7)2
m2 – 14m + 49
39m2 + 14m – 9
(3m – 1)(13m + 9)
3m – 1
3m
=
=
=
=
=
=
40(m2 + 1)
40m2 + 40
0
0
0
1
1
3

Scale 10D (0, 4, 6, 8, 10)
m
=
Low partial credit: (4 marks)
–
–
or
13m + 9
13m
=
=
m
=
0
–9
9
–
13
Any relevant first step, e.g. writes down
perpendicular distance formula and stops.
Substitutes (2, –8) correctly into equation
of a line formula.
Middle partial credit: (6 marks)
–
Substitutes correctly into perpendicular
distance formula, but fails to continue
or continues incorrectly.
High partial credit: (8 marks)
–
Finds correct quadratic equation, but fails
to finish or finishes incorrectly.
Substantially correct work with one error
and both slopes found.
–
2015.1 L.17/20_MS 49/68
(10D)
Page 49 of 67
DEB
exams
Question 5
5(a)
(25 marks)
Find the general solutions that satisfy the equation
2cos 3 = –1,
and hence, solve the equation in the domain 0° ≤  ≤ 2π.


2cos 3
=
cos 3
=
–1
1
–
2
cos 3
=
1
2
3
=
cos–1
=
π
3
Reference angle

(15D)

as cos 3 < 0
π
< 3 < π
2

3
=
π–

3
=
2π
3

or
General solution

3



∴
=
=

1
2
=
π
3
2π
+ 2nπ
3
2 π + 6 nπ
3
2π + 6nπ
9
–
–
Middle partial credit: (10 marks)
–
High partial credit: (13 marks)
–
Page 50 of 67
3π
2
3
=
π+
3
=
4π
3
3
=
3
=

=
Solutions in the domain 0° ≤  ≤ 2π
For n = 0
2π

=
or

9
For n = 1
2 π + 6π

=
or

9
8π
=
9
For n = 2
2π + 12π

=
or

9
14π
=
9
2π 4π 8π 10π 14π 16π
Solutions
=
{ ,
,
,
,
,
}
9
9
9 9 9
9
Scale 15D (0, 6, 10, 13, 15) Low partial credit: (6 marks)
2015.1 L.17/20_MS 50/68
or
π < 3 <
=
=
=
=
=
π
3
4π
+ 2nπ
3
4π + 6nπ
3
4π + 6nπ
9
4π
9
4 π + 6π
9
10π
9
4π + 12π
9
16π
9
Any correct relevant first step, e.g.
quadrants where 3 can be indicated.
Finds correct reference angle,
π
i.e. 3 = or 60°.
3
2π
4π
(40°) and
9
9
(80°), or equivalent values of 3,
2π
4π
and
, and fails to continue.
3
3
Finds both values of ,
General solution found correctly, but fails
to find all solutions in range 0 ≤  ≤ 2π.
DEB
exams
5(b)
The diagram shows two points on the graph of the function y = f (x).
y
(0, 3)
f (x)
π
( , 1)
2
x
(i)
(ii)
State whether f (x) is a sine or cosine function.
Give a reason for your answer.

Answer

Reason
(iv)
cosine function
–
–
Any 1:
maximum value of f (x) at x = 0 //
grapg of f (x) does not cross the midway line
at the y-axis // etc.
**
Accept other appropriate reasons.

Period
–
2π
or 120°
3

Range
–
[–1, 3]
Low partial credit: (2 marks)
–
One part correct.
Middle partial credit: (3 marks)
–
Two parts correct.
High partial credit: (4 marks)
–
Three parts correct.
Hence, find the equation of the graph.
(5C)
f (x)
=
=
a + b cos cx
1 + 2cos3x
f (100π)
=
f (150 ×
=
f(
=
3
Find f (100π).
Scale 5C (0, 2, 4, 5)
2015.1 L.17/20_MS 51/68
–
Write down the period and range of the graph.
Scale 5D (0, 2, 3, 4, 5)
(iii)
(5D)
2π
)
3
2π
)
3
Low partial credit: (2 marks)
–
Finds two of a, b or c correct in equation
f (x) = a + b cos px.
High partial credit: (4 marks)
–
Correct equation for f (x), but fails to find
or finds incorrect f (100π).
Page 51 of 67
DEB
exams
Question 6
6(a)
(25 marks)
Let ABC be a triangle.
Prove that, if a line l is parallel to BC and cuts [ AB ] in the ratio s : t, then it also cuts [ AC ]
in the same ratio.
(5B, 5B, 5B)
Given
Triangle ABC with line l parallel to BC
To prove
| AD |
| DB |
| AE |
| EC |
=
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
=
–
s
t
Given or To prove correct.
Diagram
A
D1
Dm =D
E1
l
E = Em
C = Em+n
Dm+n =B
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
–
Draws triangle with line l parallel to
one side.
Proof
We prove only the commensurable case.
Let l cut [ AB ] in D in the ratio m : n with natural numbers m and n.
Thus, there are points:
D0 = A, D1 , D2 , …, Dm – 1 , Dm = D, Dm + 1 , …, Dm + n – 1 , Dm + n = B,
equally spaced along [ AB ], i.e. the segments:
[ D0 D1 ], [ D1 D2 ], …, [ Di Di + 1 ], …, [Dm + n – 1 Dm +n ]
have equal length.
Draw lines D1 E1 , D2 E2 , … parallel to BC with E1 , E2 , … on [ AC ].
Then all the segments:
[ AE1 ], [ E1 E2 ], [ E2 E3 ], …, [ Em + n – 1 C ]
have the same length,
… [Theorem 11]
and Em = E is the point where l cuts [ AC ].
… [Axiom of Parallels]
Hence E divides [ AC ] in the ratio m : n
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 52/68
Partial credit: (2 marks)
Page 52 of 67
–
Misses one critical step or configures
steps in incorrect order.
DEB
exams
6(a)
(cont’d.)
Alternative Proof
Let ABC be a triangle.
Prove that, if a line l is parallel to BC and cuts [ AB ] in the ratio s : t, then it also cuts [ AC ]
in the same ratio.
(5B, 5B, 5B)
Diagram
A
s
s
X
Y
t
t
C
B
Given
Triangle ABC with line l parallel to BC
l cuts [ AB ] at X and the line [ AC ] at Y
and | AX | : | XB | = s : t
To prove
| AY | : | YC | = s : t
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
–
Given or To prove correct.
Construction
Divide [ AX ] into s equal parts and [ XB ] into t equal parts.
Draw a line parallel to BC through each point of the division.
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
–
Construction not fully complete or fully
explained.
Proof
∴
∴
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 53/68
The parallel lines make intercepts of equal length along [ AC ]
[ AY ] is divided into s equal parts
and [ YC ] is divided into t equal parts.
| AY | s
=
|YC | t
Partial credit: (2 marks)
Page 53 of 67
–
[Theorem 11]
Proof completed but one critical step
omitted e.g. justification why lines make
intercepts of equal length.
DEB
exams
6(b)
Two vertical poles AB and CD, of heights 9 m and 12 m respectively, stand on level ground,
as shown. Support cables from the top of each pole are anchored at the base of the opposite
pole. | EF | is the distance from the point of intersection of the two cables to the ground.
C
A
E
B
(i)
F
D
Prove that the triangles ABD and EFD are similar.

Scale 5B (0, 2, 5)
(5B)
Δ ABD and Δ EFD are similar as
| ∠ ABD |
=
| ∠ EFD |
| ∠ BDA |
=
| ∠ FDE |
| ∠ DAB |
=
| ∠ DEF |
Partial credit: (2 marks)
–
–
2015.1 L.17/20_MS 54/68
Page 54 of 67
... both right angles or 90°
... common angle
... remaining angle in triangle
Any correct relevant first step, e.g.
explains similar triangles.
Any correct step in showing similarity.
DEB
exams
6(b)
(ii)
Show that, no matter how far apart the two poles are located, | EF | =















But


Also, Δ BDC and Δ BFE are similar
| EF |
| BF |
=
| CD |
| BD |
| EF |
| BF |
=
12
| BD |
| EF | × | BD |
=
12| BF |
9| FD |
9
12
3
4
3
+1
4
7
4
7
4
| EF |
9
| EF |
9
=
| EF |
=
=
=
=
=
=
=
=
12| BF |
| BF |
| FD |
| BF |
| FD |
| BF |
+1
| FD |
| BF | + | FD |
| FD |
| BD |
| FD |
| FD |
| BD |
4
7
4
×9
7
36
7
Low partial credit: (2 marks)
–
–
High partial credit: (4 marks)
–
–
2015.1 L.17/20_MS 55/68
(5C)
Δ ABD and Δ EFD are similar
| EF |
| FD |
=
| AB |
| BD |
| EF |
| FD |
=
9
| BD |
| EF | × | BD |
=
9| FD |
=
Scale 5C (0, 2, 4, 5)
36
.
7
Page 55 of 67
... part 
Any correct relevant first step, e.g. states
Δ BDC and Δ BFE are similar.
States similar triangles have corresponding
sides in the same ratio or equivalent.
| BF | 3
= or equivalent, but fails
| FD | 4
to finish or finishes incorrectly.
| EF | 3
| EF | 4
Finds
= or
= .
9
7
12
7
Finds
DEB
exams
Section B
Contexts and Applications
150 marks
Answer all three questions from this section.
Question 7
(50 marks)
Table 1 below gives details of the number of births, the number of candidates who completed the
Leaving Certificate and the number of students who started college in Ireland for the first time in
five-year intervals from 1978 to 2013.
Year
Male
35,766
34,642
28,083
25,359
27,848
31,455
38,619
35,223
1978
1983
1988
1993
1998
2003
2008
2013
Births
Female
34,533
32,475
26,517
23,945
26,121
30,074
36,554
33,707
Total
70,299
67,117
54,600
49,304
53,969
61,529
75,173
68,930
Leaving Certificate Candidates
Male
Female
Total
16,381
19,423
35,804
19,719
24,139
43,858
24,556
26,603
51,159
27,977
29,253
57,230
31,333
34,589
65,922
28,532
31,004
59,536
27,015
28,528
55,543
28,133
27,444
55,577
Students started College
Male
Female
Total
6,979
4,899
11,878
8,666
7,737
16,403
10,709
9,637
20,346
14,403
13,613
28,016
16,693
18,223
34,916
17,937
19,750
37,687
18,462
21,052
39,514
19,845
21,522
41,367
(Source: Department of Education and Skills, http://education.ie)
7(a)
(i)
Represent the number of male and female candidates and the total number of candidates
who completed the Leaving Certificate in the period 1978 to 2013 on a multiple bar chart
(round your figures to the nearest thousand).
(10C)
Suitable multiple bar chart
Scale 10C (0, 4, 7, 10)
Low partial credit: (4 marks)
–
–
High partial credit: (7 marks)
(ii)
Presents total candidates only or male
candidates only or female candidates
only on a bar chart.
What trend(s), if any, do you observe from your chart? Give a reason for your answer.

Trends
–
–

Reason
–
–
–
**
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 56/68
–
Any relevant first step, e.g. one or both
axis scaled.
Attempts to round relevant figures to
nearest thousand.
Partial credit: (2 marks)
Page 56 of 67
(5B)
Any 1:
numbers of male and female candidates sitting the
Leaving Certificiate have increased substantially over
the period with both peaking in 1998 //
greater number of female candidates sitting the Leaving
Certificiate up to 2013 when the number of male
candidates exceeds the number of female candidates
for the first time over the period of the study // etc.
Any 1:
more boys than girls leaving school earlier to take up
employment opportunities or apprenticeships //
less interest from girls in jobs and occupations that
offered earlier start opportunities or apprenticeships //
less start opportunities and/or apprenticeships since the
end of the boom so more boys staying in education
and sitting their Leaving Certificate // etc.
Accept other appropriate answers.
–
Trend correct, but no reason or
incorrect reason given.
DEB
exams
7(a)
(iii)
Explain why the number of males born in 1998 has no bearing on the number of male
candidates who completed the Leaving Certificate in the same year.
Explanation
–
–
Scale 5B (0, 2, 5)
7(b)
(i)
Any 1:
male candidates sitting the Leaving Certificate in 1998
would generally have been born around 1979, 1980 or
1981 - thus those born in 1998 have nothing to do with
those sitting their Leaving Certificate in the same year //
males born in 1998 will not be sitting their Leaving
Certificate until at least 2014, 2015 or 2016 and so have
no bearing on the number sitting their Leaving
Certificate in that year // etc.
Partial credit: (2 marks)
–
Mean, x
=
=
=

standard deviation, 
=
=
Scale 5B (0, 2, 5)
(ii)
Partial credit: (2 marks)
2
 (x − x )
n
8,826⋅92 (calculator work)
–
Mean
x (excl. 1978)
=
=

Scale 5C (0, 2, 4, 5)
(5B)
 Number of candidates in each year
 Years
424,629
8
53,087⋅625
Mean or standard deviation correct.
What effect, if any, would excluding the data from 1978 have on your results in part (i)?
Refer to each of the measures of central tendency and variability to explain your answer.

2015.1 L.17/20_MS 57/68
Partial answer given, but incomplete
or insufficient.
Calculate the mean and standard deviation of the total number of candidates who
completed the Leaving Certificate over the period 1978 to 2013.

(5B)
(5C)
388,825
7
55,546⋅428571...
Explanation
–
mean would rise significantly from 53,087⋅625
to 55,546⋅428571... - which indicates that the
mean is being skewed by extreme results
(outliers), i.e. numbers of Leaving Certificate
candidates in 1978
Standard deviation
 (excl. 1978)
=
6,350⋅18 (calculator work)
Explanation
–
standard deviation would drop considerably
from 8,826⋅92 to 6,350⋅18 by removing the
outlier (1978) as it is dependent on the mean
Low partial credit: (2 marks)
–
New mean or standard deviation correct.
High partial credit: (4 marks)
–
Mean and standard deviation correct but
no explanation about how outliers affect
mean and standard deviation.
Page 57 of 67
DEB
exams
7(c)
A researcher wishes to analyse the relationship between female candidates who completed the
Leaving Certificate and female students who started college in the same year over this period.
The following data was extracted from Table 1.
Year
LC Candidates
Started College
(i)
1978
19,423
4,899
1983
24,139
7,737
1988
26,603
9,637
1993
29,253
13,613
1998
34,589
18,223
2003
31,004
19,750
2008
28,528
21,052
2013
27,444
21,522
Draw a scatter plot of the data.
(5C)
Suitable scatter plot
Scale 5C (0, 2, 4, 5)
Low partial credit: (2 marks)
–
Correct scale with at least 2 points
plotted correctly.
High partial credit: (4 marks)
–
Correct scale but not all points plotted
(1 or 2 omissions).
All points plotted but incorrect scale.
–
(ii)
Sketch the line of best fit in the completed scatter plot above.
Answer
Scale 5B (0, 2, 5)
(iii)
–
Partial credit: (2 marks)
–
–
Draws straight line but clearly not line
of best fit.
(5A)
r = 0⋅7258 (calculator work)
–
Scale 5A (0, 5)
Hit or miss.
What can you conclude from the completed scatter plot and the correlation coefficient?
Answer
–
–
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 58/68
See diagram
Calculate the correlation coefficient between female candidates who completed the
Leaving Certificate and females students who started college in the same year.
Correlation coefficient
(iv)
(5B)
Partial credit: (2 marks)
Page 58 of 67
(5B)
Any 1:
a strong positive correlation between numbers of
females who complete the Leaving Certificate and
going on to college //
shows that as more females complete the Leaving
Certificate, more of these females tend to go on to
college // etc.
–
Positive or strong positive correlation
and stops, i.e. not contextualised.
DEB
exams
Question 8
8(a)
(50 marks)
A report by the Private Residential Tenancies Board (PRTB)
claims that the average rent for one-bedroom apartments in
Dublin is €1,000 per month. In order to test this claim, a
sample of 60 one-bedroom apartments in a particular area
of Dublin was randomly selected. The mean rent for these
apartments is €1150 with a standard deviation of €400.
(i)
Using the sample of 60 apartments, find the 95% confidence interval for the mean
monthly rent of a one-bedroom apartment in this particular area of Dublin.






Scale 10C (0, 4, 7, 10)
(ii)
95% confidence interval
z-value = 1⋅96
σ
<
x – 1⋅96
N
1⋅ 96(400)
1,150 –
<
60
1,150 – 101⋅213964... <
1,048⋅786035...
<
€1048⋅79
<
<
x + 1⋅96

<
1,150 +



<
<
<
σ
N
1⋅ 96(400)
60
1,150 + 101⋅213964...
1,251⋅213964...
€1,251⋅21
Low partial credit: (4 marks)
–
Any relevant step e.g. 95% confidence
interval  z-value = 1⋅96 or defines
N = 60, σ = 400, etc.
High partial credit: (7 marks)
–
Fully substitutes into formula
σ
σ
x – 1⋅96
<  < x + 1⋅96
, but
N
N
fails to find interval or finds incorrect
interval.
Interpret your interval in this case.
(5B)
–
Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 59/68

(10C)
Partial credit: (2 marks)
Page 59 of 67
we can be 95% certain that the mean monthly rent
for a one-bedroom apartment in Dublin lies between
€1,048⋅79 and €1,251⋅21
–
Correct statement given, but not
contextualised.
DEB
exams
8(a)
(iii)
An estate agent suggests that the mean monthly rent for one-bedroom apartments in this
particular area is not representative of the whole city. Test this hypothesis using a 5% level of
significance. Clearly state your null hypothesis, your alternative hypothesis
and your conclusion.
(15D)
H0: Null Hypothesis:
The average monthly rent (for a one-bedroom apartment) in a particular area
of Dublin is representative of the whole city
H1: Alternative Hypothesis:
The average monthly rent (for a one-bedroom apartment) in a particular area
of Dublin is not representative of the whole city
z
=
=
=
=
≅
x−μ
σ
N
1,150 − 1,000
400
60
150
51⋅ 639777 ...
2⋅904737...
2⋅905
Conclusion
Since z = 2⋅905 is greater than 1⋅96, we reject H0 , the Null Hypothesis,
and accept H1 , the Alternative Hypothesis - the average monthly rent
(for a one-bedroom appartment) in this particular area of Dublin is not
representative of the whole city
–
Any relevant step e.g. states null
hypothesis.
Middle partial credit: (10 marks)
–
Finds correct z and standard error of
σ
mean (
).
N
High partial credit: (13 marks)
–
Fails to contextualise answer e.g. stops at
reject the Null Hypothesis.
Scale 15D (0, 6, 10, 13, 15) Low partial credit: (6 marks)
(iv)
Find the p-value of the test you performed in part (iii) above and explain what this value
represents in the context of the question.


Scale 5B (0, 2, 5)
2015.1 L.17/20_MS 60/68
Value
p-value
Explanation
(5B)
=
=
=
P(z ≤ –2·905) + P(z ≥ 2·905)
0·0019 + 0·0019
0·0038
–
since 0·0038 is less than 0⋅05 (level of significance), the
result is significant and we reject the Null Hypothesis
Partial credit: (2 marks)
Page 60 of 67
–
Correct value for p-value, but no
explanation or incorrect explanation.
DEB
exams
8(b)
A survey is being conducted at a busy traffic junction.
Based on a simple random sample of 300 cars, 78 were silver.
(i)
Find the 95% confidence interval for the proportion of cars
observed that were silver.

Sample Proportion
pˆ
=
=
σpˆ
=
=
=
=
=

78
300
0⋅26
pˆ (1 − pˆ )
N
0 ⋅ 26(1 − 0 ⋅ 26)
300
0 ⋅ 26(0 ⋅ 74)
300
0⋅ 000641333...
0·025324...
95% Confidence Interval




Scale 10D (0, 4, 6, 8, 10)
2015.1 L.17/20_MS 61/68
(10D)
pˆ – 1⋅96σpˆ
0⋅26 – 1⋅96(0·025324...)
0·26 – 0·0496361...
0·210363......
<
<
<
<
P
P
P
P
<
<
<
<
pˆ + 1⋅96σpˆ
0⋅26 + 1⋅96(0·025324...)
0·26 + 0·0496361....
0·309636...
95% confident that the true proportion of silver cars lies between 21·0363%
and 30·9636%
Low partial credit: (4 marks)
–
Any relevant 1st step, e.g. finds correct
value for pˆ .
Middle partial credit: (6 marks)
–
Finds correct value of the standard error
of the proportion, σpˆ .
High partial credit: (8 marks)
–
Finds 95% margin of error found
i.e. 1⋅96σpˆ = 0·0496361..., but no
confidence interval formed.
Page 61 of 67
DEB
exams
8(b)
(ii)
In order to be accurate to within 2 percentage points of the true proportion, what is the
minimum sample size necessary in the survey?


1⋅96
pˆ (1 − pˆ )
N
≤
0·02
1⋅96
0 ⋅ 26(0 ⋅ 74)
N
≤
0·02
0⋅1924
N
≤
≤

0 ⋅1924
N
N

N

Scale 10C (0, 2, 4, 5)
≤
≥
≥
≥
0⋅02
1⋅96
1
98
1
9,604
0·1924(9,604)
1,847·8096
1,848
Low partial credit: (2 marks)
High partial credit: (4 marks)
2015.1 L.17/20_MS 62/68
Page 62 of 67
(5C)
–
–
Any relevant first step, e.g. ME ≤ 0·02.
1
0⋅1924
≤ ,but fails to finish
N
98
or finishes incorrectly.
Finds
DEB
exams
Question 9
9(a)
(50 marks)
(i)
The diagram shows the configuration of fifteen
snooker balls, of radius r cm, which are packed
tightly into a rectangular box.
Calculate the internal volume of the box, in terms of r.
Volume
Scale 5C (0, 2, 4, 5)
(ii)
=
=
=
=
(5C)
l×w×h
(2r × 5) × (2r × 3) × 2r
10r × 6r × 2r
120r3
Low partial credit: (2 marks)
–
–
Any relevant first step, e.g. d = 2r.
Finds l, w or h, in terms of r.
High partial credit: (4 marks)
–
Calculates volume with one
non-arithmetic error, e.g. V = 60r3.
An alternative configuration of packing the snooker balls is shown below.
p
Using the right-angled triangle shown in the configuration, calculate p and hence,
investigate if this configuration is a more efficient method of packing the balls.





p
4r
2r
=
l×w×h
=
(2r × 5⋅5) × (2r + 2 3 r) × 2r
=
11r × (1 + 3 )2r × 2r
=
44(1 + 3 )r3
=
120.2102355... r3
>
120r3
less efficient method of packing the snooker balls
Volume

Scale 10D (0, 4, 6, 8, 10)
2015.1 L.17/20_MS 63/68
Using Pythagoras’s theorem
p2 + (2r)2
=
(4r)2
2
2
p + 4r
=
16r2
2
p
=
16r2 – 4r2
=
12r2
p
=
12 r
=
2 3 r cm
(10D)
Low partial credit: (4 marks)
–
Any relevant first step e.g. correct use of
Pythagoras’s theorem or writes down
hypoteneuse = 4r.
Middle partial credit: (6 marks)
–
Find correct value for p, but fails to
continue or continues incorrectly.
High partial credit: (8 marks)
–
Find incorrect volume, but at least two
dimensions correct and calculated
correctly with these dimensions.
Page 63 of 67
DEB
exams
9(a)
(iii)
The diagram shows a third configuration in which
the snooker balls can be packed.
Find the length of side of the triangle and hence,
calculate, in terms of r, the internal volume of the
triangular box required to pack the balls tightly
into the box.






Volume of box
=
area of equilateral triangle × height
Length of side
Angle in Δ
=
=
tan 30°
=
8r + 2x
60°
r
x
r
x
1
3
=
x
Length of side
=
=
=
Area of Δ
=
=
=
=
=
=
=
Area of Δ × 2r
3 (4 + 3 )2r2 × 2r
2 3 (4 + 3 )2r3
2 3 (16 + 8 3 + 3)r3
2 3 (19 + 8 3 )r3
(38 3 + 48)r3 or 113⋅817930...r3
–
Any relevant first step e.g. equilateral
triangle  all angles = 60°.
Middle partial credit: (10 marks)
–
Finds correct length of side, but fails to
continue or continues incorrectly.
High partial credit: (13 marks)
–
Finds correct area of triangle, but fails to
finish or finishes incorrectly.
Scale 15D (0, 6, 10, 13, 15) Low partial credit: (6 marks)
2015.1 L.17/20_MS 64/68
x
3 r cm
8r + 2 3 r
2(4 + 3 )r
=
=
Volume of Δ
r
30°
1
ab sin C
2
1
× 2(4 + 3 )r × 2(4 + 3 )r × sin 60°
2
1
3
× 2(4 + 3 )2r2 ×
2
2
3 (4 + 3 )2r2
=

(15D)
Page 64 of 67
DEB
exams
9(b)
A well-known snooker trick involves stacking balls in the shape
of a square-based pyramid. Up to thirty balls, of radius r cm,
are available to perform the trick.
(i)
Find the number of snooker balls used to form the base of
the pyramid if the largest-sized pyramid is constructed.


=
=
=
=
=
30
1
4
9
16
–
Some correct work with “square”
numbers, but base layer not correct.
Find the area of the base of the pyramid, in terms of r.
=
=
=
=
(5B)
l×w
(2r × 4) × (2r × 4)
8 r × 8r
64r2
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
–
Area calculated with incorrect side length.
Hence, by drawing a sketch, or otherwise, find the height of the pyramid, in terms of r.




Using Pythagoras’s theorem from diagram
(6r) 2
=
(3r) 2 + h2
2
36r
=
9 r 2 + h2
2
h
=
36r2 – 9r2
2
h
=
27r2
h
=
27 r
=
3 3r

Total height
Scale 10C (0, 4, 7, 10)
2015.1 L.17/20_MS 65/68
1×1
2×2
3×3
4×4
Number of balls used in bottom layer
=
4×4
=
16
Area of base
(iii)
=
=
=
=
Partial credit: (2 marks)
Scale 5B (0, 2, 5)
(ii)
Top layer
Layer 2
Layer 3
Layer 4
Total
(5B)
=
=
6r
(10C)
h
3r
3 3 r + 2r
(2 + 3 3 )r or 7⋅196152...r
Low partial credit: (4 marks)
–
Any relevant first step e.g. determines
hypotenuse = 6r or some use of
Pythagoras.
High partial credit: (7 marks)
–
Finds correct value of h, buts fails to find
or finds incorrect total height.
Page 65 of 67
DEB
exams
Notes:
2015.1 L.17/20_MS 66/68
Page 66 of 67
DEB
exams
Notes:
2015.1 L.17/20_MS 67/68
Page 67 of 67
DEB
exams
2015.1 L.17/20_MS 69/68
Page 69 of 67
DEB
exams