Edexcel GCSE Mathematics (1387) Higher Tier Model Answers

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1 Edexcel GCSE Mathematics (87) Higher Tier 004 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 Statements in italics are for information rather than a part of the answer

2 Paper (Non-Calculator), 8 June 004 Question (i) Question Set compasses to length PB and mark off an equal distance the other side of P (which I will call C). Using a slightly wider distance place point of compasses on B and mark off arcs above and below P. Place compasses on C and repeat operation. Connect the two points where these arcs intersect, to form required perpendicular. Question 6 Question Three required expressions are 7 is already prime LCM 7 66 πabc d, a, (c + d ) Question 7 Volume 0 0cm k k k (i) 4(x + ) + (x 7) Question 4x x 7x P(4 or 6) P(4) + P(6) Question 4 (c) (x + y)(x + y) x + xy + xy + 6y x + xy + 6y (d) (p + q) + (p + q) (p + q)(p + q + ) ( m 4 ) 4 m 8 6 (e) So H.C.F. t r t 4 6r t 6

3 Question 8 (i) Question 00.mm y x + 6 Gradient of BC (m) is given by 0.mm m() Question 9 m Area of triangle Area of square Therefore the side of square Therefore, perimeter Question cm 8 6cm 6cm 4 6 4cm When x - and y k 6y + x 6y x + y 6 x + k 6 ( ) (c) (i) Lines enclose area bounded by both axes, 6y + x (as shown on graph) and the line x. (, ) equation of BC is y x + 6 (c) The rule is that if opposite angles of the quadilateral equal 80 degrees then it can be circumscribed. This condition is always satisfied for rectangles. Question Each apex of triangle transforms Question (0, 0) (, ) ( 4, 0) ( 4, ) (, 4) (, 4) s (Note : Corresponding to 0 on vertical axis ) Box plot of form as shown for (c), except the box itself stretches from 6 (the lower quartile) to 4 (the upper quartile), with center line of (representing the median, corresponding to 0 on the vertical axis). Lines projecting either side extend to 9 on the left and 7 to the right. (c) Apart from straightforward comparin of items mentioned in, could al comment on the fact that the inter-quartile range is less for the girls, as is the range. Question 4 Time (t minutes) Frequency 0 < t < t 8 < t < t 0

4 Question On Jan st 000, t0, V pq 0 p 600 On Jan st 00 Question pq q q 4 q Question 6 V 600 ( ) 8 V 600 V 6. 6 Since ABC and ACB are equal, then ABC is isceles with AB AC. Since PB and PC are tangents, both radiating from P, they are of equal length. Since AP is shared by both triangles, AP B and AP C are congruent. By symmetry, angle BPC 0 degrees. Since triangle PBC is isceles, angle PBC 80 degrees. Using alternate segment theorem, angle BAC 80 degrees. Since ABC is isceles, angle ABC 0 degrees. Question 7 AC 6c 6a 6(c a) OP OA + AC OP 6a + 4(c a) 6a + 4c 4a a + 4c CB 6a OM OC + CB Therefore k (c) The area of hole as a fraction of the area of the large rectangle 8 ( + 0 ) ) ) ) 6 Therefore, area of card as fraction of large rectangle which as a percentage is Question 9 (i) Equation becomes % x x + 98 (x 7)(x 4) (x 7)(x 4) 0 OM 6c + (6a) 6c + a (a + 4c) Therefore O,P and M lie on a straight line, since (i) or x 7 0 x x 4 0 x 4 OM OP 7 n + 7

5 4 Assume that a probability of is possible 7 n + 7 (c) n + 7 which implies that n is non-integer. original assumption is wrong P (yellow) n n + 7 Therefore P(different colors) P(John/yellow)P(Mary/white) or P(Mary/white)P(John/yellow) ( 7 n + 7 n ) 4 n ( ) 7n (n + 7) 4 9 8(7n) 4(n + 7) 6n 4(n + 4n + 49) 4n 70n n n (d) Solution of above eqn., from n 4 (disregarding the other lution as unphysical) P (white) 7 n P(Mary/white) and P(John/white) 9 Question 0 (i) y sin x + y sin x Increase the amplitude (i.e. the y value) by a factor of Halve the wavelength (i.e. the x value)

6 Paper 6 (Calculator), June 004 (Note : When a number is given followed by several dots, this implies that the number is to be kept inside the calculator in full, and this full form is to be used for the entire calculation. You do not write it down in truncated form and then use this truncated form in later calcuulations - this will possibly introduce rounding errors. Question How much money, on average, do you spend on each visit to the canteen? (With relevant boxes, e.g.,.0,,.0 Question Shaded circle of radius cm, centered on Manchester Question F eb Mar Apr May 84 0% off the normal prices gives 80% of normal price Taking 0% off this 80% means a further deduction of % off normal prices giving a total deduction of 4% off normal prices Question 4 x x x So answer is between 4. and So answer is 4. to d.p. Question 0 9 No of calculations per second Question 6 Question No. of dots A constant first difference of 4 the general expression is 4n + k where k is a constant. Inspection shows k 6, general term is Question 8 Volume of the puck Mass of the puck 4n + 6 π(.8). cm π(.8).. 70g

7 6 Question 9 Using Pythagoras DG Question The volume (V) of a cone is given by V πr h DG.7m cos x 8 0 r V πh 0 π. x 6.9 therefore r... Question 0 to significant figures r.cm () () (4) - () Into () 6x y () 4x + y 9 () x 4y 66 () x + 9y 7 (4) y 9 y 4x + ( ) 9 4x 48 x Question 4 Complete first choice by inserting 0.4 at the bottom The second choice will be the same top and bottom, each of which is al identical with the first choice, i.e. 0.6 and 0.4 respectively (c) Total playing time of the CDs (0 4) ( 4.8) 90 So mean time of CDs ld 90 8mins Question since ratios are different, shapes are not similar Question Question When f8, S S α f S k f k 64 After three years, value 000( 0.) 000(0.7) Therefore S 8000 k f ( or S 8000 ) k 06.0 ( 0.) 4 (0.8) When f4 S

8 7 Question 6 Using Pythagoras Using the cosine rule So perimeter (.)(8.4) sin B 0 sin B 0 (.)(8.4) AC (.)(8.4) sin B AC AC cm to sig figs Question 7 Using Pythagoras ( ) 0 DE 60 + DE So required angle EDB is given by Question 9 Inserting y 6 into cos(edb) BD DE EDB. to d.p. x + y (x + 8) x + (x + ) x + 6x + 64 x + x + 0x + x 6x 9 0 x b ± b 4ac a x 6 ± 6 4()( 9) x , So radius of circle, to sig figs 9.9cm x + 6 x Therefore no lns. of x y 6 does not cut the curve Inserting y x into x + y x + (x ) x + 4x 8x + 4 x 8x 0 Question 8 Need to construct BDE BD is diagonal of ABCD Using Pythagoras First bracket gives (x + 7)(x ) 0 x x 7 Length of AE is given by BD BD cm and ( y 7 ) 4.8 cos 0 60 AE Second bracket gives AE 60 cos and x 0 x y () 4

9 8 Question 0 Upper bound in terms of m/s Upper bound in terms of km/h km/h Lower bound in terms of m/s Lower bound in terms of m/s Martin s speed km/h ( ) 4km/h Midway between the uper bound and lower bound (c) Age Group Total No. of people in sample Question 40 x 4 + x 40 x + x 4x 8 x 7 4x 6x 4x 9 x(x ) (x + )(x ) x x + Question Equating distance from x-axis to distance from (0, ) y (y ) + x y (y ) + x y y 4y x 4y x + 4 y 4 x +

10 Paper (Non-Calculator), 8 June 004 Question Cheese Topping Question 70 8g 70 4g Question () - (6) Inserting p 4 into (6) (80 x) p + q () p + q (6) p 8 p q q Question Question 4 (i) 0 0 p(q p ) pq p (p + ) (p ) p + 0 0p + 6 p Using results from, expression becomes Question 6 The exterior angles sum to 60 A single exterior angle of a hexagon will be Question There are several ways of doing this. From the way the line is presented, I would assume that this the method they are suggesting Choose a point O mewhere inside the angle. Set the compasses to OA and draw a circle with compass point on O, such that it cuts the line at B. Join B and O and extend to circumference of circle at C. Join A and C and this will be the perpendular to A (because angle at A will be the angle subtended by diameter BC). You can then bisect this right angle - with compass point on O, draw two arcs intersecting OC and OB. With compass points on these intersections draw two more arcs intersecting each other. Draw a line connecting this latter intersection and O. Could al use the following method, although you might not have too much space - Construct a perpendicular to the given line. Easiest to construct a perpendicular bisector. Set compasses to a width greater than half the length of the line. With the point of compasses on ends of line in turn, construct arcs above and below line that you have two sets of intersecting arcs. 9

11 0 Connect these intersections to form the perpendicular bisector. Next, set compasses to length between A and point where bisector crosses the line, Use this distance to mark off a point C the same distance along the bisector (using compasses, with its point on aforementioned point where bisector crosses line). Connect C with A. Question 8 Volume equals the area of the cross-section multiplied by length 4 7 4cm Question 0 Draw the line x : x will be the area to the right of this line Draw the line y x (i.e. a straight line thru the origin at 4 ): y x will be the area above this line. Draw the line y x + 6 (i.e. crossing the y-axis at y6 and with a negative gradient of : y x + 6 (i.e. x + y 6) will be the area below this line. Shade in area that satisfies all three inequalities above simultaneously - i.e. the area bounded by a triangle with vertices (,), (,4) and (,). (, 4), (, ), (, ), (, ) Question 9 (i) x 6 x x4 ( y 4 ) y inequality signs here indicate that points on lines are actually included within the region R Question D ut + kt ut D kt u D kt t (c) (d) (i) (t + 4)(t ) t t + 4t 8 t + t 8,, 0,,, Question < A 44 < A 00 < A 4 4 < A < A 6 60 < A 6 6 (c) Draw cumulative frequency table, using information from (d) (i) ( ) 7 9 (e) Draw a box stretching from 0 to 44. A vertical line within the box at 8 indicates the median. From both ends of the box project horizontal lines which terminates at a small vertical line at 4, to the left, and at a small vertical line at 4, to the right.

12 Question (c) BCA AC, being a diameter wii subtend a right angle at the circumference (at B), BCA ABD, DBC 90 BOA 0 this is twice BCA (from ) which is the angle AB subtends at the circumference - and the angle it subtends at center will be twice this Question 4 Question 7 Area of triangle Using Pythagoras Question x + (x) x + 4x x ( ) x 4 x cm PSQR because they are parallel and both have their end points on the same two parallel lines PQSR from the same reaning the third side is shared (the diagonal of the parallelogram) SP Q is obtuse SRQ is al obtuse But for a cyclic quadrilateral, opposite angles need to sum to 80 Question 6 Question cm Type of Proportionality y is directly proportional to x y is inversely proportional to x y is proportional to the square of x y is inversely proportional to the square of x Question 9 n (i) n + (n + ) Graph Letter D A B C Each vertical square corresponds to a unit of 0.8 Price Frequency 0 < P 40 < P < P < P 40 Range Height of Column 0 < P 6 < P < P < P using the same convention as before where each (small) vertical square corresponds to 0.8, the heights of each column would be respectively 0 squares, squares, squares, 6 squares n + n + 0n + 0 is always even, and 0n + must always be odd n ((n + )) n(n + ) n + n n(n + ) Since either n or (n + ) will be even, the whole expression will be even

13 Question 0 Question a, b (i) Area of sector πx 0, 60, 70 Arc length of this sector πx (c) This will be the same as when x 90, i.e. A πx from Question (i) P Q q p al now A πrx r A πx πx r πx r x V πr h OS OP + P Q π ( x ) h OS p + (q p) OS p + q p OS p + q OS (p + q) and since VA, from the question ( ) ( ) x πx π h 9 7 h h 7 cm RS OS OR RS (p + q) p thus RS is parallel to OQ RS q Question x + + x x (x ) + (x + ) x + x + x 4 x 4 Using (x ) (x + )(x )

14 Paper 6 (Calculator), 9 November 004 Question 40% of % of percentage of all students who went to party Question corresponds to 00cm If length is 48cm, then width cm Question x x + 4x x 4.4 to dec place must calculate expression for 4. to see whether the answer is 4. or 4.4 to one decimal place Question 4 (c) 4(x + ) ( x) 8x x 0x x 0 p 4pq p(p q) x + 7x + 6 (x + )(x + 6) Question Value (c) 8000 (.0) 96 A x(.0) 88 x You could always work out the compound interest the long way, but I would recommend mastering this method Question 6 Average Speed m/s number of figures depends on your calculator 9.4 m/s the input data has three significant figures, answer should have a maximum of three significant figures Question 7 Question tan x. 0.4 x cm to dec. place

15 4 Question 9 Using Pythagoras Question 0 Question CD CD CD CD.7 to dec. pl. This has a gradient of y r + t sin x r t sin x sin sin to sig figs r + r 4 r + r 4(r ) r + 4r 0 r + r r L : y x + gradient of L is al, giving At (,) equation of L is y x + c () + c 6 + c c 6 4 y x 4 Question No of videos watched by all 60 members No of videos watched by all boys. 40 No of videos watched by girls 68 6 So mean number watched by girls The lists must be of the same size p and q are equal Question Equating corresponding sides of similar triangles Question 4 CD 0 4 CD 0 4.cm ED ED ED 4.8 ED 7.cm x + x + 4 completing square ( x + ) ( x +. ) or x + ±. x ±. x to sig figs x to sig figs When x0, y which is not prime

16 Question Using Pythagoras on ABC AC AC 74 Using Pythagoras on ACG AG ( 74) AG to sig figs Need to calculate angle between AG and AC sin( GAC) 9. Question 8 For Upper Bound of T L.6 g 9.7 For Lower Bound of T L. g 9.8 Subtracting expression for Lower Bound from expression from Upper Bound, and multiplying by 6.8 Question to sig figs Question 6 GAC to dec pl. T α R T kr Inserting given data k(0) Question 0 4x 9 x x + (x + )(x ) (x )(x ) x + x k 0 k 6 (4 0) k (0) 40 could always use a calculator if you must When R 0 Question 7 T 40 (0) T 0 mins Scale factor volume of Y 0 ( ) 4 00 cm Need P ((Gary W ins and Gary Doesn t W in) OR (Gary Doesn t W in and Gary W ins)) (0. 0.4) + (0.4 0.) (0. 0.4) 0.49 Probability of Gary losing 0. There are only three scenarios to be considered - the symmetrical situation where they both win one and lose one, or when both games are drawn Need P ((Gary W ins and Gary Loses) OR (Gary Loses and Gary W ins) OR (Both Draws)) (0. 0.) + (0. 0.) + (0. 0.) 0.

17 6 Question x + y 00 i.e. x + y 0 is a circle based on the origin with a radius of 0 y x 4 is a straight line with a gradient of. and intersects the y-axis at or similar (c) Completing the square (d) Completing the square (6.4, 7.7), ( 4.6, 8.9) x + 6x (x + ) 9 q 9 y 4y (y ) 4 x + 6x + y 4y 87 0 (c) If height of XYZ is h Area of Y XW XY XW sin Y XW Y W h sin Y XW Area of W XZ Y W h Y W h XY XW XY XW XZ XW sin W XZ W Z h sin W XZ Since sin Y XW sin W XZ W Z h W Z h XZ XW XZ XW Y W h XY XW W Z h XZ XW Y W XY W Z XZ Y W W Z XY XZ (x + ) 9 + (y ) (x + ) + (y ) 00 x and y 0 satisfy this equation (x + ) translate x 6 by units in a negative direction, and (y ) translating y 8 by units in a positive direction Question Area of triangle 0 sin 4 60 sin m to sig figs Area of ABC Area of ACD BC h CD h Area of triangle ABC Area of triangle ACD BC h BC CD h CD