Year 11 Mathematics IAS 1.1 Numeric Reasoning Robert Lakeland & Carl Nugent Contents Achievement Standard.................................................. 2 Prime Numbers....................................................... Factors and Multiples................................................... 4 Rounding and Estimation................................................ 7 Standard Form......................................................... 12 Order of Operation..................................................... 1 Integers (+, x,, )....................................................... 17 Fractions (+, x,, )...................................................... 20 Percentages............................................................ 24 Ratio................................................................... 2 Proportion............................................................. 7 Rates.................................................................. 41 Powers................................................................. 44 Compounding Rates..................................................... 49 Practice Internal Assessment 1............................................ Practice Internal Assessment 2............................................ 6 Practice Internal Assessment............................................ 7 Answers............................................................... 8 Innovative Publisher of Mathematics Texts
2 IAS 1.1 Numeric Reasoning NCEA 1 Internal Achievement Standard 1.1 Numeric Reasoning This achievement standard involves applying numeric reasoning in solving problems. This achievement standard is derived from Level 6 of The New Zealand Curriculum, Learning Media. The following achievement objectives, taken from the Number Strategies and Knowledge thread of the Mathematics and Statistics learning area, are related to this achievement standard: Achievement Achievement with Merit Achievement with Excellence Apply numeric reasoning in solving problems. reason with linear proportions use prime numbers, common factors and multiples, and powers (including square roots) understand operations on fractions, decimals, percentages, and integers use rates and ratios know commonly used fraction, decimal, and percentage conversions know and apply standard form, significant figures, rounding, and decimal place value apply direct and inverse relationships with linear proportion extend powers to include integers and fractions apply everyday compounding rates. Apply numeric reasoning, using relational thinking, in solving problems. Apply numeric reasoning, using extended abstract thinking, in solving problems. Apply numeric reasoning involves: selecting and using a range of methods in solving problems demonstrating knowledge of number concepts and terms communicating solutions which would usually require only one or two steps. Relational thinking involves one or more of: selecting and carrying out a logical sequence of steps connecting different concepts and representations demonstrating understanding of concepts forming and using a model; and also relating findings to a context, or communicating thinking using appropriate mathematical statements. Extended abstract thinking involves one or more of: devising a strategy to investigate or solve a problem identifying relevant concepts in context developing a chain of logical reasoning, or proof forming a generalisation; and also using correct mathematical statements, or communicating mathematical insight. Problems are situations that provide opportunities to apply knowledge or understanding of mathematical concepts and methods. The situation will be set in a real-life or mathematical context. The phrase a range of methods indicates that evidence of the application of at least three different methods is required. Students need to be familiar with methods related to: ratio and proportion factors, multiples, powers and roots integer and fractional powers applied to numbers fractions, decimals and percentages rates rounding with decimal places and significant figures standard form. IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
IAS 1.1 Numeric Reasoning Prime Numbers Prime Numbers A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. A prime number has exactly two factors 1 and itself. For example 17 is a prime because it has A factor is a number that divides into only two factors 1 and 17. another number without remainder. The smallest prime number as well as the only For example 2 is a factor of 6 because 2 even prime number is 2, because it is divisible by 1 divides into 6 without remainder. and 2. A number n greater than 1 is defined as a prime number if it is only divisible by 1 and n. Positive numbers other than 1 that are not prime numbers are called composite numbers. a) Copy the numbers, 19, 2, 7, 9, 2 and circle those that are prime. Product of Primes b) List the next two prime numbers after 61. It is possible to write any positive number greater c) Draw a prime factor tree for 10. than one as a product of prime numbers. The best way to do this is to use a factor tree. At the top of the tree you start with the number you wish to write as a product of prime numbers. a), 19, 2, 7, 9, 2 You then find two numbers that multiply to give b) The next two prime numbers after 61 are 67 the number. Once one of the branches of the tree and 71. has a prime number at its branch end you stop c) Prime factor tree for 10 is as follows. simplifying that branch. 10 You continue working on each branch until only a prime number remains. If you multiply all the prime numbers at the end of each of the branches you should get the number 1 x 10 you started with. A prime factor tree for 120 is x 2 x drawn below. Prime factors are 2 x x x. 80 20 x 4 4 x 2 x 2 2 x 2 So 80 = 2 x 2 x x 2 x 2. It does not matter what two numbers you find to multiply to give 80 (i.e. 40 x 2 or 8 x or 10 x 8) you will always end up with the same prime factors at the end. IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
12 IAS 1.1 Numeric Reasoning Standard Form To enter a number in standard form on a graphics calculator we use the EXP Standard Form button (Casio 970GII) or the EE button (TI-84 Plus), We use standard form as a concise way of writing very large and very small numbers. A number in e.g. for.127 686 2 x 10 6 we enter standard form is written as a number between 1.1276862 EXP 6 Casio 970GII and 10, multiplied by a power of 10. EE Consider 127 686.2.1276862 2nd, 6 TI-84 Plus You first move the decimal point so the number has a value between 1 and 10, i.e..127 686 2 To write any number in standard form You now multiply by 1 000 000 or 10 6 to make the first move the decimal point, so that the number equal 127 686.2 number has a value between 1 and 10 then multiply by an appropriate power of 10, so the In standard form 127 686.2 =.127 686 2 x 10 6 number has the same value (found by counting the With very small numbers we work similarly. number of decimal places the decimal point has Consider 0.004 moved). Move the decimal point, so the number has a To convert a number into standard form value between 1 and 10, i.e. 4.. on your calculator use the scientific mode. Each calculator is a little You now need to divide this number by 1000 or different. multiply by 0.001 = 10 to make 4. equal 0.004. On the TI-84 Plus press MODE then choose SCI In standard form 0.004 = 4. x 10 and then the number of decimal places you want The decimal point needs to move places left from to display. 4. so the power is. On the Casio 970GII press SHIFT MENU then scroll down and select Display and choose SCI and then the number of decimal 0. 0 0 4 = 4. places you want x 10 to display. This number must always be between 1 and 10. A number displayed as 8.89E+0 means 8.89 x 10. A number displayed as 8.89E 0 In means 8.89 x 10. Write the following in standard form. a) 1 20 000 b) 0.000 6 c) 2.1 a) 1.2 x 10 6 b).6 x 10 4 c) 2.1 x 10 0 as 10 0 = 1 Write the following as an ordinary number. a) 2.4 x 10 b) 8.647 x 10 2 c).91 x 10 a) 0.0024 b) 864.7 c) 91 000 Merit Write the following in standard form, calculating the answer first if required. 6. 41 00 66. 91 67. 12.7 68. 0.04 69. 0.92 70. 7 71. 12 700 000 72. 0.000 009 6 IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
14 IAS 1.1 Numeric Reasoning 99. A blade of grass grows on average at a rate of 2.0 x 10 8 metres per second. How much will a blade of grass grow in one week in millimetres? 100. A country s national debt is $1. x 10 12. If the population of the country is 28 million, how much is owed per individual? 101. Use standard form to calculate the following information about 7 year old Martin. Round all your answers to significant figures. One estimate of the number of cells in the average human body is seventy-three point eight million million. a) Write this as an ordinary number. A person s heart beats, on average, 8 beats per minute for every minute of their life. f) How many minutes has Martin lived for? Ignore leap years and give your answer in b) Write this number in standard form. standard form. c) The population of the Earth is approximately 6 10 000 000 people. g) How many times has Martin s heart beaten? Calculate the total number of human cells for the entire population of Earth. d) The average human weighs 6. kg. What is the average mass of one cell in grams? (1000 g = 1 kg). h) Each beat of the heart pumps about 67 ml. How many litres has the heart pumped in Martin s life? e) The mass of a single hydrogen atom is 1.66 x 10 24 g. Calculate how many times heavier a single human cell is than a hydrogen atom. IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
24 IAS 1.1 Numeric Reasoning 211. Keith is a used car salesman and at the beginning of a month has 120 vehicles in stock. a) Keith sells 24 of his stock in the month. How d) Keith sells one of his cars for $27 000. The 1 purchaser puts down a deposit of and pays the balance off over 24 months at 0% interest. How much does the purchaser pay per month? many vehicles does he have left at the end of the month? 4 e) of the customers who bought vehicles one b) 1 month paid cash for them. If 20 customers paid of the vehicles are commercial vehicles, such as vans and utes, and 8 of these cost less than $10 000. How many of Keith s commercial vehicles cost less than $10 000? c) 2 of the vehicles are cars and of these cash, how many vehicles in total did Keith sell during the month? 2 f) Keith sells 24 of his stock in a month and of the remaining stock the following month. What 4 have an fraction of the original stock of vehicles remain engine capacity less than 2000 cc. How many after the two months and how many cars does cars does Keith have with an engine capacity Keith have left? greater than 2000 cc? Percentages The word percentage is made up of the prefix per meaning out of and centage from the same root as century meaning Percentages to Fractions 100. Percentage therefore means out of 100. To convert a percentage to a fraction we write By representing figures as a percentage we can the percentage as a fraction out of 100 and then easily make a comparison between two or more sets simplify the fraction if possible. of figures. Consider % To simplify a fraction on the Casio 970GII enter it into your calculator As a fraction = as a fraction and then press EXE. The 100 calculator will automatically reduce Simplify = 7 it down to its simplest form. If the simplified 20 fraction is a mixed numeral press SHIFT F D to convert it to an improper fraction. On the Decimal to a Percentage To convert a decimal to a percentage we multiply TI-84 Plus to simplify a fraction like we enter 100 by 100%. Consider 0.0 1 0 0 MATH Frac Mult. by 100% = 0.0 x 100% 1 ENTER = % To convert a decimal to a percentage Fraction to a Percentage move the decimal point two places to the right. On a Casio 970GII to To convert a fraction to a percentage we multiply convert a fraction to a percentage on the by 100%. calculator enter Consider a b/c x 1 0 0 Mult. by 100% = x 100% = 00% Simplify = 60% EXE On the TI-84 Plus to convert a fraction to a percentage enter x 1 0 0 ENTER IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
28 IAS 1.1 Numeric Reasoning Calculate the following. a) Decrease $9 by 1%. b) Increase $2 by 16.% a) Start x (100 ± Change)% = Result b) Start x (100 ± Change)% = Result 9 x (100 1)% = Result 2 x (100 + 16.)% = Result 9 x 8% = Result 2 x 116.% = Result Result = $80.7 Result = $297.08 (2 dp) a) An article is bought for $8 and sold for $70. b) Find the pre-gst amount when an article What is the percentage increase? selling for $20 includes GST of 1%. a) Start x (100 ± Change)% = Result Start x (100 ± Change)% = Result 8 x (100 ± Change)% = 70 Start x (100 + 1.0)% = 20 (100 ± Change)% = 1.2069 Start x 11.0% = 20 (100 ± Change) = 120.69 as a % 20 Start = 11.0% Change = 20.7% (1 dp) = $217.9 Achievement Answer the following percentage questions. 247. Increase $210 by 2% 248. Decrease 194 by % 249. Decrease $47.0 by 12.% 20. Decrease by 0% a shirt that costs $67.0 21. Alec has to pay a 1% surcharge on a meal costing $96.0. How much will he pay altogether? 22. A plasma screen usually retails for $999. If a purchaser pays cash they are eligible for a 12.% discount. How much does the plasma screen cost with the cash discount? 2. Due to a flu epidemic 12% of the pupils in a school are absent one day. If the school roll is normally 7 pupils, how many pupils are present? 24. A house sells for 4% above its government valuation of $48 000. How much does it sell for? 2. A school s roll has increased by 6.% over the last 10 years. If ten years ago it had a roll of 140, what is its roll now? 26. The value of a car has depreciated in value by %. If it was initially purchased for $42 00 what is it worth now? IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
IAS 1.1 Numeric Reasoning 7 Proportion Proportion A proportion is a part considered in relation to a whole or a statement of equality between two or more ratios. i.e. a b = c d Directly Proportional Problems Directly proportional problems are problems where Directly proportional problems can also a change in one quantity causes a proportional be set up as ratios, but make sure that change in another quantity. the two ratios are written in the correct order. Two quantities y and x are in direct proportion if by whatever y changes, x changes by the same In the problem on the left, proportion or multiplier. if x = the number of pens then 1. We write y x, which is read as y is directly 2 = 10. x proportional to x, this means y = kx, where k is a Solving this equation gives 1.x = 21 constant. and then x = 14 E.g. The cost of pens is directly proportional to the number of pens you buy. If two pens cost $1.0, how many pens can you buy for $10.0? First we find k, 1.0 = 2k k = 0.7 (each pen costs 7 cents) To calculate how many pens we can buy for $10.0 we divide 10.0 by the cost of a single pen 0.7 which equals 14 pens. Inversely Proportional Problems Two quantities x and y are said to be Inversely proportional problems are problems that inversely proportional if their product are similar to directly proportional problems except xy always remains constant. that when x increases y will decrease and vice versa. In the problem on the left, Two quantities y and x are inversely proportional if if x = the number of hours then their product always remain constant, i.e. xy = k or 4 men x 6 hours = 7 men x x y = k where k is a constant. x E.g. If it takes 4 men 6 hours to dig a drain, how long will it take 7 men to do the same job? First we find k, which is 4 x 6 = 24 (total number of man hours). To find how long it will take 7 men to dig the drain we divide 24 (total number of man hours) by 7 = 7 hours. A good test of an inverse proportional problem is to ask yourself, If one quantity doubles, will the other half, i.e. if x increases by a multiplier, y will decrease by the same divisor. 24 = 7x x = 7 hours The ratio of the number of men = the inverse ratio of the number of hours. i.e. 4 : 7 = x : 6 4 7 = x 6 7x = 24 x = 7 hours The more men the less time to complete the job, hence this is an inversely proportional problem. When x increases, y decreases. IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent
8 IAS 1.1 Numeric Reasoning Answers Page 6 cont... Page 9 Page 7. 0. 6.0 (1 dp) 1. a) 1, 2, 4, 7, 14, 28 47 29 101 1. 9 (1 sf) b) 1, 2,, 6, 7, 14, 21, 42 11 9 2..22 (2 dp) c) 1, 19. 9 (2 sf) 17 89 71 2. a) 17, 4, 1, 68,... 4. 2.62 ( sf) b) 21, 42, 6, 84,.... 12 (2 sf) 8. 4 = 2 + 2, = 2 +, 6 = +, c) 62, 124, 186, 248,... 7 = 2 +, 8 = +, 6. 110 (2 sf) 9 = 2 + 7, 10 = +,. a) 1,, 7, 21 12 = + 7, 1 = 11 + 2, 7. 0.1 (2 sf) and 1, 2, 4, 7, 8, 14, 28, 6 14 = 11 +, 1 = 2 + 1 8. 22.8 m (1 dp) HCF = 7 16 = 1 +, 18 = 7 + 11 19 = 2 + 17 all can. 9. 47 cm (2 sf) b) 1,,, 9, 1, 4 and 1,, 9, 1, 9, 117 14 numbers less than 20. 40..4 m (1 dp) HCF = 9 9. HCF of 448 and 616 is 41..6 L (2 sf) 6 so greatest possible length as only the 4 litres is c) 1,, 19, 9 is 6 cm. measured (the 8 parts are and 1, 2,, 6, 19, 8, 7, 114 counted). 10. LCM of 28 and 24 which is HCF = 19 168. So 168 seconds 42. 4 cm 2 (2 sf) 4. a) 12, 24, 6, 48, 60, 72,... 4. $.0 (2 sf) 20, 40, 60,... 11. 48 = 2 x 2 x 2 x 2 x 44. a) 21 m 2 LCM = 60 108 = 2 x 2 x x x b) 4 b) 6, 12, 18, 24, 0, 6, 42, 48, c) cm (2 sf) HCF = 2 x 2 x = 12 4, 60, 66,... 12. LCM of 40, 48 and 60 = 240, d) $170.9 11, 22,, 44,, 66,.. so 240 minutes (4 hours) e) $897 LCM = 66 1. HCF of 96, 144 and 224, f) No should be rounded to sf i.e 2.8 m. c) 9, 18, 27, 6, 4, 4,... so 16 pieces of chicken. Page 8 g) 18 hours. 1, 0, 4,... LCM = 4 14. 789.88 Page 10. a) prime number 1. 0.0024 4. 120 (accept 102) b) 18 16. 0.0 46. 70 (accept 77) 17. 67 00 47. 00 2 x 69 18. 0.0098 48. 9 x 2 19. 2000 49. 200 2 x x 2 20. 6.0 0. 60 c) 140 21. 0.0480 1. 100 22. 27 000 2. 20 10 x 14 2. 44. $10 2 x 2 x 7 24. 0.9 4. $10 000 2 x x 2 x 7 2..2. $100 Page 6 26. 480 6. $ 6. 8, 16, 24, 2, 40, 48, 6,... 27. 8.4 7. $12 000 14, 28, 42, 6,... 28. 60 8. 100 cm 6 seconds 29. 7220 9. $00 60. 120 km IAS 1.1 Year 11 Mathematics and Statistics Published by N New Zealand Robert Lakeland & Carl Nugent