Apply numeric reasoning in solving problems 17 Direct linear proportions Various practical problems involving rates can also be solved using proportional reasoning. If two quantities are directly proportional, then the ratio of the two quantities is constant. The ratio can be scaled up (or down) by multiplying (or dividing) both quantities in the ratio by the same number. Example Which mixture has a greater proportion of eggs: 3 eggs in a 825 g cake or 2 eggs in a 560 g cake? Solution The first mixture has eggs:cake = 3:825. Example The second mixture has eggs:cake = 2:560. 13 cups of fruit drink are made using 3 cups of cordial and 10 cups of water. Scaling up both ratios, compare 6:1 650 with 6:1 680. [double first, triple second] The ratio of cordial to water is 3:10 (using any unit of measurement). The first cake has the greater proportion of eggs. [6 eggs to smaller mixture] The table shows other quantities of fruit drink, with cordial and water in this proportion. Note: This problem could also be solved using 3 2 fractions: compare with . 825 560 1 1 with . Simplifying fractions: compare 275 280 The first fraction is larger. Cordial Water Explanation 9 30 multiplying 3:10 by 3 tablespoons tablespoons 1 1 cups 2 5 cups 10 litres 3 50 litres 3 1 litre 5 litres dividing 3:10 by 2 dividing 3:10 by 3 10 multiplying 1: by 5 3 3 Note: Each of these quantities of fruit drink is 13 10 cordial and water. 13 [each quantity has 3 parts cordial and 10 parts water making 13 parts altogether] [same numerator, larger denominator] Exercise K: Direct linear proportions 1. Complete the tables by scaling up/down the mixtures made up of two quantities in the given ratios. Remember to include units. a. cordial : water = 1:7 Cordial Fractions are useful for comparing proportions. Water 2 cups Example 10.5 tablespoons Each working day, Dave spends 3 hours in his car and 5 hours at his desk, while Kim spends 3.5 hours in her car and 5.5 hours at her desk. Who spends a greater proportion of their working day in the car? 3 litres What fraction of the mixture is water? E L P M SA S E G A P Solution 3 Dave has car:desk ratio 3:5 so he spends of his 8 day in his car. Kim’s ratio is 3.5:5.5 or 7:11 (doubling) so she 7 of her day in her car. spends 18 28 3 7 27 and [changing to equivalent = = 8 72 18 72 fractions] So Kim spends the greater proportion of time in her car. © ESA Publications (NZ) Ltd, Freephone 0800-372 266 b. milk : cream = 8:3 Milk Cream 20 mL 5 tablespoons 1 litre What fraction of the mixture is cream? Apply algebraic procedures in solving problems 69 Solving problems with quadratic equations 2. The product of two numbers is 96. The smaller number is 4 less than the larger number. Solve the equation: 96 = x(x + 4) to find the smaller number. (There are two possible answers.) 3. The formula for the sum, S, of the first n counting numbers is: When solving quadratic equations in a practical context, answers must be meaningful. For example, physical quantities such as length and area cannot be negative. Example Rod checks the price of a book in two different bookshops. He finds that the book costs $3 more in one bookshop than in the other. If the product of the prices in dollars is 180, find the cheaper price by solving the equation: x(x + 3) = 180 Solution: x represents the cheaper price (since (x + 3) is a higher price) x2 + 3x = 180 [expanding] [rearranging] x2 + 3x – 180 = 0 (x + 15)(x – 12) = 0 [factorising] x = –15 or x = 12 n (n + 1) . 2 How many numbers would need to be added to make a sum of 300? S= The cheaper price is $12 (since a price of –15 dollars is impossible). Exercise V: Solving problems with quadratic equations 1. A square room is made 2 m longer and 1 m wider. The area of the new room is 72 m2. Solve the equation: (x + 2)(x + 1) = 72 to find the area of the original square room. x 2 x 4. A box of jelly beans is 5 cm high. Its length is 2 cm longer than twice its width. Its volume is 120 cm3. Solve the equation: E L P M SA S E G A P 5x(2x + 2) = 120 1 © ESA Publications (NZ) Ltd, Freephone 0800-372 266 to find the width of the cuboid. 72 Achievement Standard 91027 (Mathematics and Statistics 1.2) Exercise W: Solving straightforward exponential equations Solving straightforward exponential equations Exponential equations may involve finding the value of an unknown exponent (index or power). It is useful to be familiar with the powers of the first few whole numbers. For example: 2n 1 2 4 8 16 32 n 0 1 2 3 4 5 1. 3n 4n 5n 1 1 1 3 4 5 9 16 25 27 64 125 81 256 625 243 1 024 3 125 If x n = x m then n = m Note: This rule does not hold if x = 0 or ±1. Example Solve: [since 34 = 81] 2. (–4)x = –64 (–4)x = (–4)3 x=3 [since (–4)3 = –64] 3. 7x + 1 7x + 1 x+1 x = 49 = 72 =2 =1 2. 6a = 36 b. 10b = 10 000 c. 2c = 128 d. 5d = 3 125 e. 1 1 = 243 3 f. (–2)f = 4 g. 7g = 1 h. (0.4)h = 0.0256 Solve for x. a. 3x + 1 = 10 b. 5x – 5 = 120 [since 49 = 72] In some equations involving exponents, there is an unknown base number. Example a. e The following general rule can be used to solve exponential equations. 1. 3x = 81 3x = 34 x=4 Find the unknown exponents. E L P M SA S E G A P 1. x3 = 125 has solution x = 5 [since 53 = 125] 343 3 5 has solution x = [since 35 = 243 2. x = 1024 4 and 45 = 1 024] c. 64 – 8x = 0 It is important to note that even powers of negative numbers are positive, eg (–2)4 = 16. This means that there are two solutions to an exponential equation such as x4 = 16, namely x = 2 or –2. d. 9x = 243 3 Note: Using the ‘plus or minus’ sign (±), this solution is written x = ±2. e. 2x + 2 = 8 f. 32x = 81 Example 1. a 4 = 625 has solution a = ±5 (ie, a = –5 or 5) 2. y 2 – 4 = 140 rearranges to y 2 = 144, which has solution y = ±12 © ESA Publications (NZ) Ltd, Freephone 0800-372 266 Achievement Standard 91028 Investigate relationships between tables, equations and graphs Mathematics and Statistics 1.3 Externally assessed 4 credits Chapter 3 Sequences of numbers 2. By adding 1 to each term of the sequence 3, 6, 9, 12, … A sequence is an ordered list of numbers (these numbers are called terms of the sequence). this linear sequence is formed: 4, 7, 10, 13, …. The general rule for this sequence is y = 3x + 1. For example, for the sequence 1, 4, 9, 16, 25, … the 1st term is 1, the 2nd term is 4, the 3rd term is 9, and so on. By observing patterns in a sequence, a rule for the values of the terms may be found. This rule is often expressed in terms of the position of the term in the sequence. y = mx + c Where x is the position of the term in the sequence, m is the steady increase, and c is a constant. Example The rule for the sequence Example Position of term, x Value of term, y For the sequence 1, 4, 9, 16, 25, … The differences between terms are 3, 5, 7, 9, … [4 – 1 = 3, 9 – 4 = 5, …] So the next difference is 9 + 2 = 11 and the 5th term is 25 + 11 = 36. Similarly the 6th term is 36 + 13 = 49, and so on. Linear sequences 2 3 4 5 ... 10 14 18 22 ... By substituting x = 1 into the rule to get 6 = 4 × 1 + c, it can easily be seen that c = 2. The rule for the sequence is therefore y = 4x + 2. Linear sequences arise in many practical situations which may involve spatial patterns of shapes. Letters relating to the objects (eg M for the number of matchsticks) may also be used in the rules. E L P M SA S E G A P A linear sequence increases (or decreases) steadily by the same amount from one term to the next. Example 1 6 is y = 4x + c [since the terms increase steadily by 4] These differences increase by 2 each time. Example 1. The linear sequence 3, 6, 9, 12, 15, … is formed by adding 3 to any term to get the next term. A general relationship between the position of the term in the sequence, and its value can be found using a table. Position of term, x Value of term, y The general rule for a linear sequence is always of the form 1 3 2 6 3 9 4 5 12 15 ... ... Joined hexagons are made with matchsticks. The table summarises the information. Number of hexagons, n Number of matchsticks, M Clearly, the value of each term is equal to 3 × the position of the term. The rule is M = 5n + c This can be written in the form y = 3x. The rule is M = 5n + 1. 1 6 2 3 ... 11 16 ... [M increases steadily by 5] Substituting n = 1 gives 6 = 5 + c, so c = 1. Investigate relationships between tables, equations and graphs 115 Finding the equation of a parabola Example By examining the features of a parabola, its equation can be found. The graph below is the same shape as y = –x2 and its x-intercepts are –2 and 4. Transformation method 10 If the coordinates of the vertex can be seen, then the transformation method can be used. 8 6 If the basic parabola y = x2 has been: 4 • Translated c units horizontally, its equation is y = (x – c)2. The parabola moves right if c > 0 and left if c < 0. 2 –4 • Translated d units vertically, its equation is y = x2 + d. The parabola moves up if d > 0 and down if d < 0. • Translated c units horizontally and d units vertically, its equation is y = (x – c)2 + d. x2 If the parabola being translated is y = – (an ‘upside down’ parabola) then the corresponding equations are y = –(x – c)2 and y = –x2 + d and y = –(x – c)2 + d. Example In the graph below the vertex of the parabola is at (–2, 3). 10 y y –2 2 The equation of the parabola is y = –(x – –2)(x – 4) which simplifies to y = –(x + 2)(x – 4) or y = (x + 2)(4 – x). To write down the equation of a parabola which has had a change of scale, see the following section on applications for examples of how to do these. Exercise Q: Finding the equation of a parabola 1. Write down the equations of the following parabolas in factored form. All have the same shape as y = x2 (or y = –x2) a. 4 8 y 3 6 2 4 1 2 –4 4 –2 2 4 –5 –4 –3 –2 –1 x x2 which simplifies to y = (x + 2)2 + 3 Factored form method E L P M SA S E G A P If the x-intercepts of the parabola can be seen then the factored form of the equation is useful. If the parabola has the same shape as y = x 2 and its x-intercepts are c and d then the equation of the parabola is y = (x – c)(x – d ). If the parabola is ‘upside down’ then the corresponding equation is y = –(x – a)(x – b). © ESA Publications (NZ) Ltd, Freephone 0800-372 266 2 x –2 The basic parabola y = has been translated –2 units horizontally and 3 units vertically. So the equation is y = (x – –2)2 + 3 –1 1 –3 –4 b. The parabola above is moved 2 units right and 3 units up. Give the equation of the resulting parabola. 126 Achievement Standard 91028 (Mathematics and Statistics 1.3) b. A table is calculated of x-values between 4.5 and 4.7, with increments of 0.01. Complete the table below. Width x 5 Length y =100 – 2x 100 – 2 × 5 = 90 Area xy 450 10 15 20 25 30 35 c. Complete the statement. The enclosure with largest area has width between d. and Use five more pairs of values for the width and length to explore this problem further. Width x Length y =100 – 2x Area xy Minimum surface area occurs between 4.63 and 4.65. A table can then be calculated with values between 4.63 and 4.65 in increments of 0.001. This shows that the minimum surface area occurs when the side length of the square is 4.64 cm to 3 sf. e. Exercise V: Optimal solutions using numerical methods 1. A farmer uses a 100 metre length of fencing to make a rectangular enclosure against an existing hedge. He wants to fence off as large an area as possible. hedge x x y a. Complete the statements. Width of enclosure with largest area is m. Largest possible area of enclosure is m2. 2. A square-based cuboid has surface area 240 cm2. E L P M SA S E G A P Using x for the width and y for the length of the enclosure, show that y = 100 – 2x. y x x What is the largest volume the cuboid can have? a. Using x for the side length of the square base, and y for the length of the cuboid, 60 x show that y = – . x 2 © ESA Publications (NZ) Ltd, Freephone 0800-372 266 152 Achievement Standard 91031 (Mathematics and Statistics 1.6) Exercise H: Angles in a circle Angles in a circle 1. Parts of a circle are labelled in the diagrams below. VÀVÕviÀiVi ÕÃ À>` ViÌÀi >ÀV ` > ÃiVÌÀ Find the angle marked x in each diagram. Give reasons for each answer. O is the centre of a circle. a. A iÌ V À` iÀ 46° O B x Ãi}iÌ Ì>}iÌ C Rules for angles in a circle The following rules (and their abbreviations) need to be known. Rule Angles at the circumference standing on the same arc are equal. Diagram Example A b. a b 30° x B 50° 28° a=b (∠s on same arc) x = 50° (∠s on same arc) 36° x D C The angle at the centre is twice the angle at the circumference. 80° x x = 160° ( at centre) (∠ at centre) Four different cases are illustrated. 120° c. A O x The angle in a semi-circle is a right angle. x a a = 90° x B 50° x = 60° ( at centre) (∠ in a semi) E C E L P M SA S E G A P 35° x = 55° ( in a semi) d. A 46° When finding unknown angles in circle problems, you may need to use any of the geometry rules you have previously learned. For example: • Parallel line rules. • Isosceles triangle rules. x O D B C • Vertically opposite angles, and so on. © ESA Publications (NZ) Ltd, Freephone 0800-372 266 Apply geometric reasoning in solving problems 177 Achievement Standard 91031 Practice test questions c. A triangular plastic track marker DEF has length DF = 21.5 cm and angle DFE = 36˚. F Question 1 a. 21.5 cm A cross-country running course is in the shape of a right-angled triangle, ABC as shown. A 655 m E 36° D Find the length EF. B 438 m C The distance AB is 655 m and the distance BC is 438 m. i. What is the length of AC? d. A post of height 3.1 m is held up by a guy rope, which makes an angle of 58˚ with the ground. guy rope 3.1 m 58° ii. b. What is the size of the angle BAC? What is the length of the guy rope? The cross-section of a spectator stand is also in the shape of a right-angled triangle, PQR, as shown. e. The side view of the sports pavillion is shaped as shown below, C P 5.36 m 19.25 m Q 67° E L P M SA S E G A P B R The length PQ is 5.36 m and the length PR is 19.25 m. Find the length QR. 9.2 m 2.9 m A D where AB = 2.9 m, CD = 9.2 m and ∠ BCD = 67˚. Find the length of the roof section, BC. © ESA Publications (NZ) Ltd, Freephone 0800-372 266 182 Achievement Standard 91035 (Mathematics and Statistics 1.10) Comparison questions This Achievement Standard requires students to investigate a given multivariate data set using the statistical enquiry cycle to make comparisons between or among groups in the population. This means that the data set will be supplied in the assessment, with the values of several variables given for each individual, eg gender, height, weight, length of foot, resting heart rate, etc. h. What is the circumference of your wrist? i. What is your popliteal* length? j. What is the length of your index finger? k. Are you left-handed, right-handed or ambidextrous? l. What is your main method of transport to school? m. How long does it usually take you to get to school? The assessment involves comparisons in which the same variable is used for two different groups (eg heights of boys are compared with heights of girls). n. What is the weight of your school bag today? o. What is your favourite learning area? p. How fast is your reaction time? Note: In this assessment, it would be incorrect to ask a relationship question using bivariate data, in which the relationship between two different variables for the same individual is investigated (eg how the height of a boy is related to the weight of a boy). q. What sport or activity do you most enjoy? r. How physically fit do you think you are? s. What is your resting pulse rate? t. How long have you had your current cellphone for? It is important to remember that the data set you are working with is usually a random sample from a larger population. The information from the sample is used to make a statement or an inference about the population. Exercise A : Statistical variables and questions *For a seated person, the popliteal length is the measurement from the underside of the leg right behind the knee to the floor. 1. For the above list, write down the variable involved in each question, and its unit or possible values (where appropriate). a. b. c. A nationwide online survey for Year 5–13 students which provides real, relevant data and classroom activities to enhance statistical enquiry across the curriculum. d. e. E L P M SA S E G A P CensusAtSchool www.censusatschool.org.nz is funded by Statistics New Zealand and the Ministry of Education. f. g. h. The following are some of the questions that are used by CensusAtSchool in its survey. i. a. Are you male or female? j. b. How old are you? c. Which country were you born in? d. Which ethnic group do you belong to? e. How tall are you? f. What is the length of your right foot? g. What is your arm span? k. l. m. n. o. p. © ESA Publications (NZ) Ltd, Freephone 0800-372 266 Demonstrate understanding of chance and data 239 The probability of the first sequence is Pascal’s Triangle P(MMFFF) = 0.4 × 0.4 × 0.6 × 0.6 × 0.6 Pascal’s Triangle is an array of numbers as shown below: 1 1 2 3 4 1 6 5 Each of the 10 sequences above has the same probability, so the chance of 2 males and 3 females out of the next 5 people is 1 3 1 = 0.42 × 0.63 1 1 1 [since p(male) = 0.4, p(female) = 0.6] 10 × 0.42 × 0.63 = 0.3456. 1 4 10 The number of ways of placing 2 males in a sequence of 5 people can also be found using Pascal’s Triangle, using the diagram above. 1 10 5 1 The numbers in each row are found by adding two adjacent numbers in the row above. For example, the left hand ‘10’ in row 5 is found by adding the left hand ‘4’ and the ‘6’ in row 4. This array of numbers can be used to show the number of pathways (without going backwards ie ‘up’) from the top point to any other point below it in the array. D 1 G 1 H 3 K 1 P 1 L 4 Q 5 R 10 Alternatively, start at A and move diagonally right for 2 places (2 males) to F, then diagonally left for 3 places (3 females) to R (10 ways). Peter is late to school one in every five school days. What is the probability that he will be late 3 times in the next 10 school days? F 1 I 3 M 6 • Example C 1 E 2 Start at A and move diagonally left for 2 places (2 males) to D, then diagonally right for 3 places (3 females) to S (which gives 10 ways). Note: This method will work only when there are only TWO possible outcomes. A 1 B 1 • Solution: P(late) = 0.2, J 1 N 4 O 1 S 10 T 5 P(not late) = 0.8 U 1 1 = 0.2] 5 [1 – 0.2 = 0.8] [ The number of combinations of Peter being late to school 3 times in 10 days is 120 (using Pascal’s Triangle – see Appendix, page 277). The diagram shows that there are The probability that Peter will be late on 3 of the next 10 school days is: • 2 pathways from A to E: ABE, ACE 120 × (P(late))3 × (P(not late))7 • 3 pathways from A to H: ABDH, ABEH and ACEH. = 120 × (0.2)3 × (0.8)7 • 6 pathways from A to M: ABDHM, ABEHM, ACEHM, ABEIM, ACEIM and ACFIM. = 0.2013 and so on. E L P M SA S E G A P Pascal’s Triangle can be used in solving probability problems. Example In a doctor’s practice, 40% of the patients are male. What is the probability that out of the next 5 people to enter the waiting room at this doctor’s surgery, 2 are male and 3 are female? Solution There are 10 ways that there can be 2 male (M) and 3 female (F) in a sequence of 5 people, as listed below. MMFFF FMFMF MFMFF FMFFM MFFMF FFMMF MFFFM FFMFM © ESA Publications (NZ) Ltd, Freephone 0800-372 266 FMMFF FFFMM Note: The first eleven rows of Pascal’s triangle are given as an appendix at the back of this workbook. Exercise P: Probabilities using Pascal’s Triangle 1. What percentage of families with 5 children will have 3 boys and 2 girls? (Assume that the birth of a boy or a girl is equally likely.) Year 11 Mathematics Index A adjacent 167 adjacent angles on a straight line 133 algebraic expression 41 algebraic fractions 45, 51 alternate angles 134 analysis 181 angle of depression 175 angle of elevation 175 angle sum of a triangle 138 angles at a point 133 arrowhead 141 asymptote 123 average 203 axis of symmetry 105, 111 B back-to-back stem-and-leaf plot 187 bar 207 base 43 base angles 139 bearings 173 BEDMAS 4, 42 BEMA 42 biased 199 bivariate data 182 C census 199 certain 223 changing the subject of a formula 59 chords 154 classes 201 clinometer 175 clusters 186 coefficient 41 co-interior angles 134 comparison questions 181 composite 1 composite bar graph 207 compound interest 36 conclusion 193 concyclic 157 conditional probabilities 236 continuous 200, 201 corresponding angles 134 cosine 167 cube root 7 cyclic quadrilateral 157 D data 181 data source 199 decay curve 123 decimal 22 decimal place 24 denominator 9 difference of two squares 48 directly proportional 17 discrete 200, 201 distribution 186, 190 distributive law 47, 49 divide 45 dot plots 185 E L P M SA S E G A P E elimination method 61 equally likely outcomes 227 equation of the line 86, 91 equilateral 138 equivalent fractions 9 estimates 25 event 223 experimental probability 223 exponent 43 exponential 5 exponential curve 122 exponential equations 72 exterior angle 138

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