Examination Revision and Solutions - PBworks

Year 10 General Mathematics 2013

Examination Revision and Solutions

Pythagoras’ Theorem

Trigonometry

Measurement

Statistics

Instructions:

We expect that you complete this revision before the final examination and strongly recommend that you commence now. Most of this revision should be completed at home.

Students are advised to review previously completed SAC's

Appropriate working must be included to gain maximum marks in the examination. Marks will be deducted in any question for poor presentation or the use of inconsistent mathematical statements. Refer to the solutions provided if necessary.

Scientific Calculator permitted.

Attempt all questions.

Give your answer to the required accuracy stated.

Marks will be deducted for missing units

PYTHAGORAS' THEOREM

Find the length of the hypotenuse, leaving your answer in surd form where appropriate: a b A

x cm 2 cm x cm

5 C ITI

5 cm •

2 Find the length of the hypotenuse, leaving your answer in surd form where appropriate:

5 cm 2 cm

8 cm x cm 0.6 cm x cm

3 Find the value of x in the following triangles, leaving your answer in surd form where appropriate:

6m

8 x m

4 Find the value of x in the following triangles, leaving your answer in surd form where appropriate:

6 cm

x cm 4.3 cm

7 7m

7 m

x m

Find the value of x in the following triangles. Where appropriate, leave your answer in surd form.

hk.N.Nn 3 m

Find the value of x in the following triangles. Where appropriate, leave your answer in surd form.

a 12 cm

2 cm 5 cm

• 17 cm x cm x cm

8 Find the value of x in the following triangles. Where appropriate, leave your answer in surd form. 9 Solve for x:

10 cm x cm

12 cm

12 Find the value of x in the following:

17 Find, correct to 2 decimal places, the value of the unknowns:

19 Find, correct to 2 decimal places, the value of the unknowns:

20 Find x, correct to 2 decimal places:

5 cm

y r8 cm

• 3 cm

23 Find the length of AB, correct to 2 decimal places. You may need to draw extra lines on your figure.

9 m rl 2 m

6m

A

29 a Is {2, 3, 5} a Pythagorean triple? Show your working.

b {6, 8, n} is a Pythagorean triple. Find n.

30 a Is {10, 15, 18} a Pythagorean triple? Show your working.

b {n, 30, 34} is a Pythagorean triple. Find n.

- -

37 A rectangle of length 14 cm has a diagonal of 19 cm. Find the width of the rectangle. 19 cm

14 cm

38 Dominic walks 15 km due east and then 10 km due north. Draw a diagram to show his course. How far is Dominic from his starting point?

41 An equilateral triangle has sides of length 18 cm. Find: a the length of one of its altitudes b the area of the triangle.

42 A sailing ship sails 46 km north then 74 km east. a Draw a fully labelled diagram of the ship's course. b How far is the ship from its starting point?

1 8 A

43 A 5 m long ladder leans against a brick wall. The feet of the ladder are 1.8 m from the base of the wall. How far up the wall does the ladder reach?

1.8 m

44 A power station P, supplies two towns A and B with power. New underground power lines to each town are required. The towns are connected by a straight highway through A and B and the power station is 42 km from this highway.

a Find the length of power line required from P to each town.

b Find the total cost of the new power lines given that each kilometre will cost $2350.

47 A cable car ascends from sea level to the top of a plateau, as shown in the diagram.

a How far does the cable car travel?

b If it travels at 3 metres per second, how long will the journey take?

A 13 km 45 kin

800 m

Three towns are situated so that B is 35 km in a direction 225°T from A, and C is 24 km in a direction 135°T from A. Find the distance from B to C to the nearest 100 metres.

A plane travels from A on a bearing of 66°T for 450 km to B, and then travels on a bearing of 156°T until it reaches C. If A and C are 600 km apart, what is the distance from B to C? A

-11— 650

A

C

51 A helicopter travels from base station S on a true bearing of 063° for 143 km for an emergency rescue. It then travels 157 km on a true bearing of 153° to an emergency hospital H. How far is the emergency hospital from the base station S?

53 The slant height of a cone is 15 cm and its height is 11 cm. Find the diameter of the base.

11 cm 15 cm

....... .C1 .....

49

50

PYTHAGORAS' THEOREM

SOLUTIONS

1 a x2 = 22 + 52 {Pythagoras} ...

x2 =4+25

... x2 = 29 x = {x >

•-. hypotenuse is 1/ cm

• x2 = 52 + 52 {Pythagoras} x2 =25+25

*. X2 = 50

x =/ {x > 0}

•.. hypotenuse is A/7) cm

2 a x2 = 52 + 82 {Pythagoras} ... x2 =25+64

x2 = 89

x =/ > 0}

... hypotenuse is Vg cm

• x2 = 22 + (0.6)2 {Pythagoras} x2 =

4+ 0.36

... x2 = 4.36 x =Jfl {x > 0}

... hypotenuse is -A L/ cm

3 a x2 = 72 +92 {Pythagoras} ... x2 = 49 + 81 ... x2 = 130

x lA/J {x > 0}

• x2 + 62 = 82 {Pythagoras} X2 + 36 = 64

x2 = 28 x = N/N {x > 0}

4 a x2 = 62 + (4.3)2 {Pythagoras} ... x2 = 36 + 18.49 ... x2 = 54.49

x = .A/519 {x > 0}

• X2 ± 72 = 102 {Pythagoras} x2 + 49 = 100

x2 = 51 x = {x > 0}

5 a

x2 +32 = 42 {Pythagoras} x2 + 9 = 16 ... x2 = 7

x = {x > 0}

b 12 ±

= (v75)2 {Pythagoras}

X2 + 8 = 15 ... x2 = 7

x = Ar7 Ix > 0}

6 a x2 + 122 = 172 {Pythagoras} • X2 + 144 = 289

x2 = 145 x {x > 0}

b x2 + 22 = 52 {Pythagoras} • X2 + 4 = 25

• x2 = 21 x = AM_ {x > 0}

8 a x2 +102 = 122 {Pythagoras} X2 + 100 = 144

x2 = 44 x = A/474 {x >

9 b x2 + x2 = 72 {Pythagoras} 2x2 = 49

x2 =

X = {x > 0}

12

4 cm

2 cm x2 ± 22 = 42 {Pythagoras} x2 + 4 = 16

x2 = 12

.*. x = {x > 0}

17 x2 = 22 + 12 {Pythagoras} x2 = 4 +1

... x2 = 5 x > 0} x 2.24 (2 d.p.)

Also, y2 = x2 + 12 {Pythagoras} y2 = 5 + 1

• y2 = 6 y = 'Vfi {y > 0} y 2.45 (2 d.p.)

19 x2 + 52 = 82 {Pythagoras} ... x2 + 25 = 64

x2 = 39 x = A/ {x > 0} x = 6.24 (2 d.p.)

Also, y2 = x2 + 32 {Pythagoras} y2 = 39 + 9

y = Vr8 {y > 0} y *. 6.93 (2 d.p.)

20 (2x) 2 = 42 +42 {Pythagoras} 4x2 = 16 + 16 4x2 = 32

x = N/g > 0} x ,* 2.83 (2 d.P.)

41 a h2 +92 = 182 {Pythagoras} h2 + 81 = 324

18 cm h2 = 243 h = th > 0} • h + 15.59 —0- 9 cm

i.e., the altitude is 15.6 cm.

b Area = base x height

1 x 18 x 15.59 • 2 140.3 cm2

74 km

46 km PIFFV-, km

start

x2 = 462 ± 742 {Pythagoras} x2 = 2116 + 5476 x2 = 7592

x = {x > 0} x = 87.13

the ship is 87.13 km from starting point.

43 X2 ± 1.82 = 52 {Pythagoras} x2 + 3.24 = 25

x2 = 21.76 x = A/21.76 {x > 0} x + 4.66

i.e., ladder reaches 4.66 m up the wall.

5 n/x m

1.8 m

42 a

23

m 2 m

6m

A 9m

AC = 9 m, BC = 2 + 6 = 8 m •

•

= 92 + 82 {Pythagoras} x2 = 81 + 64

• X2 = 145 x = A/1.5 {x > 0}

i.e., AB * 12.04 m long

29 a 22 + 32 = 4 + 9 = 13 and 52 = 25 22 + 32 52

{2, 3, 5} is not a Pythagorean triple.

62 + 82 = n2 36 + 64 = n2

n2 = 100 n = N/11:03 > 0} n = 10

30 a 102 + 152 = 100 + 225 = 325 and 182 = 324

• 102 + 152 182 {10, 15, 18} is not a Pythagorean triple.

n2 + 302 = 342 • n2 + 900 = 1156

n2 = 256 • n=/F; {n > 0}

n = 16

37

19 cm x cm 14 cm

x2 + 142 = 192 {Pythagoras} .*. x2 + 196 = 361

x2 = 165 x = {x > 0}

•

•

x + 12.8

i.e., the width of the rectangle is 12.8 cm.

38 AN

F x km 101cm

15 km x2 = 102 ± 152 {Pythagoras} X2 = 100 + 225

▪ x2 = 325 x = {x > 0} x + 18.03

i.e., Dominic is 18.03 km from his starting point.

44 a 13 km 45 km A •

42 km x km y km

X2 = 132 + 422 {Pythagoras} x2 = 169 + 1764 X2 = 1933

• x = {x > 0} • x + 43.97

i.e., 43.97 km of power line is required from P to A.

y2 = 422 ± 452 {Pythagoras} y2 = 1764 + 2025

.'. y2 = 3789 • y = > 0}

y 61.55

i.e., 61.55 km of power line is required from P to B.

b Total cost = (43.97 + 61.55) x $2350 = $247972

i.e., total cost is approximately $248000.

47 a

x ITA

800 m

650m

x2 = 6502 + 8002 {Pythagoras} x2 = 422 500 + 640 000 x2 = 1 062 500 x = •■./1 062 500 {x > 0} x+ 1031

i.e., the cable car travels 1031 m,

distance b time = —

speed

1031 .•. time = —

3 * 343.7 sec

time * 5 min 44 sec

49 225°

:16) 135°

45°45° 24 km

35 km

xkm

ZBAC = 450 + 450 = 90° • x2 = 352 + 242 {Pythagoras} • x2 = 1225 + 576

x2 = 1801 x = {x > 0} x + 42.44

i.e., the distance from B to C is 42.4 km.

51

153° 143 km 63° \

63° \ 157 km

27° \

x km

ZSRH = 63° + 27° = 90° • x2 = 1432 + 1572 {Pythagoras} • x2 = 20 449 + 24 649 • x2 = 45 098

x = V45 098 {x > 0} x + 212.4

i.e., the emergency hospital is 212.4 km from the base station.

53

11 cm 15 cm

r CM

---- ----

r2 + 112 = 152 {Pythagoras} r2 + 121 = 225 .•. r2 = 104

r = 1)/ > 0} .•. r + 10.20

2r * 20.40

i.e., the diameter of the base is 20.4 cm.

50

A

ri

600 km

LABC = 66° + 24° =90° x2 + 4502 = 6002 {Pythagoras}

x2 + 202 500 = 360 000 x2 = 157 500 x {x > 0} x+ 396.9

i.e., the journey from B to C is 396.9 km.

TRIGONOMETRY

I find: 2 find: a cos 0° a sinO° b sin90° b cos 60° C sin30° C sin20°

3 find: 4 find:

a sin60° a cos 90° b cos 50°

b sin50°

C tan45°

C tan 0°

5 Use your calculator to find, correct to 4 decimal places: a sin 43° b cos 43° c tan 43°

6 Use your calculator to find, correct to 4 decimal places: a cos 18° b sin 76° c tan 81°

7 Use your calculator to find, correct to 4 decimal places: a sin 28° b cos 16° c tan 54°

8 Use your calculator to find, correct to 4 decimal places: a tan 64° b cos 73° c sin 23°

9 In the diagram alongside, name the: • a hypotenuse b side opposite the angle marked a

c side adjacent to DCYZ. x

10 In the diagram alongside, name the: A a hypotenuse b side opposite the angle marked 0 c side adjacent to BCA.

13 For the right angled triangle alongside, find: a sin 0

P

b cos c tan0 •

14 a Find x. b Find sin 0, cos 0 and tan O.



12.7 m

• x m

21 Find, to 2 decimal places, the unknown length in:

a ,diegiAo m x m


a x cm

IriFr'21 cm

23 Find, to 2 decimal places, the unknown length in: a 2 cm

•

x cm 60'


a 1.95 km

x km.

111DPIPP- •


a

x cm

9 cm


a

12 cm

• x cm

29 Find, correct to 2 decimal places, the value of 0 if: a sin 0 = 0.654 b cos 0 = 0.1

30 Find, correct to 2 decimal places, the value of 0 if:

a tan 0 = 5.3 b sin 0 =

31 Find, correct to 2 decimal places, the value of 0 if: a cos 0 = 0 b tan 0 =

3.6 mm

Slam

28°

•

11.2m

62° x m

35 cm •

61° x cm

18.4 cm •

x cm 62°

C tan° = 1

„ 9.6 c cos iy =

11—

.2

c sin 0 = 0.95

33 Find, to one decimal place, the measure of the angle marked 0 in:

7 cm

6 cm r


13 mm

8 mm •

7.4m 3.2 m

•


•

5.7 cm 8.2 cm 13.4m

11.5 m


st

6.1 m

8.2 m

37 Find x correct to 3 significant figures:

44416 cm

23 cm

11 cm x cm m 5 m

13.6° •

38 Find x correct to 3 significant figures: 11.6 cm

x 20 cm

1711.9 cm

39 Find x correct to 3 significant figures: 4.62 m

5.1 m

.'4111111111Irm .v m

40 Find x correct to 3 significant figures:

mm

x mm

x" 11.1 cm

45 Find the height of a flagpole which casts a shadow of 9.32 in when the Sun makes an angle of 63° to the horizontal.

63° —A

-■••

9.32 m

46 A surveyor standing at A measures the angles sub-tended by two posts B and C on the opposite side of a canal. The posts are 120 m apart. If the angle of sight between the posts is 37°, how wide is the canal?

B 120m C

canal

A

47 An aeroplane moves from directly overhead to a position where the angle of elevation is 10°, in 2 4!■111ft. minutes. Find the speed of the aeroplane given -

....----- 000 m that it is constant and that the aeroplane is tray- . ----- ' .... ----- - to° elling at a constant height of 4000 m above the ground ground.

48 Two people of the same height stand at A and B at the top of two buildings of heights 335 m and 280 m. The buildings are 185 m apart. Find the angle of elevation of B from A.

49 Find the angle subtended at the centre of a circle of radius 12 cm by a chord of length 15 cm.

........... 7° :2:§§§2i ... ....... . A m===-Lin -

335 m 280m

- - - .

185 m -

52 Find the area of an isosceles triangle which has base angles of 37° and equal sides of length 15 cm.

53 From an observer 0, the angles of elevation to the bottom and the top of a flagpole are 36° and 38° respectively. Find the height of the flagpole. (Hint: Find AB and AC.)

000000 01:101100

36° A., 38.

0 200 m A

56 From point A, the angle of elevation to the top of a tall building is 20°. On walking 80 m towards the building the angle of elevation is now 23°. How tall is the building?

57 Find the bearing of X from Y if the bearing of Y from X is: a 034°

b 190° c 335°

58 Find the bearing of R from Q if the bearing of Q from R is: a 3150 b 220° c 007°

59 Find the bearing of:

a B from A

b B from C

c C from A

261°

61 An orienteer runs in a direction 042° for a distance of 7 km and then in the direction 132° for 12 km. Find: a the distance from the starting point b the bearing of the final position from the starting point.

62 Mary is in a direction 337° and a distance 23 km from Sue. Jane is in a direction 067° and a distance 35 km from Sue. Find:

a the distance of Jane from Mary b the bearing of Jane from Mary

TRIGONOMETRY SOLUTIONS

1 a cos 0° = 1

b sin90° = 1 c sin 30° = 0.5

2 a sin 0° = 0

b cos 60° = 0.5 c sin 20° * 0.34

3 a sin 60° * 0.87

b cos 50° * 0.64 C tan 45° = 1

4 a cos 90° = 0

b sin 50° * 0.77 c tan 0° = 0

5 a sin 43° * 0.6820

b cos 43° 4:0.7314

c tan43° * 0.9325

6 a cos 18° 0.9511

b sin 76° * 0.9703

c tan 81° * 6.3138

7 a sin 28° = 0.4695

h cos 16° 0.9613 c tan 54° 1.3764

8 a tan64° 4:2.0503

b cos 73° * 0.2924

c sin 23° * 0.3907

9 a ZY b XY c XY

10 a BC b AC c AC

13 a sin 0 = —r

b cos 0 = c tan 0 = —p

14 a 10 X2 ± 102 = 132 {Pythagoras} •

X2 + 100 = 169

13 .•. x2 = 69 0 X = .10 {X > 0}

b sin 0 = ' cos 0 = ' 69 tan 0 = 3-0 13 13

15 a X2 = 62 ± 112 {Pythagoras} 0 x .•. X2 = 36 + 121

6 x2 = 157 •

11 x =1,/W7 > 0}

b sin 0 = —, cos 0 = 4—, tan 0 = 1‘,/7 v157 6

16 a X2 = 12 + 12 {Pythagoras} • X2 = 1 + 1

1 1 X2 = 2

0 x=/ >

b sin° = cos0 =N/P tan0 = 1

21 a tan 47° = • 10 lOrn

10 x tan 47° = x v m 47° x = 10.72

i.e., length 10.72 m

cos 37° = —x

12.7 m 12.7 12.7 x cos 37° = x

x = 10.14 x m i.e., length * 10.14 m

22 a sin 71° = -x x cm 21

21 x sin 71° = x

• x * 19.86 IPPPIPP-21 cm i.e., length * 19.86 cm

cos 33° = -12

12 x cos 33° = x

.*. x * 10.06 12 cm • i.e., length * 10.06 cm

x cm

2 23 a sin 60° = -

x x x sin 60° = 2 2 cm

• 2

sin 60°

x 4' 2.31 x cm 60°

i.e., length * 2.31 cm

tan 28° = -3.6

x x tan 28° = 3.6 in

3.6 X M

tan 28° 28° .*. x * 6.77

i.e., length * 6.77 mm

24 a sin 41° = -x 1.95 km

1.95 41°

1.95 x sin 41° = x x km

• 1.28 • i.e., length * 1.28 km

• cos 62° = -

11.2

11.2m x x cos 62° = 11.2 62°

11.2 x m

cos 620

x * 23.86 i.e., length * 23.86 m

25 a tan 51° = -9

x x tan 51° = 9 x cm

9 .441111

tan 510 9 cm

x * 7.29 i.e., length * 7.29 cm

35 cm sin 61° = -35

x x sin 61° = 35 35 x cm

sin 610

x * 40.02 i.e., length * 40.02 cm

26 a cos 23° = -x

12 12 cm 12 x cos 23° = x

iiil • i.e., length * 11.05 cm x cm

18.4 • 18.4 cm r„,

sin = - x

x x sin 62° = 18.4 x cm 62° 18.4

sin 62°

x * 20.84 i.e., length * 20.84 cm

29 a sin 0 = 0.654 b cos 0 = 0.1 ;. 0 = sin-1 (0.654) .'. 0 = cos-1 (0.1)

* 40.84° 0 * 84.26°

c tan 0 = 1 ;. 0 = tan-1 (1)

0 = 45°

31 a cos 0 = 0 b tan 0 = A 7 0 = COS-1(0) .'. 0 = tan-1 (1) 0 = 90°

c sin 0 = 0.95 = sin (0.95)

0 * 71.81°

33 a tan 0 = •

1 7 cm 6

0 = tan-1 (i)

• 04'49.4° 6 cm

0

b sin 0 = 11 m 0

• = sin-1 (*) 0 4'27.0° • 5m

34 a cos 0 = 13

• 0 = cos- ' (i)

• 0 52.0°

b sin 0 = 7.4 0 = sin-1 ( : 24 )

.'.

0# 25.6°

35 a tan 0 = -8.2 5.7

0 = tan-1 (N)

;. * 55.2°

b cos 0 = -11.5 13.4

0 = cos ' (113)

• 0# 30.9°

30 a tan 0 = 5.3 b sin 0 = i- ... 0 = tan-1(5.3) .*. 0 = sin'()

.6 mm .*. 0 *. 79.32° .., 0 = 30°

> c cos 0 = M

(M)

.'. 0 * 31.00°

13 mm

8 mm •

7.4 m 3.2m

0 •

5.7 cm • 8 - 2 cm

13.4 m 11.5 m

•

4.62 m 44°

x m

-‘1111.

6.1 m

148.2 m

0

16 cm

cm a

x

37 a sin 13.6° = -11 11

... 11 x sin13.6° = x x 13.6° •

... x * 2.59

b tanx° = a 7 ..diellik7 5 ;. X° = tan-1 ()

35.5

38 a • 20 cm x cm

58°

20 tan 58° = -

x x x tan 58° = 20

20

x tan58°

x * 12.5

b cos 3.7° - 11-q - 17.9 . =_ cos-1(11.6\

k 17.9 / x4: 49.6

39 a sin x° = 5.1

• X ° = sin-1 ( 1.2 ) • •

5.1 x4: 49.9

b tan44° 4.62

x = 4.62 x tan44°

.•.

x4: 4.46

8.7

40 a • sin 53° = -x

8.7

x mm sin 53° 10.9

b tanx° = 2 cmtak....4.._

xo=ta,1„ ) ll.,

x. 10.2 11.1 cm

45

h m

63° • 9.32m

tan 63° = - 9.32

9.32 x tan 63° = h h 18.29

i.e., the flagpole is 18.29 m high.

120 46 tan 37° =

d x tan37° = 120 1

• d 20

. -

tan 37°

d 159.2

i.e., the canal is 159.2 m wide.

47 •

4000 m= 4 km 10° x r

ground 4

tan 10° = -x

x x tan10° = 4 4

tan 10°

x * 22.685 km

distance travelled speed =

time taken

22.685 {2 mm = to hr}

hour 30 * 680.6 km/h

48 B a° • ------------ - -- --- A ----

335 m C 185m EE 280 mr-r------ =-:41

angle of elevation is a°

BC = 335 - 280 m = 55 m

Now tan a = a = tan-1 (*) a * 16.56

i.e., the angle of elevation is 16.56°.

49

36 a sin0 = 8.2

.'. 0 = sin-1 (H)

.*. 48.1°

b tan 0 = 16 0 = tan-1 (t)

55.2°

B 120m C

d m

A

sin (g) sin-1

• 2 = 38.682.... • • 2 77.36°

i.e., the angle at the centre is 77.36°

52 15 cm

b cm 37°

a 'cm a cm

a cos 370 =

15 sin 37° =

10 a = 15 x cos 37° b = 15 x sin 37° a * 11.980 .*. b 9.027

Now area = base x height

= x 2a x b 2

= ab * 11.980 x 9.027

area * 108.1 cm2

53

0 200 m A

AB In OAB, tan 36° = —

200

200 x tan 36° = AB AB * 145.31

In AOAC, tan 38°

200 200 x tan 38° = AC

AC* 156.26

Now, BC .= AC — AB .•. BC * 156.26 — 145.31* 10.9

i.e., the flagpole is 10.9 m in height.

56 T

Ah m

20° B , A-4– gorn–s-4— xm —1.-C

h In ABTC, tan 23° = —

x ... h=xxtan23°

h In AATC, tan 20° =

x + 80

... h = (x + 80) x tan 20°

... x x tan23° = (x + 80) x tan 20° ... x tan 23° = x tan 20° + 80 tan 20°

... x tan 23° — x tan 20° = 80 tan 20° ... x(tan 23° — tan 20°) = 80 tan 20°

80 tan 20° tan 23° — tan 20°

80 tan 20° h= x tan 23°

tan 23° — tan 20°

h = 204.28

i.e., the building is 204.3 m high.

57 a

1461) 214°

34°

X

bearing of X from Y is 214°

17031 190°

10°


155°

25° •335°


58 a

135°

45°

‘73 15°

bearing of R from Q is 135°

140b 220° 40°


173 187°

7°


61 59

]

261° A -,--..-

a bearing of B from A is 058°

b 0 + 151 = 180 {co-interior angles} ... 0 = 39

0 + 261 + 0 = 360 {angles at a point} ... 39 +261 + 4) = 360

...

4) = 60

... bearing of B from C= 261+ 60 = 321°

c a + 58 + 29 +4) = 180 {angles in a Al ... a + 58 + 29 + 60 = 180

.-. a = 33 Thus 58° + a° = 91°

... bearing of C from A is 091°

62 N

NI cl' e x km

------ J

N 23 km

67° 35 km 23° &Lit

WV 337°

a LMSJ is a right angle

x2 = 232 + 352 {Pythagoras} :. x2 = 529 + 1225 ... x2 = 1754 ... x = .‘/1754 {x > 0} ... x + 41.88

.-. Mary and Jane are 41.88 km apart.

b tan0 = N ... 0 = tan-1 ()

... 0 + 56.7

But 0 + 0 + 23 = 180 {co-interior angles} .'. 0+56.7+23= 180

100.3

... Jane is on a bearing of 100.3° from Mary.

132°

N

km N 12 km

S x lat- F

a LSXF is a right angle

x2 = 72 + 122 {Pythagoras} :. x2 = 49 + 144 :. x2 = 193

... x = \ 1/ {x > 0}

... x + 13.89

.-, distance from starting point is 13.89 km.

b Now tan 0° = 1,

... 0° = tan'()

.'. 0° * 59.74°

Thus 0° + 42° + 101.7°

... bearing of final position from starting point is 101.7°.

1 Convert the following lengths:

2

3

4

a 41.6 km to m b 920 cm to m c 11 250 mm to m

Convert the following lengths: a 52 mm to cm b 0.53 m to mm c 1145 cm to km

Convert the following lengths: a 3120 mm to m b 0.076 lcm to cm c 2.3 lcm to mm

Convert the following lengths: a 46.7 cm to mm b 2500 m to km c 24 km to mm

5 Jane runs 6.4 km while Jenny walks 2600 m. How much further did Jane run than Jenny walk?

6 Due lives 2.7 km from school. He runs the first 800 m on his way home, then walks the rest of the way. How far does he walk?

9 Find the perimeter of: a a triangle with sides 12.2 cm, 11.4 cm and 10.8 cm b a rhombus with one side 7.9 cm c a circle with diameter 8 cm.

10 Find the perimeter of: a an equilateral triangle with sides 2.7 cm b a rectangle with adjacent sides 2 m and 3.2 m c a circle with radius 3 m.

11 Find the perimeter of: a an isosceles triangle with equal sides 13.4 cm and a third side 11.2 cm b a rhombus with sides 7.1 m c a circle with radius 10 cm.

13 Find the perimeter of:

5 cm

14 Find the perimeter of

120° 7 cm

16 Find the perimeter of a sector of radius 8 cm and angle 150°.

18 Find the perimeter of the given shape:

10 cm

19 Find the perimeter of the following shape:

,1400 6 cm

20 Find the perimeter of the following shape: 3 cm

(You may need to use Pythagoras' theorem.)

9 cm _I.

25 Determine the length of fencing around a 60 m by 150 m rectangular playing field if the fence is to be 15 m outside the edge of the playing field.

26 A new house is to have eight aluminium windows, identical in shape to that shown in the diagram. The outer framing costs $9.50 per metre and the inner slats cost $4.20 per metre.

Find the total cost for the framing and slats.

27 A farmer wishes to subdivide his property into six sections as shown. If he wishes to fence his entire property, as well as each section, with fencing which costs $0.34 per metre, find the total cost of the fenc-ing.

1.2 In

1.5 in

700 m

1100 m

28 A farmer wants to fence his house garden to keep out wandering cattle. The garden measures 26 in by 40 m. The fence is to have five strands of wire, and posts are to be placed every two metres.

a Calculate the perimeter of the garden. b What length of wire is needed? c Now many posts are needed? d Find the total cost of the fence if wire is $0.34 per metre and each post costs

$3.95.

31 Find the radius of a circular fish pond with circumference 5 m. (Answer to the nearest cm.)

41 Find the shaded areas: a

l4kmfl10

km

121cm

42 Find the area of the following figures: a 4 cm Atm

12 cm 43 Find the area of the following figures:

a

13 m

•1 11.. 10 m

44 Find the shaded area of the following figures: a

-411111111111111111d 5 cm 3 cm 1.14 cmi,

45 Find the area of the following figure: 10 cm

12 cm

20 cm

47 Find the area of the following figure:

9m

48 Find the area of the following figure:

6.4 m T1) 3 m

49 Find the area of: a a trapezium with parallel sides 10 cm and 6 cm, and height 12 cm b a sector of radius 10 cm and angle 240°

52 8 m

'11111111111010 m

Find: a the radius of the semi-circle b the perimeter of the figure c the enclosed area.

63 Find a formula for the area A, of the following region:

69 Find the surface area of the following solids: a 00 4 cm

cm

10 cm

Alo; cm

15 cm

71 Find the surface area of the following solids:

a

5 cm

if4r4(040c; 3 cnAidhl

6 cm 5 cm

73 Find the total surface area of the following solids a

8 cm

74 Find the total surface area of the following solids: a

..

-4- 4 cm

80 Find the radius of a sphere of surface area 500 cm 2 .

91 Find the volume of: a a sphere of radius 5.2 cm b a cone of radius 12 cm and slant height 15 cm.

92 Find the volume of: a a cylinder of height 16 cm and base diameter 40 cm b an equilateral triangular prism with triangles of side 4 cm and length 10 cm.

93 A swimming pool has the dimensions shown. a Find the area of the trapezium-shaped side. b Find the volume of water needed to fill the

pool. c Find the capacity of the pool in:

i kilolitres ii litres. d If water runs into the pool at 80 L per minute,

how long will it take to fill?

20 in 111111P1P-- 1 in

.- 2.5 m 6 111

97 A glass vase is hemispherical, as shown, with a hemi-spherical hollow for water. Find the approximate volume of glass, if the vase is slightly flattened on the bottom for stability.

100 2000 pipes with dimensions as shown are to be made of concrete. Find:

a the area of the concentric circular end b the volume of each pipe c the total weight of concrete required to make the

pipes, given that each cubic metre of concrete weighs 3.17 tonnes.

SOLUTIONS

I-I— 12 cin 0.8 cm

II 1.4m end IIII 2 m

I* 1.2 m PI

1 a

c

2 a

c

41.6 km = (41.6 x 1000) m = 41 600 m

11 250 mm = (11 250 ÷ 1000) m = 11.25 m

52 mm b = (52 ± 10) cm = 5.2 cm

1145 cm

b

= =

920 cm = (920 ± 100) m = 9.2 in

0.53m (0.53 x 1000) mm 530 mm

3 a 3120 mm = (3120± 1000) m = 3.12 m

b 0.076 km = (0.076 x 1000 x 100) cm = 7600 cm

c 2.3 km = (2.3 x 1000 x 1000) mm = 2 300 000 mm

= (1145 ± 100 ÷ 1000) Icin = 0.01145 km

A

4 a 46.7 cm b 2500m = (46.7 x 10) mm = (2500 ÷ 1000) km = 467 mm = 2.5 km

c 24 km = (24 x 1000 x 1000) mm = 24 000 000 mm

19 Perimeter = semi-circle diameter 12 cm + 2 semi-circles diameter 6 cm

=ix7rx 12 ± 2xix7rx6 = 67r + 67r = 127r * 37.70 cm

5 Jane runs 6.4 km Jenny walks 2600 m = (2600 +1000) km

= 2.6 km Difference = 6.4 - 2.6

= 3.8 km

i.e., Jane runs 3.8 km further.

20

3 cm

x cm =4 cm

-4-3 cm-P-4-3 cm-P-4-3 cm-1.-

By Pythagoras

x2 = 32 + 42 x2 = 9 + 16

x2 = 25 x =

{x > x = 5

6 Due walks 2.7 km - 800 m = 2.7 km - (800 ÷ 1000) km = (2.7 - 0.8) km = 1.9 km

i.e., Duc walks 1.9 km

9 a Perimeter b Perimeter = 12.2 + 11.4 + 10.8 cm = 4 x 7.9 cm

= 34.4 cm = 31.6 cm

c Perimeter =7r x 8 cm *25.13 cm

10 a Perimeter b Perimeter = 3 x 2.7 cm = 2 x (2 + 3.2) m

= 8.1 cm = 2 x 5.2 m

c Perimeter = 10.4 m

=2 x7rx 3m * 18.85 m

11 a Perimeter b Perimeter = 13.4 + 13.4 + 11.2 cm = 4 x 7.1 m

= 38 cm = 28.4 m

c Perimeter =2 x7rx 10 cm * 62.83 cm

13 Perimeter = 5 -F 5 -F (t) x 27r x 5

= 10 + (t) x 107r * 13.93 cm

14 Perimeter = 7+ 7 + (*) x 27r x 7

147r * 28.66 cm

16 Perimeter =8+8±( 131-g)x 27r x 8 g cm

= 16 + (*) x 167r 150°

36.94 cm

18

P = 10+10+7rd = 20 ± ir x 10 * 51.42 cm

perimeter = 9 + 5 +3 + 5 = 22 cm

25 180m

150 m

90m 60 m -0- -4-- 15 m

Length of fencing = 2 x (90 + 180) = 540 m

26 Length of outer framing for 1 window = 2 x (1.5 ± 1.2) = 5.4 m

Length of inner slats for 1 window = 2 x 1.5 +2 x 1.2 = 5.4 m

Total cost for 8 windows = 8 x (5.4 x $9.50 + 5.4 x $4.20) = $591.84

27

700 m

Length of outer fences = 2 x (1100 ± 700) = 3600 m

Length of inner fences: Now AB ± CD = 700

EF + GH = 700 IJ + GK = 1100

length of inner fences = 700 + 700 + 1100 = 2500 m

total length of fencing = 3600 + 2500 = 6100 m

total cost of fencing = 6100 x $0.34 = $2074

10 cm

44 a By Pythagoras

h2 + 9 = 25 h2 + 32 = 52 {Pyth}

.*. h cm :. h2 = 16

{h > 0} 5 cm 3 cm

... h = 4

141cnin 10 km

12 km

A = area large circle — area small circle A = 7r x 42 — ir x 22 A = 37.70 cm2

2 cmdi

4 cm

45

31 C 27rr 5 = 27rr

5 r =

r * 0.796 m r * 80 cm

41 a A = (a ±2

x h

A = (14 - F 1(:)

• 2 x 12

.• A = 12 x 12 A= 144 lcm2

Area shaded = x 5 x 4 = 10 cm2

Af% —D] 4 cmk-

Area = area large semi-circle — area small semi-circle

= x (7r x 82 ) — 1 X (7r X 42 ) 2 * 75.40 cm2

42 a 4 cm

\tri h em

/4 cm h 4 gm d 4 cm

12 cm Area = area of A + area of B

A — 10 cm

12 cm

B E20cm

47 2m

9m

21/1

Area = area of A + area of B + area of C = (9 x 2) + (4 x 5) +( x 2 x 4) =18+20+4 = 42 m2

area = (a+ b

) x h 2

/4 + 12 \ = 2)

= 8 x 3 = 24 cm2

A= () xirx62

• A = x 367r 360

A * 42.41 cm2

28

a

b

c

d

Length

Total

40 m

43 a

b

13 t AilW •

1-4=0.4.■ 0-1

Number

26 m

P = 2 x (26 -F 40) P = 132 m

of wire required = 132 x 5 {5 strands} = 660 m

of posts = 132 ±2 = 66

cost = 660 x $0.34 + 66 x $3.95 = $485.10

5m 10 m

h2 + 52 = 132 {Pythagoras} ... h2 + 25 = 169

.*. h2 = 144 ... h = 12 {h > 0}

area = base x height = (5 + 10) x 12 .=. 180 m2

A = (35) x 7r x 82

A = x 647r

A = 30.72 cm2

h2 42 = 52 {Pythagoras} = (12 + 10 + 12) x 10 + 10 x 20 h2 + 16 = 25 = 340 + 200

h2 = 9 = 540 cm2 h = 3 {h >

48 3.4m

6.4m

3.4m

3 m

Area = area square 6.4 m x 6.4 m — area triangle base 3.4 m, height 3.4 m

= 6.4 x 6.4 — x 3.4 x 3.4

69 a The figure has 2 faces of 10 cm x 8 cm 2 faces of 10 cm x 4 cm 2 faces of 8 cm x 4 cm

total surface area = 2 x 10 x 8 + 2 x 10 x 4 + 2 x 8 x 4 = 160 + 80 + 64 = 304 cm2

b The figure has 1 square base 4 triangular faces

cm =

49 a

35.18 m2

6 c

12 cm

+ b Area h

= =

=15x15

— 15 cm

15 cm total surface area

+ 4 xx 15 x 10 225 + 300 525 cm2

= (a 2

x

12 = x /6+ 10 )

2

= 96 cm2 . 10 cm

to cm b Area = (32460o) X ir X 102

240° * 209.4 cm2

52 8m •

r m

10 m r m

71 a The figure has 4 faces of 8 cm x 5 cm 2 faces of 5 cm x 5 cm

.-. total surface area =4 x8x5 + 2 x5 x5 = 210 cm2

b The figure has 1 face of 6 cm x 4 cm 1 face of 6 cm x 3 cm /ON 1 face of 6 cm x 5 cm

3 cm 4 cm 2 faces of i x 3 cm x 4 cm

a (2r)2 +82 = 102 {Pythagoras}

5 cm

The triangle with sides 3 cm, 4 cm, 5 cm is right angled as 32 + 42 = 52 {Pythagoras}

4r2 + 64 = 100 4r2 = 36 ... total surface area

r2 = 9 =6 x4+6 x 3+6 x 5+2x x 3 x 4

r = 3 {r > 0} = 24 + 18 + 30 + 12 i.e., radius = 3 m

P = 8 + 10 + (27rr) • P=18+7rx 3 • P = 27.42 m

c A = area of triangle + area of semi-circle = x 8 x 2r + 1 (7rr2 )

=8 x 3 + xirx 32

* 38.14 m2

3r.

Area = area of square — area of circle = 3r x 3r — T-7.2 =9r2 — 7rr2 = (9 — 7r) r 2

= 84 cm2

73 a Total surface area = 27rrh 27rr2 {r = 2, h 8} = 27r x 2 x 8+ 27r x 22

125.7 cm2

74 a Total surface area = 47rr2 {r = 5} = 47r x 52

314.2 cm2

80 A = 4R-r2 500 = 4n-r2

..•

r2=

47r

r= / {r > 0}

r+6.308

i.e., radius is 6.31 cm

63

91 a V = n-r3

= x ir x 5.23 3

* 589.0 cm3 15 cm

b h2 + 122 = 152 cm {Pythagoras} • .........

h2 + 144 = 225 —11.1 12 cm N-

.'. h2 81

h = 9 {h > 0}

V = iirr2 h

=ix7rx 122 x9 * 1357.2

i.e., volume is 1357.2 cm3

92 a V = area of end x height = r x 202 x 16 * 20 106 cm2

411k111 4 cm 10 cm

h2 +22 _42

{Pythagoras} Akcm h2 + 4 = 16 • • h2 = 12 2 cm

• h = N/T2 {h > 0}

V = area of end x height = 1 x4 xVf2 x 10

2

* 69.28 i.e., volume is 69.28 cm3

93 a 20 m

1m 2.5 m

Area of trapezium (2.5 + 1)

x 20 2

= 35 m2

b V = area of end x length = 35 x 6 = 210 m3

i.e., volume of water is 210 m 3

c i 1 ms = 1 kL capacity = 210 kL

ii 210 kL = 210 x 1000 L = 210 000 L

210 000 d time taken =

80 = 2625 minutes = 43i hours

97 outer radius = 6 cm, inner radius = 5.2 cm Approximate volume of glass

= outer volume — inner volume =•x 7rx63 — 1 xi x 7rx5.23

2 3 2 3

* 157.9 cm3

100

end 1.4m

1.7 m

a area of end = area of large circle — area of small circle =7rx (0.7)2 — 7rx(0.6)2 * 0.4084 m2

b volume = area of end x length * 0.4084 x 2 * 0.8168 m3

c total weight * 0.8168 x 2000 x 3.17 tonnes * 5179 tonnes

8 Families at a school were surveyed and the number of children in each family was recorded. The results of the survey are shown along-side:

STATISTICS

7 Construct a vertical column graph for the fol-lowing data:

9

Number of lollies in a packet Frequency

23 2 24 6 25 9 26 15 27 14 28 9 29 3

The following marks were scored for a test where the maximum mark was 40:

a Construct a vertical column graph of the data.

Number of children Frequency 1 5 2 28 3 15 4 8 5 2 6 1

Total 59

38 24 24 21 29 32 37 35 32 35 15 30 36 31 37 4 35 22 38 18

a Construct an ordered stem-and-leaf plot for the data.

10 The weights (in kg) of pumpkins grown by a market gardener are recorded below: 2.6 3.4 3.7 6.0 5.8 5.2 6.7 5.6 7.8 4.5 3.0 7.0 5.1 4.5 5.3 4.6 9.5 7.2 3.7 4.1

a Construct an ordered stem-and-leaf plot for the data using stems 2, 3, 4, etc.

c Copy and complete: "The modal weight was between .... and .... kg."

11 The maximum daily temperature (in °C) was recorded (correct to 1 decimal place) in Queenstown in April as follows:

14.1 16.2 18.7 19.2 20.5 20.7 21.4 22.2 18.6 19.0 18.2 18.1 16.9 16.4 17.2 16.8 15.2 15.6 14.9 15.5 17.4 17.2 17.0 17.3 18.4 16.9 18.2 18.7 20.1 20.7

a Construct an ordered stemplot for the data using sterns 14, 15, 16, etc. b Describe the distribution of the data. c What fraction of April days had maximum temperatures greater than 18°C? d What fraction of April days had maximum temperatures between 16°C and 20°C?

12 A frequency table for the heights of a football squad is given alongside.

Height (cm) Frequency 170 - < 175 1

a Explain why height is a continuous variable. 175 - < 180 7 b Construct a histogram for the data. The axes 180 - < 185 9

should be carefully labelled and include a heading 185 - < 190 11 for the graph. 190 - < 195 6

c What is the modal class? 195 200

- < - <

200 205

3 1

13 A frequency table for the ages of students at a country high school is given alongside.

a Explain why age is a continuous variable. b Construct a histogram for the data. The axes

should be carefully labelled and include a head-ing for the graph.

c What is the modal class?

Age (years) Frequency 12- 2 13- 21 14- 25 15- 24 16- 17 17- 10 18- 6 19- 1

14 Josef recorded the masses of the pumpkins that he grew on a histogram as shown:

a How many pumpkins did Josef grow? b What percentage of the pumpkins

weighed more than 4 kg? c What percentage of the pumpkins

weighed between 2 kg and 8 kg?

20 Josef's pumpkins

• ITN 15

10

5

I . 0 0 I 2 3 I 5 6 7 8 9

mass (kg)

The salaries of employees of Better Health Pharmaceuticals are shown in the histogram alongside.

a How many employees were there? b What percentage of the employees

earned more than $50 000? c What percentage of the employees

earned between $30 000 and $40 000 per year?

15 10

8

6

4

2

0

Staff salaries p.a.

16 A group of Year 10 students organised a golf competition to find out who could hit a golf ball the furthest. The results of the competi-tion are displayed using the histogram shown:

a How many students took part in the com-petition?

b What percentage of the students hit the ball from 200 m up to 210 m?

c What percentage of the students hit the ball 200 m or more?

17 Find the mean and median of the set of scores:

18 Find the mean and median of the set of scores:

0, 1, 1, 1, 2, 3, 3, 3, 5, 8

23, 24, 25, 25, 24, 25, 22

50

40

30

20

10

0

19 The following table shows the average monthly rainfall for a city.

Month2 1 F r A M T A

S 10 NID Av. rainfall (mm) 14 32 36 43 104 168 1657 104 I 67 49 I 19 I

Calculate the mean average monthly rainfall for this city.

20 The selling prices of the last 9 houses sold in a particular district were: $186000, $193000, $210000, $192000, $185000, $179000, $194000, $203000, $198000

a Calculate the mean and median selling prices of these houses. Comment on your results.

b The next house sold brought a price of $560 000. Calculate the mean and median selling prices of the 10 houses.

c Which of your results in b better represents the typical selling price? Give a reason for your answer.

Number of matches Frequency 46 1 47 1 48 7 49 11 50 18 51 9 52 3

Number of lollies in a packet Frequency

23 2 24 6 25 9 26 15 27 14 28 9 29 3

21 A manufacturer maintains that packets con-tain 50 matches. A quality control inspector tested 50 packets and found the following dis-tribution:

a Find the: i mean ii median number of matches in a

packet b Comment on these results in relation to the

manufacturer's claim.

22 A manufacturer maintains that packets contain 27 lollies. To test this a quality control inspec-tor tested 58 packets and found the following distribution:

a Find the: i mean ii median number of lollies in a

packet

b Comment on these results in relation the manufacturer's claim.

23 Niko grew a fine crop of pumpkins in his backyard. He recorded their masses in the table shown.

a Find the: i approximate mean ii modal class

iii approximate median mass of the pumpkins.

b Comment on the statement that `Niko's pumpkins usually weigh 4 kg or more'.

24 The frequency table alongside shows the number of goals scored by a squad of hockey players over the course of a season.

a Construct a column graph for this data. b Describe the distribution of the data. c Find the i mean ii median of the data. d Which measure of centre is the most suitable

for this data set?

Mass (kg) Frequency

1 - < 2 4 2 - < 3 5 3 - < 4 8 4 - < 5 10 5 - < 6 16 6 - < 7 9 7 - < 8 6 8 - < 9 2

No. of goals Frequency

0 7 1 9 2 5 3 2 4 3 5 2 6 0 7 1

124, 125, 125 find the:

5.2, 5.2, 5.6 find the:

29 For the data set 112, 113, 114, 117, 118, 120, 121, 123, a median b lower quartile c upper quartile d interquartile range

30 For the data set 3.5, 4.1, 4.3, 4.8, 4.8, 4.9, 5.0, 5.1, 5.2, a median b lower quartile c upper quartile d interquartile range

31 The time spent (in minutes) by 24 people in a queue at a hardware store, waiting to be attended, has been recorded as:

0 3.1 0 2.3 3.1 0 1.2 0 1.5 2.7 1.3 2.8 0 3.1 4.7 1.6 2.9 0.8 3.6 5.5 1.3 2.5 3.0 1.5

a Find the median waiting time and the upper and lower quartiles. b Find the range and interquartile range of the waiting time. c Supply the missing numbers:

i "50% of the waiting times were greater than minutes." ii "75% of the waiting times were less than minutes."

iii "The minimum waiting time was minutes and the maximum waiting time was minutes. The waiting times were spread over minutes."

33 Stem Leaf 3 027 4 0 1 5 6 7 8 5 1 1 2 4 5 8 8 9 6 0368 71

For the data set given, find the: a minimum value c median e upper quartile g interquartile range

maximum value lower quartile range

Scale: 7 I 1 means 7.1

34 Stem 0 1 2 3 4 5

Leaf 9 71 8 3 6 7 6 4 9 3 5 5 6 8 2 1 7 9 3 4 2 1

For the data set given, find the: a minimum value c median e upper quartile g interquartile range

maximum value lower quartile range

Scale: 4 I 7 means 4.7

For the data set given, find the: minimum value b maximum value median d lower quartile upper quartile f range interquartile range

35 Stem Leaf 0 9 a 1 56 c 2 3 4 4 5 5 7 7 8 8 9 e 3 0 1 1 2 3 4 5 5 6 6 7 8 8 9 4 0 2 3 6 6 7 7 9 g 5 0

Scale: 4 I 0 means 40

37 The boxplot given summarises the points scored by a basketball team for the --I I 1— matches played in a season.

0 10 20 30 40 50 60 70 80 a What was the: points scored by basketball team

i highest score ii lowest score? What was the median score for the season? What was the range of scores for the season? In what percentage of the matches did the team score 43 points or more? What was the interquartile range for the season? The lowest 25% of the scores were between and Is a score of 44 points in the upper 50% of the scores?

38 A boxplot has been drawn to show the distribution of marks (out of 50) in a science Marks in a Science test test for a class. 1 F-

a What was the highest mark? b Explain the significance of the aster- 0 10 2(1 1(1

isk on the boxplot. c The middle 50% of the class scored a mark between and d Find the median mark. e 50% of the students scored a mark greater than or equal to f Find the range of marks scored.

39 A box-and-whisker plot has been drawn Marks in Mathematics test to show the distribution of marks (out of 50) in a mathematics test for a particular —1 1 I—* class.

a What was the: 20 30 40 50

i highest mark ii lowest mark scored? What was the median mark for this test? What was the range of marks for this test? What percentage of students scored 30 or more for this test? What was the interquartile range for this test? The top 25% of students scored a mark between and If you scored 35 for this test, would you be in the top 50% of students in this class?

41 An English teacher recorded the number of spelling mistakes made by 30 students in a test. The results were:

4, 7, 9, 3, 1, 11, 5, 4, 6, 6, 4, 4, 4, 12, 8, 2, 3, 3, 6, 7, 8, 4, 4, 3, 5, 5, 5, 8, 7, 0 a Find the median, lower quartile and upper quartile of the data set. b Find the interquartile range of the data set. c Draw a box-and-whisker plot for the data set.

42 Donna counted the number of chocolates in some boxes of chocolates and tabulated the data as shown below:

Number of chocolates 23 24 25 26 27 28 29 30 Frequency 1 4 8 14 11 7 3 1

a Find the five-number summary for this data set. 1) Find the i range ii IQR for the data set. c Construct a box-and-whisker plot for the data set.

50 60

Time (seconds) Frequency

30 t < 35 4 35 t < 40 10 40 .“ < 45 13 45 t < 50 7 50 t < 55 5 55 t < 60 1

Score

20 x < 30 x < 40 x < 50 x < 60 x < 70 x < 80 x < 90 x <

Frequency

30 3 40 7 50 12 60 21 70 30 80 28 90 19 100 7

43 Chinwe worked in the sales and marketing department for a publisher. She recorded the number of international telephone calls she made each work day for a month, as follows:

Number of calls 0 1 2 3 4 5 6 8 15 Frequency 2 1 2 4 6 3 2 1 1

a Find the five-number summary for this data set. b Find the i range ii IQR for the data set. c Construct a box-and-whisker plot for the data set.

47 A survey was carried out on the waiting time in a supermarket queue on a Friday afternoon at 4.00 pm. The results are as shown in the table alongside. a Can we accurately give the longest waiting

time? b Draw a histogram of the data. c What is the length of each class interval? d What is the modal class? e Add further columns to help find the approximate mean. f Comment on the shape of the distribution.

48 The speeds of 100 cars passing a point in an 80 km/h limit zone were recorded as follows:

a Can we accurately give the highest speed? b Draw a histogram of the data. c What is the length of each class interval? d What is the modal class? e Add further columns to help find the approxi-

mate mean. f Comment on the shape of the distribution.

Speed (km/h) 55 - < 60 60 - < 65 65 - < 70 70 - < 75 75 - < 80 80 - < 85 85 - < 90 90 - < 95

Frequency 1 2 1

17 48 23 6 2

Time (seconds) 0 - <20

Frequency 5

20- < 40 11 40 - < 60 29 60 - < 80 12

80 - < 100 3

165 x < 170 x < 175 x < 180 x < 185 x < 190 x < 195 x <

49 The following table gives the height groups of bas-ketball players in a league. Draw a cumulative fre-quency graph of the data and use it to estimate:

a the median height of the players b the number of players shorter than 183 cm c the height below which lie the 4 shortest

players.

Frequency

170 5 175 6 180 13 185 22 190 19 195 19 200 7

50 The following table gives the lap times (in seconds) of 40 racing car drivers. Draw a cumulative fre-quency graph of the data and use it to estimate:

a the median lap time b the number of drivers whose time was less

than 52 seconds c the time in which the fastest 10 drivers com-

pleted the lap.

51 The following table gives the scores of an examina-tion achieved by a group of students: Draw a cumulative frequency graph of the data and use it to estimate:

a the median examination mark b the number of students who scored less than 63

marks c the percentage of students who scored a distinc-

tion, given that the distinction mark was 75.

STATISTICS SOLUTIONS

7 Contents of nacket

I I I I.

Stem

1 04

2

Leaf

58

3 4 5 6 number of children

2 4 4 1 9 2 3 8 2 7 5 2 5 0 6 1 7 5 8

318 means 38 Stem Leaf

04 1 58 2 1 2 4 4 9 3 0 1 2 2 5 5 5 6 7 7 8 8

3 1 8 means 38

Stem Leaf Stem Leaf 2 3 4 5 6 7 8 9

6 4707 5561 8 2 6 1 07 802

5

3

2 3 4 5 6 7 8 9

6 0477 1556 1 2 3 6 8 07 028

5 7 I 8 means 7.8

c The modal weight was between 5.0 kg and 6.0 kg.

11 a Stem 14 15

Leaf 19 256

16 2 4 8 9 9 17 0 2 2 3 4 18 1 2 2 4 6 7 7 19 02 20 21

1577 4

22 2 17 1 2 means 17.2

12 a Height may take any value, not just the integers 170 to 205 cm.

Heights of football squad

15

10

5

0 170 180 190 200

height (cm)

c 185 - < 190 cm

13 a Age can take any value, not just the integers 12 to 20 years.

30

20

10

0 12 13 14 15 16 17 18 19 20

age (years)

c 14 - < 15 years

14 a 1+4+7+12+18+15+12+8+3=80 i.e., Josef grew 80 pumpkins

b Percentage weighing more than 4 kg 18+15+12+8+3

80 x100%

= x 100% 80

=70%

c Percentage weighing between 2 kg and 8 kg 7+ 12 + 18 + 15 + 12 +8

80 x100%

= x100% 80

=90%

15 a 3+8+9+4+1=25 i.e., there were 25 employees

b Percentage earning more than $50 000/year = x 100% 25

= x100% 25

= 20%

c Percentage earning between $30 000/year and $40 000/year = A x 100%

=32%

16 a 5+17+26+42+19+7=116 i.e., there were 116 students

b Percentage who hit from 200 m up to 210 m = x 100% 116

* 22.4%

c Percentage who hit the ball 200 m or more 26 + 42 + 19 + 7

x100% 116

x 100% 116

* 81.0%

8 a Number of children in families 30 frequency -

20

10

0

9 a

10 a

sum of scores 17 mean = 21 No. of matches

46 47 48 49 50 51 52

Total

Freq.

1 1 7

11 18 9 3

50

Product

46 47

336 539 900 459 156

2483

number of scores

0+1+1+1+2+3+3+3+5+8 10

—

=2.7

2 + 3 5th 6th median i of and scores}

2 = 2.5

sum of scores 18 mean = number of scores

23+24+25+25+24+25+22 7

168 — 7 = 24

22 23 24 24 25 25 25

sum of all data values a i mean =

number of data values 2483 =

50 * 49.7

ii median is average of 25th and 26th scores 50 +50

2 =50

19 mean

20 a

median = 24 {middle score}

average monthly rainfall

14 + 32 + 36 + + 67 + 49 + 19

b These results tend to support the manufacturer's claim.

22 No. of lollies Freq. Product

23 2 46 24 6 144 25 9 225 26 15 390 27 14 378 28 9 252 29 3 87

Total 58 1522

sum of all data values i a mean =

12 955 12

* 79.6 mm per month

$186000 + $193000 + mean =

+ $198000 9

$1 740 000

193000,

9 * $193333

The ordered data set is: 179000, 185000, 186000, 192000, 194000, 198000, 203000, 210 000

number of data values 1522 = 58

* 26.2

median = $193000 {middle value}

The mean and median are almost the same which means that both are reliable measures of the centre of the distribution of this data set.

$179000 + $185000 + + $560000 b mean =

10 = $230000

. $193 000 + $194 000 median =

2 = $193500

c The new data value of $560 000 is an outlier and affects the mean only. The median gives a better representation of the selling price in this case as the median is not affected by outliers.

ii median is average of 29th and 30th scores 26 +26

— 2 =26

b These results suggest that the manufacturer's claim is incorrect. The packets contain 26 lol-lies would be more accurate.

23 Mass (kg)

1 - < 2 2 - < 3 3 - < 4 4 - < 5 5 - < 6 6 - < 7 7 - < 8 8 - < 9

Total

Frequency 4 5 8 10 16 9 6 2

60

Midpoint of interval

1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5

Product 6

12.5 28 45 88

58.5 45 17

300

30 3.5, 5.6

a

b

4.1, 4.3, 4.8, 1.4.8;(4.91 5.0, 5.1, 5.2, 5.2, 5.2,

The median is the average of the 6th and 7th scores •

4.9 + 5.0 median =

2 9.9

= 4.95 2

As the median is not a data value we split the original data into two equal groups of 6.

3.5, 4.1,14.3, 4.8; 4.8, 4.9

4.3 + 4.8 9.1 Qi = = =4.55

2 2

C 5.0, 5.1, 5.2, 5.2; 5.2, 5.6 Q3 = 5.2

d interquartile range = Q3 - Qi = 5.2 - 4.55 = 0.65

sum of all data values a i mean *

number of data values

300 5 kg

ii modal class is 5 - < 6 kg

iii median = 30th score + 31st score

2 5 + -3.5

• {27 values are <5

16 and 27 + 3.5 = 30.5} * 5.2 kg

b

24

Niko's pumpkins usually weigh 4 kg or more, since the mean * 5 kg (from a i) and the median is * 5.2 kg (from a In fact, 43 out of 60 wei2h 4 k2 or more.

No. of goals Freq. Product 0 7 0 1 9 9 2 5 10 3 2 6 4 3 12 5 2 10 6 0 0 7 1

Total 29 5 1

31 The 0,

a

2.5, (24

ordered data set is: 0, 0, 0, 0, 0.8, 1.2, 1.3, 1.3, 1.5, 1.5,11.6,

2.7, 2.8, 2.9, 3.0, 3.1, 3.1, 3.1, 3.6, 4.7, 5.5 scores)

n + 1 25 As 24 12.5 n =

2 =--=

2

12th score + 13th score median =

2 1.6 + 2.3

2

3.9 - 2

a 10

8

6

4

0 0 1 2 3 4 5 6 7

no. of goals b The data is positively skewed.

sum of all data values C i mean =

number of data values

= =, 1.86

ii 29

median is the 15th data value = 1 (the 8th through 16th data values are all 1)

d The mean is the most suitable measure of centre in this case because the mean takes into account the skewness of the data.

29 112, 113,114; 117, 118,4201 121, 123,4241 125, 125

Qi median Q3

a median = 6th score = 120

b lower quartile, Q i = 114

c upper quartile, Q 3 = 124

d interquartile range = Q3 -

= 124- 114 =10

= 1.95 minutes

As the median is not a data value we split the original data into two equal groups of 12.

0, 0,

2.3, 4.7,

0, 0, 0,10.8, 1.2;

0.8 + 1.2

1.3, 1.3, 1.5, 1.5, 1.6

2.0 = = 1.0 minutes

2

...... 3.1, 3.1, 3.6,

6.1 = = 3.05 minutes

2

Qi= 2 2.5, 2.7, 2.8, 5.5

3.0 + 3.1 t-13 2

b Range of the waiting times = 5.5 - 0 = 5.5 minutes

Interquartile range = 3.05 - 1.0 = 2.05 minutes

c i 50% of the waiting times were greater than 1.95 minutes.

ii 75% of the waiting times were less than 3.05 minutes.

iii The minimum waiting time was 0 minutes and the maximum waiting time was 5.5 minutes. The waiting times were spread over 5.5 min.

d As the median is not a data value, we split the original data into two equal groups of 11. So, Qi is the 6th score = 4.5

e Q3 is the 17th score = 5.9

I range = max. value - min. value = 7.1 - 3.0 =4.1

g interquartile range = Q3 -

= 5.9 - 4.5 = 1.4

9th score + 10th score d Qi= 2

27+ 27 2

27th score + 28th score e Q3 =

2 39+ 40

2 = 39.5

=27

33 a 3.0

b 7.1 „ n + 1 23 „

c Since n = zz, - 2 = - = 11.0 2

11th score + 12th score • median =

2 5.1+ 5.2

2 = 5.15

35 a 9

b 50 37

c Since n = 36'

n + 1 2 =

2 = 18.5


2 33 + 34

2 = 33.5

34 The

a

b

c

d

e

f

g

or ered stemplot is:

Stem Leaf

0 9

1 17

2 3 4 6 6 7 8

3 1 2 3 5 5 6 8 9

4 2 3 4 7 9

5 1 4 1 7 means 4.7

0.9

5.1 n + 1 24

Since 23, 12 n = =

37 a b

c

d

e

I

g

38 a

b

c

d

e

I

f range = max. value - min. value = 50-9 = 41

g interquartile range = Q3 - Q1 = 39.5 - 27 = 12.5

i 66 ii 13

43

range = 66 - 13 = 53

Percentage scoring 43 points or more = 50%

Interquartile range = Q -Q = 55 - 25 = 30

The lowest 25% of the scores were between 13 and 25.

Yes, since the median score is 43.

50

The asterisk represents an outlier, a mark of 5 out of 50.

The middle 50% of the class scored a mark be-tween 30 and 45. 38

50% of the students scored a mark greater than or equal to 38.

range = 50 - 5 = 45

2 2 median = 12th score = 3.3

Qi is the 6th score = 2.6

Q3 is the 18th score = 4.2

range = max. value - min. value = 5.1 -0.9 =4.2

interquartile range = Q - Qi = 4.2 - 2.6 =1.6

39 a i 49 ii 18

b 34

c range = max. mark - min. mark = 49 - 18 =31

d 75% of students scored 30 or more

e IQR = Q3 - Q1 = 41 - 30 = 11

f The top 25% of students scored a mark between 41 and 49.

g Yes, since the median mark is 34.

41 The ordered data set is 0, 1, 2, 3, 3, 3, 3,1,42; 4, 4, 4, 4, 4, 4,(51 .51 5, 5, 6, 6, 6, 7,Ct 7, 8, 8, 8, 9, 11, 12 (30 scores)

a median = —5 + 5

= 5, Qi = 4, Q3 = 7 2

b IQR = Q3 — = 7 — 4 = 3

c I I .i•

0 2 4 6 8 10 12 spelling mistakes

42 a min. value = 23, max. value = 30

Total number of scores is 49 median is the 25th score

i.e., median = 26 {there are 27 scores from 23 to 26}

12th + 13th score 25 + 25 Qi = 2 = 2 =

25

37th + 38th score 27 + 27 Q3 — 2 = 2

=27

b i range = max. value — min. value = 30 — 23 =7

ii IQR = Q3 — Qt = 27 — 25 = 2

23 24 25 26 27 28 29 30 number of chocolates

43 a min. value = 0, max. value = 15

Total number of scores is 22


2 4 + 4

2 =4

Qi = 6th score = 3, Q3 = 17th score = 5

b i range = max. value — min. value = 15-0 =15

ii IQR Q3 — Q1 = 5 — 3 = 2

•I' 1 . 11'1 • 1 . 1 . 1 • 1 P- r) 2 4 6 8 10 12 14 16

number of calls

47 a No, as we do not have exact data values, only data intervals.

Waiting time in supermarket queue

r)-' 30 A 1=1

25 ,11=1 20 —

ii 15 10

c 20 seconds

d 40 - < 60 seconds

Time (s) Freq. Midpoint of in Product

0 - < 20 5 10 50 20 - < 40 11 30 330 40 - < 60 29 50 1450 60 - < 80 12 70 840 80 - < 100 3 90 270

Total GO 2940

2940 estimated mean * —

60 * 49 seconds

f The distribution is approximately symmetrical.

b Speed of cars

>, 60 . i Fl. 40

20

0- 55 65 75 85 95

speed (km/h)

c 5 km/h

d 75 - < 80 km/h

Speed (km/h) Freq. Midpt. of int. Product

55 - <60 1 57.5 57.5 60 - < 65 2 62.5 125 65 - <70 1 67.5 67.5 70 - <75 17 72.5 1232.5 75 - <80 48 77.5 3720 80 - <85 23 82.5 1897.5 85 - < 90 6 87.5 525 90 - <95 2 92..5 185

Total 100 7810

estimated mean 7810=

78.1 km/h 100

f The distribution is slightly negatively skewed.

48 a No, as we do not have exact data values, only data intervals.

49 100 cumulative frequency

90

80

70

60

50

40

30

20

I () Fad!

0 —11111111M 165 170 175 180 185 190 195 200

height (cm) a There are 91 players in total.

the median is the 46th height. Reading from the graph, the median * 185 cm

b From the graph, approximately 37 players are shorter than 183 cm.

c From the graph, the shortest 4 players are below approximately 169 cm.

51 cumulative frequency 140

120

100

80

40

20

0 20 30 40 50 60 70 80 90 100

marks

a There are 127 students in total. the median is the 64th score.

Reading from the graph, the median * 67 marks.

b From the graph, approximately 52 students scored less than 63 marks.

c From the graph, 87 students scored less than 75 marks. 127 — 87 = 40 students scored more than 75 marks. the percentage of students who scored a distinction* —142°7 x 100% .* 31.5%.

50 45 A cumuh five frequency

52

40

35

30

25

20

15

10

5

v I I 1.1 0 30 35 40 45 50 55 60

time (seconds) a The median is the average of the 20th and the

21st times. Reading from the graph, the median * 42.5 seconds.

b From the graph, approximately 36 drivers had times less than 52 seconds.

c From the graph, the fastest 10 drivers completed the lap in approximately 38 seconds or less.

70 cumulative freluency

60

50

40

30

20

10

0 50 54 58 62 66 70 74

speed (km/h)

a There are 59 cars in total the median is the 30th data value. Reading from the graph, the median * 63 km/h

b From the graph, approximately 7 cars were driv-ing slower than 55 km/h.

c From the graph, approximately 40 cars were driv-ing slower than 65 km/h. 59 — 40 = 19 cars were driving faster than 65 km/h.

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Examination Revision and Solutions - PBworks