1

Junior Certifi cate Higher Level Maths Solutions

SEC 2013 Sample Paper 2

1. (a) Area = Length × Width

= 8 m × 4 m

= 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

8 + 2 + 2 + 2(0.25) 2πr ____

2 + 2(0.25)

length + diameter of circle + 2 buried pieces (3.14)(2) + 0.5

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

2

2. (a) 600 – (92 + 101 + 115 + 98 + 105)

= 89

(b) Based on the above results, I disagree. Each number should have a 1 __

6 or 16.7% chance of

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

_____________

600 =

304 ____

600 =

152 ____

300

3. (a) Example 1: Black jeans, White shirt, Black jumper and Boots.

Example 2: Black jeans, Red shirt, Black jumper and Flip-fl ops.

(b) 3 × 4 × 2 × 3 = 72 outfi ts

4. (a) B

S.F

NY

7

5

13

1 49

5

U

(b) 5 ___

35 =

1 __

7

(c) 9 ___

35

(d) 9 + 3 + 4 + 1 + 1 + 7 _________________

35 =

25 ___

35 =

5 __

7

5. (a) 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) +

(17 × 6) + (19 × 1) + (21 × 4) = 650

____

100 = 6.5 hours

(d) (1) He is using a much smaller sample.

(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.

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6. (a) No. of beans 17 22 23 24 25 26 29 30 31 32 33 34 35

Brand A 0 0 1 0 2 2 0 0 0 2 1 1 1

Brand B 1 2 0 2 0 0 5 0 0 0 0 0 0

Brand C 0 0 0 0 3 2 2 2 1 0 0 0 0

0

1

2

3

4

Freq

uenc

y

Number of beans

5

17 22

Brand A

Brand B

Brand C

23 24 25 29 3026 31 32 33 34 35

(b) Example answer: If I had to choose, I would buy Brand C. In Brand C, the mean number

of sweets is 276

____

10 = 27.6 sweets, compared with a mean of 24.5 sweets for Brand B

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) 93.725

______

360 × 3165 = 824 schools

Angle

______

360° × Total no. of schools

(b) Example answer: I disagree, as the fi rst pie chart represents 3,165 primary schools, but the second chart only represents 729 post-primary schools.

Both angles are approximately 45°.

Primary: 45

____

360 × 3,165 = 396 schools

Post-primary: 45

____

360 × 729 = 91 schools

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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 ?

625 = 576 + 49

625 = 625

⇒ It is a right-angled triangle.

(d) (Missing side)2 = (20)2 + (24)2

(Missing side)2 = 400 + 576

Missing side = √ ____

976

= 31.2409987

= 31.24 cm

9. (a) The sin of |∠EAB| = opposite ÷ hypotenuse which can be written as

|BE|

____

|AE| =

80 ____

120 or

|CD| ____

|AD| =

200 _________

120 + |ED|

(b) sin |∠EAB| = 80

____

120 =

200 _________

120 + |DE|

80(120 + |DE|) = 200(120)

9600 + 80|DE| = 24,000

80|DE| = 24,000 − 9,600 = 14,400

|DE| = 180 m

10. (a) ΔADB is similar to ΔAPC:

|∠BAP| = |∠PAC| told.

|∠ABD| = |∠APC| ... (x + y)

|∠ADB| = |∠ACP| ... (180 − 2x − y)

Similar by A, A, A (3 angles the same).

25 cm

24 cmBlue

7 cmRed Yellow

?

24 cmBlue

20 cmWhite

120 m80 m

A B C

D

E200 m

A

D

C

B

P

180 – 2x – y

180 – 2x – y

180 – x – y x + y

x – yx

x x

y

5

(b)

AC

P

180 – 2x – y

x + y

x

AD

B

180 – 2x – y

x + y

x

Small Δ _______

Big Δ · |AC|

____

|AD| =

|EC| ____

|BD| =

|AP| ____

|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) 1

2

34

6

(b)

A B C D

x = y

A Central symmetry

B Axial symmetry (through the x-axis)

C Translation

D Axial symmetry (through the line x = y)

13. (a)

–2

–2

2

4

6

8

10

2 4 6 8 10 12

A

B

C

D

(b) |AD| = |BC|

(2,3)(4,8) (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) E = midpoint of |(2,3)(12,9)| = ( 12 + 2 ______

2 ,

3 + 9 _____

2 ) = (7,6)

F = midpoint of |(10,4)(4,8)| = ( 10 + 4 ______

2 ,

4 + 8 _____

2 ) = (7,6)

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.

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14. (a) & (b)

A

l1

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

√ __

3 = h

sin 60° = opposite

__________

hypotenuse =

√ __

3 ___

2

sin 60° = √

__ 3 ___

2

16. The Leaning Tower of Pisa Tower of Suurhusen Church

A

55.863

3.9

A

27.37

2.47

cos A = 3.9 ______

55.863 cos A =

2.47 _____

27.37

cos A = 0.069813651 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°.

2 cm60° 1 cm

2 cm 2 cm√3

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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

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Junior Certifi cate Higher Level Maths Solutions

Sample Paper 2, no. 1

1. (a) The missing side is

(H) 2 = 92 + 122

H2 = 225

H = 15 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) Volume = pr2h − 2 __ 3 pr3

= (3.14)(3.75)(3.75)(10) − ( 2 __ 3 ) (3.14)(3.75)3

= 441.5625 − 110.390625

= 331.174375

= 331 cm3

(iii) 1 __ 3 pr2h = 331

( 1 __ 3 ) (3.14)r2(10) = 331

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

12 cm

9 cm

H

2

2. (a) 4 ___ 24

= 1 __ 6

(b) 7 ___

24

(c) Denis is correct, since P (prime) = 9 ___ 24

but P (Even) = 12 ___ 24

.

(d) 10

___ 24

= 5 ___

12 since nine numbers are greater than 10, the cost of the game, and the black is a

chance to win the maximum prize of €50.

(e) 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)°

3

6. (a)

8 cm5 cm

10 cmP Q

R

(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) 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.

(b) 3 × 4 × 2 = 24

9. (a) 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) Usually, people at a sports complex would be health-conscious and anti-smoking.

(c) Median = 38 + 44

_______ 2 = 41

Mode = 11

Mean = 756

____ 20

= 37.8

The mean amount of cigarettes smoked was 37.8 per week but some smoked much more, up to 69 per week.

4

10. (a) Grade A B C D E F

No. of students 4 7 8 4 3 4

(b) 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 ___ 5 ) m = 12 sec

= 15 m

sin A = Opposite

__________ Hypotenuse

sin A = 7.5

___ 15

A = 30°

12. (a) Tree diagram:

H34–

14–

T

H34–

14–

T

H34–

14–

T

(b) 3 ___

16 +

3 ___

16 =

6 ___

16 =

3 __

8

(c) 9 ___

16 + 1 ___

16 =

10 ___

16 =

5 __

8

HH = 3 __ 4 × 3 __ 4 = 9 ___ 16

HT = 3 __ 4 × 1 __ 4 = 3 ___ 16

TH = 1 __ 4 × 3 __ 4 = 3 ___ 16

TT = 1 __ 4 × 1 __ 4 = 1 ___ 16

5

13. (a) 2y = −x + 5

y = − 1 __ 2 x +

5 __

2 slope = − 1 __

2

(b) 3 + 2y − 5 = 0 2y = 2 y = 1 A = (3,1)

(c) Since both p and q are parallel, then slope = − 1 __ 2 . Since (0,0) → (−1,−2),

then (3,1) → (2,−1).

y + 1 = − 1 __ 2 (x − 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

|(0,0)(2,−1)|2 + |(3,1)(2,−1)|2 = |(0,0)(3,1)|2

( √ _________

(2)2 + (−1)2 ) 2 + ( √ _______________

(2 − 3)2 + (−1 − 1)2 ) 2 = ( √ _________

(3)2 + (1)2 ) 2 5 + 5 = 10

14. (a) sin 40 = h __

8

h = 8 sin 40 = 5.1 m

(b) cos 40 = A __ 8

A = 8 cos 40 = 6.128355545

2A = w = 2(6.128355545)

= 12.3 m

6

(c) The diagonal of the door is

= √ _________

(2)2 + (3)2

= √ _____

4 + 9

= √ ___

13

= 3.605551275 m

This is too small, so the sheet of wood will not fi t 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.

16. cos q = 2 ____ √

___ 13

q = cos−1 2 ____ √

___ 13

q = 56.309° tan 56.309 = 1.5

17. (a)

15 – x 12 – x

8

F H

x

24 = 15 − x + x + 12 − x + 8

24 = 35 − x x = 35 − 24

x = 11

Eleven students play both sports.

(b) 4 ___ 24

= 1 __ 6

3

2

1

Junior Certifi cate Higher Level Maths Solutions

Sample Paper 2, no. 2

1. (a) Area = 1 __ 2 (16) × h = 44 cm2

8h = 44

h = 44 ___ 8

h = 5.5 cm

(b) (i) prl = 260p

26r = 260

r = 260

____ 26

r = 10 cm

(ii) l2 = h2 + r2

(26)2 = h2 + (10)2

676 − 100 = h2

√ ____

576 = h

24 cm = h

(c) (i) Volume = 2 __ 3 p(2)(2)(2)

= 5 1 _ 3 p cm3

(ii) pr2h = 10 2 _ 3 p

(2)(2)h = 10 2 _ 3

h = 10 2 _ 3

____ 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) A corollary is a statement that follows readily from a previous theorem.

(b) Multiple possible answers. Example: A diagonal divides a parallelogram into two congruent triangles.

2

3. (a) A B O TotalMale 4 26 30 60

Female 16 25 49 90

Total 20 51 79 150

(b) 20

____ 150

= 2 ___ 15

(c) 51 + 79

_______ 150

= 130

____ 150

= 13

___ 15

(d) 26

____ 150

= 13

___ 75

(e) 16 + 25

_______ 150

= 41 ____ 150

(f) 16 + 25

_______ 90

= 41 ___ 90

4. (a) 4 × 2 × 3 = 24 choices

(b) 72

___ 24

= 3 choices

5. (a) 3, 4, 6, 6, 7 or 3, 3, 6, 6, 7 or 3, 5, 6, 6, 7

(b) 4, 6, 6, 7

6. Diagram: A

1 2

3 4

DB C

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

1 2B C

A

1

3

DB

∠1 is common to both triangles

|∠BAC| = |∠ADB| = 90°

ΔABC and ΔABD are similar.

|BC|

____ |AB|

= |AB|

____ |BD|

|AB|2 = |BC| · |BD| … Equation 1

3

Now consider the triangles ABC and ADC.

A

1 2B C

A

2

4

D C

∠2 is common to both.

|∠BAC| = |∠ADC| = 90°

So ΔABC and ΔADC are similar.

|AC|

____ |DC|

= |BC|

____ |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) Line 1; slope is −3.

(b) Lines 3 and 4; slopes are − 1 __ 3 .

(c) Lines 1 and 2, since −3 × 1 __ 3 = −1.

8. (a)

15 cm9 cm

V

S T12 cm

(b) 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°.

4

(b) cos 50 = |QR|

____ 72

|QR| = 72 cos 50 = 46.3 m

(c) First fi nd |PR|.

sin 50 = |PR|

____ 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 specifi c 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

x = − 4 1 _ 2

P = (−4 1 _ 2 ,0)

(b) Let x = 0

2 (0) – 3y = –9

3y = 9

y = 3

Q = (0,3)

(c)

–4

–3

–2

–1

1

2

3

x

y

O

Q

P

–1 1 2 3 4–2–3–4 00

5

(d) Area = 1 __ 2 (4.5) × 3

= 6.75 units2

(e)Axial symmetry in x-axis ( – 4 1 _ 2 ,0 ) (0,0) (0,–3)

Central symmetry ( 4 1 _ 2 ,0 ) (0,0) (0,–3)

Axial symmetry in y-axis ( 4 1 _ 2 ,0 ) (0,0) (0,–3)

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 7 7 9

6 2 2 5

7 2 6

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).

13. Homework = 45

____ 360

× 24 hours = 3 hours

School = 90

____ 360

× 24 hours = 6 hours

Sleeping = 120

____ 360

× 24 hours = 8 hours

Leisure = 75

____ 360

× 24 hours = 5 hours

Meals = Remainder = 2 hours

Sleeping School Homework Meals Leisure

No. of hours 8 6 3 2 5

6

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′

15.

34°P O

Q

R

S

118°

118°34°

28°

28°

28° 56°28°

62° 62°

180° – 56° = 124 – 90 = 34

(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.

16. Slope of (0,6)(4,0) = 0 − 6

_____ 4 − 0

= − 6 __

4 = −

3 __

2

Slope of p = 2 __ 3 Point (−2,−1)

y + 1 = 2 __ 3 (x + 2)

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)

1

Junior Certifi cate 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) 1 _ 2 pr 2 = (0.5) ( 22 ___ 7 ) (3.5) (3.5)

= 19.25 m2

(ii) 214.5 = 19.25 + 19.25 + (0.5) (2) ( 22 ___ 7 ) (3.5) (x)

214.5 = 38.5 + 11x

176 = 11x

16 m = x

(c) (i) Volume of cone = 1 __ 3 ( 22 ___ 7 ) (15) (15) (28)

= 6600 cm3

= 6.6 litres

6.6 ___

2 = 3.3 minutes

= 3 minutes 18 seconds

(ii) Volume of cone = ( 1 __ 3 ) ( 22 ___ 7 ) (7.5) (7.5) (14)

= 0.825 litres

Answer: No, it takes 24.75 seconds to fi ll.

This is eight times faster than the previous cone.

2. (a) x + 2y = −1

4x − 2y = −19

5x = −20

x = −4

−4 + 2y = −1

2y = −1 + 4

2y = 3

y = 1 1 _ 2

B = (−4,1.5)

2

(b) ( 4, −2 1 __ 2 ) or x + 2y + 1 = 0

(4) + 2 ( −2 1 _ 2 ) + 1 = 0

4 − 5 + 1 = 0

0 = 0

True

(c) | ( 4, −2 1 _ 2 ) ( −4, 1 1 _ 2 ) | = | ( −4, 1 1 _ 2 ) (0, 9.5) | √

__________________

(−4 − 4)2 + ( 1 1 _ 2 + 2 1 _ 2 ) 2 = √ _________________

(0 + 4)2 + ( 9 1 _ 2 − 1 1 _ 2 ) 2 √

_______ 64 + 16 = √

_______ 16 + 64

√ ___

80 = √ ___

80

True

(d) Slope of p = − 1 __ 2 , slope of q = −4 ___

−2 = +2

− 1 __ 2 × 2 = −1

Hence, p is perpendicular to q.

3. (a) A = axial symmetry through the x-axis B = central symmetry through the origin

(b)

(c) Transformations change position but lengths, angles and areas remain the same.

4.

50° 40°10 cmA B

C

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

3

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: 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.

(c) 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) Agree. Reasons: The school is all-male; also, the interests of his class are not refl ective of the interests of the older or younger students in his school.

7. (a) Fuel type, colour, model = categorical nominal data Engine size = numerical discrete data

(b) 2 × 5 × 3 × 2 = 60

8. (a) Given that the adjacent = 208 m and the angle of elevation = 68°, use tan to fi nd the height of the building but add on 80 m to get the total height.

(b) Tan 68° = Height

______ 208

Height = 208 tan 68° + 80

= 515 + 80

= 595 m

4

9. (a) Multiple possible answers:

4 5 6 7

No. of potatoes per plant8 9

(b) Mode = 6 potatoes

(c) Mean = (4 × 3) + (5 × 10) + (6 × 13) + (7 × 12) + (8 × 7) + (9 × 5)

______________________________________________ 50

= 325

____ 50

= 6.5 potatoes

(d) 12 + 7 + 5

_________ 50

× 100 = 48%

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

5

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 fi rst 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 fi ve 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) Slope = 6 __

4 =

3 __

2

(b) Slope = − 6 __ 2 = −3

(c) Multiple possible answers:

13. (a) A cyclic quadrilateral is a four-sided fi gure 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

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

(b) 6

(c) 2

2 boxes

2 boxes

4 boxes or

1 box

} }}}

6

(d) 8 ___

36 = 1 __

9

(e) 4 ___ 36

= 1 __ 9

15. (a) 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

17. Midpoint = ( 2 + 10 ______

2 ,

8 + 14 ______

2 )

= ( 12 ___ 2 , 22 ___

2 )

= (6, 11)

1

Junior Certifi cate Higher Level Maths Solutions

Sample Paper 2, no. 4

1. (a) 2pr + 8(14)

= 2 ( 22 __ 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

√ ___

44 = h

6.6 cm = h

(c) (i) pr2h − 4 __ 3 pr3

p(8)(8)(16) − (1.333) p(8)(8)(8)

1024p − 682 2 __ 3 p

= 341 1 __ 3 p cm3

(ii) pr2h = 341 1 __ 3 p

64h = 341 1 __ 3

h = 341 1 __ 3 ÷ 64

h = 5 1 _ 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).

12 cm

10 cm

2

3. 9 cm

C

A B

6 cm

14 cm

4. (a) 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) Slope of AD = 1 − 2 _____ 3 − 4

= −1 ___ −1

= 1

y − 2 = 1(x − 4)

y − 2 = x − 4

0 = x − y − 2

Answer: x − y − 2 = 0

5. 1. Online survey 2. Postal survey 3. Face-to-face survey/questionnaire 4. Telephone survey

6. (a) Grade A B C DNo. of students 12 10 7 1

(b) Multiple possible answers.

A0

2

4

6

8

10

12

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, refl ecting the days she listens, and none on the other three days.

(b) Eoghan, as all bars are the same for all seven days.

3

(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) |∠SOQ| = 96° (= 2 × 48°)

(b) |∠QRS| = 360 − 96 = 264 ÷ 2 = 132°

9. Prove that: ΔBAE ≡ ΔDCE:

Angles |∠EAB| = |∠BCD| Standing on same arc

Side |AB| = |CD| Given

Angles |∠ABE| = |∠ADC| Standing on same arc

Hence, ΔBAE and ΔDCE are congruent by ASA.

10. (a) 8 __

2 = 4

(b) 6 __

6 = 1

(c) Multiple possible answers.

c

11. (a) ( −1 + 3 ______

2 , 2 + 4 _____

2 ) = (1,3) = M

(b) 4 − 2 ______ 3 + 1

= 2 __ 4 = 1 __

2

(c) Slope = −2 Point (1,3)

y − 3 = −2(x − 1)

y − 3 = −2x + 2

2x + y = 5

(d) x − 2y = 0

2x + y = 5 (Mult 2)

x − 2y = 0

4x + 2y = 10 2 − 2y = 0

5x = 10 y = 1 N = (2,1)

x = 2

4

12. Yes

Josh ⇒ tan 32 = H ___ 40

H = 40 tan 32 ≈ 25 metres

David ⇒ tan 50° = H ___ 21

H = 21 tan 50 ≈ 25 metres

13. (a) 1 1 2 2 2 6

Green (G) G1 G1 G2 G2 G2 G6

White (W) W1 W1 W2 W2 W2 W6

Red (R) R1 R1 R2 R2 R2 R6

(b) 6 ___

18 = 1 __

3

(c) 4 ___ 18

= 2 __ 9

(d) 2 ___ 18

= 1 __ 9

(e) 360 − 45 − 90

____________ 360

= 225

____ 360

= 5 __

8

(f) P(Red) = 70% of 360°

= 252° − 45° = 207°

14. (a) cos 45° sin 45° tan 45°

1 ___ √

__ 2 1 ___

√ __

2 1

(b) To fi nd x:

cos 45 = x ____

5 √ __

2

x = 5 √ __

2 · 1 ___ √

__ 2 ⇒ 5 = x

To fi nd y (can also use Pythagoras’ Theorem)

sin 45 = y ____

5 √ __

2

y = 5 √ __

2 · 1 ___ √

__ 2 ⇒ 5 = y

To fi nd z:

tan z = 5 __ 5

tan z = 1

z = tan−1 (1)

z = 45°

5

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°