AQA GCSE Mathematics (0) Higher Tier Model Answers In general, the number of significant figures in an answer should not exceed the number of significant figures in the input data, or if this data has differing numbers of significant figures, the data with the lowest number of significant figures. Brian Daugherty Summer/Autumn 00
Paper - 0/H, 4 June 00 Question Question (n ) 80 = 6n 80n 60 = 6n 8n = 60 n = 0 Question 6 (xy ) = x (y ) = 8x y 6 Expression is approximate to Question C and D Question 4 50 5 0 = 50 00 =.5 Between 50 m and 50 m from B draw two arcs of cm and 5cm from B using compasses nearer to A than C draw a wide arc from A and from C such that they cross each other twice, once either side of a line connecting AC. Connect the points where they cross and this will be the line equidistant between A and C. more than 00m from the path Construct a line parallel to the path and cms away. Question 5 So required values of n are y 8y 4y(y ) 6 n < n <,, 0,,,, 4 a =, b = Hint: Since you know b is prime, work thru the primes from the bottom to find an appropriate value. = 8 and 8 does not divide into a so that is not the right answer. = 7 which does divide 54 by, and is a prime number. 54 5.7.45..5.5 So H.C.F. Question 7 = = 7 Something like, for example How often do you read for pleasure? with appropriate responses like, for example - daily, weekly, etc. The sample suggests that the fraction of pupils who read comics is 6 50 = 8 5 Therefore the estimate of pupils in the entire school who read comics is Question 8 8 000 = 0 5 7. 0 6 g
http://www.maccer.co.uk Question 0 No of grains of sand Question 9 6 0 4 g = 7. 06 6 0 4 = 7. 6 00 =. 0 0 4(m + ) + (m 5) 4m + + 6m 5 0m x + y = 9 () x + y = () First Spin 0.7 needs to be added to bottom branch Second Spin Likewise add 0.7 to the bottom branch of the top half. The bottom half needs 0. on the top branch and 0.7 on the bottom branch Identify the relevant branches - there is only one branch, i.e. the top one. Multiply along this branch which is the answer Question 0. 0. = 0.09 0 = 8 = (8 ) = = 4 6 44 0.5 = 6 ( ) times 6x + 9y = 7 () = 6 ( ) times 6x + 4y = (4) ( ) - ( 4) 5y = 5 y = 5 Substituting this into ( ) x + (5) = 9 x = 6 Question = Calculating the gradient from the coordinates of B and A 4 0 = 8 4 = y = x + x = y = x + (ii) x + 6x 6 (x + 8)(x ) Question 5 < t 0 : 0 < t 60 : (x + 8)(x ) = 0 T = 64 minutes Either or x + 8 = 0 x = 8 x = 0 x = 00 shoppers in total. Require the top 0 in the highest bracket, which contains 5 people. Since the bracket encompasses 0 mins, an estimate would correspond to the top 6 minutes of this period, i.e time above 60 + 4 = 64m
http://www.maccer.co.uk Question 4 40 (ii) Since opposite angles of a cyclic quadrilateral equal 80 y = 80 40 = 40 Since AD and CD are tangents, then the angles at A and C are both right angles. The other angle in both triangles equals 50, so angle ABC will equal 50 = 00 By the alternate segment theorem alternate angles gives BDA = DBC = since this is in an isosceles triangle BDC = Question 6 So Question 7 + 7 4 + 9 4 + 9 + 5 a = 5 ( + 8) + 8 + 8 + 8 0 + ( 8) 0 + ( 4) 0 + ( 4) 0 + 8 = 8 therefore DCB = 6 since opposing angles in a cyclic quadrilateral sum to 80 BAD = 64 Question 5 When W=, P=6 so W α P W = k P = k 6 k = 6 = 4 = W = P W = 5 = 5 = 5 = P P = = 7 P = 49 Probability both pick toffees = 5 0 4 9 = 0 90 Probability both pick chocolates = 0 9 = 6 90 Probability both pick mints = 0 9 = 90 Probability they pick sweets of the same type Question 8 (ii) = 0 90 + 6 90 + 90 = 8 90 = 4 45 BA = OA OB = a b MQ = MB + BQ = MB + BA = b + (a b) = a + b = (a + b)
http://www.maccer.co.uk 4 (iii) From above, both is a trapezium Question 9 OP = OB + BP = b + (a b) MQ and = a + b = (a + b) OP are parallel, so OMQP arc length = rθ = 70 60 π = 8π if s = radius of base πs = 8π s = 9cm Question 0 the same curve as original but shifted 45 degrees to the right The coordinates of P are P(5,) the same curve as original but all values doubled The coordinates of P are P(90,) Question x 0x + 8 So = (x 5) 5 + 8 = (x 5) 7 a = 5, b = 7
Paper - 0/H, 0 June 00 Question Using Pythagoras s Theorem Question x = +. x = 0.44 x =.m There are 4 shares in all, so Laura receives Question 7 of 000 4 = 7 4 000 = 500 x x + 7x Comment Too small 48 Too big.5.5 Too big.4 0.64 Too big. 8.67 Too small.5 9.48 Too small So answer is.4, to decimal place Hint : Remember to test down to the second decimal place. We had determined that the result lay between. and.4 but we needed to test.5 to decide whether it was to be rounded up or down when quoted to one decimal place. Question 4 If we consider the original figure to correspond to 00%, then the figure of 78.0 will correspond to 0%. So the original figure Question 5 = 78.0 00 = 76.50 0 Taking the mid-values of each range, the times will total to ( )+(5 6)+(7 7)+(9 8)+( 5)+( ) = 8 So estimate of mean Question 6 Question 7 7.8067 (ii) 7.4 = 8 0 = 7.6mins 6 < t 8 sin = 5 x x = 5 sin x = 9.5m x = 0m to sig figs (.8 0 5 ) (4.5 0 ).55 0 8 =.55 0 9 Hint : In standard form, there is only one number to the left of the decimal point Question 8 x 5 = x = 5 x = 8 x = 4 x + 8 < 9 x < x < 7 5
http://www.maccer.co.uk 6 Question 9 by extrapolation from the graph Question 0 (, ) a a a(a ) (ii) When a = -4.5, above expression becomes (ii) Question Question Upper Quartile = 9 mins Lower Quartile = mins 4.5( 0) = 45 (4x )(x + 5) 4x + 0x x 5 4x + 7x 5 x 5 x = x 5 = x y 5 y = y 5 ( ) = y 7 w = x + y x = w y x = w y So required limits are mins and 9 mins Question x 6 y 0 y x Question 4 Let Question 5 x = 0. 4 8 00x = 48. 4 8 00x = 48 + x 99x = 48 x = 48 99 = 6 A = C alternate angles B = D alternate angles and opposite angles at E mean opposite angles of triangles are equal Given that it is given that AB = DC Therefore triangles are congruent Note: to prove congruence only actually need to show equality of two angles and one side Question 6 (50 + 46 + 45) = 47 No - trend shown by graph is to level off and not reach 5% Various factors possible, e.g. sample size, ages, social class, ages, location, etc. Question 7 The length of line HF is given by So from DFH Question 8 Need to find AB in diagram HF = + 5 HF = 44 + 5 = 69 HF = tan DF H = 5 = 0.846 DF H =.04 AB = 8 + 5 (8)(5)(cos 8) AB = 8.70km
http://www.maccer.co.uk 7 Question 9 Question Volume of sand = π() 4 From πr h = 6π If dimensions are increased by a half Volume of the cone π( r) h = π() = 6π Volume remaining in cyclinder after inversion 6π 6π = 0π This volume will occupy a cyclindrical shape of height h, such that So x in the diagram will be Question 0 0π = π() h h = 0 9 = + = 5 cm (x + 4) x + 4x + 4x + 6 x + 8x + 6 Substituting for y in the circle equation x + (x + 4) = 6 x + x + 8x + 6 = 6 x + 8x 0 = 0 x + 4x 0 = 0 So volume of larger bottle = 7 8 Remember to insert the units Question Question π 9 4 r h 7 8 πr h 480 = 60ml x x + x = x(x ) (x + ) = (x + )(x ) x x x = x x = x = r = π(t r) r = πt πr r + πr = πt + r( + π) = πt + r = πt + + π x + 4x 0 = 0 Question 4 x = 4 ± 4 4..( 0) = 4 ± 56 = ±.74 = 5.74 or.74 5x + 4x x 9 (5x )(x + ) (x )(x + ) (5x ) (x )
http://www.maccer.co.uk 8 Question 5 The upper limit of coffee that will be dispensed could be 0.5ml The smallest size of cup could be 74.5ml The upper volume limit of cartons could be.5ml So maximum volume of liquid = 0.5 + (.5) = 7.5ml and this will be less than smallest size of cup, so liquid will never overflow
Paper - 0/H, November 00 Question x = 50 x = 5 x = 5 75 =.5 =.5 Question 4 80.75 = 40km The second stage will be 50 km This will take 0 minutes Average speed for second stage Question 5 Construct the 60 angle by = 50 0.5 = 00km/h Question The number of sticks goes up by 4 each time, so Diagram 5 has sticks i.e. 50th. diagram Question 4n + 4n + = 0 4n = 00 n = 50 5 00 = 5 00 (ii) Yes - For a fair dice, the expected values from 00 throws would be - red 50, blue 00 and green 50. The outcomes are compatible with these expected values The number of throws is too small to make definite conclusions. Using A as the center draw a wide arc thru and above AD Using the point where this arc cuts AD as a new center, and using the same radius, mark off an arc intersecting the previous arc. Connect A and the point where these two arcs intersect. Mark off 0 cms along this line To drop a perpendicular at B Using B as the center, draw an arc intersecting AD in two places Using these two intersections, use each in turn to draw two new arcs below AD which intersect with each other. Connect this latter intersection with B Question 6 After two years, the account will contain 500( + 0.) = 500(.) = 500(.) = 05 Note : If you are uncertain about the above procedure, you can always calculate the final value differently, year by year Easiest way would be 0 0 = 00 9
http://www.maccer.co.uk 0 Question 7 (x )(x + ) x + x 6x x 5x x 7x 8 (x + )(x 8) Question 8 5x + y = (5) x + 5y = (6) (5) 5x + 9y = 9 (7) (6) 5 5x + 5y = 5 (8) (8) - (7) 6y = 4 y =.5 Substituting this into (5) 5x 4.5 = 5x = 7.5 x =.5 Question 9 Question 0 x is the area to the right of, and including, the line x= y x is the area above, and including, the line y = x-, which is a line at 45 crossing the y-axis at y =. x + y 7 is the area below, and including, a line connecting the y-axis at y=7 and the x-axis at x=7. Question Using Pythagoras (ii) (ii) (iii) (iv) Question OP = + OP = 9 + = 0 OP = 0 x + y = 0 90 y y 0 = (x x 0 ) y = (x ) y = x + 9 + y = x + 0 cos x = BD 5 BD = 5 cos x (ii) BD = 5 = 0cm Using Pythagoras 0 = 6 + BC BC = 0 6 = 00 6 = 64 so 7 = x = BC = 8 64 = (4 ) = 4 So so sin y = 8 0 = 4 5 y =
http://www.maccer.co.uk Question Each bar of given width, each with height corresponding to frequency Need 0th. item (strictly speaking we need value of (0th + st)) So Question 4 Looking at t=, h =0 Trying (A) 700 + 7 50 = 75mins 0 0 = k k = 5 However this will not work for t=5, h=6.5 Trying (B) for t = 5, h = 6.5 which is valid and fot t=6, h = 90 0 = k k =.5 6.5 =.5 5 Question 6 ( + 7) 9 + 7 + 7 + 7 6 + 6 7 (ii) Substitute above result into LHS of given equation Second solution is because Question 7 ( + 7) 6( + 7) + 6 + 6 7 8 6 7 + = 0 7 + 7 + 7 = 6 5 P (red /Beth) = 4 5 P (red /Beth) = 4 5.5 = P(red/Amy and not red/beth) which again is valid So (B) fits the results 90 =.5 6 = 5. = 5 P(not red/amy and red/beth) Question 5 The angle will be double the angle subtended at the circumference 00 (ii) Because it is a cyclic quadrilateral y = 80 50 = 0 Because ABCD is a cyclic quadrilateral (ii) 9x + x = 80 x = 80 x = 5 DCA = 80 0 5 = 5 Since ACE is an isoceles triangle So EAC = 80 (5) = 50 EAD = 50 0 = 0 = 5. = 6 5 P(not red/amy and not red/beth) Question 8 Question 9 = 4 5 5 6 5 = 5 = 5 Graph A : y = (x ) Graph B : y = (x + ) Graph C : y = x Graph D : y = x + = x 6 x + 0x 8 (x + 4)(x 4) (x )(x + 4) = x 4 x
http://www.maccer.co.uk Question 0 Volume of original cone = π( )(0) 480π Volume of cone cut off (with height 5cm) = π(6 )(5) So volume of frustrum = 60π = 480π 60π = 40π For the cone πr (5) = 40π r = 40 = 6 5 r = 6cm
Paper - 0/H, 4 November 00 Question The largest angle is Question 7 60 = 6 0 Question 5 D - the increase of height in both segments is a straight line, i.e. the height is proportional to time. The change from one segment to the other is maybe smoother than expected but there is no other alternative offering two straight regions for the two separate segments of the bottle. Perimeter = πr Question Probability of vegetarian + 9 = π 4.5 + 9 = 4. + 9 =. = 0.8 0 constitutes a fifth of the ttal no. of pupils, so total no. of students = 0 5 = 600 Cylinder Question 6 y = 5 y = 5 y = Question 4 5(a c) + 4(a + c) 0a 5c + a + 8c a + c (x ) = 5x 5 x 6 = 5x 5 Question 7 x + 4 + 4x + 6 = 6(x + ) + 4(4x + ) = 4 x + 6 + 6x + 4 = 4 8x = 4 x = x = x = Using Pythagoras s Theorem OA = 6 +.5 x + 7 < x < 6 x < Therefore OA = 6 + 6.5 = 4.5 OA = 6.5 AB = 6.5.5 = 4cm
http://www.maccer.co.uk 4 Question 8 sin 48 = x 5. x = 5. sin 48 =.8cm Hint : Input data contains decimal place at most, so answer should contain no more than one decimal place From above, height of PQRS is.8 cm Area of PQRS Question 9 = 6.8.8 = 5.84cm Consider three consecutive numbers These sum to a, a +, a + a + which will always be divisible by Question 0 u = t + 5 Question corresponding to 0 on the vertical scale, the median is 00 (ii) Corresponding to 0 on the vertical scale is 9 Corresponding to 0 on the vertical scale is 06 So the inter-quartile range is 06 9 = George - who has a lower Inter-Quartile Range (ii) Brian - who has the lowest median Question Let So therefore Question 4 x = 0. 5 0x =. 5 000x = 5. 5 000x = + 0x 990x = x = 990 = 5 65 Question u = t + 5 t = u 5 7 0 9 x 0x 5 = 0 x = ( 0) ± ( 0) 4..( 5) 00 + 0 x = 5 ± 0 x = 5 ± x = 5 ± 5.477 x = 0.48, 0.48 0.0045 Hint : the index of - implies you shift the decimal point by three places - you just have to decide in what direction.7 0.75 0 5 = 0.8 0 = 8 0 Question 5 Length becomes Width becomes So percentage increase = 5 +.5 = 7.5cm 0 + = cm (7.5 ) (5 0) 5 0 00 = %
http://www.maccer.co.uk 5 Question 6 Question Area s = (0 + + 9 = 5 = 5 5 4 6 = 800 = 4.4cm Least distance would be 45m Maximum speed would be 5.5 m/s Corresponding time would be 45 5.5 = 8.s Question 7 To find the no. of kilowatt hours when costs become equal 9.60 + (n 5).0 =.5n 9.6 +.n 6.5 =.5n 0.n =. n = 5.5 Above this figure Alpha gasco becomes cheaper Question 8 Volume of solid cube Volume of square hole = 0 = 8000cm = 0 0 0 = 000cm Volume of that part of the circular hole which does not coincide with the square hole So volume reaining 5(π(4 )) = 60π = 8000 000 60π = 5497.4cm Question 9 4,57, 57,8 respectively various possibilities : sex of students, age etc. Question 0 So and x + + 5x x = (x ) + 5x(x + ) = ((x + )(x ) x + 5x + 5x = x x 6 x + 9x + 4 = 0 (x + )(x + 4) = 0 x + = 0 x = x + 4 = 0 x = 4 Question Question P(black and black) P(white and white) y = x + 4 x (x )y = x + 4 xy y = x + 4 x(y ) = y + 4 x = y + 4 y = 5 8.4 7 = 5 4 = 8. 7 = 8 Probilty of both balls being same color Question 4 Minimum point of is (0.5, -6.5) so for is (0.5, -.5) = 0 8 + 8 = 8 y = x x 6 y = x x equate the two equations x x 6 = x + x x 8 = 0
http://www.maccer.co.uk 6 Question 5 Using given formula base area height = 00 (5) height = 00 height = 00 5 = cm Length (L) of a diagonal across base is given by L = 5 + 5 = 50 L = 50 set up a right-angled triangle consisting of height, half the diagonal across the base and side x, and use Pythagoras ( ) 50 x = + x =.5cm