Mathematical Literacy Grade 12-Measurement PDF

1  Page TABLE OF CONTENTS 1. Introduction ( to be provided, generic for all study guides) 3 2. How to use this self-study guide 4-5 3. 3.1 Measurement 3.1.1...

1  Page TABLE OF CONTENTS 1. Introduction ( to be provided, generic for all study guides) 3 2. How to use this self-study guide 4-5 3. 3.1 Measurement 3.1.1 Key concepts 6-7 3.1.2 Conversion 8-16 3.1.3 Calculating Perimeter, Area and volume 17-41 3.2 Solutions to practice questions 3.2.1 Conversion Practice Questions 42-45 3.2.2 Calculating Perimeter, Area and volume Practice Questions 45-57 4. Examination guidance 58-60 5. Study and Examination Tips 61 6. References 62 7. Acknowledgements 63 2 1. Introduction The declaration of COVID-19 as a global pandemic by the World Health Organisation led to the disruption of effective teaching and learning in many schools in South Africa. The majority of learners in various grades spent less time in class due to the phased-in approach and rotational/ alternate attendance system that was implemented by various provinces. Consequently, the majority of schools were not able to complete all the relevant content designed for specific grades in accordance with the Curriculum and Assessment Policy Statements in most subjects. As part of mitigating against the impact of COVID-19 on the current Grade 12, the Department of Basic Education (DBE) worked in collaboration with subject specialists from various Provincial Education Departments (PEDs) developed this Self-Study Guide. The Study Guide covers those topics, skills and concepts that are located in Grade 12, that are critical to lay the foundation for Grade 12. The main aim is to close the pre-existing content gaps in order to strengthen the mastery of subject knowledge in Grade 12. More importantly, the Study Guide will engender the attitudes in the learners to learning independently while mastering the core cross-cutting concepts. 3 2. How to use this Self Study Guide? This study guide covers selected sections of Measurement which form part of paper 2. The topic is drawn from the CAPS Grade 10 – 12 curriculum. Selected sections are presented in the following way: o What you should know at the end of the section. o Explanation of key concepts. o Summary/Notes. o Worked examples. o Practice questions. o Solutions to practice questions. Mathematical Literacy is a highly contextualised subject. Whilst every effort has been taken to ensure that skills and concepts you will be examined on are covered in this study guide, it is in fact the context used in the examination that will determine how these skills and concepts are assessed. This study guide covers all the cognitive levels. Go through the worked examples on your own. Do practice examples on your own. Then check your answers. Read symbols and explanation table below to understand how marks are allocated. Symbol Explanation M Method M/A Method with accuracy MCA Method with consistent accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RD Reading from a table/graph/diagram SF Correct substitution in a formula O Opinion/Example/Definition/Explanation P Penalty, e.g. for no units, incorrect rounding off, etc R Rounding off NPR No penalty for rounding NPU No penalty for the units AO Answer only, if correct, full marks Reward yourself for things you get right. If any of your answers are incorrect, make sure that you understand where you went wrong, before moving on to the next section. The study guide covers both generic and subject specific examination tips. You are expected to read and understand the tips, so that you are able to study more effectively. 4  3. Measurement 3.1.1 Key Concepts TERMINOLOGY TERM MEANING Area The amount of two-dimensional space occupied by a2-D shape. The area of a shape is the size of its surface. BODMAS Brackets, of/orders (powers, squares, etc.), division, multiplication, addition, subtraction. A mnemonic (reminder) of the correct order in which to do mathematical operations. Body mass index (BMI) A number calculated from an adult’s weight and height, expressed in units of kg/m2 Breadth How wide something is. From the word “broad”. Capacity The amount of space available to hold something. OR A measure of the volume a hollow object can hold – usually measured in litres. Circle A closed cure that is everywhere the same distance from the middle point. Circumference Distance around a circle / the perimeter of a circle. Conversion A change from one system / unit to another. Cubed The power of three; multiplied by itself three times. Cubic Shaped like a cube; having been multiplied by itself three times. Cylinder A 3-dimensional object with congruent parallel sides and bases are circles. A tall shape with parallel sides and a circular cross-section – think of a log of wood, for example, or a tube. Degrees Celsius Unit used to measure temperature in most countries. Diameter A straight line passing through the centre of a circle and touching the circle at both ends, thus dividing the circle into two equal halves. Dimension A measurable extent, e.g. length, breadth, height, depth, time. Physics, technical: the base units that make up a quantity, e.g. mass (kg), distance (m), time (s). Distance How far it is from one place to another, e.g. from one town to another or from one point to another point. Growth Charts Graphs consisting of a series of percentile cures that show the distribution of the growth measurements of children. Imperial System A system of measurement using inches, pounds, feet, gallons and miles. Length The measurement between two points, in a straight line, e.g. the length of a room. Measure Using an instrument to determine size, weight etc. Measuring Determine the value of a quantity directly, e.g. reading the length of an object from a ruler or the mass of an object from a scale. Metric System A system of measurement that uses metres, litres, kilograms, etc. A measurement system, using a base of 10 (i.e. all the units are divisible by 10). Perimeter The total distance around the boundary or edge that outlines a specific shape. Pi π, the Greek letter p, the ratio of the circumference of a circle to its diameter. A constant without units, value approximately 3,142. Radius The distance from the centre of the circle to any point on the circumference of the circle. Scale An instrument that is used to measure the weight of an object. Surface Area The area of all the faces / surfaces of an object added together. Volume The amount of 3-D space occupied by an object. It is measured in cubic units. Weight An indication of how heavy an object is. 5 3.1.2 Conversion Objectives By the end of this section, leaners must be able to: : Express measurement values and quantities in units appropriate to the context Estimate lengths and or measure lengths of objects accurately to complete tasks Estimates distance and or measure distance accurately between objects/position in space using appropriate maps and scale. Measure out quantities to complete tasks Calculate the cost of a certain amount of product Calculate the values using a formulae involving mass Summary Metric conversions You need to memorise the conversions between metric units. Length Conversion factors for length 10 millimetres (mm) = 1 centimetre (cm) 1 000 millimetres (mm) = 1 metre (m) 100 centimetres (cm) = 1 metre (m) 1 000 metres (m) = 1 kilometre (km) General method: BIG unit down to a SMALLER unit MULTIPLY by the conversion factor SMALL unit up to a BIGGER unit DIVIDE by the conversion factor 6 Conversion diagram km m cm mm × 1000 × 100 ×10 We can also reverse it to find lengths in larger units: km m cm mm ÷ 1000 ÷ 100 ÷ 10 Volume Conversion factors for volume 1 000 millilitres (mℓ) = 1 litre (ℓ) 1 000 litres (ℓ) = 1 kilolitre (kℓ) Here is a visual representation of converting between units of volume: kℓ ℓ mℓ ×1 000 × 1 000 And you can also reverse it: kℓ ℓ mℓ ÷ 1000 ÷ 1000 Weight Conversion factors for weight 1 000 mg (mg) = 1 gram (g) 1 000 grams (g) = 1 kilogram(kg) 1 000 kilograms (kg) = 1 tonne (t) 7 Here is a visual representation of converting between units of weight: t kg g × 1000 ×1000 And one can also reverse it: t kg g ÷1000 ÷ 1000 Conversion involving imperial Units of Length and Distance From METRIC to IMPERIAL 1 cm = 0,3937 inches (in) 1 m = 1,0936 yards (yd.) 1 km = 0,6214 miles (mi) 1 m = 3,2808 feet (ft.) From IMPERIAL to METRIC 1 inch (in) = 2,54 cm 1 yard (yd.) = 0,9144 m 1 mile (mi) = 1,6093 km 1 foot (ft.) = 0,3048 m Cooking conversion Conversions for cooking and baking 1 cup = 250 mℓ 1 tablespoon (tbsp.) = 15 mℓ 1 teaspoon (tsp) = 5 mℓ 8 Temperature conversion 5 °C = (°F – 32 ) × 9 9 ºF = ( × ºC) + 32 5 Worked example 1 Mary needs to measure the length of a sliding door, to find out how much material she must buy to make a curtain. The curtain material cost R89.99 per metre on sale, sold only in half metres. 1.1 Mary estimates the length of a sliding door to be 1, 5 metres long (using her arm). If Mary goes to the shop with this estimate 1.1.1 How many metres of material should she buy? 1.1.2 How much would the material cost? 1.2 Mary decides to double-check her estimated measurement before she buys the material and so she uses her tape measure to accurately measure the length of a sliding door. She determines that the sliding door is actually 4, 2 m long. 1.2.1 Calculate the total length of the material in metre that she needs to buy. 1.2.2 Calculate the cost of the material. 1.2.3 If she pays with R405, how much change will she get and why that amount? Solutions 1.1.1 1.5 m 1.1.2 Cost = R 89,99 + ½ (R89,99) = R89,99 + R44,995 = R134,99 OR Cost = R89.99 x 1.5 = R 134.99 1.2.1 4,5 m 1.2.2 Cost = 4 × R89,99 + (R89,99) = R359,96 + R44,995 = R404,96 OR Cost = R 89.99 x 4.5 = R 404.96 1.2.3 She will get R0, 10 change since we don’t have R0, 05 and R0, 01 circulating in our economy anymore. 9 Worked example 2 Francis Mohapanele Primary school has two classrooms. The two classrooms need to be sanitised. A 25 litre container of sanitizer costs R575, 00. The school needs to buy 75 containers. 1.1 Calculate the total cost of the sanitizers? 1.2 How many millilitres of sanitizers will they buy in total? Solutions 1.1 R575,00 × 75 = R43 125 1.2 = 75 × 25 = 1 875 ℓ 1 l = 1000 # ℓ = 1 875 × 1000 m ℓ = 1 875 000 m ℓ Worked example 3 It is recorded that Javier Sotomayor of Cuba is the current men’s record holder in long jump with a jump of 245 cm whereas Stefka Kostadinova of Bulgaria is the women ’s record holder with a jump of 2090 mm. 3.1 Determine with calculations who is the longest jumper between the two. 3.2 Calculate how many feet (ft) did Javier Sotomayor jump if 1m = 3, 2808 feet. 3.3 Calculate how many inches(in) did Stefka Kostadinova jump if 1 inch = 2,54 cm. Solutions 3.1 Javier Sotomayor: 245 cm To convert cm to mm : we multiply by 10 ( Bigger unit to smaller unit) 245 cm × 10 = 2450 mm Javier Sotomayor is the longest jumper. 3.2 Stefka Kostadinova jumped 2090 mm We need to convert mm to m To convert from mm to m : we divide by 1000 (Smaller unit to bigger unit) 2090 mm ÷ 1000 = 2.09 m 2,09 m = 2,09 × 3,2808 = 6,86 ft ( Bigger unit to smaller unit) 3.3 Javier Sotomayor jumped 245 cm. To convert from cm to inches : we divide cm by 2,54 (Smaller unit to bigger unit) 245 cm ÷ 2,54 = 96,46 in 10 Worked example 4 Three friends bought sugar tins with different prices shown below. A :125 000 mg B: 500 g C :0,75 kg R 9,50 R 32,40 R 39,50 [adapted from answer series grade 12] Which sugar tin is the most economical? Solutions 125 000 mg = 125 000 ÷ 1000 = 125 g A : Cost per gram = R 9,50 ÷ 125 g = R 0 ,076/g Cost per gram = R 32, 40 ÷ 500 g = R 0,0648/g B : Convert kg to g = 0,75 kg × 1000 = 750 g C : Cost per gram = R 39,50 ÷ 750 g = R 0,052/g Therefore C – 0, 75 kg of sugar cost less than the other two bottles. 11 Practice Questions Question 1 John is preparing for a union meeting that he will host on Friday afternoon. Including himself, the union consist of 8 members. He decided to make the Vetkoek with curried mince. TABLE 1: RECIPE OF VETKOEK WITH CURRIED MINCE Vetkoek Curried mince 2 cups plain flour 250 g beef mince 7g instead yeast 2 large potatoes 1 cup lukewarm water 1 large onion 1 teaspoon salt 1 large carrot 1 teaspoon sugar 2 large tomatoes Sunflower oil 2 tablespoon chutney 1 tablespoon olive oil 1 cube beef stock Frying temperature for vetkoek (Fat Cakes) is 350°C for 30 minutes. [adapted from foodleclub.com/vetkoek-and-curried-mince] 1.1 1 John bought kg of plain flour. Will the flour be enough for the 2 vetkoek with curried mince? (Hint: 1 cup = 120 g) (4) 1.2 Convert the frying temperature of 350ºC to ºF. 9 The following formula may be used: ºF = ( × ºC) + 32 (2) 5 1.3 John fries the vetkoeks for 30 minutes. If he puts the pan in the oven at 08:43, at what time must he take them out of the oven? (2) 12 Question 2 Jabu owns a supermarket in Barkly west. He buys his stock from a wholesaler in Kimberley. Below is some of the stock he buys monthly. 10 kg Tastic rice 5 kg Iwisa Samp 155 g of Lucky Star pilchards Cost Price: R174,99 Cost Price: R53,49 Cost Price: R12,99 Selling price : R192,49 Selling Price : R58,84 Selling Price R 14,30 [adapted from Northern Cape June Paper 1 Nov 2019] Use information above to answer the questions that follow: 2.1 Convert 155 g to kg. (2) 2.2 Determine the profit he will make if he sells a tin of Luck Star pilchards. (3) 2.3 He buys a 5 kg Iwisa samp and repacks the samp into 125 g packets. Determine how many packets will be able to get from one pack of 5kg samp. (3) 2.4 If the 10 kg Tastic rice is divided into 8 smaller packets. Calculate the selling price of ONE small packet. (2) Question 3 Nkgono sews dresses for children. The material costs R120, 50 per metre and she needs 3 metres of material to make a dress for a 6-year old, 4, 5 metres to make a dress for a 7- year old and 5 metres to make a dress for 10-year old. The embroidery cotton costs R15, 25 for a roll of 4 metres. She uses 3 rolls of cotton per dress. Use above information to answer the questions that follows 3.1 How many metres of material will she need to make the above 3 dresses? (2) 3.2 What will the material cost for the three dresses? (2) 3.3 Calculate in centimetre the length of the embroidery cotton that Nkgono is going to use when sewing one dress. (3) 3.4 Calculate the total amount that she will pay for the embroidery cotton? (2) 3.5 What is the total cost of a dress for a 10 year old? (3) 13 Question 4 Jappie will win a club cycling cup if he is able to log at least 600 km of cycling distance in eight months period. He cycles the Vineyard Race in January (70 miles); the Ocean-to- Ocean Race in March (130 000 yards); the Karoo Fun Race in April (888 500 feet); and the Charity Fun Sprint in June (5 832 154 inches). NOTE: 1 km = 0,6214 miles 1m = 3,281 feet 1 yard = 0,9144 metres 1 inch = 2,54 cm Use the information above to answer the questions that follow 4.1 Calculate in km the distance cycled in Vineyard Race. (2) 4.2 Calculate in km the distance cycled in Ocean to Ocean Race. (2) 4.3 Calculate in km the distance cycled in Karoo Fun Race. (3) 4.4 Calculate in km the distance cycled Charity Fun Sprint. (3) 4.5 Verify with calculations whether Jappie will be able to win the club cycling cup. (3) 3.1.3 CALCULATING PERIMETER, AREA AND VOLUME Objectives In this topic learners need to : Solve problems and complete tasks/project Determining and/or calculating appropriate quantities of materials needed to complete a task. Calculate the cost of materials needed to complete a task. Determine the required budget for a given project. Make a choice regarding costs and/or quantities and/or materials used in order to complete the task within a given budget. 14 Summary Formula Shape Rectangle Perimeter =2 × length + 2 × width ℓ Area = Length × width w Square Perimeter = 4 × length or 4 × side side × side = side2 s s Triangle Perimeter = Length 1 + length 2 + length 3 Area = ½ base × perpendicular height s s h s b 15 Circle Circumference = π × (2 × radius) C OR r Circumference = π ×diameter Area = π × #adius2 Volume Shape Volume Diagram ℓ Rectangular box V=ℓ ×b×h b h r Cylinder V = π × r2 × h h 16 WORKED EXAMPLES Worked example 1 Uncle Joe bought a house and decided to do renovations in the lounge area. He plans to change the wall panels on one of the walls in this room. Below is the plan of the lounge Diagram Plan of the wall Wall to be decorated Measurements of the wall (8 m ´ 4, 5 m) Use the above information to answer the questions that follow 1.1 Calculate the perimeter of the part of the wall that needs to be decorate The following formula may be used: Perimeter = 2 x (Length ´ Width) 1.2 Calculate the area of the wall that is going to be decorated with wall panels. The following formula may be used: Area = Length ´ Breadth 1.3 Uncle Joe is decorating the wall with the ceiling panels Ceiling panel is 1 m2 Part of the ceiling (8 m ´ 9,5 m) 17  1.3.1 How many wall panels will be needed for the wall? Give your answer to the nearest whole number. The following formula may be used: Area of the wall to be decorated Wall panels needed = 1,5 m 2 1.3.2 The area of the part of the ceiling that is going to be decorated is 45 m2. Determine how many packs of ceiling panels must uncle Joe buy if there are FOUR pieces in a pack 1.3.3 The colour of the wall panels is white and uncle Joe’s lounge is orange. He bought the same colour paint to paint the wall panels. The paint that he wants to use can cover approximately 9 m2 per litre. How many litres of paint must he buy? Solutions 1.1 P = 2 ´ (8 + 4,5) = 25 m 1.2 A = 8 ´ 4.5 = 36 m2 1.3.1 36 ÷ 1.5 = 24 panels 1.3.2 45 ÷ 1= 45 pieces 45 ÷ 4 = 11,25 packs = 12 packs 1.3.3 36 ÷ 9 = 2,72222 = 4 litres of paint 18 Worked example 2 The following pond is made of rectangular shape at the center and two semi-circular shapes on the either side as shown below. The centre circle (has a radius of 3 m) in the middle indicates where the statue will be erected. The area around the pond is paved by bricks 24 m 32 m Use information above to answer questions that follow. 2.1 A protective fence must be erected around the pond. Determine the perimeter to be covered by the fence. 19  This formula may be used: Perimeter of rectangle = 2( l + b) 2.2 Currently, the quantity and cost of paint is as follows Quantity Cost 5 litre R 475,00 10 litres R580,00 20 litres R998,00 [source:builders.co.za] 2.2.1 One litre of paint covers one coat of paint on 3m2 of the wall. Determine the number of litres of paint to cover all the inside walls of the pond. 2.2.2 Determine the tin size (tin of paint) that is the cheapest per litre 2.2.3 Provide an advice on the size of the tin that will be most economical Solutions 2.1 Perimeter = 2 (l + b) = 2 (32 + 24 ) = 2 (56) = 112 m 2.2.1 1litre = 3m2 Litres = 38,78 m2 = 12,94litres = 12,94litres = 13 litres 2.2.2 5 litres = R475.00 : 1L = 475 ÷5 = R95 10 litres = R580 : 1L=580÷10 = R58 20 litres = R998 : 1L= 998 ÷20 = R49,99 Therefore 20 litres tin is cheaper 2.2.3 13 litres required 1 x 10 litres and 1 x 5 litres 10 litre x 1 = R580 1 x 5 litres = R475 Total = R1055 20 litres = R998.00 Therefore it is economical to purchase the 20L tin 20 Worked example 3 Julia buys some material to make dish cloths and tea towels which she uses at her catering project. The material she buyshas a fixed width of 120 cm and is cut into any required length. She cuts the material into rectangular pieces of 30 cm × 45 cm to make the tea towels and 30 cm × 30 cm to make the dish cloths, as shown in the photographs below. Use information above to answer the questions that follow. 3.1 Calculate the area of the material needed to make ONE tea towel. You may use the formula: Area of a rectangle = length × breadth 3.2 Julia wants to make a decorative border on some of the tea towels Calculate the perimeter of a tea towel. You may use the formula: Perimeter of a rectangle = 2(length + breadth) 21 3.3 To make the dish cloths and tea towels she uses the cutting layout as shown in the sketch Cutting layout for dish cloths and tea towels using dotted lines: Use the layout above to answer the questions that follow. 3.3.1 Determine the minimum length of material that Julia will need if she wants to make exactly 4 dish cloths and 4tea towels. 3.3.2 Determine the maximum number of dish cloths and tea towels she can make if she has a piece of material that is exactly 180 cm long. You need to make sure that no material is wasted. 3.3.3 Determine the maximum number of dish cloths and tea towels she can make if she has a piece of material that is exactly 180 cm long. You need to make sure that no material is wasted. The edges of the dish cloths are stitched, making the corners round as shown in the diagram To calculate the area of the finished dish (x is the distance from the side of the cloth, she uses the following formula: square where she starts to make the 2 2 Area (in cm ) = 900 – x (4 – π) corners round.) Calculate the Area of ONE finished dish cloth if x = 3 and using π = 3, 142. 22  3.4 Julia buys the material at R45, 00 per metre. She uses the table below to help her calculate the cost of the material required. Length of material ( in metres) 0 2 4 5 7 B 10 Cost ( in rand) 0 90 180 A 315 360 450 Use information above to answer the questions that follow. 3.4.1 Complete the following formula: Cost of material = R 45, 00 × …….. 3.4.2 Calculate the missing values A and B. 3.4.3 Draw a line graph that represents the relationship between the length and the cost of the material. Solutions 3.1 Area of a rectangle (Tea towel) = length × breadth = 30cm × 45 cm = 1350 cm2 3.2 Perimeter of a rectangle = 2(length + breadth) = 2 (30 cm + 45 cm) = 2 (75 cm) = 150 cm 3.3.1 30 cm + 45 cm = 75 cm 3.3.2 180 cm = 2 × 75 cm + 30 cm She can make 8 tea towels and 12 dish cloths 3.3.3 Area (in cm2 ) = 900 – (3)2 (4 - 3,142) = 900 – 7,74 = 892,26 3.4.1 Cost of the material = R45,00 × length of material (in meters) 3.4.2 A = R 45,00 × 5 =R 225 For calculating B: Cost of material =R 360 Length of material = B According to the formula: Substitute what you have i.e. R 360 =R 45,00 × Length of the material (B) Therefore B = 360 ÷ R 45,00 = 8 meters Cost of material for different lenght in metres 600 400 COST (R) 200 0 0 2 4 6 8 10 12 LENGTH (M) 23 Worked example 4 Rachel sells hair products. One of the Picture of cylindrical jar of aqueous product that is in high demand is aqueous cream cream used for cleansing and moisturising, as shown in the picture alongside. The aqueous cream is sold in 100 ml jars. The cylindrical jars are filled with cream to a height of 4, 5 ml. 4.1 Calculate in cm the diameter of the jar. You may use the formula: Volume of a cylinder = & × ( radius)2 × height, where & = 3,142 NOTE:1 mℓ = 1 cm3 Solution Volume of a cylinder = π ⨉ (radius)2 ⨉ height 100 ml = 3,142 ⨉ (radius)2 ⨉ 4,5 100 cm3 =14,139 (radius)2 100 cm3 ÷ 14,139 = 14,139 (radius)2 ÷ 14,139 7, 072635971 = (radius)2 (7,072635971 =r 2,659442793 = radius Diameter= 2,659442793 ⨉ 2 = 5, 32 cm 24 Practice questions Question 1 Mr Masigo is working as an intern at Bata shoe company in Durban. The company designs boxes and other containers for shoes. He is required to work on a design for a shoe box, as shown in the diagram Picture of the Bata shoe box Dimensions of the Bata shoe box: length = 320 mm; width = 205 mm; height =111 mm Note: The dimensions of the base are the same as the dimensions of the top (lid). Area = Length × width [ source:www.boxesonline.co.za] Use the above information to answer the question that follow. Determine how much cardboard (the surface area) is needed to make one Bata shoe box in cm2. NB: The Shoe box has six faces. The overlap Is not taken into account when forming a box. (5) 25 Question 2 Ten years ago, John bought a rectangular prism-shaped ottoman and two matching cubic-shaped ottomans1. He wants to refurbish each of them by having the side surfaces (excluding the top and bottom) repainted. He will also employ an upholsterer2 to re-cover the top of each ottoman and to attach cylindrical-shaped legs to the base of each ottoman. Each cubic-shaped ottoman will have 4 legs, while the rectangular prism-shaped ottoman will have 6 legs. 400 mm 190 cm DIMENSIONS: Rectangular ottoman Length = 190 cm Width = 400 mm 400 mm Height = 400 mm 400 mm Cubic-shaped ottomans 4 mm Side = 400 mm mm PICTURE OF A LEG DIAGRAM OF A LEG DIMENSIONS OF A LEG Diameter = 75 mm 8 cm Total height = 8 cm 75 mm 1 Ottoman: a piece of furniture like a large box with a soft top, used as a seat 2 Upholsterer: someone whose job it is to cover furniture with material Use information above to answer the question that follow 2.1 Determine the total number of legs for the ottomans John has to purchase (3) 2.2 Calculate the radius of the ottoman's leg. (2) 2.3 Calculate, in centimetres, the total height (including the legs) of ONE cubic-shaped ottoman (2) 2.4 Calculate, in cm2 , the total surface area of the side surfaces of all three Ottomans that need to be painted. You may use the following formulae: Area of a rectangle = length ´ width Area of a square = side ´ side (5) 26 2.5 John bought a one-litre tin of luxurious silk paint to paint the side surfaces. The paint has a spread rate of 9 m 2 per litre. 1-litre tin of pain [Source: www.farmcity.co.za]. Calculate, in millilitres, the amount of paint needed to paint ALL the ottomans with TWO coats of paint (4) 2.6 The tin has an inner radius of 3.3 cm. Calculate the height (in cm) of the paint in the tin, if 1 litre = 1 000 cm3. You may use the following formula: Volume Height = (3) 3,142 ´ (radius) 2 Question 3 Dr Seroto the head of Maths, Science and Technology has to produce rectangular display boards when organising a science fair. The external dimensions of each board are 52 inches by 40 inches. The front surfaces of the boards must be spray – painted with one layer of non- reflective white paint. NOTE: 1 mℓ of the paint covers a surface area of 50 cm2 1 inch = 25 mm Use information above to answer the questions 3.1 Determine, showing ALL calculations, whether 8 litres of the paint would be enough to spray paint 30 display boards. The following formula may be used: Area = length × width (10) 27 3.2 Dr Seroto has a choice of two cylindrical containers (as shown below) for display purposes. A decorative label with a 1 cm overlap will be placed right around the container completely covering the curved surface only. CYLINDER - A CYLINDER - B Diameter = 30 cm Diameter = 40 cm Height = 30 cm Height = 20 cm Verify, showing ALL calculations, whether cylinder B would require less material to make the decorative labels. The following formula may be used: Curved surface area of cylinder = π × diameter × height, π = 3,142 (7) 3.3 Dr Seroto travelled from his office directly to the school 45 km away. He travelled at an average speed of 100 km per hour and arrived at the school at 11:20. Verify, showing ALL calculations, whether Dr Seroto left his office at exactly 10:50. The following formula may be used: Distance = average speed × time (5) 28 Question 4 4.1 Use the above information to answer the questions that follow. 4.1.1 Give, in simplified form, the ratio of distance AD to distance CB. (2) 4.1.2 The perimeter of ABCD is 125, 92 m. Calculate the distance CD (2) 4.1.3 Write down the length of the radius of the pool. (2) 4.1.4 A fence will be erected along the curved side AB at a cost of R97, 56 per (2) running metre. Calculate the total cost of erecting the fence 4.2 Anju uses a sedan vehicle to travel. The fuel consumption of her vehicle is 8, 4 litres per 80 km travelling at an average speed. Use the above information to answer the questions that follow 4.2.1 Calculate (to the nearest km) the distance her vehicle can travel using 60 litres of petrol. (3) 4.2.2 Anju spends 1 hour 45 minutes on a particular day driving between two workstations that are 196 km apart. Determine the average speed of the vehicle. You may use the following formula: Average Speed = distance ÷ time (3) 29 4.3 The dimensions (in centimetres) of Anju’s rectangular bag are given below. Rectangular Bag Dimension of the bag without the handle [Source : Amazon.com] Use the above information to answer the questions that follow. 4.3.1 Calculate the volume (to the nearest litre) of ONE bag excluding the handle. You may use the following formula: Volume = length × width × height (4) QUESTION 5 Below is a drawing of a big cardboard box, A. Smaller boxes of chalk are packed into this cardboard box. The smaller box, B, are in the shape of a cube (all the sides are equal in length). The sketches are not drawn to scale. A 50 cm B 9 cm 40 cm 9 cm 96 cm Use information above to answer the questions that follow. 30 5.1 Calculate the area of the shaded rectangular side of the cardboard box A. Use the following formula Area = length × breadth (2) 5.2 Calculate the volume of cardboard box (B). Use the following formula Volume = length × breadth × height (3) 5.3 Determine how many boxes of chalk (B) can be packed into the larger cardboard box (A). (5) 5.4 If one small chalk box (B) contains 100 pieces of chalk, determine how many pieces of chalk one big cardboard box (A) holds. (2) 31 Question 6 Paul is residing in Kimberley under Sol Plaatje municipality. He is planning to build himself a swimming-pools in such a manner that one part is deeper than the other part. A sketch drawing of a similar swimming- pool is given below. The dimensions are clearly marked on the figure. ABHG, DCEF and BCEH are rectangles. Picture of the swimming pool Diagram of the swimming pool G 13 m F 5m 2.2 m E A D H 4m C B Use information above to answer the questions that follow. 6.1 Calculate the area of ABCD Use the Formula : (4) Area of ABCD = ½ (AD)(AB + DC) 6.2 Name another side of the pool with the same shape and size as that of ABCD (2) 32 6.3 Calculate the area of the following sides of the pool. Use the Formula: Area = length x breadth 6.3.1 ABHG (2) 6.3.2 CEFD (2) 6.4 All 4 side walls of the swimming-pool are to be tiled. The length and width of each tile is 1600 mm and 1200 mm respectively. 6.4.1 Calculate the total area (m) of all 4 side walls of the swimming-pool (3) 6.4.2 Calculate the area of on one tile in square metres (m2). (4) 6.4.3 Calculate the number of tiles Paul will need to finish the tiling of the 4 side walls? (Ignore the waste in cutting.) (2) 6.4.4 CTM sells tiles in boxes of 24. How many boxes of tiles Paul has to buy? Give your answer to nearest whole number. (3) 6.4.5 A box of tiles cost R159, 90/m2. What is the total cost of the tiles for the above job? (2) 6.5 The swimming-pool is to be filled with water. The formula to determine the volume is: Volume = Area of ABCD x AG Calculate the capacity (volume) of the swimming-pool 6.5.1 In cubic metres (m3) (2) 6.5.2 In kilolitres (kℓ) (Hint :1 m3 = 1 kℓ) (2) 6.6 The Sol Plaatje municipality charges R6, 65 per kilolitre (kℓ). Calculate the cost of water needed to fill the pool. (2) 6.7 What is the depth at the centre of the pool? (2) You may use the following formula: Depth = ½ (AB + DC) 33 Question 7 Kwazi bought a 750 ml can of motor oil. He is aware that the can of oil is not filled to its fullest capacity. Below is a photograph of the can of motor oil and a diagram showing the external dimensions of the cylindrical can. Photograph of the can of motor oil Diagram of the cylindrical can of motor oil 10, 5 cm 12, 5 cm NOTE: 1 cm3 = 1 ml Calculate: 7.1.1 The actual capacity (volume) of the can You may use the following formula: Volume of a cylinder = π × ( radius )2 × height ,where π = 3,142 (4) 7.1.2 The volume of the empty space in the can when it is filled with 750 ml of motor oil. (2) 7.1.3 The height of the motor oil in the can You may use the following formula: !"#$%& "( %")"* "+# Height of motor oil in can = , × (/01234)! where π = 3,142 (3) 34 7.2 Mr Koba owns a refuse removal company that removes refuse from building sites. His refuse containers are rusted and Mr Koba wants to repaint the exterior sides of the containers. Each container is an open – top right prism with two sides that are trapezium- shaped and two slanted sides that are rectangular. The photograph below is of one of his refuse containers that he delivers to building sites. The dimensions of a container are shown on the diagram below. Use the above diagram to answer the following questions 7.2.1 Calculate the area (in mm2) of the triangular part marked Y on the diagram. You may use the following formula: 6 (2) Area of a triangle = × 2345 × 657869 7 7.2.2 Use the formula below to calculate the total exterior area (in m2) of the trapezium side of the container if the area of the rectangular part X is equal to 2 088 000 mm2. You may use the following formula: Area of trapezium side of the container = 2 × (area of Y) + (area of part X) NOTE: 1 m2 = 1 000 000 mm2 (5) 7.2.3 The total exterior surface area of the container, excluding the top and the bottom base, is 11,676 m2. Hence, calculate the area of ONE of the slanted rectangular sides of the container if they are identical in size. (3) 35 7.3 To paint the exterior sides of the container it must first be rustproofed with a coat of Optirustbusta. It is then painted with Optimetalcoat. The technical consultant recommends that two layers of Optimetalcoat must be applied. Mr Koba has 25 identical containers, each having an exterior area of 11,676 m2 that need to be repainted as described above. Optimetalcoat is sold in 5 litre tins. One tin of Optimentalcoat will cover 25 m2. 7.3.1 Calculate the total area of all the containers that will be coated with Optirustbusta. (2) 7.3.2 Determine the total number of coats of Optimetalcoat that will be needed for all the containers. (2) 7.3.3 Mr Koba estimates that he needs to paint a total area of 585 m2.Calculate the minimum number of tins of Optimetalcoat that he will need to order, based on his estimation. (3) 36 Question 8 8.1 Simone uses the local swimming pool to give swimming lessons. The rectangular pool has a shallow section, C, a deep end, A, and a sloping section, B as shown in the various views below. The capacity of section of B of the swimming pool is 300 m3. You may use the following formula: Volume of a rectangular prism = length × width × height NOTE: 1 gallon = 3,785 litres 1 m3 = 1 000 litres 8.1.1 Show, with calculations, that the maximum capacity of the swimming pool is 765 m3. (5) 8.1.2 Calculate the volume of water (in gallons) required to fill the swimming. Calculate the volume of water (in gallons) required to fill the swimming pool to 94% of its capacity. (4) 8.1.3 The pool must be topped up with 135 000 ℓ of water due to water loss. The pool is filled with water at a constant rate of 2 350 litres per hour. Simone says it will take (5) 8 exactly 29 days to top up the pool. Verify, showing ALL calculations, if her statement is valid. 37 Question 9 9.1 Koos lives in Pelican Road in Port Elizabeth. He is making a pentagonal post box for a house as shown in the diagrams below: The front and rear ends of the post box are regular pentagons with side lengths equal 270mm. The bottom, the top and sides of the post box are rectangles with a length 360mm and a breadth of 270mm. [source: Free state June Paper 2 June 2021] Use the information above to answer the questions that follow: 9.1.1 Calculate the perimeter of ONE of the pentagonal ends of the post box. (2) 9.1.2 Calculate the total surface area (in m2) of the post box (excluding the openings for the newspaper and letter), if the following are given: SHAPE AREA Pentagon 0,13m2 Letter opening 0,017m2 Newspaper opening 0,013m2 The following formulae may be used: Area of a rectangle = length × breadth Total surface area of post box = 5 × areas of rectangles (side) + area of front + area of back (5) 38  9.1.3 A newspaper folded into a cylindrical shape has a diameter of 12 cm. The area of the newspaper opening of the post box is 0,013 m2. Show, with calculation, whether the folded newspaper will fit in the newspaper opening of the post box. The following formula may be used: Area of a circle = π × radius2 and π = 3,142. (5) 3.2 SOLUTIONS TO PRACTICE QUESTIONS 3.2.1 Conversion Quest Solution Explanation L 1.1 2 cups = 120 g × 2 üüMA 2MA multiplying by 2 L2 = 240 g 1 kg = 500 güA 1A simplification 2 John bough more than enough flour üA 1A conclusion (4) 1.2 9 L2 ºF = ( × 350) + 32 üSF 5 1 SF substituting = 662 ºFüA 1 A simplification (2) 1.3 Time = 8:43 + 0:30 üMA 1MA adding L2 = 9:13üA 1A simplification (2) 39 Quest Solution Explanation L 2.1 1 kg = 1 000 g L1 ? = 155 g Quantity in kg = 155 ÷ 1000üMA 1MA dividing by 1 000 = 0,155 kg üA 1A amount in kg OR OR 155 g = 155 g × 0,001 kgüMA 1MA dividing by 1 000 = 0.155 kgüA OR 1A amount in kg 155 g = 155 ÷ 1000üMA = 0,155 kgüA OR 1MA multiply by 0,001 1A amount in kg (2) 2.2 Profit = R14,30 – R12,99üRT üM 1RT correct values L1 = R1,31üCA 1M subtracting values 1CA simplification (3) 2.3 Number of packet: L1 = 5 kg ×( 1000 ÷ 125 g) üMAüM 1MA multiply by 1 000 = 5× 8 1M dividing by 125g =40 packets üCA 1CA simplification OR OR 5 ×1000 = 5000 g ÷125 )üMAüM 1MA multiply by 1 000 = 40 packets üCA 1M dividing by 125g 1CA simplification. (3) 2.4 Selling price = R192,49 ÷ 8üMA 1MA dividing correct value by 8 L1 = R24,06üCA 1CA simplification (2) 40 Quest Solutions Explanation L 3.2 Material needed = 3m + 4,5 m + 5 m üM 1M addition L1 = 12,5 m üA 1 A answer (2) 3.2 The cost for the 3 dresses= Length of material ×price L1 = 12,5× R120,50üM = R 1506,25üCA (2) 3.3 The length of embroidery = Length of one roll × 3üMA 1M multiplication L2 = 4m × 3 1C conversion = 12m × 100üC 1A answer = 1200 cmüA (3) 3.4 Total amount = No. of dresses × 3 cotton rolls × price L2 = 3 × 3 × R 15,25üMA 1 MA = R 137,25 üA 1A answer (2) 3.5 Total cost for 10 yrs. old dress L2 = (Length of material × price) + (3 rolls of cotton ×price) = ( 5 ×R 120,50) + (3 ×R 15,25) üSF 1SF substitution =R 602,25 + R 45,75 üM 1M adding = R 648, 25üA 1A answer (3) 41 Quest Solution Explanation L 4.1.1 Distance cycled in Vineyard Race L2 = 70 ÷ 0,6214ü MA 1MA conversion =112,65 km üA 1A simplification (2) 4.1.2 Distance cycled in Ocean to Ocean Race L2 = 130 000 yards × 0,9144 = 118 872 m üMA 1MA conversion m = 118 872 ÷ 1 000 = 118,87 km üA 1M simplification (2) 4.1.3 Distance cycled in Karoo Fun Race: 888 500 feet L2 = 888 500 ÷ 3,2808üMA 1MA division 3,2808 = 270 818,09 m ü A 1A 270 818,09 m = 270 818 ÷ 1000 = 270,82 km ü C 1C Simplification (3) 4.1.4 Distance cycled Charity Fun Sprint: 5 832 154 L2 inches = 5 832 154 × 2,54 üMA 1MA conversion cm = 14 813 671,16 cm üA 1A simplification = 14 813 671,16 ÷ 100 000 üM 1M diving by 100 000 = 148,14 km (3) 4.1.5 Total distance covered L2 = 112,65 + 118,87 + 270,82 + 148,14 üMA 1MA addition = 650,43 km üA 1A simplification Therefore, Jappie will be able to win the club cycling cup üO 1O explanation (3) 42 Calculating Perimeter, Area and Volume Quest Solution Explanation L 1. l = 320 ÷ 10 = 32 cm w = 205 ÷ 10 = 20,5 cm üC 2C converting to cm h = 111 ÷ 10 = 11,1 cm 2 × (l × w ) + 2 × ( w × ℎ ) + 2(> × ℎ ) =2 × ( 32 × 20,5 ) + 2 × (20,5 × 11,1 ) + 2 ×(32 × 11,1) = 1 312 + 455,1 + 710,4 2MA calculating surface area = 2 477,5 cm2 üA üMA 1A simplification ü (5) Quest Solution Explanation L 2.1 Legs of ottomans L1 2 cubic ottomans × 4 legs = 8 legs/pote üA 1A number of legs 1 retangular ottoman × 6 legs = 6 legs = 8 + 6 üMA 1MA adding 6 legs = 14 legs üCA 1CA total number of legs (3) 2.2 75 mm üMA 1MA concept of radius L1 Radius = 2 1A simplification = 37,5 mm üA (2) 2.3 Total height L1 (400 ÷ 10 ) + 8üC 1C converting to cm = 40 + 8 üM 1MA finding the height = 48 cm (2) 43 2.4 8 square sides × (40 × 40 ) L2 = 12 800 cm2üA 1A simplification 2 rectangular side × (40 × 190 ) = 15 200 cm2 üA 1A simplification 2 square × ( 40 × 40 ) = 3 200 cm2 üA 1A simplification Total area to be painted = 12 800 cm2 + 15 200 cm2 + 3 200 cm2 üM 1M adding all values = 31 200 cm2 üMA 1MA finding total area (5) 2.5 31 200 ÷ 10 000 = 3,1 m2 üM 1C converting from cm2 to L2 3,1 × 2 = 6,2 m2 üM m2 6,2 m2 ÷ 9 = 0,69 üM 1M area for 2 coats = 690 millilitres üA 1M divide by spread rate 1A answer in millilitres (4) 2.6 Volume L2 Height = π ´ (radius) 2 1C conversion from ml to cm3 1000 cm 3 üC = 1SF substitution 3,142 ´ (3,3)2 üSF 1CA simplification = 29,22576848…. cm üCA (3) 44 Quest Solutions Explanation L 3.1 Area of display 4 = length × width = 52 inches × 40 inches üSF 1SF substituting correctvalues 1C converting inches to mm = 52 × 25 mm ÷ 10 × 40 × 25 mm ÷ 10 üC 1C converting mm to cm 2 = 130 cm × 100 cm = 13 000 cm üCA 1CA area of one display Total area of 30 displays =13 000 cm2 × 30 üM 1M multiplying by 30 = 390 000 cm2üCA 1CA calculating total area Amount of whiteboard paint needed 1M working with ratio = 390 000 cm2 ÷ 50 cm2üM 1CA calculating paint used üCA = 7 800 ml ÷ 1 000 = 7,8 litres üC 1C converting to liters Therefore 8 ℓ will be enough. üO OR 1O conclusion 8 liters of paint can cover OR 8 ℓ × 1 000 üC = 8 000 m ℓ × 50 üM 1C converting to mℓ 1M multiplying by 50 = 400 000 cm2 üCA 1CA area Area of display = length × width 1SF substitution = 52 inches × 40 inches üSF 1C conversion to mm = 52 × 25 mm ÷ 10 × 40 × 25 mm ÷ 10üC 1C conversion to cm = 130 cm × 100 cm üC 1CA one display board area = 13 000 cm2 ü Total area of 30 displays 1M multiplying by 30 = 13 000 cm2 × 30 üM 1CA total area = 390 000 cm2 üCA 1O conclusion 8 ℓ is enough. ü (10) 45 3.2 Total Surface Area of cylinder A 4 = π × diameter × height = 3,142 × 30 × 30 ü SF 1SF correct values 1CA = 2 827,80 cm2 ü CA calculating area Total Surface Area of decorative sticker for cylinder A = 2 827,80 cm2 + (1 × 30) cm2 ü M 1M adding area of overlap = 2 857,80 cm2 ü CA 1CA calculating area of sticker Total Surface Area of cylinder B = π × diameter × height = 3,142 × 40 × 20 = 2 513,60 cm2 ü CA 1CA area of cylinder B Total Surface Area of decorative sticker for cylinder B = 2 513,60 cm2 + (1 × 20) cm2 = 2 533,60 cm2 ü CA 1CA area of sticker B Correct, B will require less ü O 1O conclusion OR Total Surface Area of sticker for cylinder A OR = [(π × diameter) + 1] × height ü M 1M formula 1M for adding 1 to üM circumference = [(3,142 × 30) + 1 ] × 30 ü SF 1SF substitution = 2 857,8 cm2 ü CA 1CA calculating area Total Surface Area of sticker for cylinder B = [(π × diameter) + 1] × height = [(3,142 × 40) + 1] × 20 ü SF 1SF correct values = 2 533,6 cm2 ü CA 1CA calculating area Correct, B will require less üO 1O conclusion (7) 46 3.3 Distance = average speed ⨉ time 4 45 km = 100 km⨉ time üSF 1SF substitution Time = 0,45 hoursüA 1A time in hours = 27 minutes üA 1Atime in minutes Dr Seroto left home at 27 minutes before 11:20 1CA simplification = 10:53 üCA 1O conclusion He did NOT leave at 10:50 üO OR OR 1A time in hours Time diff. = 11:20 – 10:50 = 30 min = 0,5 hours 1SFsubstitution1CA üA distance 1O comparing 1O Distance =100 km/h × 0,5 h üSF üCA conclusion = 50 km more than 45 km üO Dr Seroto did NOT leave at 10:50 but a bit (5) laterüO Question 4 Quest Solutions Explanation L 4.1.1 AD: CB = 10,9 : 9,45 üM 1M ratio form L1 = 218 : 189 üCA 1CA (2) 4.1.2 CD = 125,92m – ( 57,5 + 10,9 + 9,45) ü MA 1MA subtracting all lengths L1 = 48,07m üCA 1CA length (2) 4.1.3 Radius = 4,73 m ÷ 2 üM 1M dividing by 2 L1 = 2,365 m üA 1A simplification (2) 4.1.4 Total Cost = R 97,56/m × 57,5 m ü MA 1MA multiply cost by correct L1 = R 5 609,70 üCA distance 1CA simplification (2) 4.2.1 Distance = 60 litres ÷ 8,4 litres × 80 km 1MA multiply by 80 L2 üüMA 1MA divide by 8,4 = 571,43 = 571 km ü R 1R distance (3) 47 4.2.2 Average speed = 196 ÷ 01h45 üSF 1SF to hours L2 = 196 ÷ 1,75 üC 1C correct values = 112 km üCA 1CA Average speed (3) 4.3.1 Volume = 53,34 cm × 17,78 cm × 42,32 cm 1SF correct substitution L3 üSF = 40 135,66 cm3 üCA 1CA volume = 40 135,66 litres ÷ 1000ü MA 1MA dividing by 1 000 = 40 litres ü R 1R volume in litres (4) Quest Solution Explanation L 5.1 Area = 50 cm × 40 cm PSF 1SF substitution L2 = 2 000 cm2 PCA 1CA area (2) 5.2 Volume B = 9 cm × 9 cm × 9 cm PSF 1SF substitution L2 = 729 P cm3 PCA 1CA volume of B 1A unit (3) 5.3 Length = 96 cm ÷ 9 cm = 10 PA 1A length Width = 40 cm ÷ 9 cm = 4 PA 1A width L3 Height = 50 cm ÷ 9 cm = 5 PA 1A height Nr of chalk boxes = 10 × 4 × 5 PM 1M = 200 chalk boxes PCA 1CA number of boxes (5) 5.4 Pieces of chalk: 100 × 200 PM 1M multiplying L1 = 20 000 pieces of chalk PCA 1CA number of pieces (2) 48 Quest Solution Explanation L 6.1 ½ (13 m)(4 m + 2,2 m) ü üSF 1SF for 13 m L2 1SF for 4 m and 2,2 m = 6,5 m x 6,2 m üS 1S simplification = 40,3 m2 üA 1A answer (4) 6.2 GHEF üü 2A answer L1 (2) 6.3.1 Area of ABHG = 4 m x 5 m üSF 1SF substitution L2 = 20 m2 üCA 1CA simplification (2) 6.3.2 Area of CEFD = 5m x 2.2m üSF 1SF substitution L2 = 11 m2 üCA 1CA simplification (2) 2 2 2 6.4.1 (40,3 m x 2) + 20 m + 11 m üMAüMA 1MA multiply by 2 L2 = 111,6 m2 üCA 1MA adding 1CA: simplification OR OR (40,3 m + 40,3 m ) + 20 m2 + 11 m2 üüMA 2 2 2MA adding 40,3m2 = 111,6 m2 üCA 1CA: simplification (3) 6.4.2 L = 1 600 ÷ 1000 = 1,6 müR 1R converting length to m L2 W = 1 200 ÷ 1000 = 1,2 m üR 1R Converting w to m Area = 1,6 X 1,2 = 1,92 m2üCA 1CA simplification (3) 6.4.3 111,6 m2 ÷ 1,92 m2 üMA 1MA: dividing by 1,92 = 59 tiles üCA 1CA simplification (2) 6.4.4 59 ÷ 24 üMA 1MA dividing by 24 = 2,45 boxes üA 1A answer ≈ 3 boxes üR 1R rounding 6.4.5 Cost = 3 × R159,90 üMA 1MA multiplying L1 = R479,70 üA 1A simplification (2) 6.5.1 Volume = 40,3 × 5 üMA 1MA multiplying L1 = 201,5 m3üA 1A simplification (2) 6.5.2 1 m3 = 1 kl 2A answer L1 = 201,5 klüüA (2) 6.6 Cost= R6,65 × 201,5üMA 1MA multiplying = R 1 339,98üA 1A simplification (2) 6.7 ½ (4 m + 2,2 m) üMA 1MA adding L1 = ½ (6,2) = 3,1 m üA 1A simplification (2) 49 Quest Solutions Explanation L 7.1.1 Radius =10,5 cm ÷ 2 = 5,25 cm ü Mü SF 1M radius Volume of cylinder = 3,142 × 5,252 × 12,5 cm3 1SF substitution L2 = 1082,517 cm3 = 1082,5 cm3ü CA 1CA volume (3) 7.1.2 Volume of empty space = 1082,5 – 750 cm3ü A 1M subtracting 750 L3 = 332,5 cm3 ü CA 1CA volume (2) 7.1.3 :;< =>" 1SF substitution Height of motor oil in can = ?,8A9 ×( ;,7; =>)! ü L2 SF 1A simplification :;< =>" = üA 1CA height BC,C