GCSE. Edexcel GCSE Mathematics A November 2005 Mark Scheme. Edexcel GCSE. Mathematics A 1387

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1 GCSE Edexcel GCSE Mathematics A 187 November 005 Mark Scheme Edexcel GCSE Mathematics A 187

2 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information please call our Customer Services on , or visit our website at November 005 Publications Code All the material in this publication is copyright Edexcel Ltd 005

3 1 (a) B for.88468( ) (B1 for or.047 or.88(...)) (b).88 1 B1ft ( to sf any answers to (a) that have sf ) (1) (8) (97) (44) (75) (96) (1) 1.8.8() (59) (915) (a) 1 share = 10 (=70) Total = '70' (+4+4) = B for trial between 1.6 and 1.7 inclusive (B1 for a trial between 1 and inclusive) B1 for a different trial between 1.6 and 1.7 exclusive B1 (dep. on at least one previous B1) for 1.7 NB trials should be evaluated to at least 1 dp truncated or rounded. 770 M1 for association of 10 with M1 for = (10 ) (+4+4) oe A1 cao (b) Average speed = 10 (hrs 40 mins) 10 = km/h = M1 uses speed = distance/time M1 (dep) evidence of converting time to a single unit. A1 for 77 to 81 (SC B for answer of 87.5) B1 for km/h (or other valid unit if consistent with their value)

4 4 (a) Reason 1 B1 for 'radius is perpendicular to tangent' oe (b) tan x = (= 0.464) 4.9 M1 for tan (x =) or sin (x =) 5.8 ' OP ' or P = tan -1 (0.464) = 4.89 cos (x =) 1.5 ' OP ' 5.8 M1 for tan oe correct use of inverse A1 for 4.9 (or better) (SC M1M1A0 for either 0.44(4..) or 7.6(5..) seen) (c) OP = OP = ( = ) = 1.78 PC = OP M1 for M1 for OR M1 for cos'4.9' = or sin'4.9' = OP OP M1 for OP = 1.5 cos' 4.9 or 5.8 sin'4.9' A1 for 1.7 to 1.8 B1ft (dep on OP >1.5) for adding or subtracting 5.8

5 5 (a) eg x =, y = ; x = 0; y = Any correct pairs of integers B for two correct pairs (B1 for one correct pair) (b) (1,1) (1,) (,1) B for three correct points (B for two correct points, B1 for one correct point) B If more than pts marked, mark best three then deduct 1 mark for each additional point SC B for indicating the correct region 6 8 = CE BC 1.5 = or or CE 8 = 9 6 BC 1. = 6 95 CE = BC = (i) 1 (ii) M1 for scale factor or 6 9 or 8 6 or or oe A1 cao for 1 A1 cao for 9 6 or 8 7 eg eqn(1) 4 then subtract eqn() 1y = 65 or eqn() 7 then subtract eqn(1) 5 1x = 9 eg 4x + 5 '5' = 1 x = y = 5 4 M1 correct full process to eliminate either x or y (condone one error) A1 cao either y = 5 or for x = M1 (dep. on 1st M1) for correct substitution of their found value into one of the eqn's A1 cao (both needed) 4

6 8 x + 4x + 4 = (x + )(x + ) x + 4x + 4 = 0, so (x + ) = 0, so x = is the only value of x that satisfies the equation oe Lisa is correct. See working column B for a complete solution (B1 for verifying that x = is a root or for factorising oe) 9 (a) Increase = (=88) % increase = '88' = or 14. or M1 for (=88) M1 for '88' OR 708 M1 for M1 for '114.19(...)' 100 A for 14 or 14. or (A1 for unrounded or truncated answer) (SC if A0 award B1 for an answer rounded or given to dp, 1dp, sf or nearest whole number) (b) (c) M1 for recognizing that (100 14)% is equivalent to = M1 (dep.) for 100 oe '100 14' A1 cao M1 for adding three consecutive numbers and dividing by A1 cao 5

7 10 (a) 8 5, , 0.5, (= 190, 60, 975, 00) Σfx = = 4095 Mean Σfx / Σf = 4095/ M1 for fx with x within intervals (including ends) at least two consistently M1 (dep) for fx consistently using midpoints M1 (dep on 1st M) for use of Σfx / Σf A1 for 7. cao (b) Frequency density (number of pictures per cm ) e.g. Width 0 to 10 height of rectangle.8(k) Width 10 to 5 height of rectangle.4(k) Width 5 to 40 height of rectangle (k) Width 40 to 60 height of rectangle.(k) Bars with correct heights, widths, label and scaling B for 4 rectangles with correct widths and heights (B1 for rectangles with correct widths and heights) B1 for correct label or key and consistent scaling (SC if 0/ award M1 if clearly using area or freq. density) 11 Vertices at (, 1), (, 4), (, 1) See working column B1 for all sides 1 B1 for correct orientation with two vertices almost correct B1 cao 1 5x 15 = 4y xy 5x + xy = 4y + 15 x(5 + y) = 4y + 15 x = 4 5 y M1 for expanding into four terms three of which are correct y M1(indep) for rearranging correctly to isolate x terms M1(indep) for factorising x from terms with one factor involving y A1 cao for final answer x = 4 5 y y oe 6

8 1 (a) D = kt 0 = k(40) k = 0/1600 (= ) 0 D = t M1 for D = kt seen or implied( k 1) 1600 M1(dep) for substitution or sight of k = 0/40 oe 0 A1 for D = t oe k = 0.018(75) truncated or rounded 1600 (b) M1 for ' ' 64 k ( ) k 1 seen (c) ( t = ) 1 ( 0 /1600 ) t = 640 = M1 for 1 ' k ' ( k 1) A1 for 4.4 to 5.9 (ignore 5.) 14 (a) 'minimum possible diameter is twice minimum possible radius' oe minimum possible diameter = = 8.cm See working column M1 for 'minimum possible diameter is twice minimum possible radius' or seen A1 for 8. cao (b) upper bound, in cm, for radius of outer circle is lower bound, in cm, for radius of inner circle is area, in cm, of shaded region = πr πr = π(15.85) π(14.15) = 51π k = 51 4 B1 for or 789.( ) seen B1 for or 69.0(1 ) seen M1 for using πr πr A1 cao (accept final answer left as 51π) 7

9 15 1 (1,5) on curve so pq = 5 (4,0) on curve so pq 4 = 0 4 pq = pq ( q ) = 5 q = 0 so q= 64 = 4 and p 5 5 = = q 4 q = 64 p = 1.5 q = 4 B1 for pq= 5 B1 for pq 4 = 0 B1 cao for both p = 1.5 oe and q = 4 (B for correct answer with no working) 16 (i) (b a) B1 for (b a) oe (ii) OD = OA + 4AB = a + 4(b a) 4b a M1 for OA + 4AB or OA + 'AC' oe A1 cao for 4b a 17 (5 + x ) (5 x ) 5 (5 + x ) 5 5 x B1 for (5+x)(5 x) B1 for 5(5+x) B1 cao for 5 x oe 5 8

10 18 (a) (19x + 8)(x 8) x = 8 x = 8/19 M1 for either ( ax + b )( cx + d ) with ac = 19 and bd = 4 or for a clear attempt to use b=± 14, c=± 4 ± a b b 4 ac A1 for either (19x + 8)(x 8) or for x = 14± A1 for 8 and 8/19 oe (accept 1.47 or better) with a= 19, (b) red = n blue = n + white = n + (n+) n + (n + ) + [n + (n + )] = 4n + 4 = 4(n+1)* Proof 1 B1 for n + (n + ) + [n + (n + )] (c) n 14 1 n = 4( n + 1) 4( n + 1) 81 n n = 4( n + 1) 4( n + 1) 81 81n(n + 4) = 14 16(n + 1) 4n + 4n = 4(n + n + 1) 4n + 4n = 4n + 448n + 4 Proof 5 M1 for multiplying two fractions A1 for n 1 n 4( n + 1) 4( n + 1) B1 for correct expansion of (n + 1) M1 for a valid method to eliminate fractions from an algebraic expression A1 complete proof oe 19n 14n 4 = 0* 9

11 (d) from (a) n = 8 so 4(n + 1) = 6 Proof 1 B1 for substituting n = '8' into 4(n + 1) or 8,10,18 seen (e) P(different colours) = 1 [P(RR)+P(BB)+P(WW)] = OR P(different colours) = [P(RB)+P(RW)+P(BW)] = OR P(different colours) =P(RR )+P(BB')+P(WW') = M1 for [P(RR)+P(WW)+P(BB)] or [P(RB)+P(RW)+P(BW)] 16 or [P(RR )+P(BB )+P(WW )] Allow algebraic fractions M1 (dep) for 1 [P(RR)+P(WW)+P(BB)] or [P(RB)+P(RW)+P(BW)] or P(R) [1-P(R)]+P(B) [1-P(B)]+P(W) [1-P(W)] Numerical values required 101 A1 cao for oe or 0.6( ) 16 0 SC B for oe or 0.65(1 ) 15 0