# Gcse revision cards checked 190415

Maths Teacher en Community Maths School
19 de Apr de 2015
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### Gcse revision cards checked 190415

• 1. N1 - Percentages (core) ‘Percent’ means ‘out of 100’. We can convert a fracƟon to a percentage e.g. or change a percentage to a fracƟon e.g. or a percentage to a decimal e.g. Finding a percentage (Calculator) A) Write percentage as a decimal B) MulƟply decimal and original amount. e.g. 14% of 260 = 0.14 x 260 = 36.4 Finding a percentage (Non-calculator) 10% = total amount divided by 10 20% = 2 x 10% 5% = 10% ÷ 2 etc. e.g. Find 35% of 320 10% of 320 = 32 5% = 16 (10% ÷ 2) 30 % = 3 x 32 = 96 so 35% = 96 + 16 = 112 Increase 75 by 12% Decrease 75 by 12% 1.12 x 75 = 84 0.88 x 75 = 66 (112%) (88%) N2 - Percentage proﬁt/loss Percentage Proﬁt/Loss/Increase/Decrease Percentage Triangle: Cover up the secƟon you want to ﬁnd the value of. Example 1: A man buys a car for £400 and sells it for £700. Calculate the percentage increase. We want the percentage so 700 ÷ 400 = 1.75 1.75 is a 75% increase. Example 2: A man buys a house for £120000 and sells it for £90000. Calculate the percentage loss. We want the percentage so 90000 ÷ 120000 = 0.75 0.75 is a 25% decrease or loss.
• 2. N3 - Factors, mulƟples and primes etc. The SQUARE NUMBERS are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ……. A FACTOR of a number is a whole number (integer) which will divide exactly into that number. E.g. the factors of 12 are: 1, 2. 3. 4, 6 and 12 A MULTIPLE is a number in the relevant ‘Ɵmes table’ e.g. the mulƟples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, ….. A PRIME NUMBER is a number with exactly 2 factors e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, ……. 2 is the only even prime number. One has only 1 factor so it is NOT a prime number. The HIGHEST COMMON FACTOR (HCF) is the largest number which is a factor of 2 or more numbers. At foundaƟon level it is probably easiest to write out a list of factors and ﬁnd the largest one. E.g. ﬁnd the HCF of 12, 20 and 28 Factors of 12 are: 1, 2, 3, 4, 6 and 12. Factors of 20 are: 1, 2, 4, 5, 10 and 20. Factors of 28 are: 1, 2, 4, 7, 14 and 28. The highest number that is in every list is 4 so the HCF of 12, 20 and 28 is 4. LEAST (or LOWEST) COMMON MULTIPLE (LCM) is the lowest number which is a mulƟple of 2 or more numbers. E.g. ﬁnd the LCM of 12 and 16 MulƟples of 12 are 12, 24, 36, 48 , 60, …… MulƟples of 16 are 16, 32, 48, 64, 80 so the LCM of 12 and 16 is 48. N4 - Prime factor decomposiƟon A PRIME FACTOR of a number is a factor of the number which is also a PRIME NUMBER. E.g. the factors of 20 are 1, 2, 4, 5, 10 and 20 but the prime factors of 20 are 2 and 5 Any ‘non-prime’ number (except 1) can be wriƩen as a product of its prime factors. 153 Find 2 numbers which mulƟply to give 153. If the number chosen is a prime then that branch is ﬁnished 3 51 Repeat unƟl you are leŌ with only prime numbers 3 17 153 = 3 x 3 x 17 or 32 x 17 (must write the prime factors down as a product - the product is the result of a mulƟplicaƟon). A tougher quesƟon might be ... The prime factors of a number are 2, 2 and 5. List all the factors of this number. Easiest way is to ﬁnd the number, in this case 20 (2 x 2 x 5) and then list the factors: 1, 2, 4, 5, 10 and 20
• 3. N5 - Indices FoundaƟon Ɵer Rule 1: xa x xb = xa+b Example: 34 x 35 = 34+5 = 39 Rule 2: xa ÷ xb = xa-b Example: 340 ÷ 315 = 340-15 = 325 Rule 3: (xa )b = xab Example: (72 )4 = 72x4 = 78 Rule 4: x0 = 1 Examples: 70 = 1 3.20 = 1 Higher Ɵer Rule 5: Examples: Rule 6: Examples: Mixture of rules 5 & 6 N6 - FracƟons TOP TIP: Every Ɵme you ﬁnd a ‘mixed number’, change it to ‘top-heavy’ fracƟon. AddiƟon/subtracƟon Can only add/subtract fracƟons with a common denominator. MulƟplicaƟon Simply mulƟply the numerators and denominators Division ‘Flip’ the divisor and mulƟply ‘Special case’ Any whole number will need to be wriƩen with a denominator of ‘1’ when mulƟplying or dividing. FracƟons of an amount - Find three-ﬁŌhs of 35. One-ﬁŌh of 35 = 7 (35 ÷ 5) so three-ﬁŌhs of 35 = 21 (3 x 7). Two- thirds of a number is 12. What is the number? } 6 12 66 If 2 boxes are equal to 12, then 1 box is 6 So original number is 18 (3 x 6).
• 4. N7 - Rounding and EsƟmaƟon. When a number is rounded, it has a similar value. The key is to look at the digit aŌer the value you are rounding to and then use the normal rules of rounding i.e. 5 or more rounds up and 4 or less rounds down. Round 3570 to the nearest 100 3570 = 3600 to nearest 100 Round 3570 to the nearest 1000 3570 = 4000 to nearest 1000 Round 2.0348 to 1 decimal place 2.0348 = 2.0 (to 1 d. p.) Round 2.0348 to 2 decimal places 2.0348 = 2.03 (to 2 d. p.) When rounding to one signiﬁcant ﬁgure, ﬁnd the ﬁrst signiﬁcant ﬁgure (ﬁrst digit which isn’t a zero) and follow the normal rules of rounding. E.g. Round 213 to 1 sig. ﬁg. First sig. ﬁg. is the 2 which is in the 100’s column so round to the nearest 100 - 213 = 200 (to 1 sig ﬁg) E.g. Round 0.0273 to 1 sig. ﬁg. First sig. ﬁg. is the 2 which is in the 100th’s column so round to 2 d. p. - 0.0273 = 0.03(to 1 sig ﬁg) When you are esƟmaƟng an answer you must show your working. Round all the numbers to 1 signiﬁcant ﬁgure and then perform the calculaƟon. EsƟmate Watch out for square roots etc. Don’t round them to 1 sig. ﬁg. - instead round to the nearest sensible value. E.g. N8 - RaƟo Ratios are used when we talk about comparing quantities, they are set out using the form x:y pronounced "x to y". They can be simplified in a similar way to fractions e.g. 12:15 = 4:5 Amit is 12 years old. His brother, Arun, is 9. Their grandfather gives them £140, which is to be divided between them in the ratio of their ages. How much does each of them get? Firstly work out the ratio of their ages. In this case it is 12:9 which simplifies to 4:3 This means that the £140 is to be broken down into 7 parts (4 parts and 3 parts) ALWAYS CHECK that the amount you finish with adds up to give the initial amount 80 + 60 = 140 which is what we started with. A recipe to make lasagne for 6 people uses 300 grams of minced beef. How much minced beef would be needed to serve 8 people? In this case you need to find out how much beef would be needed for one person: 300 grams ÷ 6 people = 50 grams per person So if you are making lasagne for 8 people, you would need 50 grams x 8 people = 400 grams of Beef £140 If 7 boxes are worth £140, then each individual box is worth £20 (140 ÷ 7) Amit gets £80 (4 x 20) Arun gets £60 (3 x 20) 20 20 20 20 20 20 20
• 5. N9 - BODMAS Remember that most scienƟﬁc calculators use BODMAS by default so if it is a quesƟon on a calculator paper, use your calculator to answer or check your answer. Not all mathemaƟcal operaƟons are equal BRACKETS POWERS DIVISION MULTIPLICATION ADDITION/SUBTRACTION (addiƟon and subtracƟon are equally important) The higher up the list an operaƟon comes, the more important it is and so you do that part ﬁrst. E.g. 3 + 8 x 2 = 19 (‘x’ then ‘+’) NOT 22 (‘+’ then ‘x’) Three common types of quesƟons (but none of them will menƟon BODMAS) a) Work out the value of 8 + 6 ÷ 2 (answer is 11 NOT 7) a) Place the operators or numbers in an expression to give the correct answer E.g. 3 … 5 … 2 … 4 = 23 3 x 5 + 2 x 4 = 23 or … + … x … = 9 5 + 1 x 4 = 9 b) Place brackets to make the ‘sum’ correct. E.g. 3 x 7 + 2 - 6 = 21 3 x (7 + 2) - 6 = 21 N10 - Venn Diagrams Venn Diagrams are useful ways of presenƟng data. You are given the following as part of the formulae sheet at the start of the Methods 1 paper. A A’ (not A or the complement of A) A∩B (intersecƟon or things in both set A and set B) AUB (union or things in either set A or set B or both) Problem solving with Venn diagrams. 34 cannot sing or dance so they are in neither set. This leaves us with 66 people to ﬁt on the diagram (100 - 34) Total number of people which sing or dance (and some will do both) is 96 (46 + 50). This means that 30 will be in the intersecƟon (96 - 66) . If 46 people can sing (including those who can dance as well), then 16 can only sing (46 - 30) and similarly 20 can just dance. Number of people who can sing but not dance = 16. Probability someone chosen can just sing = 16 /100 = 4 /25 34 dancesing 3016 20
• 6. N11 - Standard (Index) form Standard (index) form is a way of wriƟng very large and very small numbers. Most scienƟﬁc calculators have a buƩon for entering in standard form EXP EE x 10x Numbers in standard form are always wriƩen in the same ‘standard’ format. a x 10b 1 ≤ a <10 b is an integer (whole number) Examples of numbers in standard form 2300000 = 2.3 x 106 (don’t just count the zeros) 0.00306 = 3.06 x 10-3 HIGHER TIER ONLY MulƟplying Dividing AddiƟon and subtracƟon (3 x 104 ) x (1.5 x 109 ) (3 x 104 ) ÷ (1.5 x 109 ) No short cuts— write as normal numbers (3 x 1.5) x (104 x 109 ) (3 ÷ 1.5) x (104 ÷ 109 ) Then add or subtract 4.5 x 1013 2 x 10-5 Change into std. form if required IMPORTANT—you must know how your calculator displays standard form. N12 - Non-calculator arithmeƟc 367 135 + 502 1 1 Don’t forget to ‘carry’ 364 135 + 229 5 1 Cannot take ‘5 from 4’ so need to ‘borrow’ from the 10s This changes our ‘4’ into ‘14’ 48 × 138 × 100 30 8 40 4000 1200 320 8 800 240 64 4 0 0 0 1 2 0 0 3 2 0 8 0 0 2 4 0 6 4 + 6 6 2 4 1 1 4.8 × 1.38 Ignore the decimal points and mulƟply as before 48 × 138 = 6624 so 4.8 × 1.38 = 6.624 (3 digits in the quesƟon aŌer a decimal point so there will be 3 aŌer the decimal point in the answer) 4.8 × 1.38 = 6.624
• 7. N13 - Using a calculator FracƟon Square root Brackets Change your answer into a decimal Cube Power Root ‘Yellow’ funcƟons above the keys are found by pressing ‘shiŌ’ ﬁrst Allows you to enter a negaƟve number Square ‘Show your working’ (on a calculator paper) does not mean that you shouldn’t use your calculator. ScienƟﬁc calculators will always use BODMAS so make sure you use brackets correctly e.g. input using brackets (3 + 12) ÷ (4 + 1) or a fracƟon using the fracƟon buƩon. S1 - Averages 7 3, 5, 3, 7, 2, 11, 9 Mode (most common) = 3 Range (highest value - lowest value) = 11 - 2 = 9 Median (middle value) A) arrange in size order B) Find the middle value (s) 2, 3, 3, 5, 7, 9, 11 Median is 5 Mean (‘average’) = 3 + 5 + 3 + 7 + 2 + 11 + 9 = 5.7 (7 numbers altogether) 3, 5, 3, 7, 2, 11 Mode (most common) = 3 Range (highest value - lowest value) = 11 - 2 = 9 Median (middle value) A) arrange in size order B) Find the middle value (s) 2, 3, 3, 5, 7, 11 Median is (3 + 5) ÷ 2 = 4 Mean (‘average’) = 3 + 5 + 3 + 7 + 2 + 11 = 5.2 (6 numbers altogether) 6
• 8. S2 - Averages (grouped data) The following table shows the heights of students in a class. State which is the modal group and calculate an esƟmate of the range, mean and median. Mode (most common) is obviously 1.3 < h ≤ 1.5 Est. range = highest poss. value - smallest poss. value = 1.75 — 1.2 = 0.55 Height, h Frequency 1.2 < h ≤ 1.3 3 1.3 < h ≤ 1.5 11 1.5< h ≤ 1.6 10 1.6 < h ≤ 1.7 5 1.7 < h ≤ 1.75 2 EsƟmated mean requires an extra couple of columns EsƟmated mean = 46.35 ÷ 31 = 1.50 m (to 2 d.p.) Median will only be on HIGHER paper. Est. median will be the 16th value which is the second piece of data (out of the 10 pieces) in the 1.5 < h ≤ 1.6 group. Est. median is 2/10 of the way through 1.5 < h ≤ 1.6 group 2/10 x 0.1 = 0.02 so esƟmated median is: 1.5 + 0.02 = 1.52 Width of 1.5 < h ≤ 1.6 group Height, h (metres Frequency 1.2 < h ≤ 1.3 3 1.3 < h ≤ 1.5 11 1.5< h ≤ 1.6 10 1.6 < h ≤ 1.7 5 1.7 < h ≤ 1.75 2 Mid-point 1.25 1.4 1.55 1.65 1.725 Group total 3 x 1.25 =3.75 11 x 1.4 = 15.4 10 x 1.55 = 15.5 5 x 1.65 = 8.25 2 x 1.725 = 3.45 Total 31 46.35 S3 - ScaƩergraphs ScaƩergraphs are used to look for relaƟonships between two sets of data. If there is a link (correlaƟon) then a ‘line of best ﬁt’ can be drawn and used to predict a value. Use the scaƩergraph to esƟmate height of someone 187 cm tall. About 20 cm This diagram shows posiƟve correlaƟon as the line of best ﬁt goes ‘uphill’ . NegaƟve correlaƟon slopes in the opposite direcƟon i.e. ‘downhill’. If no line can be drawn then there is no correlaƟon. CorrelaƟon can be further graded as weak or strong depending on proximity of the points to the line.
• 9. S4 - More staƟsƟcal diagrams Stem and leaf Nice way of displaying discrete informaƟon. If data is ordered then allows you to easily ﬁnd median, quarƟles etc. Weights of dogs at vets 0 3, 3, 4, 8 1 0, 1, 2, 2, 2 Key (very important) 2 3, 5 1 0 means 10 kg Mode (most common) = 12 kg Range = 25—3 = 22 kg Median = 11 (11 pieces of data so we need 6th) Higher content Lower quarƟle = 4 kg (3rd of the 11 data values) Upper quarƟle = 12 kg (9th of the values) Remember to ﬁnd which piece of data is the median (total pieces of data + 1) ÷ 2 and ‘÷ 4’ for quarƟles We have 11 pieces of data so (11 + 1) ÷ 2 means median is the 6th piece of ordered data. Pie chart InformaƟon below shows frequency of diﬀerent eye colours in a class. Use the informaƟon to draw a pie chart. Find out how many students in total (45) and then calculate angle which represents one student. 360° ÷ 45 = 8° (1 person = 8°) Now add on new column to table, work out angles, draw and LABEL correctly. Colour Freq . Blue 37 Brown 6 Green 2 Colour Freq. Blue 37 Brown 6 Green 2 angle 37 x 8 = 296 6 x 8 = 48 2 x 8 = 16 S5 - Data handling cycle
• 10. S6 - QuesƟonnaires You may be asked to design or criƟque a quesƟon in a quesƟonnaire as part of your exam. Below is a list of common, do’s and don’ts. Don’t ask leading quesƟons e.g. ‘Most intelligent people like dogs. Do you like dogs?’ Always give a Ɵme frame in the quesƟon e.g. ‘How many curries do you eat in a week?’ Don’t have overlapping answer boxes e.g. 0 1 - 3 3 - 5 Make sure all answers are available e.g. include an ‘other’ response. Choose your audience carefully e.g. don’t ask a quesƟon about exercise to people at a gym unless they are your target audience. The more responses you get; the beƩer your results. Don’t ask embarrassing quesƟons - you may not like the responses! F1 - Risk The hazard of an event will depend on the effect of the event and will be defined in the question. You should know the rule Risk factor of event = P(event) x hazard factor The risk factor can be shown graphically using the graph shown below: The vertical scale is the probability of the event which goes from 0 to 1. The horizontal scale is the hazard of the event. This is not a standard figure and in this case is given on a scale of 0 to 100. 0 would be no hazard and 100 would be a total hazard. Example of exam quesƟon A ‘Micromort’ is the 1 in a million chance of an activity causing death. It is estimated that a motorcyclist has 1 micromort per 6 miles travelled. It is estimated that a car driver has a 1 micromort per 250 miles travelled. Mr Green rides 2500 miles per year on his motor- cycle and 15 000 miles per year in his car. How many times more likely is he to have a fatal crash on his motorcycle than in his car? Risk on motor bike = 2500 ÷ 6 = 416.66 micromorts Risk in car = 15000 ÷ 250 = 60 micromorts 416.66666 ÷ 60 = 6.9444444 Almost 7 times more likely to have a fatal crash on motor bike.
• 11. F2 - Spreadsheets All formulae must start with ‘=‘. MathemaƟcal signs used will be: + Addition – Subtraction * Multiplication / Division ^ Power ( ) Brackets Only funcƟon students will be expected to know is ‘SUM’ E.g. = SUM(E1:E5) total of the values in those 5 cells AVERAGE is NOT required. Euan is ordering some engineering parts. This spreadsheet shows some of the orders. The formula used in cell D3 is = B3 * C3 The formula used in cell E3 is = E2 + D3 These formulae are copied down to cells D7 and E7. (a) (i) What formula will be in cell D7? = B7 * C7 (ii) What value will be in cell D7? £117.50 (b) (i) What formula will be in cell E7? = E6 + D7 (ii) What value will be in cell E7? £541.10 F3 - Flow charts Flow charts are a diagrammaƟc representaƟon of a set of instrucƟons which must be followed. Flow charts are made up of diﬀerent boxes, which each have diﬀerent funcƟons. Terminator InstrucƟon (used at start/end of ﬂow chart) Data input/output Decision Nice acƟvity at: hƩp://www.cimt.plymouth.ac.uk/projects/mepres/book8/
• 12. F4 - Annual Equivalent Rate (AER) If you have a sum of money to invest for a number of years, the best rate to compare is the Annual Equivalent Rate (AER) i.e. the annual rate of interest you will get on your savings. For the Annual Growth Account (interest paid once at the end of the year) this is the same as the gross % rate, but for the other accounts in which interest is added more than once per year, the AER is higher than the gross % rate because the interest is compounded. Formula will be given in the quesƟon r = gross interest rate (as %age NOT decimal) n =number of Ɵmes that interest is paid. Example—The annual rate of interest is 3.2%. Interest is calculated and paid monthly. What is the AER? r = 3.2 n = 12 (12 months in a year) F5 - R.P.I. and C.P.I. Retail Price Index (RPI) A measure of how the cost of living increases or decreases over Ɵme. It measures average change in the price of goods and services MONTHLY. In January 1987 the RPI was set at 100. In September 2006 the RPI was 200. This meant that the cost of goods and services had doubled. Consumer Price Index (CPI) This is slightly diﬀerent to the RPI in that it measures the change in price of goods and services purchased by households annually. In 2005 the CPI was set at 100. In 2010 the CPI for fuel bills was 130. In 2005 the average bill was £676 per annum. So fuel bill in 2010 has risen by 30% 1.3 x £676 = £878.80 in 2010
• 13. P1 - Basic Probability Probability is how likely something is to happen. The total probability will always be ‘1’ or 100% and no probability can be greater than 1. You can write probabiliƟes as fracƟons, decimals or percentages but NOT raƟos. Some quesƟons will ask you to use words but you will lose marks if you use words when numbers are expected. 0 0.5 1 Impossible even chance Certain Probability of something not happening is: 1 - probability of it happening. E.g. probability of it raining is 0.3 so probability of it not raining = 1 - 0.3 = 0.7 If I have 3 yellow counters and 2 blue counters the probability of choosing a blue counter is 2 /5 The probability of John winning at chess is 0.4 and the probability of a draw is 0.25; ﬁnd the probability of him losing 1 - (0.4 + 0.25) = 0.35 P2 - RelaƟve frequency RelaƟve Frequency This is simply a way of esƟmaƟng probability using the results of an experiment. The table below shows the result of 100 rolls of a die. The die looks biased (‘un-fair’) as ‘6’ comes up much more oŌen than the other numbers while ‘1’ comes up very liƩle. The following table shows the relaƟve frequencies as well This means that the esƟmated probability of ‘rolling a 4’ would be 15/100 or that you would expect to get a 4, ‘ 15 Ɵmes out of every 100 rolls’. This means that if you rolled the die 300 Ɵmes (3 Ɵmes as many as before), you would expect to get 45 ‘4’s’ (3 Ɵmes as many as before). EXAM TIP—The greater the number of trials (rolls, picks etc.) the beƩer the esƟmate of the probability is . Number on die 1 2 3 4 5 6 Frequency 6 14 16 15 17 32 Number on die 1 2 3 4 5 6 Frequency 6 14 16 15 17 32 RelaƟve frequency 6_ 100 14_ 100 16_ 100 _15_ 100 _17_ 100 _32_ 100
• 14. P3 - Sample Space Diagrams These are diagrams to help visualise a problem. Values are added to a table and then the outcomes are counted to show probabilities for different events. Here is one that has been filled out for you. We consider the two following spinners. I spin them and I multiply the outcomes. * spinner1/ spinner2 2 3 6 1 2 3 6 2 4 6 12 2 4 6 12 3 6 9 18 What is the probability of geƫng the score 6? What is the probability of geƫng the score 4? What is the probability of geƫng an even number? To work out these probabilities, count up how many outcomes appear in the table and put it into a fraction with the total number of outcomes in the denominator. So the probability of getting a 6 is 4/12 as there are 4 “6’s” and 12 outcomes possible. 4/12 simplifies to 1/3 Probability of getting a 4 = 2/12 = 1/6 Probability of getting an even number = 10/12 = 5/6 * Sample Space Diagrams are straight forward provided you carefully read what the question is asking you to do. A1 - Algebraic NotaƟon and Terminology An expression is a collecƟon of numbers, leƩers and operators e.g. 3n, 2y - 1 or 5 - w An idenƟty (≡) is something which is true for all values e.g. 3(y + 1) ≡ 3y + 3 An equaƟon is a number of expressions which are equal to some value e.g. 3y - 7 = 4 or 5w + 3 = 8 - w A formula is a special type of equaƟon which shows the relaƟonship between variables e.g. d = 2r y + y + y = 3y y × y × y = y³ a + 3a + b = 4a + b (avoid ’×’ sign in algebra) (cannot be simpliﬁed further) w ÷ 3 = w /3 ‘4 more than m’ is m + 4 ‘3 less than m’ is m - 3 (avoid ’÷’ sign in algebra) ’3 lots of y’ is 3y n² + 4n + 2n - 8 = n² + 6n - 8 3a × 4b = 12ab (cannot be simpliﬁed further) a × b² = ab² 3ab × 4b = 12a²b (ab)² = a²b²
• 15. A2 - Expanding and Factorising. Expanding Brackets This just means mulƟplying out the expression. 3(y + 4) = 3y + 12 Expand and simplify 3(5y + 2) - 2(y - 6) 3(5y + 2) - 2(y - 6) = 15y + 6 -2y + 12 = 13y + 18 (watch out for double negaƟves) Double brackets are slightly more involved: (x + 3)(x - 8) So (x + 3)(x - 8) = x2 + 3x - 8x - 24 = x2 - 5x - 24 x +3 x x2 +3x -8 -8x -24 Factorising This is the process of puƫng an expression back into brackets. E.g. x2 + 3x = x(x + 3) Always try to take out the highest common factor. E.g. 12y2 + 15y factorises to 3y (4y + 5) And not 3(4y2 + 5y) or y(12y + 15) You may be asked to expand and then factorise - if so beware of ‘double negaƟves’ 3(2x + 4) - 2(x + 3) = 6x + 12 - 2x - 6 = 4x + 6 = 2(2x + 3) But 3(2x + 4) - 2(x - 3) = 6x + 12 - 2x + 6 = 4x + 18 = 2(2x + 9) A3 - Solving EquaƟons Solving equaƟons is all about using ‘inverses’ to isolate the term you are looking to ﬁnd the value of. Solve the following equaƟons: 3y - 7 = 8 k /2 + 3 = 14 (+7) 3y = 15 (+7) (- 3) k /2 = 11 (-3) (÷3) y = 5 (÷3) (x2) k = 22 (x2) 7w - 4 = 2w + 11 3v + 28 = 12 - v 3(2d - 5) = 39 (-2w) 5w - 4 = 11 (-2w) (+v) 4v + 28 = 12 (+v) 6d - 15 = 39 (+4) 5w = 15 (+4) (-28) 4v = - 16 (-28) (+15) 6d = 54 (+15) (÷5) w = 3 (÷5) (÷4) v = -4 (÷4) (÷6) d = 9 (÷6)
• 16. A4 - Seƫng up and solving equaƟons The rectangle below has a perimeter of 48 cm. 4y - 1 2y + 4 Set up an equaƟon and solve to ﬁnd the value of y. Perimeter is the distance around a shape so: P = 2y + 4 + 4y - 1 + 2y + 4 + 4y - 1 = 12y + 6 But we know from the quesƟon that the perimeter is 48 so: 12y + 6 = 48 12y = 42 y = 3.5 The three angles in a certain triangle are: m + 10, m - 30 and 2m - 20 Calculate the size of each angle. Angles in a triangle add up to 180° so: m + 10 + m - 30 + 2m - 20 = 180 4m - 40 = 180 4m = 220 m = 55 SubsƟtute this value into our expressions to calculate the angles. 55 + 10 = 65; 55 - 30 = 25; 2x55 - 20 = 90 So angles are 25°, 65° and 90° A5 - InequaliƟes Algebraically, inequaliƟes are solved in the same way as equaƟons, as long as you don’t mulƟply or divide by a negaƟve number. 2x + 15 = 7 2x + 15 > 7 2x = - 8 2x > - 8 x = - 4 x > -4 (this means that x is any value bigger than –4, not just integers) On a number line, < or > are represented by and ≤ or ≥ are represented by So - 1 ≤ x < 3 is represented by …… If x is an integer, then x could be –1, 0, 1 or 2 HIGHER TIER ONLY In a similar way, when graphing inequaliƟes, < or > should be a dashed line and ≤ or ≥ is a solid line. Make it clear which region is ‘acƟve’ on your graph —shading OUT is oŌen easiest to see. Find the region, R, enclosed by: x + 3y ≤ 24 x > 2 y ≤ 8R
• 17. A6 - Coordinates On both the foundaƟon and higher papers, you will be expected to plot coordinates in all 4 quadrants. If you are asked to ﬁnd the mid-point, then it will be the point in between your 2 points. Remember, that when ploƫng points, you always plot in alphabeƟcal order i.e. the x value is before the y value. Point A is at (2, 6) Point B is at (6, 2) The midpoint of AB is (4, 4) On the higher paper, you may be required to look at 3D coordinate sys- tems. This will mean that there is an extra axis (called z). You sƟll plot and read points in alphabeƟcal order (x. y. z) but make sure you look at the labels to make sure you know which axis is which. A) If F is (10, 4, 8), then what is C? B) Find the midpoint of CF. A) C is at (0, 4, 0) B) C (0, 4, 0) and F (10, 4, 8) so the midpoint is: (5, 4, 4) (0 + 10) ÷ 2 (4 + 4) ÷ 2 (0 + 8) ÷ 2 A7 - Straight line graphs. To draw any straight line graph from its equaƟon, choose a values for x or y (usually x) and calculate the corresponding values for x or y. This will make a set of coordinates. When you have three points on your line, plot them on your axes and draw a line through them. E.g. for the line y = 3x + 1 when x = 0, y = 3 x 0 + 1 = 1 so we have the point (0, 1) when x = 2, y = 3 x 2 + 1 = 7 so we have the point (2, 7) For more complicated lines such as 2x + 3y = 12 try to ﬁnd out what happens when x = 0 and then when y = 0 When x = 0 2 x 0 + 3y = 12 → 3y = 12 → y = 4 so we have the point (0, 4) When y = 0 2x + 3 x 0 = 12 → 2x = 12 → x = 6 so we have the point (6, 0) The equaƟon of any straight line can be wriƩen as: y = mx + c gradient y - intercept (where the line crosses the y axis) So the line with an equaƟon of y = 3x + 1 has a gradient of 3 and crosses the y axis at 1. 2y = 3x - 7 is the same as y = 1.5x - 3.5 so has a gradient of 1.5 and crosses the y axis at -3.5 Parallel lines have the same gradient (are as steep as each other). So the line parallel to y = 3x + 1 and passing through 5 on the y axis will have an equaƟon of y = 3x + 5
• 18. A8 - Solving equaƟons with graphs. The soluƟon to the pair of simultaneous equaƟons: x + 3y = 8 and y = 8x - 7 could be found by drawing the 2 lines on the same graph and reading oﬀ the coordinates of the point that they cross at to ﬁnd the values of x and y. This is also true for graphs which are not of straight lines (i.e. curves). If you are asked to draw the curve, you will need to calculate more than the 3 points you need for a straight line - remember to draw a smooth curve through them and not to use a ruler. The following PP quesƟon shows you how to solve quesƟons involving curves. Here is the curve of y = x2 - 4x + 1 Use the graph to solve: A) x2 - 4x + 1 = 0 B) x2 - 4x + 1 = 2 A) This is the equaƟon of our graph when y = 0 so just read oﬀ the values where the curve crosses the x axis (or y = 0). So x = 0.2 or x = 3.8 B) As above but y = 2 so draw line y = 2 and read oﬀ so x = -0.3 or x = 4.3 REMEMBER - we are only ﬁnding the x values. TOP TIP - These are very badly answered quesƟons (less than 10% of candidates will get any marks) so make sure you can answer these. A9 - Subject of Formulae The subject of a formula is the part of the formula to the leŌ of the equals sign e.g. for V = IR the subject is V and for v2 = u2 + 2as the subject is v2 Rearranging formulae follows all the usual rules of algebra. Make t the subject of v = u + at (-u) v - u = at (-u) (÷a) (÷a) so Make r the subject of Make m the subject of Ft = mv - mu Because the leƩer we are looking for appears twice, we will need to factorise. Ft = m(v - u) These are the same skills that you need for solving equaƟons. Higher only
• 19. A10 - Sequences A sequence is just a paƩern of numbers linked by a rule. Each number in the sequence is a term and 4th term will be the 4th number in the sequence. At foundaƟon level, you will oŌen be asked to write the next term (or terms) in a sequence and explain the rule for conƟnuing the sequence. Don’t over complicate this - the answer could be as simple as ‘add 6’. E.g. What is the next number in this sequence and why? Explain how you know that 1002 isn’t a number in this sequence. 23, 27, 31, 35, 39, ….. Next term is 43 and the rule is ‘add 4’. 1002 (even) will not be in this sequence as all the numbers in the sequence are odd. Occasionally you may be asked to ﬁnd the nth term of a sequence. This is someƟmes called the ’posiƟon-to-term’ rule. This allows you to calculate the value of a term from it’s posiƟon (n) in the sequence. E.g. 3, 5, 7, 9, 11, …. The rule is ‘add 2’ so the nth term will start ‘2n’. What do we need to do to 2 to get our ﬁrst term (3) - add 1 so nth term is 2n + 1 or you could say, what would be the number before our 1st term (1) so our nth term is sƟll 2n + 1 The nth term can be used to ﬁnd any other term in the sequence e.g. 10th term in our sequence is 2 x 10 + 1 = 21 You could be given the nth term and asked to ﬁnd some terms in the sequence. E.g. Find the ﬁrst 3 terms in the sequence with the nth term 3n - 2 1st term, n = 1 3 x 1 - 2 = 1 2nd term, n = 2 3 x 2 - 2 = 4 3rd term = 3 x 3 - 2 = 7 so sequence starts 1, 4, 7 A11 - Trial and Improvement Trial and Improvement is a way of solving equaƟons by guessing a soluƟon, trying the value and then using the ’answer’ obtained to make a ‘beƩer guess’. Example: Solve x² + 3x - 6 = 200 giving your answer correct to 1 decimal place First guess (13) is too high (>200) so guess lower 2nd guess is too low so answer must be between 12 and 13 When you have your answer within a pair of consecuƟve 1 d p numbers, you must ‘split the diﬀerence’. Answer is between 12.9 and 12.95 so x = 12.9 (to 1 d p) x x²+ 3x - 6 High/low 13 202 High 12 174 Low 12.5 187.75 Low 12.8 196.24 Low 12.9 199.11 Low 12.95 200.5525 High
• 20. A12 - Travel Graphs Distance - Time graph Gradient of the line gives you the speed. The steeper the line; the faster you are travelling. Horizontal line means you are staƟonary. Speed/Velocity - Ɵme graph Gradient of the line gives the acceleraƟon. The steeper the line; the greater the acceleraƟon. Horizontal line means travelling at a constant speed. Area underneath the graph is the distance travelled.Gradient = verƟcal change ÷ horizontal change So for gradient = 8 ÷ 4 = 2 G1 - Basic angle facts. Acute angle - less than 900 . Obtuse angle - bigger than 900 and less than 1800 . Reﬂex angle - bigger than 1800 and less than 3600 . A right angle is 90° (is not acute or obtuse) Angles on a straight line add up to 1800 Angles at a point add up to 3600 Opposite angles are equal Angles in triangles add up to 1800 In diagram below a = c and b = d Scalene triangles have 3 diﬀerent sides and 3 diﬀerent angles. Isosceles triangles have 2 equal sides and 2 equal angles and are very common exam quesƟons. Equilateral triangles have 3 equal sides and all the angles are 600 .
• 21. G2 - Angles in Parallel lines Alternate angles are equal DO NOT call ‘Z’ angles Co-interior angles add up to 1800 Corresponding angles are equal. You may be asked to ﬁnd the value of an angle and then say why it has that value. The examiner is looking for words such as alternate, corresponding, opposite etc. REMEMBER - parallel lines are denoted by using arrows. Perpendicular lines are at right angles to each other. G3 - Symmetry A shape has line symmetry if it can be folded exactly in half - the more ways it can be folded then the more lines of symmetry it has. Lines of symmetry are usually drawn as dashed lines. 4 lines of symmetry 2 lines of symmetry RotaƟonal symmetry tells us how oŌen our shape will ﬁt exactly onto its original posiƟon during a full turn. We talk about orders of rotaƟonal symmetry. All shapes have rotaƟonal symmetry of at least order 1. It is acceptable to say that a shape with rotaƟonal symmetry of order 1 has no rotaƟonal symmetry. RotaƟonal symmetry RotaƟonal symmetry of order 1 of order 5
• 22. G4 - Angles in polygons A regular polygon has all its sides the same length and all its angles the same size. The total of all the exterior angles in any polygon is 3600 . This means that for any regular polygon you can calculate the number of sides easily by using the formula: 3600 ÷ exterior angle = number of angles E.G. A regular polygon has an exterior angle of 200 . How many sides does the polygon have? 3600 ÷ 20 = 18 sides The sum of the interior angles increases as the number of sides increases. Sum of interior angle = (no. of sides - 2) x 180 so for an octagon (8 sides), sum of interior angles = (8 - 2) x 180 = 6 × 180 = 10800 If you know the exterior angle, you can calculate the interior angle (Interior + exterior = 1800 ) Interior angle Exterior angle Interior + exterior = 1800 For any polygon G5 - TessellaƟon, Congruence and Similarity Shapes tessellate when they together without ANY GAPS. An alternaƟve descripƟon is that all the angles at the point they join add up to 1800 . The shapes do not have to be the same. If asked to show a shape tessellates, make sure that you draw enough of them (may give a number to draw in the quesƟon) Shapes are congruent when they are exactly the same size and shape (although they may be reﬂecƟons or rotaƟons of each other). In the example below B and D are congruent. Similar shapes are enlargements of each other i.e. all the sides are a common factor larger in one shape than the other. In the example above, A and B are similar as all the lengths in B are twice those in A.
• 23. G6 - Parts of a Circle G7 - ReﬂecƟon/RotaƟon There are 4 transformaƟons to deal with at GCSE level. ReﬂecƟon The main thing to remember when reﬂecƟng is that the image will be the same distance from the mirror line as the object (but facing the opposite direcƟon). Example bellows shows a reﬂecƟon in the line y = - x RotaƟon To describe a rotaƟon you must state: A) the fact that it is a rotaƟon B) The size and direcƟon of the turn C) The centre of rotaƟon - the point that you will be spinning your tracing paper about. Here is a rotaƟon of 1800 (or a half turn) about the point (0, 0). The direcƟon is not important when dealing with half turns. NB - the point (0, 0) has its own special name - the origin and this may used in the exam.
• 24. G8 - TranslaƟon/Enlargement TranslaƟon A translaƟon is just a movement leŌ/right and/or up/down. They are oŌen described using column vectors. Top number is how far across + to right and - to leŌ. BoƩom number is how far up/down with + up and - down So our example represents a transformaƟon of 1 to the leŌ and 6 down. Enlargement If you are asked to enlarge a shape by a scale factor simply draw the same shape with all the sides that many Ɵmes larger. However the if you are given a centre of enlargement to work from you need to measure the distance from the centre of enlargement and then mulƟply this by the scale factor. The green triangle has been enlarged by a scale factor of 2 from the point marked 0. If you measure the length OA you will ﬁnd that OA’ is twice (because the scale factor is 2) as long (and similarly for the other points). If you are given two shapes and need to describe the enlargement; divide the length of the side on the image by the corresponding side length on the original to ﬁnd the scale factor. Then draw in the ‘ray lines’ as above - where they cross will be the centre of enlargement. G9 - Pythagoras’ Theorem This is used to ﬁnd a missing side in a right-angled triangle. The side opposite the right-angle is the longest side and is called the hypotenuse. Pythagoras’ Theorem says that for any right-angled triangle: Finding the hypotenuse (longest side, c) Finding a shorter side.
• 25. G10 - Area and Perimeter PERIMETER is the distance around a shape and is measured in cm, m etc. AREA is the space inside a 2D object and is measured in ‘square units e.g. cm2 , m2 etc. b Trapezium area is given on formula sheet ba Kite G11 - Area and Circumference of circle Circumference is just the perimeter of a circle. We calculate the circumference using the formula: C = πd where π = 3.14 and d is the diameter. Example: Calculate the circumference of this circle 5 cm We have the radius so need to ﬁnd the diameter. d = 2r so d = 2 x 5 = 10 C = πd = 3.14 x 10 C = 31.4 cm To ﬁnd the area of a circle, we use the formula: A = πr2 where π = 3.14 and r is the radius NB - Only r is squared i.e. A = π x r x r Example Calculate the area of the circle. 5 cm A = πr2 A = 3.14 x 52 A = 3.14 x 25 A = 78.5 cm2 Always check that you have been given the radius. If you are given the diameter instead, the radius will be half of the diameter.
• 26. G12 - Volume and Surface area - cuboids Volume is the space inside a 3D shape and is measured in ‘cubed units’ e.g. cm3 , m3 . Surface area is the total area of all the sides of a 3D shape. It is measured in ‘square units’ e.g. cm2 . The formula for the volume of a cuboid is: Volume = length x width x height Find the volume of this cuboid. Volume = 10 x 3 x 7 = 210 cm3 10 cm 7 cm 3 cm The surface area means you have to ﬁnd the area of each face and add them all together. Remember for a cuboid there will be pairs of idenƟcal faces. Calculate the surface area of the cuboid. Surface area = 2 x (10 x 3) + 2 x (3 x 7) + 2 x (10 x 7) = 242 cm2 3 cm 10 cm 7 cm G13 - Volume of Prisms A PRISM is a 3D shape. The cross-secƟon (end of the shape) is the same throughout. Prisms are usually named by the shape of the cross-secƟon e.g. This is a triangular prism. The following is given in the formula sheet: Example Find the volume of this shape. 5 cm 12 cm4 cm Volume = 10 x 12 = 120 cm3 x
• 27. G14 - ProperƟes of quadrilaterals G15 - Loci Loci (pl. locus) - posiƟon(s) an object can be in. Equidistant from 2 points - perpendicular bisector Equidistant from 2 lines - angle bisector Equidistant from a point - circle Equidistant from a line - at least a parallel line
• 28. G16 - Bearings There are 3 key facts you need to remember about bearings: 1) Always wriƩen as 3 ﬁgures (so 075° not 70°) 2) Always measured clockwise 3) Always measured from North. Make sure that you read the quesƟon carefully and measure from the correct point. The bearing from A to B is 120° (clockwise from North) The bearing of A from B is 300° (we have the 60° angle but must measure in a clockwise direcƟon so angle we want is 360 - 60 (full turn = 360°)) To calculate a ‘reverse’ bearing i.e. the bearing needed to return to your start point: A) if original bearing is less than 180° then add 180° B) if original bearing is more than 180° then subtract 180° G17 - Conversions You must know the diﬀerence between metric and Imperial units. Metric units are the ones you currently use while Imperial are ‘old school’. You must learn these as they will not be given in the exam. Metric mm = millimetres, cm = cenƟmetres, m = metres, km = kilometres, g = grams, kg = kilograms, ml = millilitres 1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m 1 kg = 1000 g 1 tonne = 1000 kg 1000 ml = 1 litre Imperial 12 inches = 1 foot 3 feet = 1 yard 16 ounces = 1 pound 14 pounds = 1 stone 8 pints = 1 gallon Metric ↔ Imperial 1 inch ≈ 2.5 cm 1 mile ≈ 1.6 km (or 5 miles ≈ 8 km) 1 kg ≈ 2.2 pounds 1 yard ≈ 0.9 m 4.5 litres ≈ 1 gallon 1 litre ≈ 1.75 pints