Junior Certificate Higher Level Maths Solutions
SEC 2013 Sample Paper 2
1. (a)
Area = Length × Width
=8m×4m
= 32 m2
(b)
Total length needed
= 5(semicircles) + length
= 5(3.14 × 2) + 8
= 31.4 + 8
= 39.4 metres
(c)
Length: = 12.5 m
Width: 6.78 m
since
since
2πr
____ + 2(0.25)
2
(3.14)(2) + 0.5
8 + 2 + 2 + 2(0.25)
length + diameter of circle + 2 buried pieces
Area = length × width
= 12.5 × 6.78
= 84.75 m2
(d)
Volume = Cylinder ÷ 2
πr2h
= ____
2
(3.14)(2)(2)(8)
= ____________
2
= 50.24 m3
(e)
Area of bed = (8 – 0.4) × (4 – 0.4)
= 7.6 × 3.6
= 27.36 m2
Volume of soil = 27.36 × 0.25 = 6.84 m3
Cost of topsoil = 6.84 × 0.75 × €80
= €410.40
1
2. (a)
600 – (92 + 101 + 115 + 98 + 105)
= 89
(b)
1
Based on the above results, I disagree. Each number should have a __ or 16.7% chance of
6
occurring. Currently, 5 has a 14.8% chance of occurring, whereas 3 has a 19.2% chance
of occurring.
(c)
Answer: 152
Reason: The probability of getting an even number from six hundred throws is
101 + 98 + 105 ____
304 ____
152
_____________
=
=
600
3. (a)
600
300
Example 1: Black jeans, White shirt, Black jumper and Boots.
Example 2: Black jeans, Red shirt, Black jumper and Flip-flops.
(b)
4. (a)
3 × 4 × 2 × 3 = 72 outfits
U
NY
B
1
9
3
7
5
(b)
5 __
1
___
=
(c)
9
___
(d)
5. (a)
35
4
5
1
S.F
7
35
9_________________
+ 3 + 4 + 1 + 1 + 7 ___
25 __
5
=
=
35
35 7
No. of hours 0–2 2–4 4–6 6–8 8–10 10–12 12–14 14–16 16–18 18–20 20–22
No. of students 11 31 18 13 11
3
1
1
6
1
4
(b)
2−4 hours
(c)
(1 × 11) + (3 × 31) + (5 × 18) + (7 × 13) + (9 × 11) + (11 × 3) + (13 × 1) + (15 × 1) +
650
(17 × 6) + (19 × 1) + (21 × 4) = ____ = 6.5 hours
100
(1) He is using a much smaller sample.
(d)
(2)
His sample consists of First Year boys only and ignores other years and the
opinions of girls.
(3)
His survey is being conducted after the mid-term break, when students would have
had more free time and probably spent many more hours on social networking
sites.
2
6. (a)
No. of beans 17
Brand A
0
Brand B
1
Brand C
0
5
22
0
2
0
23
1
0
0
24
0
2
0
25
2
0
3
26
2
0
2
29
0
5
2
30
0
0
2
31
0
0
1
32 33 34 35
2 1 1 1
0 0 0 0
0 0 0 0
Brand A
Brand B
Frequency
4
Brand C
3
2
1
0
(b)
17
22
23
24
25
26 29 30 31
Number of beans
32
33
34
35
Example answer: If I had to choose, I would buy Brand C. In Brand C, the mean number
276
of sweets is ____ = 27.6 sweets, compared with a mean of 24.5 sweets for Brand B
10
and 29.1 for Brand A. The reason I didn’t pick Brand A, even though it has a greater
mean, is because Brand C has a range of 6 (31 – 25) unlike Brand A (35 – 23) and
Brand B (29 – 17), which both have a range of 12, double that of Brand C.
This means there is a greater difference in the number of sweets between the biggest and
smallest packages.
When buying sweets, I’d expect a consistent number of sweets in any brand package
I buy.
⇒ 1 choose Brand C.
7. (a)
(b)
93.725
______
× 3165 = 824 schools
360
Angle
______
× Total no. of schools
360°
Example answer: I disagree, as the first pie chart represents 3,165 primary schools, but
the second chart only represents 729 post-primary schools.
Both angles are approximately 45°.
45
Primary: ____ × 3,165 = 396 schools
360
45
Post-primary: ____ × 729 = 91 schools
360
3
8. (a)
Answer: No
Reason: No two lengths are equal in measure; hence, an isosceles triangle with two sides
of equal measure cannot be constructed.
(b)
In a parallelogram, opposite sides are of equal length, but no two strips are of equal
measure. Hence, they cannot be used to form a parallelogram.
(c)
Use Pythagoras’ theorem:
Is (25)2 = (24)2 + (7)2 ?
25 cm
7 cm
Red
625 = 576 + 49
Yellow
625 = 625
24 cm
Blue
⇒ It is a right-angled triangle.
(d)
(Missing side)2 = (20)2 + (24)2
?
(Missing side)2 = 400 + 576
20 cm
White
____
Missing side = √976
= 31.2409987
24 cm
Blue
= 31.24 cm
9. (a)
The sin of |∠EAB| = opposite ÷ hypotenuse
which can be written as
|BE| ____
80
____
=
|AE|
(b)
120
D
|CD|
200
or ____ = _________
|AD| 120 + |ED|
200
80
sin |∠EAB| = ____ = _________
120 120 + |DE|
80(120 + |DE|) = 200(120)
9600 + 80|DE| = 24,000
12
80|DE| = 24,000 − 9,600 = 14,400
|DE| = 180 m
10. (a)
200 m
E
0
m
A
80 m
B
ΔADB is similar to ΔAPC:
C
A
|∠BAP| = |∠PAC| told.
x x
|∠ABD| = |∠APC| ... (x + y)
180 – 2x – y
|∠ADB| = |∠ACP| ... (180 − 2x − y)
Similar by A, A, A (3 angles the same).
180 – x – y
y
B
x
x–y
P
D
4
C
x+y
180 – 2x – y
P
(b)
B
x+y
x+y
180 – 2x – y
x
C
A
180 – 2x – y
D
|AC| ____
|EC| ____
|AP|
Small Δ ____
_______
·
=
=
Big Δ
|AD|
|BD|
|AB|
Cross multiply
|AC| ⋅ |BD| = |AD| ⋅ |PC|
11. Three angles of a triangle sum to 180°.
a + a + 73 + 60 = 180
2a = 47°
a = 23.5°
a + b + 73 = 180
b = 180 − 73 − 32.5
b = 83.5°
a + 60 + y + 73 = 180°
23.5 + 60 + y + 73 = 180
y = 180 − 156.5
y = 23.5°
12. (a)
4
1
3
2
5
x
A
(b)
x=y
A
B
C
D
A Central symmetry
B Axial symmetry (through the x-axis)
C Translation
D Axial symmetry (through the line x = y)
13. (a)
10
C
D
8
6
4
B
A
2
–2
2
4
6
8
10
12
–2
(b)
|AD|
=
(2,3)(4,8)
_______________
|BC|
(10,4)(12,9)
_________________
√(4 − 2)2 + (8 − 3)2 = √(12 − 10)2 + (9 − 4)2
_________
_________
√(2)2 + (5)2 = √(2)2 + (5)2
______
______
√4 + 25 = √4 + 25
___
___
√29 = √29
(c)
(
(
)
)
12 + 2 3 + 9
E = midpoint of |(2,3)(12,9)| = ______, _____ = (7,6)
2
2
10 + 4 4 + 8
F = midpoint of |(10,4)(4,8)| = ______, _____ = (7,6)
2
2
Since the midpoint of both diagonals |AC| and |BD| is (7,6)
⇒ the diagonals bisect each other.
(d)
No, we cannot. We would have to prove that opposite sides and opposite angles are equal.
6
14. (a) & (b)
l1
A
l2
15. (a)
Answer: False/incorrect
Reason: tan 60° = 1.732, which is greater than 1.
(b)
Answer: True/correct
Reason: sin 30° = 0.5, but sin 60° = 0.8.
(c)
False/incorrect
Reason: cos 30° = 0.866, but cos 60° = 0.5.
(d)
(2)2 = l2 + h2
3 = h2
__
2 cm
2 cm
√3 = h
√3
__
opposite
√3
sin 60° = __________ = ___
hypotenuse
2
60° 1 cm
2 cm
__
sin 60° =
16. The Leaning Tower of Pisa
√3
___
2
Tower of Suurhusen Church
27.37
55.863
A
A
3.9
2.47
3.9
cos A = ______
55.863
cos A = 0.069813651
2.47
cos A = _____
27.37
cos A = 0.090244793
A = cos−1(0.069813651)
A = cos−1(0.090244793)
A = 86°
A = 84.822°
The most tilted tower is the tower of Suurhusen Church, which makes an angle of 84.822°.
7
17. (a)
x + 4x + 90 = 180
5x = 90
x = 90 ÷ 5 = 18°
Hence, the other angle = 4(18) = 72°
Answers: 18°, 72°
(b)
Slope = tan x
= tan 18°
= 0.324919696
= 0.325
8
Junior Certificate Higher Level Maths Solutions
Sample Paper 2, no. 1
1. (a)
The missing side is
9 cm
(H)2 = 92 + 122
H2 = 225
H = 15 cm
H
12 cm
Total perimeter = 12 + 12 + 21 + 15 = 60 cm
(b)
(i)
Total surface area = Curved surface area of cylinder + Surface area of hemisphere
= 2prh + 2pr2
= 2(3.14)(3.75)(10) + 2(3.14)(3.75)(3.75)
= 235.5 + 88.3125
= 323.8125 cm2
= 323.81 cm2
(ii)
2
Volume = pr2h − __pr3
3
2
= (3.14)(3.75)(3.75)(10) − __ (3.14)(3.75)3
3
= 441.5625 − 110.390625
()
= 331.174375
= 331 cm3
(iii)
1 2
__
pr h = 331
3
1
__
(3.14)r2(10) = 331
3
10.4666666r2 = 331
()
r2 = 31.62420384
___________
r = √31.62420384
r = 5.62
r = 6 cm
(c)
(i)
Volume = 10 × 10 × 8 = 800 cm3
(ii)
44% of 800 cm3 = p(r2)(8)
352 cm3 = 25.14285714r2
14 = r2
___
√14 = r
3.74 cm = r
1
2. (a)
(b)
4 __
1
___
=
24
6
7
___
24
(c)
9
12
Denis is correct, since P (prime) = ___ but P (Even) = ___ .
24
24
(d)
10 ___
5
___
=
since nine numbers are greater than 10, the cost of the game, and the black is a
(e)
24 12
chance to win the maximum prize of €50.
Multiple possible answers. But the lower the entry fee the better one’s chances of scoring
more than the fee. However, the charity must break even at least, and have a little reserve
in case the wheel stops on the maximum prize.
3. A = Axial symmetry in the y-axis
B = Central symmetry
C = Translation
4. (a)
Categorical nominal
(b)
Categorical nominal
(c)
Numerical discrete
(d)
Numerical continuous
(e)
Categorical nominal
(f )
Numerical continuous
(g)
Categorical nominal
5. Diagram:
B
A
C
X
Given: A triangle with interior angles A, B, C and with exterior angle X.
To prove: X = A + B
Construction: None
Proof: X + C = 180° (straight angle)
X = 180° − C
Also A + B + C = 180° (by theorem)
A + B = 180° − C
Hence
X = A + B since both equal to (180 − C)°
2
6. (a)
R
5 cm
8 cm
10 cm
P
Q
(b)
52°
(use protractor)
(c)
30°
(use protractor)
(d)
The three angles of a triangle add up to 180 degrees, with the smallest angle opposite the
smallest/shortest side.
7. (a)
It is an isosceles triangle, since |OA| = |OD| (both radii).
(b)
Δ AOD → Δ COB
(c)
|∠COB| = 110°, since |∠OCB| = |∠OBC| (isosceles Δ) = 35°
180 − 35 − 35 = 110° (three angles of Δ)
8. (a)
(b)
9. (a)
If one event has m number of ways or outcomes and a second event has n number of ways or
outcomes, then the number of total outcomes possible by combining the events is m × n.
3 × 4 × 2 = 24
Multiple possible answers. Examples:
1. How many cigarettes do you smoke in a day?
none ❒,
less than 10 ❒,
more than ten ❒
2. What is your opinion of smoking?
3. When are you most inclined to smoke?
4. Do you agree with the banning of smoking in public places?
(b)
(c)
Usually, people at a sports complex would be health-conscious and anti-smoking.
38 + 44
Median = _______ = 41
2
Mode = 11
756
Mean = ____ = 37.8
20
The mean amount of cigarettes smoked was 37.8 per week but some smoked much more,
up to 69 per week.
3
10. (a)
(b)
Grade
A
B
C
D
E
F
No. of students
4
7
8
4
3
4
Multiple possible answers. Line plot shown below:
A
B
C
D
E
F
Grade
(c)
Multiple possible answers. In a line plot, the grades can be clearly laid out on the
horizontal axis and the number of dots indicates the frequency.
(d)
The modal grade is C, but in general 19 students got a grade A, B or C. However,
7 students did not pass.
11. Hypotenuse ⇒ 4,500 m = 60 mins
75 m = 60 sec
75
___
m = 12 sec
5
= 15 m
Opposite
sin A = __________
Hypotenuse
7.5
sin A = ___
15
( )
A = 30°
12. (a)
Tree diagram:
3
–
4
3
–
4
3
–
4
T
(b)
(c)
1
–
4
H
1
–
4
H
T
H
1
–
4
T
3 3 9
HH = __ × __ = ___
4 4 16
3 1 ___
3
HT = __ × __
=
4 4 16
3 ___
3
1 × __
=
TH = __
4 4 16
1 = ___
1
1 × __
TT = __
4 4 16
3 ___
6 __
3
3 ___
___
+
=
=
16 16 16 8
9 ___
10 __
5
1 ___
___
+
=
=
16 16 16 8
4
13. (a)
(b)
(c)
2y = −x + 5
5
1
y = − __x + __
2
2
1
slope = − __
2
3 + 2y − 5 = 0
2y = 2
y=1
A = (3,1)
1 . Since (0,0) → (−1,−2),
Since both p and q are parallel, then slope = − __
2
then (3,1) → (2,−1).
1
y + 1 = − __(x − 2)
2
2y + 2 = −x + 2
x + 2y = 0
(d)
⊥ to q so slope = 2.
y − 1 = 2(x − 3)
y − 1 = 2x − 6
2x − y − 5 = 0 ⇒ m
(e)
Find B.
2x − y = 5
x + 2y = 0
4x −2y = 10
x + 2y = 0
5x = 10
x=2
2(2) − y = 5
4−5=y
−1 = y
B = (2,−1)
|OB|2 + |AB|2 = |OA|2
2 + |(3,1)(2,−1)|2 = |(0,0)(3,1)|2
|(0,0)(2,−1)|
_______________
_________
(√
(2)2
+
5
14. (a)
) + ( √(2 −
2
(−1)2
3)2
+ (−1 −
+
5
=
h
sin 40 = __
8
h = 8 sin 40 = 5.1 m
(b)
_________
) ( √(3)2 + (1)2 )2
2
1)2 =
A
cos 40 = __
8
A = 8 cos 40 = 6.128355545
2A = w = 2(6.128355545)
= 12.3 m
5
10
(c)
The_________
diagonal of the door is
= √(2)2 + (3)2
2
_____
= √4 + 9
___
3
= √13
= 3.605551275 m
This is too small, so the sheet of wood will not fit in the door.
15. Mean = 6 ⇒ Total = 30.
Biggest number − smallest number = 4.
So the two missing numbers add up to 11.
The missing numbers are 4 and 7.
2
___
16. cos q = ____
√13
2
___
q = cos−1 ____
√13
q = 56.309°
tan 56.309 = 1.5
17. (a)
F
H
15 – x
x
12 – x
8
24 = 15 − x + x + 12 − x + 8
24 = 35 − x
x = 35 − 24
x = 11
Eleven students play both sports.
(b)
4 __
1
___
=
24
6
6
Junior Certificate Higher Level Maths Solutions
Sample Paper 2, no. 2
1. (a)
1
Area = __(16) × h = 44 cm2
2
8h = 44
44
h = ___
8
h = 5.5 cm
(b)
(i)
prl = 260p
26r = 260
260
r = ____
26
r = 10 cm
(ii)
l2 = h2 + r2
(26)2 = h2 + (10)2
676 − 100 = h2
____
√576 = h
24 cm = h
(c)
(i)
2
Volume = __p(2)(2)(2)
3
1
= 5_3 p cm3
(ii)
2
pr2h = 10_3p
2
(2)(2)h = 10_3
2
10_3
____
h=
4
h = 2.7 cm
(iii)
Total surface area = 2prh + pr2 + 2pr2
= 2p(2)(2.7) + p(2)(2) + 2(p)(2)(2)
= 10.8p + 4p + 8p
= 22.8p cm2
2. (a)
(b)
A corollary is a statement that follows readily from a previous theorem.
Multiple possible answers. Example: A diagonal divides a parallelogram into
two congruent triangles.
1
3. (a)
A
4
16
20
Male
Female
Total
(b)
20 ___
2
____
=
(c)
130 ___
13
51 + 79 ____
_______
=
=
(d)
13
26 ___
____
=
(e)
16 + 25 ____
41
_______
=
(f)
16 + 25 ___
41
_______
=
4. (a)
(b)
5. (a)
(b)
150
B
26
25
51
O
30
49
79
Total
60
90
150
15
150
150
15
75
150
150
150
90
90
4 × 2 × 3 = 24 choices
72
___
= 3 choices
24
3, 4, 6, 6, 7
or
3, 3, 6, 6, 7
or
3, 5, 6, 6, 7
4, 6, 6, 7
6. Diagram:
A
3 4
1
2
C
D
B
Given: A triangle ABC in which A is 90°
To prove: |BC|2 = |AB|2 + |AC|2
Construction: Draw AD ⊥ BC and mark in the angles 1, 2, 3, and 4
Proof: Consider the triangles ABC and ABD.
A
A
3
1
B
2
C
1
B
D
∠1 is common to both triangles
|∠BAC| = |∠ADB| = 90°
Δ ABC and ΔABD are similar.
|AB|
|BC| ____
____
=
|AB| |BD|
|AB|2 = |BC| · |BD| … Equation 1
2
Now consider the triangles ABC and ADC.
A
A
4
1
2
2
C
B
D
C
∠2 is common to both.
|∠BAC| = |∠ADC| = 90°
So ΔABC and ΔADC are similar.
|AC| ____
|BC|
____
=
|DC|
|AC|
|AC|2 = |BC| · |DC| … Equation 2
Adding Equation 1 and Equation 2 we get
|AB|2 + |AC|2 = |BC| · |BD| = + |BC| · |DC|
= |BC|(| BD| + |DC |)
= |BC| · |BC|
|AB|2 + |AC|2 = |BC|2
or
|BC|2 = |AB|2 + |AC|2
7. (a)
(b)
(c)
Line 1; slope is −3.
1.
Lines 3 and 4; slopes are − __
3
1
Lines 1 and 2, since −3 × __ = −1.
3
8. (a)
V
15 cm
9 cm
S
(b)
12 cm
T
Testing Pythagoras’ Theorem:
(15)2 = (12)2 + (9)2
255 = 144 + 81
255 = 255
True; therefore, the triangle is right-angled.
9. (a)
180 − 90 − 50 = 40°, since three angles in a triangles add up to 180°.
3
(b)
(c)
|QR|
cos 50 = ____
72
|QR| = 72 cos 50 = 46.3 m
First find |PR|.
|PR|
sin 50 = ____
72
|PR| = 72 sin 50
|PR| = 55 m
The man was already standing 127 m above sea level, therefore
P = 55 + 127 = 182 m above sea level.
10. (a)
Multiple possible answers.
1. Many people may be busy or working between the hours of 5 pm and 7 pm.
2. A phone survey may be limited to landlines or a specific mobile phone network, so
others, including those without phones, cannot take part.
(b)
No, as people generally get internet and phone packages together. So those who don’t
have phones often don’t use the internet and are not represented here.
(c)
Multiple possible answers.
A door-to-door questionnaire or a street survey where people are randomly selected at
various times of the day.
11. (a)
Let y = 0
2x – 3(0) + 9 = 0
2x = −9
1
x = − 4_2
1
P = (−4_2,0)
(b)
Let x = 0
2 (0) – 3y = –9
3y = 9
y=3
Q = (0,3)
(c)
y
3 Q
2
1
0O
–4 –3 –2 –1 0 1
–1
P
2
3
4 x
–2
–3
–4
4
(d)
1
Area = __(4.5) × 3
2
= 6.75 units2
(e)
( –4_12,0 ) (0,0) (0,–3)
( 4_12,0 ) (0,0) (0,–3)
( 4_12,0 ) (0,0) (0,–3)
Axial symmetry in x-axis
Central symmetry
Axial symmetry in y-axis
12. (a)
Multiple possible answers.
Stem
Leaf
2
5
3
6
7
9
4
3
3
6
7
7
8
9
9
5
0
1
2
3
4
6
6
6
6
2
2
5
7
2
6
7
7
9
Key: 7|2 = 72 km
(b)
Multiple possible answers. Examples:
A stem-and-leaf plot clearly displays all the raw data, whereas a line plot or bar chart
would be too cluttered, as the range in data is too large.
(c)
Given that the speed limit is 50 km/hr, more than half of all drivers break this limit. Also,
from the graph we can see that:
•
The modal speed is 56 km/ hr.
•
The median speed is 51.5 km/hr.
•
The range in speeds is 51 km (between 25 and 76 km/hr).
45
13. Homework = ____ × 24 hours = 3 hours
360
90
School = ____ × 24 hours = 6 hours
360
120
Sleeping = ____ × 24 hours = 8 hours
360
75
Leisure = ____ × 24 hours = 5 hours
360
Meals = Remainder = 2 hours
No. of hours
Sleeping
School
Homework
Meals
Leisure
8
6
3
2
5
5
14. X = Z, since both are alternate angles.
The three angles of a triangle add up to 180°.
180 = 70 + 2X + X
110 = 3X
36°40′ = X
Hence, X = Z = 36°40′
180° – 56° = 124 – 90 = 34
15.
R
S
34°
28°
118°
62°
62°
°
8
11
28°
28° 56°
28°
P
34°
Q
O
(a)
|∠QPR| = 34°, since PSRQ is a cyclic quadrilateal and all angles in each triangle add
up to 180°.
(b)
|∠QPS| = 28°+ 34° = 62° from the diagram above.
0−6
6
3
16. Slope of (0,6)(4,0) = _____ = − __ = − __
4−0
4
2
2
Point (−2,−1)
Slope of p = __
3
2
y + 1 = __(x + 2)
3
3y + 3 = 2x + 4
0 = 2x − 3y + 1
Answer: 2x − 3y + 1 = 0
17. l = 180 − 124°
l = 56°
(since isosceles Δ)
a = 180 − (56 − 56)
a = 180 − 112
a = 68° (3 angles in a Δ)
6
Junior Certificate Higher Level Maths Solutions
Sample Paper 2, no. 3
1. (a)
By Pythagoras’ theorem, the length of the side of the smaller square (x) is
x2 = 42 + 32
x2 = 16 + 9
___
x = √25
x=5
Area = 5 × 5 = 25 cm2
(b)
(i)
( )
22
1 2
_
pr = (0.5) ___ (3.5) (3.5)
2
7
= 19.25 m2
(ii)
( )
22
214.5 = 19.25 + 19.25 + (0.5) (2) ___ (3.5) (x)
7
214.5 = 38.5 + 11x
176 = 11x
16 m = x
(c)
(i)
( )
1 22
Volume of cone = __ ___ (15) (15) (28)
3 7
= 6600 cm3
= 6.6 litres
6.6
___ = 3.3 minutes
2
= 3 minutes 18 seconds
(ii)
( )( )
1 22
Volume of cone = __ ___ (7.5) (7.5) (14)
3 7
= 0.825 litres
Answer: No, it takes 24.75 seconds to fill.
This is eight times faster than the previous cone.
x + 2y = −1
2. (a)
4x − 2y = −19
5x
= −20
x
= −4
−4 + 2y = −1
2y = −1 + 4
2y = 3
1
y = 1 _2
B = (−4,1.5)
1
(b)
( 4, −2__12 ) or x + 2y + 1 = 0
(
1
)
(4) + 2 −2 _2 + 1 = 0
4−5+1=0
0=0
True
| ( 4, −2 _12 ) ( −4, 1_12 ) | = | ( −4, 1_12 ) (0, 9.5)|
__________________
_________________
1
1 2
1
1 2
_
_
_
2
2 + 9_
(−4
−
4)
+
1
+
2
=
(0
+
4)
−
1
(
)
(
2
2
2
2)
√
√
(c)
_______
_______
√64 + 16 = √16 + 64
___
___
√80 = √80
True
(d)
3. (a)
−4
1 , slope of q = ___
= +2
Slope of p = − __
−2
2
1 × 2 = −1
− __
2
Hence, p is perpendicular to q.
A = axial symmetry through the x-axis
B = central symmetry through the origin
(b)
Transformations change position but lengths, angles and areas remain the same.
(c)
4.
C
50°
A
40°
10 cm
B
5. Multiple possible answers, Examples:
(a)
1. Relevant data can be collected.
2. Unbiased.
(b)
1. Expensive to gather
2. Time consuming to go through
2
6. Multiple possible answers, Examples:
(a)
Questionnaire
Interview
(b)
Questionnaire: Disadvantages: 1. Time-consuming to gather data.
2. Expensive to have data analysed.
Advantages: 1. Questions you require answered.
2. Relevant data for his school.
Interview:
(c)
Advantages: 1. Further detail can be obtained.
2. Allows one to explain the purpose of the survey.
Disadvantages: 1. Students may be inclined to lie.
2. It is very time-consuming.
Examples:
1. Do people have access to the internet?
2. How much time do students spend on the internet?
3. What are the most popular websites?
(d)
Examples:
1. Do you use the internet?
2. How many hours per week do you spend on the internet?
3. For what purpose do you use the internet?
4. What websites do you use frequently?
(e)
7. (a)
(b)
8. (a)
(b)
Agree. Reasons: The school is all-male; also, the interests of his class are not reflective of
the interests of the older or younger students in his school.
Fuel type, colour, model = categorical nominal data
Engine size = numerical discrete data
2 × 5 × 3 × 2 = 60
Given that the adjacent = 208 m and the angle of elevation = 68°, use tan to find the
height of the building but add on 80 m to get the total height.
Height
Tan 68° = ______
208
Height = 208 tan 68° + 80
= 515 + 80
= 595 m
3
9. (a)
Multiple possible answers:
4
(b)
(c)
(d)
5
6
7
8
No. of potatoes per plant
9
Mode = 6 potatoes
(4 × 3) + (5 × 10) + (6 × 13) + (7 × 12) + (8 × 7) + (9 × 5)
Mean = ______________________________________________
50
325
= ____
50
= 6.5 potatoes
12 + 7 + 5
_________
× 100 = 48%
50
10. Multiple possible answers:
(a)
Similarity:
Both played 12 games. Both scored more than 130 runs but less than
200 runs per match.
Difference: The Mayo team scored between 133 and 159 runs in most of their games,
whereas the Galway team’s scores were more dispersed, as they achieved
higher scores of 175 to 190 runs more frequently than Mayo.
(b)
Mayo range
= 183 − 133 = 50
Galway range = 190 − 131 = 59
Galway have a greater range between their best and worst game.
(c)
Mayo modal runs = 152
Galway modal runs = 175
Galway’s frequent score (175 runs) is higher than Mayo’s frequent score (152 runs).
(d)
Mayo median
= 149.5
Galway median = 158
Galway have a greater median (or middle value).
(e)
Mayo interquartile range = 159.5 − 139.5
= 20
4
Galway interquartile range = 177.5 − 145.5
= 32
Galway have a greater interquartile range in their scores.
In conclusion, Patrick is incorrect.
11. The youngest child is 8, so that is the first digit in the sequence.
The middle number (or 3rd number) must be 13, the median.
The range is 17, so the largest number is 8 + 17 = 25.
Another age to: include is 15.
Since the mean is 14, the sum of all five children’s ages amounts to 70. Hence, the missing
child’s age is: 70 − 25 − 8 − 13 − 15 = 9
Answer: 8, 9, 13, 15, 25
12. (a)
(b)
(c)
6 3
Slope = __ = __
4 2
6
Slope = − __ = −3
2
Multiple possible answers:
13. (a)
4 boxes or
}
}
}
}
2 boxes
1 box
2 boxes
A cyclic quadrilateral is a four-sided figure that touches the inside of a circle at all four of
its vertices on the circle.
(b)
|∠PRT| = 90° (half the angle at the centre of the circle, by theorem)
(c)
|∠RPT| = 180 − 90 − 56 = 34° (alternate angle to |∠PRS| = 34°)
14. (a)
1
2
3
4
5
6
(b)
6
(c)
2
1
0
2
3
4
5
6
2
2
0
3
4
5
6
3
3
3
0
4
5
6
4
4
4
4
0
5
6
5
5
5
5
5
0
6
6
6
6
6
6
6
0
5
(d)
(e)
15. (a)
8 __
1
___
=
36 9
4 __
1
___
=
36 9
Fact 1: The three angles of a triangle add up to 180 degrees.
50° + x° + y° = 180°
Fact 2: The exterior angle of a triangle is equal to the sum of the two interior
opposite angles.
135° = 50° + y°
Fact 3: The angles in a straight angle add up to 180°.
x° + 135° = 180°
(b)
x = 180° − 135°
x = 45°
(Straight angle)
y = 180 − 45° − 50°
y = 85°
(3 angles in a triangle add up to 180°)
16. 5x + 3y + 4 = 4x + y + 8
⇒
x + 2y = 4
8x − y − 2 = 3x + 2y + 5
⇒ 5x − 3y = 7
x + 2y = 4
5x − 3y = 7
3x + 6y = 12
10x − 6y = 14
13x = 26
x =2
2 + 2y = 4
2y = 2
y=1
Answer: x = 2, y = 1
(
2 + 10 8 + 14
17. Midpoint = ______ , ______
2
2
22
12 , ___
= ___
2 2
(
)
)
= (6, 11)
6
Junior Certificate Higher Level Maths Solutions
Sample Paper 2, no. 4
1. (a)
2pr + 8(14)
( )
22
= 2 __
7 (14) + 8(14)
= 88 + 112
= 200 cm or 2 metres
(b)
(i)
2prh = prL
2(5)(12) = 10L
120 = 10L
12 cm = L
(ii)
(12)2 = h2 + 102
144 − 100 = h2
12 cm
___
√44 = h
6.6 cm = h
(c)
(i)
(ii)
10 cm
4
pr2h − __ pr3
3
p(8)(8)(16) − (1.333) p(8)(8)(8)
2
1024p − 682__p
3
1
= 341__ p cm3
3
1
pr2h = 341__ p
3
1
__
64h = 341
3
1
h = 341__ ÷ 64
3
1
h = 5 _3 cm
2. (a)
A theorem is a statement or rule that you can prove by following a certain number of
logical steps or by using other axioms or theorems that you already know.
(b)
Multiple possible answers. Example: Vertically opposite angles are equal in measure.
(c)
The converse of a theorem is the opposite or reverse of the theorem.
(d)
Multiple possible answers. Example: If two angles are equal in a triangle, then the sides
opposite these angles are equal (the triangle is isosceles).
1
C
3.
9 cm
6 cm
A
4. (a)
14 cm
B
A = (4,2), B = (6,2), C = (7,1), D = (3,1)
(b)
Vertices: (−4,2), (−6,2), (−7,1), (−3,1)
(c)
Vertices: (4,−2), (6,−2), (7,−1), (3,−1)
(d)
Vertices: (−4,−2), (−6,−2), (−7,−1), (−3,−1)
(e)
1 − 2 −1
Slope of AD = _____ = ___ = 1
3 − 4 −1
y − 2 = 1(x − 4)
y−2=x−4
0=x−y−2
Answer: x − y − 2 = 0
5. 1.
2.
3.
4.
6. (a)
(b)
Online survey
Postal survey
Face-to-face survey/questionnaire
Telephone survey
Grade
No. of students
A
12
B
10
C
7
D
1
Multiple possible answers.
12
10
8
6
4
2
0
A
B
C
D
(c)
Multiple possible answers. Examples: A stem-and-leaf plot only displays numerical data,
whereas a bar chart is the most common means of displaying categorical data.
(d)
The modal grade is A, occurring 12 times. Out of 30 students, 29 got a grade A, B, or C.
(e)
The mean is obtained by adding numerical data, but the data provided is categorical data.
7. (a)
Rachel, as only four bars are visible, reflecting the days she listens, and none on the other
three days.
(b)
Eoghan, as all bars are the same for all seven days.
2
(c)
Paul, as the graph shows more music was listened to at the weekend.
(d)
Sarah, as she only listens to her iPod from Monday to Friday, i.e. school days.
8. (a)
(b)
|∠SOQ| = 96° (= 2 × 48°)
|∠QRS| = 360 − 96 = 264 ÷ 2 = 132°
9. Prove that: Δ BAE ≡ ΔDCE:
Angles
|∠EAB| = |∠BCD|
Standing on same arc
Side
|AB| = |CD|
Given
Angles
|∠ABE| = |∠ADC|
Standing on same arc
Hence, Δ BAE and Δ DCE are congruent by ASA.
10. (a)
(b)
(c)
8
__
=4
2
6
__
=1
6
Multiple possible answers.
c
11. (a)
(b)
(c)
−1 + 3 _____
2+4
= (1,3) = M
,
( ______
2
2 )
2 1
4 − 2 __
______
= = __
3+1
4
Slope = −2
2
Point (1,3)
y − 3 = −2(x − 1)
y − 3 = −2x + 2
2x + y = 5
(d)
x − 2y = 0
2x + y = 5
x − 2y = 0
4x + 2y = 10
5x
= 10
(Mult 2)
2 − 2y = 0
y=1
N = (2,1)
x=2
3
12. Yes
H
Josh ⇒ tan 32 = ___
40
H = 40 tan 32 ≈ 25 metres
H
David ⇒ tan 50° = ___
21
H = 21 tan 50 ≈ 25 metres
13. (a)
1
G1
W1
R1
Green (G)
White (W)
Red (R)
(b)
(c)
(d)
(e)
(f)
1
G1
W1
R1
2
G2
W2
R2
6 __
1
___
=
18 3
4 __
2
___
=
18 9
1
2 __
___
=
18 9
360
− 45 − 90 ____
225 __
5
____________
=
=
360
360 8
P(Red) = 70% of 360°
= 252° − 45°
= 207°
14. (a)
(b)
cos 45°
1__
___
√2
sin 45°
1__
___
√2
tan 45°
1
To find x:
x__
cos 45 = ____
5√2
__
1__
x = 5√2 · ___
⇒5=x
√2
To find y (can also use Pythagoras’ Theorem)
y__
sin 45 = ____
5√2
__
1__
y = 5√2 · ___
⇒5=y
√2
To find z:
5
tan z = __
5
tan z = 1
z = tan−1 (1)
z = 45°
4
2
G2
W2
R2
2
G2
W2
R2
6
G6
W6
R6
15. 4 × 3 × 2 = 24
16. 2a = 180 − 58 ⇒ 2a = 122 ⇒ a = 62°
q = 71° (alternate angle)
b = 180 − 71 − 62 = 47°
Answer: a = 62°, q = 71°, b = 47°
5