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REMEMBER!!!!

REMEMBER!!!!. Have a quick look at the paper first, making small notes if necessary, then answer what you know you can do first. Always attempt EVERY question no matter what, showing as much working as possible.

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REMEMBER!!!!

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  1. REMEMBER!!!! • Have a quick look at the paper first, making small notes if necessary, then answer what you know you can do first. • Always attempt EVERY question no matter what, showing as much working as possible. • Remember to use the formulae sheet provided, and if unsure of a question look at this, as it may trigger something in your mind. • Check all of your answers if you have time at the end.

  2. Formulae to Remember:- The Formulae NOT given in exam which you must know include: To calculate the Area of a Sector Area of a circle: A = r2 To find Arc Length Circumference: C = D = 2r Inter Quartile Range (measure main 50% spread in Boxplots): IQR = Q3 – Q1 Semi-Inter Quartile Range SIQR = 1/2(Q3 - Q1)

  3. Standard Form/Scientific Notation (a x 10n) The final value must bebetween 1 and 10 [i.e. 99 x 104must be written as 9.9 x 105] 4230000 = 4.23 x 106 13280000 = 1.328 x 107 0.000975 = 9.75 x 10-4 2 million million = 2 x 1012 1.2 x 103 = 1200 7.425 x 106 = 7425000 2.91 x 10-2 = 0.0291 5.23 x 10-5= 0.0000523

  4. More complex calculations • A massive container holds4.23 x 106 litres • A spoon holds1.52 x 10-2litres Question:- How many spoonfuls fill the container? give your answer to 3 sf & in standard form Answer:- (use EXP button on calculator for ease) 4.23 x 106 = (4.23 EXP6) = 278289473.684 =2.78 x 108 1.52 x 10-2 (1.52 EXP-2) spoonfuls

  5. Significant Figures Round the figure to a certain number of important or significant digits. It is NOT rounding to a certain amount of decimal places. E.g. If you have to represent 2323 to 1sf you would write 2000 (NOT 2). As the original number is in thousands, your rounded figure should still be in thousands. 14243 to 1sf is 10000 16.932 to 1 sf is 20 14243 to 2sf is 14000 16.932 to 2 sf is 17 14243 to 3sf is 14200 16.932 to 3 sf is 16.9 1.2374 to 1 sf is 1 0.00237 to 1 sf is 0.002 1.2374 to 2 sf is 1.2 0.00237 to 2 sf is 0.0024 1.2374 to 3 sf is 1.24 0.00237 to 3 sf is 0.00237 • NBIn the last example zeros are not significant at the front of a number, the first significant digit must be one or more.

  6. Conversion: • Remember to pay attention to the units required and to how many dec plcs/sig figs! • 60mins  1 hr • to convert into hours we must therefore  60 (secs to mins  60) (SDT!!!) • 1cm3 = 1ml & 1000cm3 = 1000ml = 1litre • 1m3 = 1m x 1m x 1m = 100cm x 100cm x 100cm = 1,000,000cm3

  7. Fractions ADDING/SUBTRACTING:- the denominatormust be the same e.g. 2 + 5 = 2(y-1) + 5y = 2y-2 + 5y – 5 = 7y – 2 y (y-1) y(y-1) y(y-1) y(y-1) MULTIPLICATION:- Multiply straight over and thensimplify e.g. a x 2b = 2ab = b 4c a 4ac 2c DIVISION:- Change to multiply  turn fraction upside down e.g. ab a = abx3c= 3abc= b 9c 3c 9c a9ac 3 • Substitution **Remember (-3)2= 9 NOT –9 and follow rules of BODMAS! • BODMAS Brackets Order Divide Multiply Add then Subtract

  8. Algebraic Fractions • Remember to follow through from previous part of factorising question if given an algebraic fraction, then simplify wherever possible: x2 – 8x + 7 = (x – 7) (x – 1) = (x – 1) x 2 - 49 (x – 7) (x + 7) (x + 7)

  9. Profit/Loss (Appreciation/Depreciation) can be calculated as a percentage by:-Difference in Value x 100% = % Profit or Loss Original Value

  10. To make into a percentage.Given 12 out of 30 pupils are girls, what is this as a percentage of all pupils?12 out of 30 can be written as 12 = 4 = 40 30 10 100{NB easier to make into a percent if over 10 or 100}

  11. Compound Interest Shortcut Method [£ x (1.00 + %)y] Appreciation:- The value increases by itself (100%) + the % increase e.g 5% increase in value is (100% + 5%) = 105% so multiplying factor  1.05 Depreciation:- The entire value decreases (100%) - % decrease e.g 5% decrease in value is (100% - 5%) = 95% so multiplying factor  0.95

  12. Example of Compound Interest Ex: Calculate compound interest when investing £720 at a rate of 5.7% over 3 years? An increase in 5.7% over 3 years can be written as (1.00 + 0.057)3 (1.057) 3 £720 x (1.057) 3 = £850.27 (to the nearest penny) [Only use this method if same percent change each year.]

  13. Reversing Values (i.e. Going Backwards to find original value** DIFFICULT) After a 10% increase a house is worth £88 000. Find its original value? Original value = 100% • House is now valued at (100% + 10%) = 110% = 1.10 House Value 110%  £88 000 1.10  £88 000 1  £88 000 = £80 000 1.10

  14. A car depreciates in value by 12% over the year and is now £9680. • What was its value at the start of the year? • Original value = 100% • Car is now valued at (100% - 12%) = 88% = 0.88 • Original Car Value • 88%  £9 680 • 0.88  £9 680 • 1  £9 680 = £11 000 • 0.88

  15. Circles 1 Arc Length is part of the circumference, so use A x D 360 2 Area of a Sector is related to ‘Area’, so useA x r2 360 3 In the circle a line, or tangent, to the circle creates an angle of 90o 4 Any triangle in a semi-circle with an angle touching the circle has an angle of 90o 5 Use radii, right angled triangles and Pythagoras theorem when given ‘Chords’ in a circle. 1 & 2 3 4 5 r A

  16. Chords • Always use the radius in this question to create a right angled triangle. • Using Pythagoras or basic trig you may then find the unknown side • If finding the height of a tunnel take unknown + radius • If finding depth below take radius - unknown

  17. Make into a right angled triangle • Bottom depth will be  H = (radius – x) • Height will be  H = (radius + x) • If to find radius need to use pythagoras & solve algebraically as a Quadratic problem x r r r (r-x) x r r x

  18. Make into a right angled triangle • Given AB = 8cm and r = 5cm find the depth of water below chord AB? • x2 = r2 – (OB)2 Depth below AB: x2 = 52 – 42 D = (radius – x) = 25 – 16 = 5cm - 3cm = 9 = 2cm depth x = √9 = 3cm r x A B 5cm x A B 4cm 8cm

  19. Make into a right angled triangle C D x • Given CD = 12cm and r = 10cm • Find the water level Height at CD? • x2 = r2 – (OD)2 Water level at CD : x2 = 102 – 62 H = (radius + x) = 100 – 36 = 10cm + 8cm = 64 = 18cm high x = √64 = 8cm r r r 12cm 6cm C D x 10cm 10

  20. Make into a right angled triangle x E F • Given EF = 12cm and x = 3cm • Find the radius, r? • r2 = (r - 3)2 + 62 r2 = r2 – 6r + 9 + 36 6r = 45 r = 45 = 7.5cm 6 r r 12cm 3 6cm E F (r - 3) r cm

  21. Linear Relationships : y = mx + c Line cuts Y-axis at (0,c) thus can find value of ‘c’ by looking at graph Gradient is ‘m’ which can be found by Gradient m = Vertical = Rise = Difference in y’s = y2 - y1 Horizontal Run Difference in x’s x2 – x1

  22. If asked to draw a line given an equation: • Make a small grid with x values 0, 1, 2 and plot line i.e. if y = 3x - 4 Y = 3x – 4 • x 0 1 2 thus can plot 3 coords (0, -4); (1, -1) & (2, 2) y -4 -1 2 and extend to draw a line • Special Cases:-Gradient = 0 Gradient Undefined Remember Positive Gradients  Uphill (Positive) Negative Gradients Downhill (Negative) & Parallel lines have the same gradient

  23. Algebra: Opening Brackets • Always use a FOIL to help multiply two brackets (x + 2)(x + 6) (x - 4) 2= (x -4)(x - 4) = x2 + 2x + 6x + 12 = x2 - 4x - 4x + 16 = x2 + 8x + 12 = x2 - 8x + 16 If a larger bracket is to be opened then it may be easier to use a grid:- (x + 5)( x2 + 8x + 12) x2 + 8x + 12 = x3 + 8x2 + 5x2 + 12x + 40x + 60 x x3 + 8x2 + 12x = x3 +13x2 + 52x + 60 +5 5x2 + 40x + 60 x3 + 13x2 + 52x + 60

  24. Understanding the signs x2 + 5x + 4 All signs positive so either = (x + ?)(x + ? ) 1 x 4 or 2 x 2 to get 4 at end. = (x + 1)(x + 4) As 5 in middle => must use 1 and 4 x2 – 14x + 13 13 is positive, BUT -14x in middle = (x - 1)(x - 13) => 2 negatives multiplied together to get +13 Prime number so can only be -1 x -13 which gives -14 in middle x2 + 2x – 3 -3 implies must be 2 different signs + / - to give -3 = (x + 3)(x - 1) -3 x 1 or 3 x 1. As +2x in middle must be 3 and -1 x2 – 6x – 7 -7 implies 2 different signs to make -7 => Either -7 x 1 or +7 x -1 = (x - 7)(x + 1) As -6x in middle the larger number is influencing a negative result Thus -7 and 1 shall give -6x in middle

  25. Summary x2 + ?x + ? => Both + & sum gives positive middle value x2 – ?x + ? => Both – & sum gives negative middle value x2 + ?x – ? => 2 different signs & difference gives positive middle value (Large part is Positive) x2 – ?x – ? => 2 different signs & difference gives negative middle value (Largest part is Negative)

  26. ALWAYS FOLLOW THE SAME PROCEDURE EACH TIME WHEN FACTORISING:- Is there a COMMON FACTOR? i.e. 2x - 12x = 2x(x – 6) Is there a DIFFERENCE OF TWO SQUARES? 4a – 25b2 = (2a) – (5b) = (2a + 5b)(2a – 5b) Can it FACTORISE (Is it a trinomial in 3 parts?) x – 8x + 7 = (x – 7)(x – 1) ELSEif to solve to A SPECIFIC NUMBER OF DECIMAL PLACES OR SIGNIFICANT FIGURES use the QUADRATIC FORMULA

  27. Solving a Quadratic Problem Example: Solve the following x2 – 8x + 7 for x x2 – 8x + 7 = 0 (Set = 0 if not given) (x – 7)(x – 1) = 0 (Factorise quadratic) (x – 7) = 0 & (x – 1) = 0 x = 7 & x = 1

  28. Quadratic Formulae Use when factorisation isnot possibleand asked to a specific number of decimal places/significant figures ax2 + bx + c = 0 and a  0 where a, b and c are constants from the quadratic x = -b +/- (b2 - 4ac) 2a (******* formula given on worksheet ********)

  29. Sketch the graph of f(x) = (x - 4) 2 - 1 • Positive 'a'  Happy U  Minimum Tpt • Vertex (b,c) Minimum Tpt at (4, -1) • Symmetry at x = 4 • Cuts y-axis at x=0, y = (0 - 4)2 - 1 = (16) - 1= 15 So cuts at (0,15) • Cuts x-axis at y=0, (x - 4) 2 - 1 = 0 x2 – 8x + 16 – 1 = 0 x2 – 8x + 15 = 0 (x – 3)(x – 5) = 0 x – 3 = 0 & x – 5 = 0 x = 3 & x = 5 So cuts at (3, 0) & (5, 0) Having obtained all information it is then possible to sketch the quadratic graph.

  30. Volume of Solids Volume of a Cone : V = 1r2h [Easier to multiply values 3 Volume of a Sphere: V = 4r3 together and  by 3 at the end] 3 *** 2 Formulae NOT Given to you Vol of a Cylinder: V = r2h & Vol of ANY Prism : V = Ah (Main shape x height) **** Make sure answer to the correct number of decimal **** places OR significant figures, else marks will be lost!

  31. Simultaneous Linear Equations : 3 Methods • 1. Graphical: This solves for the values of x and y where both lines have same value (i.e. they cross/intersect). It is necessary to sketch two lines for this and the point of intersection is the where the solution can be found. • 2. Substitution: Place one equation into the other and solve using substitution. e.g y = 2x and y = -x + 12 y =3x and 7x + 2y = 39 If y = y using the method of substitutionThis time subst y = 3x into other equation 2x = -x + 12 7x + 2y = 39 2x + x = 12 7x + 2(3x) = 39 3x = 12 7x + 6x = 3 x = 4 13x = 39 x = 3 Then find y using either equation given at start.

  32. Simultaneous Linear Equations 3. Elimination: Whenever presented with TWO EQUATIONS and you need to find the values of x and y, p and q etc Use the system of SIMULTANEOUS EQUATIONS • ALWAYS eliminating one variable to solve for one, then substitute this into the equation to find the other. • SAME SIGNS => SUBTRACT EQUATIONS • DIFFERENT SIGNS => ADD EQUATIONS TOGETHER • Then use this value to find the other unknown choosing ANY equation.

  33. Statistics • You must know how to interpret, construct and present the following:- • Bar and Line graphs • Stem and Leaf Plot and Pie charts • Frequency tables, diagrams and polygons • Cumulative Frequency diagrams • Box Plots; Dot Plots and 5-Figure Summary • Probability and Relative Frequency (same idea as probability) • Scatter Diagram • Standard Deviation

  34. Boxplots • You must ORDER DATA & find the 5-figure Summary:- • L or Min - minimum value • Q1 - 1st/lower quartile • Q2 - median (or middle value) • Q3 - upper/3rd quartile • H or Max - maximum value NOT given in the exam so MUST be remembered! To find the spread of the main 50% of your information use:- • The IQR (Inter Quartile Range) = (Q3 - Q1) • SIR (semi-inter quartile range) is SIQR =1/2(Q3 - Q1)

  35. Stem and Leaf Plots Remember to puta key for the data • Draw a vertical line with the main number group being the ‘stem’, and the remaining smaller units of the number being the ‘leaves’. • Place in numerical order from the lowest to highest value and represent every piece of data, even if a number is repeated more than once.

  36. Dot Plots • Draw a horizontal scale from the lowest to highest value • Horizontal scale must be in numerical order. • Represent each piece of data above the scale with a dot • If repeated values place a dot above the previous data value entered.

  37. Scatter Diagrams Refer to“GCSE Maths Revision book” this is a good example to follow. Ensure line going through data cuts y-axis to find y intercept, c. Try to have same amount of data above & below line, and about same distance from line.

  38. Standard Deviation:Method 1 • Create a table with 3 columnswith the following headings: _ _x (x – x) (x – x)2 _ _ x (x - x) (x - x)2 _ • & mean of x being x bar, x = x  n

  39. Standard Deviation:Method 2 • Create a table with 2 columnswith the following headings: x x2  x  x2 • (∑ means the total sum of) • NB:- If presented with Standard Deviation data in the format • x2 = 12345 and x = 123 use the 2nd formula given on the front sheet without creating a table.

  40. Similar Shapes • Similar triangles are Equiangular • have the same angles & sides are in proportion • Scale Factor (SF) • Similar Areas. Area units are cm2 (Scale Factor) 2 • Similar volumes have all dimensions in proportion i.e Volume units are cm3 (Scale Factor) 3

  41. Separate Similar Triangles • When can’t see the problem clearly make a new sketch by separating the triangles. • This will make the problem far more clear.

  42. Redraw complicated shapes • Use z-shape angle rules & separate for simplicity C A B α θ * * B θ α A C C * * θ α θ α E E D D

  43. Ratio and Proportion Ratio If Michael, Lisa and Stephanie share a bag of 81 sweets in the ratio 2:3:4, how many sweets does each person get? • Add parts together to find total amount of shares 2 + 3 + 4 = 9 • Take value and divide by total shares to find 1 share 81  9 = 9 • When you have found 1 share  9 sweets here, you can then substitute back into original ratio 2 : 3 : 4 2 x 9 : 3 x 9 : 4 x 9 18 : 27 : 36 • Check by adding together 18 + 27 + 36 = 81 sweets

  44. Problem Solving with Sequences If given a sequence e.g. 3, 7, 11, 15,… find the formula for the sequence: E.g. nth term 1 2 3 4 5 …. N Sequence 3 7 11 15 19 … 4N – 1 +4 +4…. +4 (4*N ; As increases by 4 each time)

  45. Variation Direct Variation When both increase/decrease at same rate i.e. as one halves the other halves y = kx or y  x Inverse As x increases  y decreases i.e.Time for job to be completed reduces as manpower increases. y = k or y 1 x x

  46. Surds 3 x 3  2 x 50 12 3 = 9 = 100 = 4 = 3 = 10 = 2 72  72 + 50 - 98 = 36 x 2 = (36 x 2) + (25 x 2) – (49 x 2) = 62 = 62 + 52 – 72 = 42 • Can multiply/divide straight away • You can only add/subtract if a ‘common surd’ exists . • Remember Rationalise  Do not leave as surd on denominator

  47. Surds Simplify and RATIONALISE:Find the area of the rectangle: 5 3 + 1 = (5 3 + 1) x (4 3 – 7) (3 + 1) (23 – 5) (43 + 7) (43 + 7) (43 - 7) = (3 x 2 3) - 53 + 2 3 - 5 = 2 x 3 – 5 - 3 3 = 2033 -35 3 + 43 – 7 = 1- 3 3 1633 -283 + 283 – 49 = 20x3 - 313 - 7 16x3 – 49 = 60 – 7 - 313 48 – 49 = 313 – 53 -1 = 53 - 313 (23 – 5) (3 + 1)

  48. Indices ax x ay = a x+y ax ay = ax-y a2 x a3 = a5 a7 a4 = a3 a3/2 x a5/2 = a8/2 = a4 a5/2 a1/2 = a4/2 = a2 a x/y = yax 82/3 = 3 (82)= 364 = 4 or 82/3 = (38)2 = 22 = 4 a0 = 1 and a-1 = 1 a • Easy to remember that a ‘x’ sign is a slanted ‘+’ sign so if multiplying weADD the powers • If ‘’ or ‘/’ think of ‘-‘ within symbol i.e. when dividing indices we SUBTRACT the powers

  49. Indices b1/2( b1/2 + b-3/2) y7 x y-4 a-4/5 x a-3/5 = (b1/2 x b1/2) + (b1/2 x b-3/2) y3 a-2/5 = b1/2+1/2 + b1/2- 3/2 = b2/2 + b-2/2 = y3 = a-7/5 = b + b-1 y3 a-2/5 or b + 1 b = y3-3 = a-7/5 –(-2/5) = y0 = a-7/5+2/5 = 1 = a-5/5 = a-1 or 1 a

  50. Right Angled Triangles • SOH CAH TOA and Pythagoras only in Right Angled Triangles. • Pythagoras if sides only  c2 = a2 + b2 • If angles and sides use SOH CAH TOA Sin x = O; Cos x = A; Tan x = O H H A

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