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Mathematics Higher Tier

Mathematics Higher Tier. Number. GCSE Revision. Higher Tier - Number revision. Contents : Calculator questions Long multiplication & division Best buy questions Estimation Units Speed, Distance and Time Density, Mass and Volume Percentages Products of primes

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Mathematics Higher Tier

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  1. Mathematics Higher Tier Number GCSE Revision

  2. Higher Tier - Number revision Contents : Calculator questions Long multiplication & division Best buy questions Estimation Units Speed, Distance and Time Density, Mass and Volume Percentages Products of primes HCF and LCM Indices Standard Form Ratio Fractions with the four rules Upper and lower bounds Percentage error Surds Rational and irrational numbers Recurring decimals as fractions Direct proportion Inverse proportion Graphical solutions to equations

  3. 9.26.1 7.5 8 – 8.5 x 103 3.4 x 10-1 2.5 3 1.72 + 5.22 6.31 1 1 3.6 2.3 – 1000 4 Calculator questions Which buttons would you press to do these on a calculator ? 1.7 + 2.8 2.3 – 0.2 2.5 + 4.1 3.5 1.56

  4. Long multiplication Use the method that gives you the correct answer !! Question : 78 x 59 70 8 50 3500 400 9 630 72 Total = 3500 + 400 + 630 + 72 Answer : 4602 Now try 84 x 46 and 137 x 23 and check on your calculator !!

  5. 23 times table 23 46 69 92 115 138161 184207230 29 227 68 23 2987 Long division Again use the method that gives you the correct answer !! Question : 2987  23 2 9 1 6 22 20 Answer : 129 r 20 Now try 1254  17 and check on your calculator – Why is the remainder different?

  6. 87p 34p 400g 150g OR 78p 32p OR 2.1L 0.95L Best buy questions Always divide by the price to see how much 1 pence will buy you Beans Large 40087 = 4.598g/p Small 15034 = 4.412g/p Large is better value (more grams for every penny spent) Milk Large 2.178 = 0.0269L/p Small 0.9532 = 0.0297L/p Small is better value OR (looking at it differently) {Large 782.1 = 37.14p/L Small 320.95 = 33.68p/L}

  7. If you are asked to estimate an answer to a calculation – Round all the numbers off to 1 s.f. and do the calculation in your head. DO NOT USE A CALCULATOR !! 4899  46 1906  44 377  19 6. 10. 4. 7. 5. 2. 8. 1. 9. 3. 360 x 87 4.89 x 6.01 1.92 58 x 21 47 x 22 399 x 31 58 x 21 7.12 x 39.2 0.87 Estimation e.g. Estimate the answer to 4.12 x 5.98  4 x 6 = 24 Always remember to write down the numbers you have rounded off Estimate the answer to these calculations

  8. x 1000 x 1000 x 1000 x 100 x 100 x 100 x 10 x 10 x 10 km kl kg g l m cg cl cm mg ml mm ÷ 1000 ÷ 1000 ÷ 1000 ÷ 10 ÷ 10 ÷ 10 ÷ 100 ÷ 100 ÷ 100 Units 1 inch  2.5 cm 1 yard  0.9 m 5 miles  8 km 2.2 lbs  1 kg 1 gallon  4.5 litres Learn these rough conversions between imperial and metric units Learn this pattern for converting between the various metric units Metric length conversions Metric weight conversions Metric capacity conversions

  9. D T S = S = S = S = D T D T D T S = T = 65 = 600 . 65 17 . 0.417 600 . T 45 = D . 120 Speed, Distance, Time questions Speed, Distance and Time are linked by this formula To complete questions check that all units are compatible, substitute your values in and rearrange if necessary. • Speed = 45 m/s • Time = 2 minutes • Distance = ? 2. Distance = 17 miles Time = 25 minutes Speed = ? 3. Speed = 65 km/h Distance = 600km Time = ? 45 m/s and 120 secs 17 miles and 0.417 hours 45 x 120 = D S = 40.8 mph T = 9.23 hours D = 5400 m

  10. M V D = D = D = D = M V M V M V V = D = 12 = 5000 . 800 564 . 12 564 . V 8 = M . 6000 Density, Mass, Volume questions Density, Mass and Volume are linked by this formula To complete questions check that all units are compatible, substitute your values in and rearrange if necessary. • Density = 8 g/cm3 • Volume = 6 litres • Mass = ? 2. Mass = 5 tonnes Volume = 800 m3 Density = ? 3. Density = 12 kg/m3 Mass = 564 kg Volume = ? 8 g/cm3 and 6000 cm3 800 m3 and 5000 kg 8 x 6000 = M D = 6.25 kg/m3 V = 47 m3 M = 48000 g ( or M = 48 kg)

  11. £600 Simple % Percentage increase and decrease A woman’s wage increases by 13.7% from £240 a week. What does she now earn ? Increase: New amount: 13.17% of £240 240 + 31.608 = 271.608 Her new wage is £271.61 a week 13.17 100 x 240 = 31.608 30% = 75% = 10% = 1% = Percentages of amounts 25% = 45% = 5% = (Do these without a calculator) 85% = 20% = 50% = 2% =

  12. % Frac Dec 83% 56% 0.04 0.17 19 20 9 50 1 2 28% 4 25 0.92 Simple % Fractions, decimals and percentages 50% Copy and complete: 0.5

  13. ÷ 1.05 ÷0.83 x 1.05 x 0.83 Original amount Original weight £660 Original weight Original amount 6 ton. £660 6 ton. Reverse % e.g. A woman’s wage increases by 5% to £660 a week. What was her original wage to the nearest penny? Original amount = 660 ÷ 1.05 = £628.57 e.g. A hippo loses 17% of its weight during a diet. She now weighs 6 tonnes. What was her former weight to 3 sig. figs. ? Original weight = 6 ÷ 0.83 = 7.23 tonnes

  14. x 1.065 x 1.065 x 1.065 x 1.065 x 1.065 £12000 ? x 0.77 x 0.77 x 0.77 x 0.77 £40000 ? Repeated % This is not the correct method: 12000 x 0.065 = 780 780 x 5 = 3900 12000 + 3900 = £15900 e.g. A building society gives 6.5% interest p.a. on all money invested there. If John pays in £12000, how much will he have in his account at the end of 5 years. He will have = 12000 x (1.065)5 = £16441.04 This is not the correct method: 40000 x 0.23 = 9200 9200 x 4 = 36800 40000 – 36800 = $3200 e.g. A car loses value at a rate of approximately 23% each year. Estimate how much a $40000 car be worth in four years ? The car’s new value = 40000 x (0.77)4 = $14061 (nearest $)

  15. 40 5 3 2 2 2 2 3 315 105 7 5 10 20 35 Products of primes Express 40 as a product of primes 40 = 2 x 2 x 2 x 5 (or 23 x 5) 630 Express 630 as a product of primes Now do the same for 100 , 30 , 29 , 144 630 = 2 x 3 x 3 x 5 x 7 (or 2 x 32 x 5 x 7)

  16. HCF Consider the numbers 20 and 30. Their factors are: 1, 2, 4, 5, 10, 20 and 1, 2, 3, 5, 6, 10, 15, 30 Their highest common factor is 10 Expressing 2 numbers as a product of primes can help you calculate their Highest common factor HCF e.g. Find the highest common factor of 84 and 120. 84 = 2 x 2 x 3 x 7 Pick out all the bits that are common to both. 120 = 2 x 2 x 2 x 3 x 5 Highest common factor = 2 x 2 x 3 = 12 Expressing 2 numbers as a product of primes can also help you calculate their Lowest common multiple LCM Consider the numbers 16 and 20. Their multiples are: 16, 32, 48, 64, 80, 96 and 20, 40, 60, 80, 100 Their lowest common multiple is 80 LCM e.g. Find the lowest common multiple of 300 and 504. Pick out the highest valued index for each prime factor . 300 = 22 x 3 x 52 504 = 23 x 32 x 7 Lowest common multiple = 23 x 32 x 52 x 7 = 12600

  17. 10-4 43 Indices Evaluate: 115 91/2 3-2 75  73 163/2 190 2-3 2-1 36-1/2 25 x 21 811/4

  18. Standard form Write in Standard Form Write as an ordinary number 9.6 0.0001 4.7 x 109 8 x 10-3 3600 0.041 1 x 102 5.1 x 104 56 x 103 0.2 7 x 10-2 8.6 x 10-1 8900000000 9.2 x 103 3.5 x 10-3 Calculate 3 x 104 x 7 x 10-1 without a calculator Calculate 4.6 x 104÷ 2.5 x 108 with a calculator Calculate 1.5 x 106÷ 3 x 10-2 without a calculator

  19. ? : 6 ? : 12 ? : 10 14 : ? 1 : ? 7:2 ? : 12 49 : ? 21 : ? 0.5 : ? 2100 : ? ? : 1 Ratio Equivalent Ratios Splitting in a given ratio Total parts = 12 Anne gets 2 of 600 = £100 12 £600 is split between Anne, Bill and Claire in the ratio 2:7:3. How much does each receive? Basil gets 7 of 600 = £350 12 Claire gets 3 of 600 = £150 12

  20. Fractions with the four rules + – × ÷ Learn these steps to complete all fractions questions: • Always convert mixed fractions into top heavy fractions before you start • When adding or subtracting the “bottoms” need to be made the same • When multiplying two fractions, multiply the “tops” together and the “bottoms” together to get your final fraction • When dividing one fraction by another, turn the second fraction on its head and then treat it as a multiplication

  21. 14 3 14 3 14 3 3 2 3 2 2 3 = = =  +  28 6 9 6 = + 28 9 37 6 = = 6 3 1 6 1 9 = = Fractions with the four rules 4⅔ + 1½ 4⅔ 1½

  22. A journey of 37 km measured to the nearest km could actually be as long as 37.49999999…. km or as short as 36.5 km. It could not be 37.5 as this would round up to 38 but the lower and upper bounds for this measurement are 36.5 and 37.5 defined by: 36.5 < Actual distance < 37.5 Upper and lower bounds 320 7cm e.g. Write down the Upper and lower bounds of each of these values given to the accuracy stated: 9m (1s.f.) 8.5 to 9.5 2.40m (2d.p.) 2.395 to 2.405 85g (2s.f.) 84.5 to 85.5 4000L (2s.f.) 3950 to 4050 180 weeks (2s.f.) 175 to 185 60g (nearest g) 59.5 to 60.5 e.g. A sector of a circle of radius 7cm makes an angle of 320 at the centre. Find its minimum possible area if all measurements are given to the nearest unit. ( = 3.14) Area = (/360) x  x r x r Minimum area = (31.5/360) x 3.14 x 6.5 x 6.5 Minimum area = 11.61cm2

  23. e.g. This line has been measured as 9.6cm (to 1d.p.). Calculate the maximum potential error for this measurement. If a measurement has been rounded off then it is not accurate. There is a an error between the measurement stated and the actual measurement. % error The exam question that occurs most often is: “Calculate the maximum percentage error between the rounded off measurement and the actual measurement”. Upper and lower bounds of 9.6 cm (1d.p.)  9.55 to 9.65 Maximum potential difference (MPD) between actual and rounded off measurements  0.05 Max. pot. % error = (MPD/lower bound) x 100 = (0.05/9.55) x 100 = 0.52%

  24. Simplifying roots Tip: Always look for square numbered factors (4, 9, 16, 25, 36 etc) e.g. Simplify the following into the form a b 20 4 x 5 2 5 8 4 x 2 2 2 45 9 x 5 3 5 72 36 x 2 6 2 700 100 x 7 10 7

  25. Calculate the length of side x in surd form (non-calculator paper): 14 x 6 Surds A surd is the name given to a number which has been left in the form of a root. So 5 has been left in surd form. SIMPLIFYING EXPRESSIONS WITH SURDS IN A surd or a combination of surds can be simplified using the rules: M x N = MN and visa versa M ÷ N = M/N and visa versa Tips: Deal with a surd as you would an algebraic term and always look for square numbers 12 4 x 3 2 3 135 ÷ 3 45 9 x 5 3 5 5(5 + 20) 5 + 100 15 (3 – 1)2 (3 – 1) (3 – 1) 3 – 3 – 3 + 1 4 – 23 LEAVING ANSWERS IN SURD FORM Pythagoras  (14)2 = (6)2 + x2 14 = 6 + x2 x = 8 Answer: x = 22

  26. 52 16  3 20 3 3 20 53  Rational and irrational numbers Rational numbers can be expressed in the form a/b. Terminating decimals (3.17 or 0.022) and recurring decimals (0.3333..or 4.7676..) are rational. Irrational numbers cannot be made into fractions. Non-terminating and non-recurring decimals (3.4526473… or or 2) are irrational. State whether the following are rational or irrational numbers: 1/5 2.3/5.7 2.7 What do you need to do to make the following irrational numbers into rational numbers: 6 2

  27. Recurring decimals as fractions Learn this technique which changes recurring decimals into fractions: Express 0.77777777….. as a fraction. Let n = 0.77777777….. so 10n = 7.77777777….. so 9n = 7 so n = 7/9 Express 2.34343434….. as a fraction. Let n = 2.34343434….. so 100n = 234.34343434….. so 99n = 232 so n = 232/99 Express 0.413213213….. as a fraction. Let n = 0.4132132132….. so 10000n =4132.132132132….. and 10n = 4.132132132…… so 9990n = 4128 so n = 4128/9990 n = 688/1665

  28. Direct proportion If one variable is in direct proportion to another (sometimes called direct variation) their relationship is described by: p  t Where the “Alpha” can be replaced by an “Equals” and a constant “k” to give : p =kt e.g. y is directly proportional to the square of r. If r is 4 when y is 80, find the value of r when y is 2.45 . Write out the variation: y  r2 Possible direct variation questions: Change into a formula: y = kr2 Sub. to work out k: 80 = k x 42 g  u3 g = ku3 k = 5 c  i c = ki So: y = 5r2 s  3v s = k3v And: 2.45 = 5r2 t  h2 t = kh2 r = 0.7 x  p x = kp Working out r:

  29. Inverse proportion If one variable is inversely proportion to another (sometimes called inverse variation) their relationship is described by: p  1/t Again “Alpha” can be replaced by a constant “k” to give : p =k/t e.g. y is inversely proportional to the square root of r. If r is 9 when y is 10, find the value of r when y is 7.5 . Write out the variation: y  1/r Possible inverse variation questions: Change into a formula: y = k/r Sub. to work out k: 10 = k/9 g1/u3 g = k/u3 k = 30 c  1/i c = k/i So: y = 30/r s  1/3v s = k/3v And: 7.5 = 30/r t  1/h2 t = k/h2 r = 16 (not 2) x  1/p x = k/p Working out r:

  30. y y = x2 + x – 6 x -3 2 Graphical solutions to equations If an equation equals 0 then its solutions lie at the points where the graph of the equation crosses the x-axis. e.g. Solve the following equation graphically: x2 + x – 6 = 0 All you do is plot the equation y = x2 + x – 6 and find where it crosses the x-axis (the line y=0) There are two solutions to x2 + x – 6 = 0 x = - 3 and x =2

  31. y = 9 – x y y = x2 – 2x – 11 x -4 5 Graphical solutions to equations If the equation does not equal zero : Draw the graphs for both sides of the equation and where they cross is where the solutions lie e.g. Solve the following equation graphically: x2 – 2x – 11 = 9 – x Plot the following equations and find where they cross: y = x2 – 2x – 20 y = 9 – x There are 2 solutions to x2 – 2x – 11 = 9 – x x = - 4 and x = 5

  32. e.g. Solve the following equation using the graph that is given: x3 – 4x + 5 = 5x + 5 y y = x3 – 8x + 7 x If there is already a graph drawn and you are being asked to solve an equation using it, you must rearrange the equation until one side is the same as the equation of the graph. Then plot the other side of the equation to find the crossing points and solutions.

  33. y y = x3 – 8x + 7 -3 y = x + 7 x 0 3 Rearranging the equation x3 – 4x + 5 = 5x + 5 to get x3 – 8x + 7 : x3 – 4x + 5 = 5x + 5 Add 2 to both sides x3 – 4x + 7 = 5x + 7 Take 4x from both sides x3 – 8x + 7 = x + 7 So we plot the equation y = x + 7 onto the graph to find the solutions Solutions lie at –3, 0 and 3

  34. State the graphs you need to plot to solve the • following equations describing how you will find • your solutions: • 3x2 + 4x – 2 = 0 • 7x + 4 = x2 – 4x • x4 + 5 = 0 • 0 = 8x2 – 5x • 2x = 9 • 6x3 = 2x2 + 5 • If you have got the graph of y= 4x2 + 5x – 6 work • out the other graph you need to draw to solve each • of the following equations: • 4x2 + 4x – 6 = 0 • 4x2 + x - 2 = 7 • 4x2 – 3x = 2x • 3x2 = – 5 Solve this equation graphically: x3 + 8x2 + 3x = 2x2 – 2x

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