Grade Boundaries for June 2001 Examinations
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Grade Boundaries for June 2001 examinations
The table below gives the lowest raw marks for the award of the stated uniform marks (UMS):
Module UMS 120 105 90 75 60 45 Pure Mathematics P1 79 71 62 53 45 37 JUNE 2001 Pure Mathematics P2 71 62 53 45 37 29 AS/A LEVEL MATHEMATICS Pure Mathematics P3 78 70 62 54 46 38 Pure Mathematics P4 68 59 52 45 38 31 NUMERICAL ANSWERS Mechanics M1 82 73 63 53 43 33 Mechanics M2 73 63 54 46 38 30 6405 Pure Mathematics P1 Mechanics M3 66 58 50 42 35 28 6406 Pure Mathematics P2 Mechanics M4 73 63 54 45 37 29 6407 Pure Mathematics P3 Statistics T1 69 61 53 46 39 32 6408 Pure Mathematics P4 Statistics T2* 100 91 81 71 61 51 6409 Mechanics M1 Statistics T3 73 65 56 48 40 32 6410 Mechanics M2 Statistics T4* 102 89 73 57 42 27 6411 Mechanics M3 6412 Mechanics M4 Decision Mathematics D1 82 76 70 64 59 54 6413 Statistics T1 *All marks out of 100, except T2 and T4 which are out of 125. 6414 Statistics T2 6415 Statistics T3 Module 80 70 60 50 40 6416 Statistics T4 Pure Mathematics P1 57 49 41 33 26 6417 Decision Mathematics D1 Pure Mathematics P2 56 48 40 32 25 Pure Mathematics P3 56 48 41 34 27 6671 Pure Mathematics P1 Mechanics M1 54 47 40 33 26 6672 Pure Mathematics P2 Mechanics M2 62 54 46 38 31 6673 Pure Mathematics P3 Mechanics M3 52 45 38 31 25 6677 Mechanics M1 Statistics S1 55 49 44 39 34 6678 Mechanics M2 Statistics S2 60 52 45 38 31 6679 Mechanics M3 Decision Maths D1 58 50 43 36 29 6683 Statistics S1 6684 Statistics S2 All new units marks out of 75. 6689 Decision Mathematics D1 Grade Boundaries June 2001 2 23
6405 Pure Mathematics P1 6689 Decision Mathematics D1
1. x = – 4, y = 3½ 2. (a) GC, FD, FG; DE, BC, GA (b) £214 000 2. ½e2 + e–1 – 1½ 3. AGBCDEFCFGBA –½ 3/2 –3/2 3 ½ 1 3. (a) x – x + x – 1 (b) 1 + ½x + /2x (c) 4 /16 4. (b) SCFET 38 km 4. (a) d = 5 (b) 59 5. (b) 3.96; 4 tapes x 3 5. (a) f–1(x) = (b)(i) f–1(x) > 1 (ii) x > 2 x 2 6. (b) (i) 6 (ii) 10 (c) x = 0.746 (d) 2.64 7. (c) maximum profit = £15
6. (c) f(2.3) = –0.033… f(2.32) = 0.034…
7. (a) 2.95 (b) 7.0, 23.7 (c) 2.78 to 3.14
1 1 8. (b) y = /3x + /3 (c) 4, 12 (d) 10
y 9. (a) A(1, 2) B(2½, 1.33) (b) (c) (3, –6) (4½, –3.99)
O 2 x 22 3
6684 Statistics S2 6406 Pure Mathematics P2
1. (a) (i) census, electoral register x3 x3 9 1. ln 2x c (ii) sample survey, list 3 (b) Poisson 2. (a) 0.8965 (b) 0.478
2. (a) 0.407 (b) 0.066 (c) 0.3495 3. (a) k = 1.25 (b) 3500
3. Insufficient evidence of change 4. (b) (x + 1)(3x –1)(x – 2) (c) 0.70, 3.69, 0.18
4. (a) 0.3222 (b) 0.3578 (c) 0.3264 8x 5. y2 = x 2 5. (a) 0.0902 (b) 0.0166 (c) 0.4457 6. (a) (3x2 – 2)e–x – (x3 – 2x)e–x 1 2 6. (a) f(x) = 27 (3x + 12x) (b) 2 (d) 2.25 (c) –1.45312…, –1.45496…,–1.45488…, –1.45489…–1.455 (e) F(2.25) = 0.517
1 7. (b) (¾ + 2) 7. (a) 0.2 (b) 0.5 (c) 12 (d) 0.6296 (g) 0.256 8. (a) 42.8° (b) 6.12 cm (c) 5.66 cm (d) 47°
9. (a) P = 2, Q = 1½, R = ½ (b) (–1½, 2) maximum (c) (–¼, –½)
10. (c) p (7.7, 8.6) q (1.4, 1.6) (d) –1.10 4 21 6407 Pure Mathematics P3 6683 Statistics S1 3 2 27 4 135 6 1. 1 – /2x + /8x – /16x + … 1. (a) 43, 4 2. z = 1 + i w = + i (b) Mean unchanged – one value is 8 above and the other is 8 below. 1 3. –5 < x < /3 2. (a) 1 270 600 (b) 0.976 4. (a) 1.18 (b) 1.32 (b) As height above sea level increases, temperature decreases. 3 243 5 5. (b) 3y + 9y + /5y + … 3. (a) 0.1056 (b) 11.2 6. (a) R = 25, = 73.74° (b) 4.0°, 143.5° 4. (a) 0.2 (b) 0.6 (c) 0.3 (d) 4.9 (e) 8.04
7. (i)(a) OR = p + q PQ = q – r (b) 69° (ii) Rhombus 21 5. (b) 0.61 (c) 0.21 (d) 0.47 (e) 47 9. (i) y sech x = 2 cosh x + c (ii) y = Ae–6x + Bex – 5 – 6x 6. (a) 30, 42, 46 (b) All have same median and same IQR. Alan - negative skew; Diane – positive skew; Gopal – symmetrical; any other sensible comment.
7. (b) y = 19.4 – 0.968x (c) b for every extra hour of practice 1 ( 0.968) less error will be made, a without practice 19/20 errors will be made. (d)(i) Yes – all points reasonably close to the line. (ii) No – more likely to be
20 5
6679 Mechanics M3 6408 Pure Mathematics P4
1 t –1 = /3 1 2 1. (a) v = 13 – 3e 6 (b) v = 11.2 ms (c)13 1. (a) (b) /3a (a,/6) 3 2. (a) cos = 4 (c) u = 1.4 2. z = –(½ ln 3)i 3. (a) 0.589 ms –1 (b) 20.8 cm 3. (b) 0.11 (c) 0.989 4. (b) x = 2gR2 2 2gR u 4. (a) Centre (0, –½) radius ½ (b) arg(w) = – 5. (b) 65.5 4
1 2 1 2 5. (b) 26 6. (b) 2 m (h + 2g), 2 m (h – 2g) 7. (b) = –2, = –1 (c) 2 a 7. (c) g (d) 1 12 g 8. (b) 15 (72 – 8)
2 9. (b) q = –(p + ) (c) 2 p 3 6 19
6409 Mechanics M1 6678 Mechanics M2
1. (a) 2t2 – 14t + 20 (b) 2 or 5 1. 20 m s2
2. (a) –6i + 9j (b) 10.8 N (c) 33.7° 2. (a) 6 cm (b) 22.6
3. (a) 5.07 N (b) 7.18 N (c) 17.7 N 6 3 1 3. k = (1.88) 4. (a) 0.375 m s–2 (b) 10.3° 5
1 u 4. (a) (25i – 5j) m s 5. (a) v = (3k – 2) (b) 4 (b) 32.9 m 5 (c) 51 m
6. (a) 1200 N (b) (400 – X) N (c) (2X + 1200) N 5. (a) 1.45 (b) 1.36 m (e) 300 2 7. (a) 4a (b) 3.95a 6. (b) u (c) e (c) assuming centres of mass are at mid-points 3 v 1 3 7. (a) t = 4 (b) 18 12 t 8. (b) 12 (c) 9 s (d) 72 m 8
O 6 T
9. (a) 15.9 m (b) 64.4 (c) Down (d) 27.8 m 18 7 6677 Mechanics M1 6410 Mechanics M2 1. (a) 1.75 m s1 (b) 0.75 Ns 1. 20 cm 2. (a) 7.55 N (b) 14.8 u 2. (b)(i) (2e – 1) (ii) reversed 3. (a) 138.5 m (c) 5200 N 3
4. (b) 40 N (c) 0.28 3. 2a
5. (a) 0 (b) 3750 N (c) 3125 N (d) 1.6 m 4. (a) 28.3 N (b) 74.8 N
1 2 2 6. (a) 0.4 m s (b) 0.09 m s , decreasing 5. (b) 3
3 7. (a) 6i + 8j (b) 2 hours 6. (b) 3mg cos – 2 mg (c) 60°
(c) 6i 4j (d) 159.4 2 2 3 7. (b) (c) (d) 9 3 3
2k 2k 8. (b) v2 = u 2 x d
3m 9. (c) (32a2 – 5g) 25 8 17
6411 Mechanics M3 6673 Pure Mathematics P3
1. (a) (5, 3) (b) 7 1. 0.42 Nm 2. (a) a + b = 2 (b) a = 3, b = 1 61 27 2. (a) a = –9, b = 13 (b) 5 i – 5 j 4. (a) 1 – 3x + 9x2 27x3 1 ku 2 1 2 ln1 (c) 0.98058 3. (b) 2 mu 2k g 5. (b) 1.38 (c) 2.05 4. (b) 1.2 m 6. (b) (5,0, 1) (d) 1.5 km 7. (d) 2.5 e 2kt 1 7. (b) x = 4e 2kt 2
1 (c) as t , x 4
8. (a) y2 = 4 (9 – x2) x2 (c) 18 (d) 36 16 9
6672 Pure Mathematics P2 6412 Mechanics M4
1. y – 4 = 2x 1. 3 m s–1
2. (a) 2.422 2. (a) a = cos , b = sin (b) n = (–sin )i + (cos )j
3. 134 volts 3. (a) 2j – 6k Nm (b) 5i + 3j + k N (c) r = – 2i + (5i + 3j + k) 5. 118.4
–1 6. (a) 1.177b (b) 16 years 4. –103i + 10j km h
2 at 2000at 37J (c) As < 1 P = < 2000 6. (b) 4 at 4 at 40m
3e tan 7. (b) y > 0 (c) 6 7. (b) –u sin (c) 3 2 tan 3 8. (a) A = 64, B = 160, C = 20 (b) 2 8. (c) Equilibrium (d) Stable
9 9. (a) 5.39, 1.19 (b) 5.39, 1.19 10 (c) 20.4 C, 4.55 (d) 01 00, 08 30 10 15
6413 Statistics T1 6671 Pure Mathematics P1
1. (a) To make inferences about a population when it is too large 1. (a) 3 (b) 1.2 for complete enumeration (b) List/register of all students 2. (b) 13.3, 103.3, 193.3, 283.3 (c) Allocate each student a number. Use random numbers to select sample. 8 8 3. (b) k ≤ 0, k ≥ (c) 0, 25 25 2. (a) 0.4 (b) 0.618 (c) 0.382 (d) Not practical – come off production line and boxed 4. (a) £2 450 (b) £59 000 (c) 30 sequentially
31 1 3 1 8 3. (a) A and B are independent (b) A and B are mutually 5. (a) 7 32 (b) 3 x - 2x x 3 exclusive 5 (c) A and B are mutually exclusive (d) 0 (e) /12 3 6. (b) (12 + 3 ) cm (b) (18 + 2 ) cm (c) 12.9 mm 4. (a) Q = 52, Q = 65, Q = 78 1 2 3 7. (a) (3, 0) (b) (1, 4) (c) 6 3 (1,4) (d) Same median, range of males = range of females, different 4 skewness 1 8. (a) 2 (b) 6 (c) 4 5 (d) 10 (e) 2x + y – 16 = 0 5. (a) Events occur independently/singly in continuous space or (f) (4,8) time/at a constant rate (b) > 10 (c)(i) 0.134 (ii) 0.109 (d) 0.0359
6. (a) 0.3 (b) 0.5 (c) 0.3 (d) 3.1 (e) 1.29 (f) 18.2
7. (b) 0.75 (c) 0.139 (e) 18
8. (b)(i) 0.268 (ii) 0.436 (c) 37.1 cm 14 11
6417 Decision Mathematics D1 6414 Statistics T2
1. (a) There are no negative entries in the objective row 1. (a) Construct a sampling frame using 3 digit numbers, choose (b) P = 33 – (z + r + s) – At present z, r, and s are all zero. If they 15 numbers from random number tables, identify increase P will increase. Hence P is maximised. corresponding club members. (c)(i) P = 33 (ii) x = 3, y = 1, z = 0 2. (a) Class 1 A, C, F Class 2 B, G Class 3 D, I, Class 4 E Class 5 H (b) Split sampling frame into three strata 5 classes required. Water skiers 10, Yacht owners 4, Power boat owners 1. (b) Class 1 B, G Class 2 E, I, C Class 3 H, A Class 4 D, F Proceed as above but select from each subgroup. 4 classes required. (c) Use stratified, as all groups represented. 3. (a) A B C (b) 17 119 34 85 (c)(i) At each stage the larger of the 2. 0.0681. No evidence to reject service manager’s claim. 85 34 51 two numbers is replaced by the 51 34 17 poitive difference (ii) It must be zero 3. (a) 0.0922. No evidence of association between nationality 17 34 –17 (iii) It finds the Highest Common and preference. 17 17 0 Factor of A and B (b) Data collection assumed to be random – without it test not 4. (a) LA P SL NY 2450 strictly valid. NY SL P LA (b) NY – C – D – P – LA 4. (a) (12.8, 13.6) (b) 0.308. Manufacturer’s claim supported.
5. (b) A B E G (c) 21 5. (a) X < 8.34 (b) 0.392 to 0.399 (d) latest time at end of D – earliest time at start of D – length of D 15 – 10 – 4 = 1 day (e) A B D G 3 days 6. (b) 1.08. B(5, 0.4) is a good model
6. (a)(i) maximum flow along SACDT = 50 7. (a) 0.488 kg (b) 0.0271 (ii) maximum flow along SBT = 100 (c) –0.729. Sample consistent with supplier’s claim (c) SADT 70 SACCBT 30 SADCBT 20 Maximum flow 270 (d) No assumption needed as Central Limit Theorem states 7. (c) y 14 is redundant. Also x 0, y 0 are redundant that the distribution of the sample mean weight of the bags of (d) (5, 9) (6, 6) (6,7) (6, 8) rice is approximately normal for large n from any population. (e) C = 140x + 45y (f) £1105 5 lorries 9 vans 8. (b) Sxx = 1900, Syy = 161, Sxy = 541 (c) 0.977 (d) 0.6694. Evidence of positive correlation 8. (a) A, B, F, G, E, D, C (b) 778 km (c) 589 eg ABCGDGEGFA revisited vertices: D and E (e) y = 3.94 + 0.285x (f) x = 80, y = 26.7 (e) 470 (f) 482 km ABCDEGFA 12 (c) In spite of using magnitude of differences still no evidence to reject H0. 13 6415 Statistics T3 6414 Statistics T4 1. (a) Latin Square Design (b) Oven 1. (b) t5( 2 + 3 t2)10 (c) 0.0425 1 2 3 4 5 5 C1 D1 D2 D3 D4 Chef C2 D2 D3 D4 D1 2. t = –1.604. There is insufficient evidence of weight loss while C3 D3 D4 D1 D2 sleeping. C4 D4 D1 D2 D3 3. (a) F = 1.689. There is insufficient evidence of a difference in 2. (a) 0.75 (b) 0.729 M M the variances of birth weights. 0.35 0.25 A (b) Assume birth weights follow a normal distribution. (c) Boys and girls weights must be independent for test to be 0.65 0.15 M valid. Validity might be affected if twins present since their A 0.85 A weights might not be independent. 3. (a) p = 1.2, q = –0.8 (c) Acceptable model. Residuals in horizontal band about zero with no obvious pattern. 4. (a) (1.6799…, 3.7639…) (b) Since z is in the confidence interval the claim is justified. 4. (a) Poisson with = 7 H0: = 7, H1: > 7 (b) x 12 (c) 0.303 (d) Power when µ = 16 much greater than when µ = 10. Test is more 5. (a) 0.1157 (b) 0.27908 (c) 0.04044 (d) 24, 120 discriminating the higher the rate – poor discrimination when µ has shown little increase. (e) 0.1814
5. (a) y = 6.41 + 1.81x (b) 2.999. Evidence to support the chemist’s claim. 6. (a) F3,12(0.05) = 3.49. Not sufficient evidence of a difference in means 6. (a) Both requirements are satisfied (b) Both requirements are satisfied (b) Enables effect of heating systems and house types to be (c) Cost of each scheme evaluated. Likely to take cheaper scheme. isolated and estimated; can remove influence of house types. (c) Ratio = 4.6439. Evidence of a difference in mean 7. (a) 6.67. No evidence to reject H0: no reason to doubt = 0.1 (b) –3.73. Evidence to reject H0 and conclude that the operator is justified expenditure. in his doubt about the mean. (d) Analyses show that house type affects mean expenditure. (c) Length of lines drawn are normally distributed. Use of non-parametric Suggest a mixture of systems 1 and 2 depending on house test based on medians. type since they are the cheapest two.
8. (a) H0: median = 50, H1: median 50 0.0107, evidence to reject H0 and accept median 50 (b)(i) 0.0547. No difference in median scores. 7. (a) 0.99297 (b) 0.09071 (d) r = 7.943 s = 7.256 (ii) No reason to reject H0: no difference in median scores (e) 18.657. Exponential is not a good fit. (f) 0.1991