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GCSE Mathematics Revision

GCSE Mathematics Revision. C/D borderline. C/D borderline Choose your session!. Session 11 Session 12 Session 13 Session 14 Session 15 Session 16 Session 17 Session 18 Session 19 Session 20. Session 1 Session 2 Session 3 Session 4 Session 5 Session 6 Session 7 Session 8

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GCSE Mathematics Revision

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  1. GCSE Mathematics Revision C/D borderline

  2. C/D borderlineChoose your session! • Session 11 • Session 12 • Session 13 • Session 14 • Session 15 • Session 16 • Session 17 • Session 18 • Session 19 • Session 20 • Session 1 • Session 2 • Session 3 • Session 4 • Session 5 • Session 6 • Session 7 • Session 8 • Session 9 • Session 10

  3. Back to sessions Session 1 • Expand and simplify 5(3x + 4) + 6(2x – 2) • A circle has radius 3cm. Write the area and circumference of the circle in terms of π. • a) Find the nth term of the sequence 4, 7, 10, 13, 16, ... b) Hence, or otherwise, find the 50th term in this sequence. 4. Simplify 5a2b3 x 4a4b2 Ans. 1 Ans. 2 Ans. 3 Ans. 4

  4. Expand and simplify 5(3x + 4) + 6(2x – 2) Expand the first bracket 5(3x + 4) = 15x + 20 Expand the second bracket 6(2x – 2) = 12x – 12 Replace the brackets with their expansions 15x + 20 + 12x – 12 Simplify by collecting like terms 27x + 8 Back

  5. A circle has radius 3cm. Write the area and circumference of the circle in terms of π. Recall the formula for the area of a circle A = πr2 Replace the r with the value for the radius A = π x 32 Simplify A = 9 π Circumference formula C = πd Replace the d with the value for the diameter C = π x 6 Simplify C = 6 π Back

  6. a) Find the nth term of the sequence 4, 7, 10, 13, 16, ... b) Hence, or otherwise, find the 50th term in this sequence. • Find the difference between the terms. • In this case 3 • The rule is linked to the three times table and starts with 3n • Find out what needs to be added/subtracted to 3n to make the rule work • 3n + 1 • The rule is 3n +1 • For the 50th term the value of n is 50 • Substitute • 3 x 50 + 1 = 151 Back

  7. Simplify 5a2b3 x 4a4b2 Consider numbers first: 5 x 4 = 20 Consider the a’s: a2 x a4 = a6 Consider the b’s: b3 x b2 = b5 Put it all back together 20a6b5 Back

  8. Back to sessions Session 2 • Find the volume of the shape below 2. Solve the equation 4x + 5 = 15 • Write 60 as a product of prime numbers • Increase £120 by 30% 3cm 5cm 6cm Ans. 1 Ans. 2 Ans. 3 Ans. 4

  9. 1. Find the volume of the shape below 3cm 5cm 6cm Recall the formula for the volume of a prism Volume of a prism = area of cross section x length Find the area of the cross section (triangle) 6 x 3 = 9cm2 2 Multiply this by the length 9 x 5 = 45cm3 Back

  10. 2. Solve the equation 4x + 5 = 15 We need to get the x’s on their own Subtract 5 from each side 4x = 10 Now divide by the number before x x = 10 4 x = 2.5 Back

  11. 3. Write 60 as a product of prime numbers Draw a factor tree to help 60 30 2 15 2 5 3 So the answer is 2 x 2 x 3 x 5 Which simplifies to 22 x 3 x 5 Your tree may look different Back

  12. 4. Increase £120 by 30% To increase by 30% we have to start by finding 30% of the amount and then add it on. 10% of £120 = £12 30% of £120 = £12 x 3 = £36 Now add it on: £120 + £36 = £156 Back

  13. Back to sessions 5 x 107 3 x 1012 Session 3 Ans. 1 Ans. 2 Ans. 3 Ans. 4 Put these fractions into order, starting with the smallest Find the area of the rectangle below. Leave your answer in standard form. Calculate the size of an interior angle of a regular octagon. Factorise 8x + 12

  14. Put these fractions into order, starting with the smallest Start by changing the fractions to equivalent ones with the same denominator. Is there a number that is a multiple of 8, 20, 4, 5 and 40? 40 The new fractions you get are Putting them in order gives Back

  15. 5 x 107 3 x 1012 Area of a rectangle = length x width Area = 3 x 1012 x 5 x 107 Multiply the numbers 3 x 5 = 15 Multiply the indices 1012 x 107 = 1019 Put your answers back together 15 x 1019 Adjust the answer to put it in standard form 1.5 x 1020 Find the area of the rectangle below. Leave your answer in standard form. Back

  16. Calculate the size of an interior angle of a regular octagon. First calculate the exterior angle of an octagon. An octagon has eight sides. Exterior angles add to 360o All exterior angles on a regular shape are equal. 360o ÷ 8 = 45o Exterior and interior angles lie on a straight line so must add to 180o Interior angle = 180o – 45o = 135o Back

  17. Factorise 8x + 12 Start by finding a common factor. Is there a number that goes into 8 and 12? 4 Take this value outside the bracket, then work out what needs to go inside the bracket to make it work when expanded. 4(2x + 3) Back

  18. Back to sessions Session 4 • Solve the inequality 3x – 2 < 4. Show your solution set on a number line. • One angle in an isosceles triangle is 40o. Find the possible sizes of the other angles. • David worked out that 21.082 x 9.86 = 438.15 (2dp) Explain why David must be wrong. 4. Find the average speed of a car which travels 75 miles in one and a half hours Ans. 1 Ans. 2 Ans. 3 Ans. 4

  19. For a number line: Draw a circle at the key value Shade it in if it can equal the value, leave open if it can’t equal the value Draw an arrow to show the correct direction -3 -2 -1 0 1 2 3 4 Solve the inequality 3x – 2 < 4. Show your solution set on a number line. Solve inequalities in the same way you solve equations. Add two to each side 3x < 6 Divide by 3 x < 2 Back

  20. Angles in a triangle add to 180o 40o yo 40o 40o xo xo y = 100o x = 70o 2. One angle in an isosceles triangle is 40o. Find the possible sizes of the other angles. What do we know about isosceles triangles? 2 sides the same and two angles the same Sketch the two different types of triangle Back

  21. 3. David worked out that 21.082 x 9.86 = 438.15 (2dp) Explain why David must be wrong. Use estimation to find an approximate answer Round each number to one significant figure 202 x 10 400 x 10 = 4000 The answer is approximately 4000 so cannot possibly be 438.15 Back

  22. Formula method: Distance = speed x time 75 = speed x 1.5 75 = speed 1.5 Speed = 50mph Scaling method: 75 miles in 1 and a half hours 25 miles in half an hour 50 miles in one hour 50mph Easier to use when calculators are allowed Find the average speed of a car which travels 75 miles in one and a half hours Back

  23. Back to sessions Session 5 7m 2m • Find the value of x in the diagram below • Work out 7.562 + 9.52 14.36 – 3.61 a) write the full calculator display b) give your answer to two decimal places • Write the equation of a line parallel to y = 5x + 2 • Rearrange the equation in question 3 to make x the subject x Ans. 1 Ans. 2 Ans. 3 Ans. 4

  24. 7m 2m x Recognise the skill required: Pythagoras’ Theorem 1. Find the value of x in the diagram below Recall Pythagoras’ Theorem a2 + b2 = c2 (where c is the hypotenuse) Substitute the values x2 + 22 = 72 Simplify x2 + 4 = 49 Solve x2 = 45 x = √(45) = 6.71m (2 d.p.) Back

  25. = 6.202195349 2. Work out 7.562 + 9.52 14.36 – 3.61 a) write the full calculator display b) give your answer to two decimal places Either enter this into your calculator as it appears in the question or work out the numerator and denominator first: a) 66.6736 10.75 b) 6.20 Back

  26. Write the equation of a line parallel to y = 5x + 2 What would be true about any line parallel to the one given? It would have the same gradient What is the gradient of the line given? 5 (the number attached to x when the equation is in the form y = mx + c) Possible answers: y = 5x + 1 y = 5x + 10 y = 5x – 2 etc. Back

  27. 4. Rearrange the equation in question 3 to make x the subject Equation is: y = 5x + 2 We need to get the x on its own on one side of the equation Subtract 2 from both sides y – 2 = 5x Divide both sides by 5 y – 2 = x 5 Back

  28. Back to sessions Session 6 • Factorise x2 + 5x – 14 • Hence solve the equation x2 + 5x – 14 = 0 • What is the reciprocal of 0.3? • The length of a rectangle is 3 more than its width. a) Show that the perimeter of the rectangle can be written as 4x + 6 b) The perimeter is 20cm. Find the dimensions of the rectangle Ans. 1 Ans. 2 Ans. 3 Ans. 4

  29. 1. Factorise x2 + 5x – 14 Recognise how many brackets this will factorise into. ( )( ) Place an x into each bracket (x )(x ) Find two numbers that multiply to give -14 and add to give +5 (x + 7)(x – 2) Back

  30. 2. Hence solve the equation x2 + 5x – 14 = 0 Hence means we can use our last answer. We have just shown that x2 + 5x – 14 can be written as (x + 7)(x – 2) Therefore (x + 7)(x – 2) = 0 One of the brackets must be equal to zero. So either x = -7 or x = 2 Back

  31. What is the reciprocal of 0.3? A reciprocal is what we need to multiply the number by to give an answer of one. So 0.3 x ??? = 1 Turn the decimal into a fraction 0.3 = 3 10 The reciprocal will simply be the fraction flipped upside down: 10 3 Back

  32. x x + 3 Width = 3.5cm Length = 6.5cm 4.The length of a rectangle is 3 more than its width. a) Show that the perimeter of the rectangle can be written as 4x + 6 b) The perimeter is 20cm. Find the dimensions of the rectangle Sketch the rectangle a) The perimeter is the total distance around the outside of the shape: x + x + 3 + x + x + 3 Which simplifies to 4x + 6 b) Solve the equation 4x + 6 = 20 Subtract 6 from each side 4x = 14 Divide each side by 4 x = 3.5 Back

  33. Back to sessions Session 7 10 m x m 25o • Find the size of the side marked x • A ball bounces to 80% of its previous height. I drop the ball from 2m, calculate the height after two bounces • Calculate the area of a circle with radius 4.5cm. Give your answer to 3 significant figures. • £1 = $1.65 Use this to find the cost in pounds of sunglasses bought in America for $96.97 Ans. 1 Ans. 2 Ans. 3 Ans. 4

  34. 10 m x m 25o 1. Find the size of the side marked x Recognise the question as trigonometry: SOH CAH TOA Labelling the sides gives: Back

  35. First Bounce 10% of 2m = 20cm 20% of 2m = 40cm New height = 2m – 40cm = 1.60m Second bounce 10% of 1.60m = 16cm 20% of 1.60m = 32cm New height = 1.60m – 32cm = 1.28m 80% as a decimal is 0.8 Height after first bounce 2 x 0.8 = 1.6m Height after second bounce 1.6 x 0.8 = 1.28m 2. A ball bounces to 80% of its previous height. I drop the ball from 2m, calculate the height after two bounces Either find 20% and take it from the original or find a multiplier. Back

  36. Calculate the area of a circle with radius 4.5cm. Give your answer to 3 significant figures. Recall the formula for the area of a circle: A = πr2 Substitute the value of r and calculate: A = π x 4.52 = 63.61725124 Round to 3 significant figures A = 63.6cm2 Back

  37. 4. £1 = $1.65 Use this to find the cost in pounds of sunglasses bought in America for $96.97 To convert dollars into pounds we divide by 1.65 So $96.97 ÷ 1.65 = £58.77 (rounded to d.p. for money) Back

  38. Back to sessions Session 8 • Expand and simplify (x + 2)(x – 4) • Solve the equation 7x – 5 = 3x + 27 • You are given the formula v = u + at Find the value of v when u = 3, a = 4 and t = 2 • Make t the subject of the formula in question 3 Ans. 1 Ans. 2 Ans. 3 Ans. 4

  39. Write all of the terms out in full: x2 + 2x – 4x – 8 Collect like terms: x2 – 2x – 8 x2 2x - 4x - 8 Expand and simplify (x + 2)(x – 4) Use a suitable method for expansion e.g. grid method x +2 x - 4 Back

  40. 2. Solve the equation 7x – 5 = 3x + 27 Use a suitable method e.g. Do the same thing to both sides to keep the equation balanced add 5 to each side 7x = 3x + 32 subtract 3x from each side 4x = 32 divide both sides by 4 x = 8 Back

  41. 3. You are given the formula v = u + at Find the value of v when u = 3, a = 4 and t = 2 Substitute the values in for the correct letter. Remember that at means a x t v = 3 + (4 x 2) v = 3 + 8 v = 11 Back

  42. 4. Make t the subject of the formula in question 3 The formula was v = u + at subtract u from each side of the equation v – u = at divide each side by a v – u= t a Back

  43. Back to sessions Session 9 x 3.2m • Find the size of the angle marked x • Find the length of the side marked y • A circular pond has diameter 5m, find the area of the pond’s surface 4. The shapes below have the same area. Find the value of h 12m y m 7.2m 7.6m 2.5cm h cm Ans. 1 Ans. 2 Ans. 3 Ans. 4 8cm 8cm

  44. x 3.2m hyp adj 12m opp 1. Find the size of the angle marked x Recognise the question as trigonometry SOH CAH TOA Label the sides Back

  45. y m 7.2m 7.6m Find the length of the side marked y Recall Pythagoras’ Theorem a2 + b2 = c2 (where c is the hypotenuse) Substitute the values 7.62 + 7.22 = y2 Simplify 109.6= y2 Solve y = √(109.6) = 10.47m (2 d.p.) Back

  46. 3. A circular pond has diameter 5m, find the area of the pond’s surface Recall the formula for the area of a circle A = πr2 Replace the r with the value for the radius A = π x 2.52 A = 19.63m2 (2 d.p.) Back

  47. Find the area of the parallelogram A = base x height A = 8 x 2.5 = 20cm2 Recall the formula for the area of a triangle: A = base x height 2 20 = 8 x h 2 Solve the equation: 40 = 8h h = 5cm 2.5cm h cm 8cm 8cm 4. The shapes below have the same area. Find the value of h Back

  48. Back to sessions 40o xo Session 10 Ans. 1 Ans. 2 Ans. 3 Ans. 4 Solve the equation x2 + 6x – 16 = 0 A straight line has a gradient of 2 and passes through the point (0,3). What is the equation of the line? 60 = 22 x 3 x 5. Write 600 as a product of primes. Find the size of the angle marked x

  49. 1.Solve the equation x2 + 6x – 16 = 0 As it already equals zero, we begin by factorising the left hand side (x + …)(x – …) = 0 (x + 8)(x – 2) = 0 One of the brackets must equal zero, so we give two solutions From the first bracket x = -8 from the second bracket x = 2 Back

  50. 2.A straight line has a gradient of 2 and passes through the point (0,3). What is the equation of the line? Straight line graphs are in the form y = mx + c where m is the gradient and c is the y-intercept Gradient = 2 y-intercept = +3 Equation: y = 2x + 3 Back

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