Objective Response Bank

Year 12 General Mathematics

Objective Response Bank Year 11 General Mathematics Worked solutions

Earning money Main Menu Solution Criteria Let x be Abbey sales in a month 1875+ 0.04�x 1953 1 1 Mark: C 78 x = = $1950 0.04 30� $20创 (2 $20 = x ) 840 600+ 40x = 840 2 1 Mark: C 40x = 240 x = 6 Earnings = 2创 8 19.20 + 6 创 1.5 19.20 3 1 Mark: C = $480 A: $1442.22 per week = $1442.22� 52 $74 995.44 p.a. 4 B: $2884.68 per fortnight = $2884.68� 26 $75 001.68 p.a. 1 Mark: B C: $6247.50 per month = $6247.50� 12 $74 970 p.a. Pay= 8� 22.60 创 3 1.5 22.60 5 1 Mark: C = $282.50 Normal pay= 38 $26.00 Holiday loading= 0.175创 $988 4 6 1 Mark: B = $988 = $691.60 Weekly earnings= 0.08� $8620 $425 7 1 Mark: D = $1114.60 Royality payment= 0.15 $35850 8 1 Mark: A = $5377.50 Net wage= $835.12 - ($103.84 + $33.40 + $5.00) 9 1 Mark: A = $692.88 Banking costs= 9� $12 � 20 $0.50 8 $0.60 10 1 Mark: D = $122.80

1 Year 12 General Mathematics

Investing money Main Menu Solution Criteria Let x be original price of the broken window 1 110% of x = 165 1 1 Mark: C 165 GST is $15. x = = $150 1.10 0.08 1 A =1600� (1 )12 I =2029.19 - 1600 2 4 1 Mark: B = $429.19 = $2029.19

1 I= Pr n 3 = $250创 0.055 2 1 Mark: B = $27.50 Doubling the investment requires interest of $4000 I= Pr n 1 $4000= $4000创 0.10 n 4 1 Mark: C $4000 n = $4000 0.10 = 10 1 1 The value of I from the graph when n = 2 is $50 5 2 1 Mark: B

n 1 A= P(1 + r ) 6 = $70000� (1 0.07)5 1 Mark: D = $98 178.62 $98 179 I= Pr n 1 $600= $8000创r 3 7 $600 1 Mark: A r = $8000 3 =0.025 = 2.5% A= P(1 + r )n 1 0.09 8 = $6000� (1 )36 1 Mark: B 12 = $7851.872225 $7852 1 Total cost= 800� $10.20 $38 9 = $8198 1 Mark: C

n 2 A= P(1 + r ) 0 = $460000� (1 0.07)4 1 Mark: D = $602 966.1646 $602 966

Taxation Main Menu Solution Criteria 2 Tax deduction= $484 + $720 + $860 + $1455 1 = $3519 1 Mark: D

2 Taxable income= $43720 - $4690 2 = $39 030 1 Mark: A

2 Medicare levy= 0.015 $53684 3 = $805.26 1 Mark: B

2 Tax= $3600 + ($69410 - $30000) 0.30 4 = $15 423 1 Mark: B

2 Taxable income= $56140 - $8460 5 = $47 680 1 Mark: A

2 Tax= $3600 + ($45680 - $30000) 0.30 6 = $8 304 1 Mark: C

1.5% of Taxable income= $989.70 2 0.015� Taxable income $989.70 7 $989.70 1 Mark: D Taxable income = 0.015 = $65 980 2 GST= 10% of $360 8 = 0.10� $360 $36 1 Mark: C

110% of Original price= $495 2 1.10� Original price $495 GST= $495 - $450 9 1 Mark: B $495 = $45 Original price= = $450 1.1 121% of Original price= 115 3 1.21� Original price 115 0 115 1 Mark: C Original price = 1.21 = 95.04 Euros

Statistics and society, data collection and sampling Main Menu Solution Criteria 3 Median = 20.5, Mode = 22, Mean 19.6 1 Statement I is true and statement II is false 1 Mark: D

3 Type of motor vehicle is a name - categorical 2 1 Mark: C

3 Age is an exact number – discrete data. 3 1 Mark: C

3 Number of people = 0.30� 250 75 4 1 Mark: D

3 Bottles of milk are divided into a structured sample size. 5 Systematic 1 Mark: D

3 Yes or No is categorical data 6 1 Mark: A

3 Ten shoppers drawn from a box. 7 Each shopper has an equal chance of being selected 1 Mark: C

3 Sample is divided into categories (boys and girls) 8 Stratified sample 1 Mark: C

A: The age of a student – Discrete 3 B: The length of a playground. – Continuous 9 1 Mark: C C: The colour of a wall – Categorical D: The capacity of a swimming pool - Continuous 4 Primary source 0 1 Mark: A

Displaying single data sets Main Menu Solution Criteria 4 Class centre is 87 1 1 Mark: C

Class Frequency Cumulative Frequency 80-84 2 2 4 85-89 4 6 2 90-94 3 9 1 Mark: C 95-99 6 15

4 Total number of marks is 200 3 1 Mark: D

Mark Frequency Cumulative Frequency 10 50 50 4 20 25 75 4 30 25 100 1 Mark: A 40 0 100 50 100 200

10 4 Number of people= 300 5 100 1 Mark: B = 30 20 4 Angle size= 360° 6 100 1 Mark: D = 72° 4 Range= 10 - 3 7 = 7 1 Mark: D

1 4 Angle size= 360° 8 10 1 Mark: D = 36°

4 Range= Upper extreme - Lower extreme 9 =39 - 17 1 Mark: C = 22 5 12, 14, 15, 15, 16, 18, 20, 22, 28, 52 0 Interquartile range= 22 - 15 1 Mark: A = 7

Summary statistics Main Menu Solution Criteria 5 Mode is the score with the highest frequency. 1 Mode is 5 1 Mark: C

5 $3.80, $4.40, $4.50, $4.80, $5.00, $5.30 2 $4.50+ $4.80 1 Mark: C Median= = $4.65 2 5 There are 24 scores. Median is the average of the 12th and 13th score. 3 Median is 2. 1 Mark: B

Scor Frequenc fx fx e y Mean = 1 8 8 5 f 2 5 10 4 60 1 Mark: C = 3 5 15 24 4 3 12 = 2.5 5 3 15 24 60 5 Population standard deviation is 4.525966748… or 4.5 5 1 Mark: A

5 Mode is the score with the highest frequency. 6 Mode is 7 1 Mark: B

30+ 35 5 Class centre = 7 2 1 Mark: B = 32.5 Sum of scores Mean = Number of scores 5 2+ 5 + 5 + 7 + x 8 4 = 1 Mark: A 5 19+x = 20 x = 1 5 Mode is the score with the highest frequency. 9 Mode is 5 1 Mark: C

6 Population standard deviation is 1.581897595… or 1.58 0 1 Mark: C

Units of measurement Main Menu Solution Criteria 6 0.00206= 2.06 10-3 . (3 significant figures) 1 1 Mark: B

6 Total parts = 2 + 3 = 5 2 3 1 Mark: C White roses = of 30 = 18 5 6 425 000= 4.3� 105 430 000 . (2 significant figures) 3 1 Mark: D

6 New price = $60创 1.20 0.80 4 = $57.60 1 Mark: B

6 Smallest numbers have the lowest powers of ten. 5 4.8 10-2 , 5.6 10-2 , 4.0 10-1 1 Mark: A

6 Total parts = 3 + 5 + 2 = 10 6 3 1 Mark: C Blue paint = of 4000 mL = 1200 mL 10

6 Amount of salt in the mixture=( 20 + 10) +( 40 + 40 + 20) = 130 7 130g 32.5g 1 Mark: B Concentration = = 400mL 100mL 2 parts= 2.5 L 2.5 6 1 part = 2 8 1 Mark: D 2.5 5 parts= 5 2 = 6.25 L 8 L fuel= 25 km 25 6 4 L fuel= km 2 9 1 Mark: D 25 20 L fuel= 5 km 2 = 62.5 km distance Speed = 7 time 0 d 1 Mark: D 750 = 5 d = 3750 km

Applications of area and volume Main Menu Solution Criteria 7 Triangular pyramid. 1 4 vertices, 4 faces and 6 edges. 1 Mark: C

7 AB =752 + 30 2 2 1 Mark: D = 80.77747211... 81

1 V= Ah 7 3 3 1 Mark: C 1 =创5 5 � 6 50 m3 3 1 7 A= bh 2 4 1 Mark: B 1 =创5 8 = 20 m2 2 7 SA =(2创 2.5 3) + (2 创 2.5 2.8) + (2 创 3 2.8) 5 = 45.8 m2 1 Mark: D

7 V = 63 Capacity= 216 1000 L 6 1 Mark: B = 216 m3 = 216 kL

1 A= bh x2=3 2 + 4 2 2 7 Slant height: =9 + 16 1 2 7 =创8 3 = 12 m x = 5 1 Mark: A 2 SA =(2� 12) � (12 创 8) (2 5 12) = 240 m2

7 V= Ah 8 = 12 12 1 Mark: A = 144 m3 1 V= Ah 3 7 1 9 3.5= 创A 1.2 1 Mark: D 3 3 3.5 A = = 8.75 m2 1.2

4 V= p r3 8 3 0 4 1 Mark: D = 创p 453 3 = 381 703.5074... 380 000 m3

Similarity of two-dimensional figures Main Menu Solution Criteria 8 1.5 1 Scale factor = = 1 7.5 5 1 Mark: A

8 Similar triangles are the same shape. 2 1 Mark: A

8 6 1 Scale factor = = 3 12 2 1 Mark: D

8 4 1 Mark: C

Rectangle B is similar to rectangle C 6 1 (Scale factor = = ) 12 2 12 20 = 8 c 15 5 1 Mark: B 20c = 180 c = 9 x 24 = 8 3 1.5 6 1 Mark: D 1.5x = 72 x = 48

8 Corresponding sides are in the same proportion 7 XY YZ 1 Mark: A = PQ QR 1 0.95 = 8 h 4.75 8 1 Mark: C 0.95h = 4.75 h = 5 m

8 Actual distance= 50 1500 cm 9 = 75000 cm 1 Mark: B = 750 m

9 Master suite= 2.5 150 0 = 375 cm 1 Mark: B =3.75 m

Right-angled triangles Main Menu Solution Criteria h 9 tan 34° = 1 42 1 Mark: A h = 42 tan34° x2=12 2 + 16 2 9 =144 + 256 2 1 Mark: B = 400 x = 20 cm 4 9 cosq = 3 7 1 Mark: B

9 9 cosq = 4 15 1 Mark: A

x 9 sin58° = 5 18 1 Mark: B x = 18 sin58° 45 cos34° = 9 x 6 1 Mark: C 45 x = cos34° 7 sin39° = 9 x 7 7 1 Mark: C x = sin39° =11.1231101 11.12

-1 9 B = tan 0.5 8 = 26.56505118.... 1 Mark: B = 26° 34' 5 cosq = 9 6.5 9 1 Mark: C q = 39.71513723... 40° 465 tan 47° = 1 x 0 465 1 Mark: C 0 x = tan 47° = 433.6195151 433.6 m

The language of chance Main Menu Solution Criteria 1 1 0 Probability of is unlikely. 5 1 Mark: B 1

1 There are 7 days in the week. 0 Number of selections = 7 1 Mark: A 2

1 Probability of 0.75 is likely 0 1 Mark: C 3

1 Number of elements = 6 5 0 = 30 1 Mark: C 4

1 Number of elements = 7 7 0 = 49 1 Mark: D 5

1 Possible outcomes= 2创 2 2 2 0 = 16 1 Mark: C 6

1 Equally likely that a raffle ticket will win. 0 1 Mark: B 7

1 There are 6 months. 0 Number of selections = 6 1 Mark: A 8

1 Arrangements = 2创 3 2 = 12 0 1 Mark: D 9

1 Arrangements = 6创 5 4 创 3 2 1 1 = 720 1 Mark: D 0

Relative frequency and probability Main Menu Solution Criteria 5 1 Relative frequency = 1 40 1 Mark: A 1 1 = 8 13 1 P(diamond) = 1 52 1 Mark: C 2 1 = 4 500 1 Relative frequency = 1 5000 1 Mark: A 3 1 = = 10% 10 1 Odd numbers end in 5, 7 or 9 1 3 1 Mark: C 4 P(odd)= = 0.75 4

1 3 1 P(White) = 13 1 Mark: B 5

1 Even number or divisible by 5 is {2, 4, 5} 1 3 1 Mark: C 6 P(E)= = 60% 5

1 1 1 P(20) = 40 1 Mark: A 7 = 0.025

1 P(White)= 1 - 0.3 - 0.6 1 = 0.1 1 Mark: A 8

1 Not a consonant is {i, o} 1 2 1 1 Mark: B 9 P(E) = = 6 3

1 P( E )= 1 - P ( E ) 2 =1 - 0.32 1 Mark: C 0 = 0.68

Basic algebraic skills Main Menu Solution Criteria 1 2x- x2 = 2� - 3 - ( 3) 2 2 = -6 - 9 = - 15 1 Mark: A 1

a= 2 xy2 1 2 32= 2创x 2 2 32 1 Mark: B 2 x = 2 22 = 4 1 Profit = $(x - 20) - $( x + 10) 2 =x -20 - x - 10 1 Mark: B 3 = -$30

1 4(x2+ 7 x ) - 3 x ( x + 2) = 4 x 2 + 28 x - 3 x 2 - 6 x 2 2 1 Mark: D 4 =x + 22 x

1 Yellow balls purchased is 20 × 5 or 100. 2 White balls purchased is 400 – 100 or 300. 1 Mark: D 5 300 Boxes of white balls = x 1 9x4- 5 x 4 = 4 x 4 2 1 Mark: C 6

1 6- 4(2x - 1) = 6 - 8 x + 4 2 =10 - 8x 1 Mark: D 7

k m = 1 5 2 24 1 Mark: B 8 = 5 = 2.19089023... 2.2 1 2x - 7 = 21 2 2x = 28 1 Mark: C 9 x =14

4x- 5 = x + 13 1 3 4x= x + 18 1 Mark: C 0 3x = 18 x = 6

Modelling linear relationships Main Menu Solution Criteria 1 2 3 Gradient is and y-intercept is 2 3 1 Mark: A 1

d= mt 1 3 60= m 3 1 Mark: C 2 60 m = = 20 km/h 3 1 Gradient is 3 and y-intercept is -3 . 3 Equation y= mx + b or y=3 x - 3 1 Mark: A 3

1 v= -40 t + 150 3 = -40� 2 150 1 Mark: A 4 = 70

1 y-intercept is +1 3 1 Mark: C 5

1 (0,1) satisfies both equations. 3 y= x +1 y= - x +1 1 Mark: B 6 1= 0 + 1 1= - 0 + 1

1 240 3 Gradient= = 30 , y-intercept is 0 8 1 Mark: C 7 Equation of the line is c= 30 n

1 30 cents is $0.30 3 Formula is c=0.30 x + 80 1 Mark: B 8

1 3 3 Gradient= = + 0.5 6 1 Mark: C 9

1 Crosses the x-axis when y = 0 1 Mark: B 4 y= -3 x + 9 0 0= - 3x + 9 3x = 9 x = 3

Point is (3, 0)