Link to examining board: http://www.edexcel.com/migrationdocuments/qp%20current%20gcse/june%202009/1380_3h_que_20090518.pdf At the time of writing you will be able to download this paper for free from the website. These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or an organisation and would like to purchase these solutions please contact Chatterton Tuition for further details. Question 1 Walk Car Other Total Boy 15 25 note 1 14 54 Girl 22 note 2 8 note 4 16 46 note 3 Total 37 33 note 5 30 note 6 100 Note 1: 54 (15 + 14) = 54 29 = 25 Note 2: 37 15 = 22 Note 3: 100 54 = 46 Note 4: 46 (22 + 16) = 8 Note 5: 25 + 8 = 33 Note 6: 14 + 16 = 30 Check that total of bottom row makes 100 (37 + 33 + 30 = 100 ) b) There were 100 children in total and 37 of these walked to school so the probability that this child walked to school is Question 2 group all the terms 2x + 8y b) 2 compasses would cost 2 x c pence = 2c 4 rulers would cost 4 x r pence = 4r total cost would be 2c + 4r www.chattertontuition.co.uk 0775 950 1629 Page 1
Question 3 x -2-1 0 1 2 3 Y -11-7 note 1-3 1 note 2 5 note 3 9 Note 1: (4 x -1) 3 = -4 3 = -7 Note 2: (4 x 1) 3 = 4 3 = 1 Note 3: (4 x 2) 3 = 8 3 = 5 b) Question 4 substituting P with 50 50 = 4k -10 rearranging with k on the left hand side 4k 10 = 50 add 10 to both sides 4k = 60 divide both sides by 4 k = 15 b) substituting n with 2 and d with 5 y = (4 x 2) (3 x 5) y = 8 15 = -7 www.chattertontuition.co.uk 0775 950 1629 Page 2
Question 5 The point 0 is the origin which is the coordinate (0,0) b) Shape P has moved 3 units to the right and 1 unit down. The is known as a vector translation where the vector is given by Question 6 the shape is a rectangle so the opposite sides are equal in length to each other so we have 4x + 1 = 2x + 12 b) subtracting 2x from both sides (so that we only have x terms on one side of the equation) 2x + 1 = 12 subtract 1 from both sides of the equation 2x = 11 divide both sides by 2 x = 5.5 cm c) now we know that x = 5.5cm, we know the length of each side of the rectangle. The shorter side is length 5.5cm, the longer side is (4 x 5.5) + 1 = 22 + 1 = 23 cm Perimeter (length all the way around the rectangle) is 5.5 + 23 + 5.5 + 23 = 57cm www.chattertontuition.co.uk 0775 950 1629 Page 3
Question 7 3.22 x 4.8 is roughly 3 x 5 = 15 so our exact answer will be 15.456 (using the same digits as given) Also if we know that 322 x 48 = 15456 then we need to move the decimal point 2 places to get 3.22 and then a further one place to get 4.8. At the moment the decimal point is after the 6 (15456.) so it will now need to be after the 5 (15.456) b) 0.322 x 0.48 is roughly 0.3 x 0.5 = 0.15 so our exact answer will be 0.15456 Also we need to move the decimal point 3 places to get 0.322 (from 322) and a further 2 places to get 0.48 (from 48) so we are moving it 5 places in total to 0.15456 (from 15456) c) 15456 4.8 is roughly 15000 5 = 3000 so our exact answer would be 3220 Also If we first rearrange our original given equation so that 15456 48 = 322 and see how this compares to our new equation 15456 4.8 15456 has been unchanged but we are now dividing by 4.8 instead of 48 so our answer will be 10 times bigger than before. Answer is 3220 Question 8 divide both sides by 2 x 2 = 36 square root both sides x = 6 b) 72 2 36 6 6 2 3 2 3 so 72 = 2 x 2 x 2 x 3 x 3 = 2 3 x 3 2 www.chattertontuition.co.uk 0775 950 1629 Page 4
Question 9 b) Question 10 1 litre = 1000 mililitres multiply by 40 40 litres = 40,000 mililitres Water leaks out at the rate of 125 mililitres per second. We need to divide 40,000 by 125. If we first put this calculation as a fraction then we may be able to cancel it down:, =, =, = 320 seconds Question 11 62.5 (this is the smallest it can be and still be rounded to 63 (to the nearest centimetre)) b) 63.49 (this is the largest it can be and still be rounded to 63 (to the nearest centimetre)) (63.5 would have also been acceptable in this type of question) www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 12 The first condition leads to us drawing a circle of radius 4cm with B as the centre The second condition leads to us drawing the angle bisector of angle A (which gives the locus of the points that are equidistant from AB and AC) Both must be drawn using a compass and ruler, and the examiners will be looking for correct construction arcs. Question 13 Which of the following types of magazines have you read in the past month (place a tick in the relevant box(es))? None Education Fashion Sport Other b) How many magazine s have you read in the past month (place a tick in only one box)? 0 1 3 4 6 7 9 More than 9 www.chattertontuition.co.uk 0775 950 1629 Page 6
Question 14 round each number to just one significant figure. in order to get rid of the decimal on the bottom multiply the top and the bottom by 100 = = 700 x 40 = 28,000 Question 15 6.4 x 10 4 (as we have effectively moved the decimal point 4 places) b)1.56 x 10-5 (we have made 156 smaller by 100 so to compensate we need to make -7 bigger by 100) Question 16 Find something that goes into both terms. This is 2x. 2x(2x 3y) b) Find two numbers that multiply to give -6 and add to give +5. The two numbers are +6 and -1. (x + 6)(x 1) www.chattertontuition.co.uk 0775 950 1629 Page 7
Question 17 b) to find the median we find half of the total frequency. Half of 120 is 60 so draw a line across from 60 on the vertical axis until it meets the curve. Then drop the line down to meet the x axis. This meets the x axis at 240. The line is shown above in red dashes. 240 is an estimate of the median c)the women had a lower median than the men (205 compared to 240) so the women spent less money during their summer holidays than the men did. Question 18 A tangent will always make a right angle where it meets the radius. Hence triangle OAD is a right angled triangle with angle OAD = 90⁰. Angles in a triangle add up to 180⁰ so angle AOD = 180 (36 + 90) = 180 126 = 54⁰ b) i) angle ABC = half of angle AOC = ½ of 54⁰ = 27⁰ ii) angle at centre is twice the angle at the circumference www.chattertontuition.co.uk 0775 950 1629 Page 8
Question 19 we simply look to see where the two lines meet. This will give us the solutions to the simultaneous equations: x =2, y = 3 b) Any line which is parallel to y = ½x + 2 will have the same gradient as this line. The gradient is ½. We now need to find the y intercept. This is straightforward since they have given us the coordinate (0,4) which is the point where the line meets the y axis. The y intercept is 4. Our new line is y = ½x + 4 Question 20 subtract t from both sides (so that we have t on just one side of the equation and the side where it will be positive) 2t + 1 12 subtract 1 from both sides 2t 11 divide both sides by 2 t 5.5 b) t has to be less than 5.5 but also has to be a whole number. The largest it can be is 5. Question 21 M L 3 we can replace with = as long as we now have a constant k M = k x L 3 substituting in the values of l and m we are given, 160 = k x 2 3 160 = k x 8 k x 8 = 160 divide both sides by 8 k = 20 we have M = 20 x L 3 When L = 3 M = 20 x 3 3 M = 20 x 27 = 540 www.chattertontuition.co.uk 0775 950 1629 Page 9
Question 22 First we must calculate the frequency density for each class of data. Frequency density = frequency class width Insert two new columns Length (x) minutes frequency Class width Frequency Density 0 x 5 4 5 0.8 5 x 15 10 10 1 15 x 30 24 15 1.6 30 x 40 20 10 2 40 x 45 6 5 1.2 We can see that the highest frequency density is 2 so we can now decide on the scale (4cm to every 1 of frequency density) (we use most of the space given but also have a scale that is easy to plot) www.chattertontuition.co.uk 0775 950 1629 Page 10
Question 23 1 st game 2nd game win 0.5 0.3 win draw 0.5 0.2 lose 0.3 draw 0.5 0.3 win draw 0.2 lose 0.2 0.5 win lose 0.3 draw 0.2 lose b)he will win both games if he wins the first game and wins the second game 0.5 x 0.5 = 0.25 Question 24 to prove that any two triangles are congruent we must prove one of the following: all three sides are the same length (SSS) two sides and the angle in between are the same (SAS) two angles and one side are the same (AAS) or Right angle, hypotenuse and one other side (RHS) this is the one I will use side AB = side AC (as triangle is equilateral) hypotenuse side AD = side AD (as same line) other side we know the triangles are right angled (given in question) right angle These three points prove that the triangles ADC and ADB are congruent b) As ADC and ADB are congruent so side BD = side DC = ½ of BC Now BC = AB (as triangle ABC is an equilateral triangle) So BD = ½BC = ½AB www.chattertontuition.co.uk 0775 950 1629 Page 11
Question 25 u = 2½ = v = 3⅓ = = 1 u = 1 = 1 x = = 1 v = 1 = 1 x = + = + = + = = multiply both sides by f 1 = multiply both sides by 10 10 = 7f divide both sides by 7 = f f = 1 b)subtract from both sides of the equation - multiply both sides of the equation by u 1 = - multiply both sides by f f = u - multiply both sides by v www.chattertontuition.co.uk 0775 950 1629 Page 12
fv = uv uf factorise the left hand side to get u on its own fv = u(v f) divide both sides by (v f) = u Question 26 we start with y = f(x), the curve has been translated (moved) 4 places in the positive x direction so the new curve is y = f(x - 4) note that when the transformation effects the x it effects it in the opposite way to what you would think if the original equation was y = x 2 then the new equation would be y = (x 4) 2 b) the curve has had two transformations. It has been stretched by a scale factor of ½ parallel to the x axis (squashed), and stretched by a scale factor of 3 parallel to the y axis If you found this paper helpful then visit www.chattertontuition.co.uk/maths-revision-papers where you will find plenty more. It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by Edexcel. In addition these solutions may not necessarily constitute the only possible solutions. www.chattertontuition.co.uk 0775 950 1629 Page 13