Student s Name: Teacher s Name: 10 Year 10 Mathematics, 2009 Algebra Use straightforward algebraic methods and sketch and interpret features of linear graphs Time: 20 minutes. Check that you have entered your name and your teacher s name in the top left hand corner of this paper. You should answer ALL the questions in this paper. You should show ALL working. Check that this paper has pages 2 5 in the correct order and none of these pages is blank. YOU MUST HAND THIS PAPER TO THE SUPERVISOR AT THE END OF THE EXAMINATION. For Plot and interpret simple graphs. Solve linear equations. Describe simple patterns. Criteria with Merit Interpret linear graphs. Find terms in patterns. Carry out more complex algebraic manipulations. Solve equations. with Excellence Solve algebra problems using graphs and manipulation. Overall Level of Performance (all criteria within a column met)
Algebra 2 QUESTION ONE Make sure you show ALL relevant working for each question. Graphic Designer Trish contracts out her time as a graphic designer on a per hour basis. The equation she uses to calculate her charge for a job is given by C = 60H + 30, where H is the number of hours worked and C the amount in dollars charged. (a) How much does Trish charge if she works on a job for 5 hours? (b) Solve the equation 60H + 30 = 570 to find out how many hours Trish would have worked on a job she charges $570 for. (c) Trish charges a consultation fee for any job she works on irrespective of the number of hours she spends on it. What is the consultation fee? (d) Trish s charge out rate is drawn on the graph below. What is the gradient of the line drawn? Charge ($) 600 550 500 450 400 350 300 250 200 150 100 50 1 2 3 4 5 6 7 8 9 10 Hours (H) (e) On the axes above draw the line C = 50H + 70, the charge out equation of another graphics designer.
Algebra 3 (f) For what length of time would the two graphic designers charge the same amount? QUESTION TWO Trish buys art paper by the ream. The more reams she buys the cheaper the price per ream. Study the table below. Reams (N) 1 2 3 4 5 Price per ream (P$) 25 23.5 22 20.5 19 (a) How much will it cost Trish in total for 5 reams of art paper? (b) How much will Trish pay per ream if she buys 8 reams? (c) Give the rule for the price per ream (P), if N reams are purchased. (d) Trish received a bill for $117 from the art paper supplier. How many reams of art paper must she have purchased and at what price per ream. Show your working. (e) What is wrong with this pattern of pricing?
Algebra 4 QUESTION THREE (a) Expand and simplify 2x(x + 2) 3(x 1) (b) Factorise p 2 11p + 24 (c) Simplify (4x 2 y) 2. (d) Make k the subject of the equation k 4 5 = q. (e) Give the equation of the line drawn on the axes below. y 12 10 8 6 4 2-8 -6-4 -2 0 2 4 6 8 x -2-4 -6-8
Algebra 5 QUESTION FOUR The sum of Trish and her mother is 78 years. Trish s mum is three years less than twice Trish s age. How old are Trish and her mother? Show ALL working. An answer alone is not sufficient. You must form an equation and solve it for excellence.
Algebra 6 Assessment Schedule Evidence Statement Question No. Merit Excellence Justification Plot and interpret simple graphs. Solve linear equations Describe simple patterns. Interpret linear graphs. Find terms in patterns. Carry out more complex algebraic manipulations. Solve equations. Solve algebra problems using graphs and manipulation. 1 (a) $330 A1 1 (b) 9 hours A2 1 (c) $30 A1 1 (d) 60 M1 1 (e) Line through (0, 70) and (9, 520) 1 (f) 4 hours A1 2 (a) $95 A1 2 (b) $14.50 A1 2 (c) P = 1.5N + 26.5 M1 2 (d) 9 reams at $13.00 = $117 2 (e) If you purchased enough reams the price would get as low as $0 or even negative. A2 M1 E (Any equivalent or valid explanation) 3 (a) 2x 2 + x + 3 A2 3 (b) (p 8)(p 3) M2 3 (c) 16x 4 y 2 A2 3 (d) k = 4q + 20 M2 3 (e) y = 2x + 1 M2 4 x + (2x 3) = 78 3x 3 = 78 3x = 81 x = 27 Trish is 27 Mother is 51 Sufficiency 6A with at least 3A1 and 2A2 plus 2M1 and 2 M2 Merit plus 1E A2 M2 E
Student s Name: Teacher s Name: 10 Year 10 Mathematics, 2009 Measurement Solve problems involving measurement of everyday objects Time: 20 minutes. Check that you have entered your name and your teacher s name in the top left hand corner of this paper. You should answer ALL the questions in this paper. You should show ALL working. Check that this paper has pages 2 5 in the correct order and none of these pages is blank. YOU MUST HAND THIS PAPER TO THE SUPERVISOR AT THE END OF THE EXAMINATION. For Solve simple measurement problems. Use measurements in calculations and conversions. Criteria with Merit Solve measurement problems. with Excellence Plan, carry out and evaluate a measuring task. Overall Level of Performance (all criteria within a column met)
Measurement 2 QUESTION ONE Make sure you show ALL relevant working for each question. Model Boats Mark loves boats. He draws, makes and sails them. Measure the height of the mast to here. 3.7 m 2.05 m 5.1 m 7.65 m 0.85 m 6.35 m (a) This is an accurate scale drawing of one of Mark s boats. Calculate the area of the boat s sail. Total area of sail = ] (b) The hull is best represented by a trapezium. The area of a trapezium is A =h a + b g 2. Calculate the area of the boat s hull. Area of hull = (c) The diagram above does not have the height of the mast on it. It is drawn to scale where 1 mm on the diagram represents 50 mm on the boat. Use this scale to find the height of the mast in metres. Height of the mast (m) =
Measurement 3 QUESTION TWO Mark dreams of competing in the Sydney to Hobart yacht race. This race starts on Boxing Day and can be completed in less than 2 days. The race distance is 1170 km. (a) At sea distances are measured in nautical miles. 1.86 km = 1 nautical mile. The race distance is 1170 km. How many nautical miles is 1170 km. 1170 km = nautical miles (b) A winner of the Sydney to Hobart yacht race started at 1300 hours on Boxing Day (26th December) and finished the race at 0730 on the 28th December. How many hours is the winner s time? Answer must be in hours only. Time (hours) = (c) Use your answer in (b) to calculate the speed in km/h of the winner in the Sydney to Hobart yacht race. Speed (km/h) = QUESTION THREE 2.6 m (a) Mark wants to estimate the volume of a boat. To do this he needs to calculate the cross-sectional area. Calculate the area of the semi circle on the right. Later you will use the calculation to find the volume of the boat. Area of full circle =
Measurement 4 QUESTION THREE cont... (b) The bow is more like half a cone, 2.6 m in diameter (radius = 1.3 m) and 1.25 m high. Use the formula V = 3 1 πr2 h to find the volume of a full cone. Length of bow 1.25 m 2.6 m Volume of full cone = QUESTION FOUR (a) Mark needs to estimate the volume of the entire hull. The hull consists of half a cylinder, 1.3 m in radius and 6.35 m long, with half a cone for the bow as described in 3 (b). In 3 (a) and 3 (b) you completed some calculations that will help you. If you have not completed these calculations then assume the cross-sectional area of the semi circle to be 2.5 m 2 and the cone volume to be 2.4 m 3. Calculate the volume of the hull. 1.25 m 2.6 m 6.35 m Volume of half cylinder = Volume of the entire hull = (b) What are potential or real problems in estimating the volume of the hull using the approach above?
Measurement 5 QUESTION FIVE Mark is also looking at different hull cross sections. The model for the existing hull is a semi-circle 1.3 metres in radius. The alternative has a cross-section of an equilateral triangle with each of its three sides 2.6 metres long. 2.6 m 2.6 m 2.6 m 2.6 m Scale 10 mm / 1 metre. Existing hull cross-section Alternative hull cross-section Both hulls will be the same length (6.35 m). Calculate the volume of the alternative hull, explaining how you obtained the height of the triangle. Volume =
Measurement 6 Assessment Schedule Evidence Statement Question No. Merit Excellence Justification Use measurements in calculations and conversions to solve simple problems. Calculation and applications of perimeter and area of simple composite shapes in context. Time calculations and applications. Area and circumference of circles. Calculation of speed. Conversion of time measurements into decimals. Solve measurement problems. Applications of harder composite shapes. Applications of surface area of prisms, pyramids. Applications of volumes of cylinders, cones, and pyramids. Conversion between units of area and volume. Applications of speed. Plan, carry out and evaluate a measuring task. 1 (a) Area = 13.2 m 2 A Any correct rounding 1 (b) Area = 5.95 m 2 A 1 (c) 103 mm / 5.15 m Accept 5.1 to 5.2 m A Accept 5150 mm ± 50 mm 2 (a) 629 nautical miles A 2 (b) 42.5 hours A 2 (c) 27.5 km/h A 3 (a) Semi circle = 2.65 m 2 A 3 (b) Vol. cone = 2.21 m 3 M 4 (a) Vol. 0.5cyl = 16.86 m 3 Vol. hull = 17.97 m 3 4 (b) May not be semi circular, diameter along hull may vary etc. Any two valid points. 5 h = 2.25 m Alt. area = 2.925 m 2 Alt. hull = 18.57 m 3 Sufficiency 4A plus 2M or 3M Merit plus E M E M Height via Pythagoras, scale drawing or trig.
Student s Name: Teacher s Name: 10 Year 10 Mathematics, 2009 Number Solve number problems Time: 20 minutes. Check that you have entered your name and your teacher s name in the top left hand corner of this paper. You should answer ALL the questions in this paper. You should show ALL working. Check that this paper has pages 2 3 in the correct order and none of these pages is blank. YOU MUST HAND THIS PAPER TO THE SUPERVISOR AT THE END OF THE EXAMINATION. For Solve number problems. Criteria with Merit Solve number problems in context. with Excellence Solve number problems in context involving several steps or reversing processes. Overall Level of Performance (all criteria within a column met)
Number 2 Make sure you show ALL relevant working for each question. QUESTION ONE Class Trip (a) Calculate 2 1 3 + 4 and write your answer as a decimal. 1 3 2 + 4 = (b) There are 24 students in 10Sy. 15 members of the class are going on a trip to Australia in the holidays. What percentage are going on the trip? Percent on trip = % (c) Steven has a paper round. He is paid $65 a week and he works six days a week. His paper round takes him one and a half hours. How much does he earn per hour? wage/h = $ /h (d) Steven is to get a 5% increase for his paper round. At present he is paid $65 per week. What will be his pay per week after the increase? New pay per week = $ (e) The cost of the trip to Australia is $1200. Steven has arranged to share the cost with his parents. Divide $1200 in the ratio of 1 to 2. 1:2 of $1200 = $ to $ (f) 45% of the cost of $1200 is for the airfare. How much is the airfare? Airfare = $ (g) Convert number 5.012 x 10 6 to expanded form. 5.012 x 10 6 = QUESTION TWO (a) Steven has allowed for NZ$250 in spending money. The exchange rate between New Zealand and Australia is NZ$1.17 = A$1. How much spending money will Steven have in Australian dollars? Spending money in Australian dollars = A$ (b) Steven s parents are keen to impress upon him how expensive this trip is. Steven has calculated that he will be away for 8.64 x 10 5 seconds. How many seconds will Steven get for each dollar of the $1200 it will cost? Convert your answer to minutes per dollar. Minutes per dollar = 1 (c) The airfare is expected to cost 45% of the $1200 cost. Accommodation will cost 4 and transport 1 6 of the $1200. The rest will be spent on food. The trip is 10 days in length so how much will Steven have per day for food? Amount for food per day = $
Number 3 QUESTION THREE The class have been planning the trip for a year. Shortly after they started raising money the original cost increased by 4%. Two months ago this interim price was increased by a further 10% to get the final cost of $1200. What was the original cost (to the nearest dollar) when the class first planned the trip? QUESTION FOUR In Australia, Steven sees an entertainment system for A$400. His older brother is impressed with the price and suggests that if Steven buys it and brings it back, he will pay $2 for every $1 that Steven pays. The exchange rate is NZ$1 = A$0.85. Calculate how much Steven would have to pay in New Zealand dollars if he agrees.
Number 4 Assessment Schedule Evidence Statement Question No. Merit Excellence Justification Solve number problems. Numbers into standard form. Rounding of numbers. Percentage problems. Sharing a quantity in a given ratio. Solving word problems involving integers and decimals. Solve number problems in context. Standard form. Increasing or decreasing by a given percentage. Calculation of percentage change. Ratio. Fractions. Solve number problems in context involving several steps or reversing processes. 1 (a) 1.25 A 1 (b) 62.5% A 1 (c) $7.22 A must be 2 dp 1 (d) $68.25 A 1 (e) $400 : $800 A 1 (f) $540 A 1 (g) 5 012 000 A 2 (a) $213.67(8) M 2 (b) 12 minutes or 12 min. 0 sec. 2 (c) $160 / 10 = $16/day M 3 Note incorrect answer 1200 1.14 = $1053 for Merit (1200 1.1) 1.04 = $1048.95 To nearest dollar for E = $1049 4 A$400 = NZ$470.59 A$400 = NZ$470.59 Sufficiency 4A plus 2M or 3M Steven 3 1 = $156.86 Accept $157 Merit plus 1E or 2E M M/E M/E
Student s Name: Teacher s Name: 10 Year 10 Mathematics, 2009 Trigonometry Solve right-angled triangle problems Time: 20 minutes. Check that you have entered your name and your teacher s name in the top left hand corner of this paper. You should answer ALL the questions in this paper. You should show ALL working. Check that this paper has pages 2 4 in the correct order and none of these pages is blank. YOU MUST HAND THIS PAPER TO THE SUPERVISOR AT THE END OF THE EXAMINATION. For Find simple unknown lengths of right-angled triangles using trig ratios and Pythagoras. Criteria with Merit Find unknowns of rightangled triangles from words and diagrams using trig ratios and Pythagoras. with Excellence Model 2-D situations to find unknowns of a right-angled triangle using trig ratios or Pythagoras. Overall Level of Performance (all criteria within a column met)
Trigonometry 2 QUESTION ONE Make sure you show ALL relevant working for each question. The Olympic Pool A B 25 m D 50 m C An Olympic pool has a width of 25 m and length 50 m. (a) Calculate the length of the diagonal of the pool, labelled x in the diagram. A 25 m x (b) The Olympic pool at the Beijing Olympics in 2008 was 3 metres deep. An underwater camera positioned at the bottom of the pool at one end is aimed at the surface at the far end. What is the vertical angle of view of the camera, labelled X in the diagram? D 50 m C X 50 m 3 m (c) The seating around a pool is tiered to allow better visibility for spectators. The angle of elevation of the seats is 54 and the length of seating space available is 25 m. Find the horizontal space taken up by the seating, labelled x in the diagram. 25 m 54 x
Trigonometry 3 (d) The start blocks used at the Olympics are 45.0 cm in height at the back and 30 cm in height at the front. The width of the start blocks is 52.5 cm. What is the angle of slant of the blocks, labelled X in the diagram? 45.0 cm X 30.0 cm 52.5 cm (e) A diving pool next to the main pool has a 5 metre diving platform. It is strengthened by a diagonal brace at an angle of 52 to the diving board. Calculate the length of the diagonal brace, labelled x in the diagram. 52 x 5.0 m (f) To the side of the pool is a presentation area for the awarding of medals. It has an isosceles triangular frame with a base of 12.0 m and side lengths 7.3 m. Find the vertical height of the triangular frame, labelled x in the diagram. 7.3 m x 2 1 3 12.0 m
Trigonometry 4 QUESTION TWO A tourist is standing 20.0 m from the base of the swimming pool complex taking photos. The angle from the ground, where they are standing, to the top of the flag pole is 55.4 and the angle from the ground, where they are standing, to the top of building is 51. Find the height of the flag pole, labelled x in the diagram. Show ALL your working and explain each step of your calculation. x 55.4 51 20.0 m
Trigonometry 5 Assessment Schedule Evidence Statement Question No. Merit Excellence Justification Find simple unknown lengths of right-angled triangles using trig ratios and Pythagoras. Find unknowns of right-angled triangles from words and diagrams using trig ratios and Pythagoras. Model 2-D situations to find unknowns of a right-angled triangle using trig ratios or Pythagoras. 1 (a) 55.9 m (1 dp) A 1 (b) 3.4 (1 dp) A 1 (c) 14.7 m (1 dp) A 1 (d) 15.9 (1 dp) A/M 1 (e) 6.3 m (1 dp) A/M 1 (f) 4.2 m (1 dp) A/M 2 Height of building is 20 x tan 51 = 24.6979... Height to top of flag pole is 20 x tan 55.4 = 28.9916... Height of flag pole = (28.9916 24.6979) = 4.3 m (1 dp) Sufficiency 2 or 3 of A plus 2M or 3M Merit plus E A M (both) E