Link to past paper on AQA website: www.aqa.org.uk The associated question paper is available to download freely from the AQA website. To navigate around the website, choose QUALIFICATIONS, GCSE, MATHS, MATHEMATICS, MATHEMATICS A (LINEAR), KEY MATERIALS. Question 1 350 represents 100%. We want to reduce this by 18% to 82% (100 18 = 82). So if we divide 350 by 100 we will get 1% and then multiply up by 82 to get 82% x 82 = 287 Alternatively: 18% of 350 = x 350 = 63 350 63 = 287 Question 2 a) 5 6 = 15625 b) 5 = 5 5 2 = 25 5 3 = 125 5 4 = 625 5 5 = 3125 and so on Each of these numbers end in 5 The reason that all the numbers end in 5 is because 5 x 5 is 25 so when you multiply any number that ends in 5 by 5 you will get another number that ends in 5. eg 625 x 5 = 3125 www.chattertontuition.co.uk 0775 950 1629 Page 1
Question 3 Plan Side elevation Front elevation www.chattertontuition.co.uk 0775 950 1629 Page 2
Question 4 a) 2Q is two quarter circles the radius of the circles is the same as the width of the rectangle the length of the rectangle is the same as two radii 2 quarter circles would fit inside one rectangle with some spare space so 2Q R b) we have 3 rectangles and 4 quarter circles area = 3R + 4Q cm 2 c) if we take off another quarter circle then we would have R 2Q and this would look like half of the shape we want. We need 2 x (R 2Q) = 2R 4Q Alternatively If we make a square surrounding this shape of area 2R and then cut away a quarter circle in of the four corners we would have the desired shape 2R 4Q Question 5 a).. = 1.897058824 b) the numbers given were all to 2 decimal places so 2 decimal places would be appropriate for our answer 1.90 Question 6 subtract 3x from both sides 4x 9 = 5 add 9 to both sides 4x = 14 divide both sides by 4 x = 3.5 www.chattertontuition.co.uk 0775 950 1629 Page 3
Question 7 Subject French German Spanish Total Ratio 5 1 3 9 Degree 200⁰ note 1 40⁰ note 2 120⁰ note 3 360⁰ The total of the ratios is 9, 360 9 = 40⁰ Note 1: 5 x 40 = 200⁰ Note 2: 1 x 40 = 40⁰ Note 3: 3 x 40 = 120⁰ as a check make sure that the total degrees add up to 360: 200 + 40 + 120 = 360⁰ Question 8 We can draw a line going North East from A (this will be on a bearing of 045⁰). Draw another line from B going on a bearing of 310⁰ from B. We always measure bearings by going clockwise from the North. This is the same as 50⁰ anti-clockwise. Where these two lines meet is our point C. www.chattertontuition.co.uk 0775 950 1629 Page 4
Question 9 a) The first and second CD will cost 7.99 but the third will cost 3.99 under the special offer. The fourth CD will cost 7.99. The post and packing charge for 4 CDs is 3.99. Total cost to Amy is 7.99 + 7.99 + 3.99 + 7.99 + 3.99 = 31.95 b) if you buy 6 CDs together then 2 of these will be half price (at 3.99). The post and packing charge will be free. Total cost is (4 x 7.99) + (2 x 3.99) = 39.94 The total cost previously was 18.97 + 31.95 = 50.92 Saving = 50.92-39.94 = 10.98 Question 10 x x 3 + 5x Comment 5 5 3 + (5 x 5) = 150 Too high 4 4 3 + (5 x 4) = 84 Too low 4.5 4.5 3 + (5 x 4.5) = 113.625 Too low 4.7 4.7 3 + (5 x 4.7) = 127.323 Too low 4.8 4.8 3 + (5 x 4.8) = 134.592 Too high 4.75 4.75 3 + (5 x 4.75) = 130.922 Too high We can see that the answer lies between 4.7 and 4.8. I have then tried 4.75 and found this to be too high. That means that solution is closer to 4.7 than to 4.8 x = 4.7 (to 1 decimal place) www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 11 a) the common difference is 5 so we know that the nth term involves 5n writing out the sequence with the 5 x table below we can see that we need to just subtract 1 each time to get from the 5 x table to our sequence 4 9 14 19 24 n 5 10 15 20 25 5n -1-1 -1-1 -1-1 So our nth term is 5n 1 b) 0.5, 1 and 1.25 are all terminating decimals. That means that the decimal places do stop, they don t go on forever 4 th term will be = = 1.4 (this is a terminating decimal) 5 th term will be 6 th term will be = = = 1.5 (this is a terminating decimal) = 1.571428571428 (this is NOT a terminating decimal) 6 th term is 1.5 7 1 4 2 8 and this is the first one that is not a terminating decimal (it recurs) www.chattertontuition.co.uk 0775 950 1629 Page 6
Question 12 a) we can draw this line by plotting three points and joining them up or by using gradients and intercepts plotting 3 points x -3 0 3 y (2 x -3) + 1 = -5 (2 x 0) + 1 = 1 (2 x 3) + 1 = 7 Plot these 3 coordinates (-3, -5), (0, 1), (3, 7) and join the 3 points with a straight line Gradients and intercepts y = mx + c where m is the gradient and c is the y intercept In this case the intercept is 1 (where the graph crosses the y axis) and the gradient is 2 (level of steepness of the graph) As the gradient is 2 this means we go up 2 for every 1 we go across www.chattertontuition.co.uk 0775 950 1629 Page 7
Question 13 a) Score 1 2 3 4 5 Frequency 19 37 21 12 11 Relative frequency b) Yes they do suggest the spinner is biased. We would expect to get around 20 of each number (⅕ of 100 is 20). So the relative frequency for 2 is a bit high and for 4 and 5 it is a bit low. Question 14 To find the shaded area we start with the area of the square and then subtract the area of the two quarter circles Area square 6 x 6 = 36 cm 2 Area quarter circle Area of full circle = πr 2 so area of quarter circle is πr 2 4 π x 3 2 4 = 7.069 cm 2 shaded area = 36 (2 x 7.069) = 36 14.138 = 21.9 cm 2 Question 15 Boys Girls Total Left-handed 3 note 6 4 note 4 7 note 2 Right-handed 11 note 7 12 note 5 23 note 3 Total 14 note 1 16 note 1 30 Note 1: there are 2 more girls than boys and the total is 30 so boys must be 14 and girls 16 Note 2: we are told that there are 7 left-handed pupils Note 3: 30 7 = 23 Note 4: a quarter of the girls are left-handed. A quarter of 16 is 4. Note 5: 16 4 = 12 Note 6: 7 4 = 3 Note 7: 14 3 = 11 Check that all columns and rows add up correctly www.chattertontuition.co.uk 0775 950 1629 Page 8
Question 16 in order to subtract (or add) fractions you must have a common denominator (number on the bottom) the common denominator here could be 6 multiply the first fraction by 3 on the top and the same on the bottom multiply the second fraction y 2 on the top and the same on the bottom = 3 = 3 multiply both sides by 6 3(x + 3) 2(x 2) = 18 expand the brackets 3x + 9 2x + 4 = 18 group terms x + 13 = 18 subtract 13 from both sides x = 5 to check put this value of x back into the original equation = = 4 1 = 3 Question 17 we are given the figure of 324.80, this represents the bill including VAT so is 116% we want to know what 100% would be so we divide by 116 to get 1% then multiply by 100 324.80 116 x 100 = 280 Question 18 a) a two point moving average is appropriate because the data comes in twice a year b) (82300 + 4700) 2 = 43500 www.chattertontuition.co.uk 0775 950 1629 Page 9
Question 19 a) as the ball is replaced this means that the probabilities for the second ball will be the same as the probabilities for the first ball First ball Second ball 0.6 Blue 0.6 Blue 0.4 White 0.6 Blue 0.4 White 0.4 White b) we could have blue and blue or white and white with probability when we say and we multiply and when we say or we add (0.6 x 0.6) + (0.4 x 0.4) = 0.36 + 0.16 = 0.52 www.chattertontuition.co.uk 0775 950 1629 Page 10
Question 20 Closer to AB than to AD we need to construct an angle bisector on A. Open the compass and with the point of the compass on A mark a point on both AB and AD. These points will be the same distance away from A. Then from these points mark an arc. Where these two arcs meet draw a straight line joining this point to A. Closer to D than C we need to construct the perpendicular bisector of CD. Open the compass to more than half of the length of that line and from each end draw an arc above and below the end. Where these arcs meet join them up with a straight line. www.chattertontuition.co.uk 0775 950 1629 Page 11
Question 21 a) we have a right angled triangle so can use basic trigonometry (SOHCAHTOA) labelling all the sides from the point of view of the angle we have opposite (0) and adjacent (a) opposite = 4 cm and adjacent = 13 cm SOHCAHTOA so use TOA tan y = = take the inverse tan of both sides (tan -1 ) y = tan -1 ( ) = 17.1⁰ b)often with these questions you are using Pythagoras Theorem on more than one triangle we are trying to get AG so we will consider the right angled triangle AEG we have length AE (4 cm) but not EG however we can work out EG from the right angled triangle EFG (where EF is 5 cm, and FG is 12 cm) E 5 cm G 12 cm F by Pythagoras: EG 2 = 12 2 + 5 2 = 144 + 25 = 169 square root both sides EG = 13 cm Now we have right angled triangle AEG A 4 cm G 13 cm E so again by Pythagoras: AG 2 = 13 2 + 4 2 = 169 + 16 = 185 square root both sides AG = 13.6 cm www.chattertontuition.co.uk 0775 950 1629 Page 12
Question 22 the negative scale factor means that the shape gets inverted coordinate (2, 1) is 2 units to the right of the centre and 2 units up from the centre. After applying a scale factor of -½ this point will now be 1 unit to the left and 1 unit down coordinate (4, 1) is 4 units to the right of the centre and 2 units up from the centre. After applying a scale factor of -½ this point will now be 2 units to the left and 1 unit down coordinate (2, 5) is 2 units to the right of the centre and 6 units up from the centre. After applying a scale factor of -½ this point will now be 1 unit to the left and 3 units down a check on the enlargement is given by drawing tramlines that join corresponding points (shown as red dashed lines). These tramlines should meet at the centre of enlargement Question 23 a) x 6 + -2 = x 4 b) x 8 --4 = x 12 c) everything inside the brackets is to the power of 3 3 3 x x 2 x 3 x y 3 = 27x 6 y 3 www.chattertontuition.co.uk 0775 950 1629 Page 13
Question 24 For each category of mass we can calculate the frequency Frequency = frequency density x class width From this we can then work out the mean. We multiply each midpoint of the class width by the frequency. We then divide the total of these by 100 to get the mean. Mass (grams) Class width Frequency density Frequency Midpoint Frequency x midpoint 16 20 4 3 4 x 3 = 12 18 216 20 22 2 8 2 x 8 = 16 21 336 22 24 2 11 2 x 11 = 22 23 506 24 25 1 23 1 x 23 = 23 24.5 563.5 25 26 1 19 1 x 19 = 19 25.5 484.5 26-30 4 2 4 x 2 = 8 28 224 Total 100 2330 Mean = 2330 100 = 23.3 Question 25 Let R 0.0 37 = 0.037037037037 there are 3 repeating numbers so we multiply R by 1000 (if there had been two it would have been by 100) 1000R = 37.037037037 remember 1R = 0.037037037 1000R 1R = 999R = 37 999R = 37 divide both sides by 999 R = 37 goes into the top and the bottom so this can be simplified to this was difficult to know but given this is a calculator paper we could use our calculators in Maths mode and put in the fraction www.chattertontuition.co.uk 0775 950 1629 Page 14
finally we can check our answer by dividing 1 by 27 Question 26 We have to give our answers to two decimal places (this implies that the equation won t factorise). We need to use the quadratic formula (given in the formulae sheet) a = 1, b = -2, c = -6 or = 3.65 or -1.65 Put both of these back into the original equation to check that they work. They don t give 0 due to rounding but they do give a very small number close to 0. Question 27 ABCD forms a cyclic quadrilateral. A property of cyclic quadrilaterals is that opposite angles add up to 180⁰. let angle ABC = x⁰ then angle ADC = 180 x angles on a straight line add up to 180⁰ angle ADE = 180 angle ADC angle ADE = 180 (180 x) = 180 180 + x = x so angle ADE = angle ABC www.chattertontuition.co.uk 0775 950 1629 Page 15
Question 28 a) 6.05 cm this is the same as saying to the nearest tenth or 0.1 of a cm b)the area of the triangle will be as big as possible if all the measurements are as big as possible length AB 7.05 cm length BC 8.05 cm angle B 62.5⁰ area of triangle = ½absinC where C is the angle between the two sides a and b (as given in formulae sheet) area = ½ x 7.05 x 8.05 x sin 62.5⁰ = 25.2 cm 2 www.chattertontuition.co.uk 0775 950 1629 Page 16
Question 29 We have to use the substitution method. The elimination method won t work as one equation has x and y, the other has x 2 and y 2 substitute the equation for x into the second equation (3 + 2y) 2 + 2y 2 = 27 expand (3 + 2y)(3 + 2y) + 2y 2 = 27 9 + 4y 2 + 6y + 6y + 2y 2 = 27 group terms 6y 2 + 12y + 9 = 27 subtract 27 from both sides 6y 2 + 12y 18 = 0 divide by 6 y 2 + 2y 3 = 0 factorise (y 1)(y + 3) = 0 if two things multiply to give 0 then either one of them must equal 0 If y 1 = 0, then y = 1 If y + 3 = 0, then y = -3 now we have y, we can substitute this value back into the first equation to get x when y = 1 x = 3 + (2 x 1) = 3 + 2 = 5 when y = -3 x = 3 + (2 x -3) = 3 + -6 = 3 6 = -3 x = 5, y = 1 or x = -3, y = -3 to check put these values into the second equation: 5 2 + (2 x 1 2 ) = 25 + 2 = 27 (-3) 2 + (2 x (-3) 2 ) = 9 + (2 x 9) = 9 + 18 = 27 www.chattertontuition.co.uk 0775 950 1629 Page 17
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