GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6
No 1(a) (b) 2(a) (b) 3(a) (b)
Working
√5.89 = 2.4269 = 3.08 3.08
Answer 108, 147 3n²
Mark 1 2
0.787965(006)
2
0.79 or 0.8 Line
1 1
7 hrs of sunshine →
2
Notes B1 cao B2 oe; accept 3n × n, 6n2 ÷ 2 etc) (B1 for anything that demonstrates recognition of a quadratic function, for n = 3n2, or 3x2) B2 cao (B1 for √5.89 or 2.4269) B1 ft (round to 1 or 2 significant figures) B1 line of best fit within guides on overlay, and at least 8cm long. B1 ft (dep on a single straight line with positive gradient) B1 ft (dep on a singe straight line with positive gradient)
200 donkey rides → 4 2.4 2.3 2.35 2.36 2.37 2.38 2.39 2.385
2.39
4
B2 trial 2.35 < x < 2.40 (ans shown) (B1 trial 2.3 ≤ x ≤ 2.4) B1 different trial 2.385 ≤ x ≤ 2.388 B1 x = 2.39 (dep on at least one B1) NB: embedded solution –B1
M1 C = ½ × π × d or 0.5 × π × 9 A1 14.1 − 14.14 B1 (dep on M1) “ ” +9 ft B1 for 37.27 – 37.3 SC: give B2 for anything between 14.1− 14.14 M1 5² + 8.5² or 25 + 72.25 or 97.25 M1(dep) √97.25 or √ (25 + 72.25) A1 cao 9.86 − 9.862
16.22 14.47 15.33 15.50 15.68 15.86 16.04 15.95
5
C=½ × π × d = 0.5 × π × 9 = 14.137 +9 = 23.1…
23.1-23.14
3
6
5² + 8.5² = 25 + 72.25 = 97.25 √(5²+8.5²) = √97.25 =
9.86 − 9.862
3
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 No 7 (a) (b) (c)
Working
Answer x5 2(2x + 3) x² + x − 6
Mark 1 1 2
(d)
2x5y6
2
(e)
3(a−2b)(a+2b)
3
1.7 × 1012
3
x=2 y = −4
3
8.786
5
= x² −2x + 3x − 6
8 9
10
11(a)
(b)
2x – 6y = −20 x + 3y = −10 2x – y = 8 6x – 3y = 24 7x = 14 7y = −28 OR x=2 y = −4 y = −4 x=2 Cos 35 = DB 9 DB=9 Cos35 = 9 × 0.819 = 7.372 Tan50= DC “7.372” DC=”7.372”Tan50= 14 ÷ 4 = 3.5; 21 ÷ 3.5 = 4 or 21 × = 14 9 × 3.5 =
6
2
31.5
2
Notes B1 cao B1 cao B2 cao (B1 for at least 3 out of 4 terms with signs correct or 4 terms ignoring signs) B1 for either x5 or y6 B1 cao B1 (a − 2b)(a + 2b) or 3(a2 − 4b2) B2 (3a − 6b)(a + 2b) or (a − 2b)(3a + 6b) B3 cao B3 cao (B2 8.83...×108or digits 17(31745…)) (B1 digits 883 (19)) M1 correct process to eliminate either x or y (condone one arithmetic error) M1 (dep on 1st M1) for correct substn of their found value A1 cao x = 2, y = −4 M1 Cos35=DB/9 M1 DB = 9Cos35 A1 DB = 7.37 – 7.4 DC M1 Tan50 = "7.372" A1 8.78 – 8.79 M1 21 ÷ (14 ÷ 4) OR finding a correct scale factor A1 cao M1 9 × “3.5” consistent with (a) A1 cao
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 No
Working 4 or 9 ÷ = 14 3.25 × 1, 3.75 × 2, 4.25 × 4, 4.75 × 7, 5.25 × 19, 5.75 × 2 = 3.25, 7.50, 17, 33.25, 99.75, 11.50 172.25 ÷ 35 = 4.921….
£4.92
4
(b)
(i) 1, 3, 7, 14, 33, 35 (ii)
cf Graph
4
(c)
Uq→27th→”£5.25” Lq→9th→”£4.70” Range = £5.25 − £4.70 = £0.55
£0.55
2
x = - -7 ±√((−7)2 – 4×2×4) 2×2 x = 7 ±√17 4
2.78 to 2.781, 0.719 to 0.72
3
450,175,90
3
cols = 16,60,2
3
12(a)
13
14(a) (b)
Answer
Mark
Notes
M1 fx using values within intervals (including ends), at least 4 consistently M1 (dep) fx using midpoints consistently M1 (dep on 1st M1) fx ÷ 35 A1 cao B1 cao B2 (dep on at least 3 correct cumulative additions in (i) or a cumulative table finishing with 35) for points plotted at interval upper ends (B1 for no more than two incorrect points, or correct apart from consistent plotting within the class interval but not at upper end) B1 (dep on non linear cumulative frequency points) for a smooth curve or straight line segments through the points from the first point B1 (dep on non linear cumulative frequency graph) ft graph for LQ or UQ, indicated by reading the LQ from 8.5 to 9.5, and the UQ from 26.25 to 27.5, and ½ square for rate of pay B1(dep on non linear cumulative graph) ft graph “£0.55” M1 subs into formula; allow sign errors in the 7. A1 for reduction to correct surd form A1 for both solutions M1 Use of frequency density or area A2 450, 175, 90 (A1 for two of the above) M1 Use of frequency density or area A2 height (1st col) =16, height (2nd col) = 60, height (7th col) = 2
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 No
Working
Answer
15(a)
Mark 3
(b) 4
16
2+3–1 6x
17(a)
E = kv² k = E/kv² = 15480/6² = 430
(b)
18
19
2 3x
4
E = 430v²
3
Notes (A1 for two of the above) note the last column stops at 150 M1 tangent drawn to the curve at t = 1.5 M1(dep) correct expression for the gradient at t = 1.5, with a least one of height or base correct A1 answer between 5 and 6 M1 distance = area under curve from 0 to 6 M1 valid attempt to find area by e.g use of trapezium rule (min 3 strips) or counting squares of min size 10 × 10 A2 94 to 98 inclusive (A1 90< area <94, 98< area <102) M1 any correct common denominator for all 3 terms A1 correct numerator for 3 terms 2 A1 for 3x M1 E = kv² M1(dep)
8707.5 = 430v² v² = 8707.5 ÷ 430 v = √20.25 =
4.5
3
P(WBWB)+P(WBBW)+P(BWWB) +P(BWBW) 2 4 1 3 = 4×( × × × ) 6 5 4 3
8 30
4
21 + 29 + 31 + 33 + 33 + 39 = 31 6
5.41 − 5.42
3
15480 62
A1 cao 8707.5 M1 from E = kv2 "430" M1 (dep) √20.25 A1 cao M1 the 4 correct cases recognised M1 a multiplication of 4 fractions (accept with rep.) 1 A1 for any single case correctly found ( ) 15 A1 cao (oe) B1 cao for mean = 31 M1 correct subs into the formula
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 No
Working
Answer
Mark
var=
Notes A1 5.41-5.42
212 + 29 2 + 312 + 332 + 332 + 39 2 − 6 × 312 6 5942 − 31²) 6 = √(990.33 – 961) = √29.3333 = 5.416 OR (“−10”)2 + (“−2”)2 + “0”2 + “2”2 + “8”2 = 176 176 sd= = 5.42 6 1 Lower: × 229.5² × 145.5 = 2554507 3 1 Upper: × 230.5² × 146.5 = 2594527 3 Difference: Upper-Lower = 40020.0833
sd =
20(a)
(b)
21(a)
OR M1 (“−10”)2 + (“−2”)2 + “0”2 + “2”2 + 22“”+ “8”2 A1 5.41 − 5.42
(
1 × (x + 0.5)² × (y + 0.5) 3 1 Lower: × (x − 0.5)² × (y − 0.5) 3 Difference: 1 (x²y + 0.5x² + xy + 0.5x + 0.25y + 0.125 3 −x²y + 0.5x² + xy −0.5x + 0.25y − 0.125) = Upper:
171 × π × 12² = 214.88 360
40000
3
x 2 2 xy 1 + + 3 3 12 3
215
3
M1 k × 145.5 × 229.5² (k>0) M1 k × 146.5 × 230.5² (k>0) A1 40000 to 40020
M1 correct algebraic expressions for Upper, Lower bounds for x & y M1 for k(x + 0.5) ²(y + 0.5) – k(x – 0.5) ²(y – 0.5) 1 x 2 2 xy 1 ( 4 x 2 + 8 xy + 1) or any A1 or + + 12 3 3 12 correct 3 term equivalent M1 sight of M1
171 oe 360
171 × π × 12² 360
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 No
(b)
Working 171 × 2 × π × 12 = 35.814 360 2 × π × R = “35.814” R = “35.814” ÷ 2π = 5.7 h² = 12² − “5.7”² = 111.51 h = √111.51 = 10.55
Answer
10.6
Mark
6
Notes A1 214.7 − 215 171 M1 × 2 × π × 12 360 A1 for 35.814 M1 for 2 × π × R = “35.81” A1 for 5.7 M1(dep) for h² = 12² − “5.7”² A1 10.55 − 10.6 OR M1 πr × 12 = “215” "215" M1 r = π ×12 A2 for 5.7 M1(dep) for h² = 12² − “5.7” ² A1 10.55 – 10.6
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 6 GCSE MATHEMATICS 1385/6 NOVEMBER 2002: Paper 6 Modifications to the mark scheme for Modified Large Print (MLP) papers. Only mark scheme amendments are shown where the enlargement or modification of the paper requires a change in the mark scheme. In all other cases apply the existing mark scheme. The following tolerances should be accepted on marking MLP papers, unless otherwise stated below: Angles: ±5º Measurements of length: ±5 mm NB: Q14 (the Histogram) has had the scale on the axis changed and reproduction to A3 size. Whilst this is effective for MLP papers, the limit is A4 for Braille papers, which may cause some problems for Braille candidates. Accordingly, once the Braille paper has been marked, please post it to the GCSE Maths office for their response to this particular question to be looked at. Q12: scatter graph The axes scales and labelling have been changed. The tramlines on the overlay, however, the existing tramlines will need to be transferred on to the differently scaled diagram for MLP papers. Q21
A different scale has had to be used for the graph (1.5 cm grid).