Edexcel GCSE Mathematics (1387) Higher Tier Model Answers

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1 Edexcel GCSE Mathematics (387) Higher Tier 003 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 5 (Non-Calculator), 4 June 003 Question Question 0% of = 00 Therefore value of car after year = 00 = 0800 Now 0% of 0800 = 080 Therefore value of car after years = = 970 Question 3 7r + = 5(r 4) 7r + = 5r 0 r = r = Question 5 Constant first difference of 5 the required expression is of the form 5n + k where k is a constant On inspection k =. So expression is Question 6 D has vertices 5n + (0, 0), (3, 0), (0, ) rotation by 80 (anti-clockwise or clockwise) about (0,) Question 7 Draw bisector of BC through A Draw an arc of 5cm from A (using compasses) Shaded area will be above bisector and to the left of arc. Question 8 x + y + z xyz : Length : V olume Question 4 xy + yz + xz : Area, 0, Insert lines corresponding to x = x = y = y = x + giving the required points, as follows (, ), (, 0), (, ), (0, 0), (0, ), (, ) Question 9 A short question could be on the lines : list, in order, your three favourite genres of food you like to eat in a restaurant? (any attempt to include all possibilities and not lead the respondents on will be satisfactory.. His family is not representative of the population as a whole. The question is directing the respondents to a particular answer

3 Question 0 Distance travelled = = = km ( 0 5 ) + (3 0 4 ) = (0 0 4 ) + (3 0 4 ) = = km Question (x + y) = (x + y)(x + y) = x + xy + y Expression can be stated as ( ) = 5 = 5 Question 5 Length of arc So total perimeter Question 6 Question 7 40 = 40 π radians 360 = 9 40 π = π 360 = (8 + π) cm 4 0 = 4 = 4 = = (6 ) 3 = 4 3 = 64 F α x Question Angle ACB = 7, because FE is a tangent to the circle, and therefore angle ACE = 90. Angle BAC = 63, because () the diameter AC subtends an angle of 90 at the circumference and () from we know that angle ACB is 7. Question 3 When x=3, F=4 so F = k x 4 = k 9 k = 36 F = 36 x F = 36 4 = 9 3q 4 q 5 q 3 (p 3 ) 3 = p 9 = 6q9 q 3 = 6q6 Question 8 64 = 36 x x = x = 6 8 = 3 4 Question 4 lower quartile = 5 upper quartile = 77 Box plot consisting of left hand side of box at 5 and right-hand side at 77. A line representing the median is drawn at 67. Lines are extended to display the range - to the left to 3 and to the right to 8. ( ) ( 5 3 ) = 5 3 = =

4 3 Question 9 0 x x Between 30 and 40,.5 units high Between 40 and 70, 0 units high Question 0 (3x + 4) 3(4x 5) = 6x + 8 x + 5 = 6x + 3 P( green) ( 5 = ) 0 = 3 5 = 375 = P (G, G, NG) + P (G, NG, G) + P (NG, G, G) So total probability ( = ) 0 = 3 3 = 96 ( xy 3 ) 5 = = 5 x 5 y 5 = 3x 5 y 5 = 660 Question 3 = n n + n (n )(n + ) n + = (n ) n n AB = 6b 6a = 6(b a) EF = 6a = (n ) (n ) Question EB = 6b = b Multiply all coordinates by.5, so that apexes transform so BC = 6a (, ) ( 3,.5) so (, 3) ( 3, 4.5) EX = b + ( 6a) = b 3a (3, ) ( 4.5,.5) Question BY = 6(b a) = 4(b a) 3 P( black) = P (B, B, NB) + P (B, NB, B) + P (NB, B, B) EY = b + 4(b a) P( red) ( 3 = ) 0 = 3 63 = 89 = b + 4b 4a = 6b 4a = 4 (b 3a) 3 = P (R, R, NR) + P (R, NR, R) + P (NR, R, R) so EX and EY lie along the same line

5 4 Question 4 (5, 4) (, 9) (, 4) (iv) (, 4) The general form of equation is y = ax + bx + c At (0,0) At (,-4) 0 = a(0) + b(0) + c c = 0 4 = a() + b() 4a + b = 4 () By symmetry, also crosses x-axis at (4,0), giving 0 = a(4) + b(4) 6a + 4b = 0 () 4 ( ) - () and so 4b = 6 b = 4 4a + ( 4) = 4 4a = 4 a = f(x) = x 4x

6 Paper 6 (Calculator), 0 June 003 Question Volume of cylinder = πr h = π(4 )(0) = 60π = 503cm Consider a right-angled triangle of sides 0 cm and 8 cm Hypotenuse = = = 64 Hypotenuse = cm So a pencil of 3cm cannot fit inside the cylinder Question Question 4 Volume of cuboid Therefore = x x (x + ) = x 3 + x x 3 + x = So answer to d.p. is 5.9 Question 5 HCF LCM Question 3 60 = 30 = 5 = = 48 = 4 = 3 = 4 6 = 5 3 = 3 = = = 480 Area Question 6 = πr = π7.5 = 88.4cm 5 = x + x = 4 x = 8 y = x + 3 for example (any constant on the end will suffice) Median = hald the 0th ans st items, which will fall in the class 50 < C 00 One extra item at will raise the median to be the st item, which lies in the same class as before = 6500 y = x + x = y x = (y ) 5

7 6 Question 7 (3) (4) 3 Add (5) and (6) therefore Question 8 Perimeter x 3y = (3) 5x + y = 8 (4) 4x 6y = (5) 5x + 6y = 54 (6) 9x = 76 x = 4 (4) 3y = 3y = 3 y = CD 4.8 = 0 6 CD = 48 6 = 8cm fracca4.5 = 0 6 CA = 45 6 = 7.5 Question For a fair dice, we would expect 00 sixes. 00 sixes is too divergent a result, so dice is not fair. Left-hand side of tree requires a probability of 5/6 o n the bottom branch. The right-hand side will have two branches from each node. Each set of branches will mirror those on the left-hand side with a probability of /6 for a six and 5/6 for a not six. P(no six) so P(at least one six) Question Volume of large cone Volume of cone A 6 6 = 36 = = 5 36 = 5 36 = 36 = 3 πr h = 3 π(7.5) (30) = 56.5π = 3 π(.5) (0) Question 9 Question 0 = = 9.8cm y = (3 08 )( 0 7 ) (3 0 8 ) + ( 0 7 ) = = = y = tan 38 = AB 8.5 AB = 8.5 tan 38 AB = 6.64cm Volume of B = 0.8 3π = 50π 50 3 π = 54. 6π = = 700 to 3 sig figs (Note: Although I have laid out intermediate figures here, I would in reality do it all in one go, storing figures in my calculator right through the calculation. This comment applies to other questions on this paper, as well ) S = πd h + d S = 4π d (h + d ) S = 4π d h + 4π d 4 4π d h = S 4π d 4 S h = 4π d d S h = 4π d d

8 7 If d and l are lengths related to the smaller frustrum, the surface area of the larger frustrum is given by S = π ( 3d ) ( ) 3h + ( ) 3d ( ) 3d 9 = π 4 (h + d ) ( ) ( ) 3d 3 (h = π + d ) = 9 ( πd ) (h 4 + d ) Therefore Surface Area of larger frustrum Question 3 Area of trapezium Rearrange Using formula = = 0.5cm 4 x + 0 x = 400 x + 0x = 400 x + 0x 400 = 0 Question 5 (a ) (b ) = 4a a a+ (4b b b+) = 4a 4a + 4b + 4b = 4a 4a 4b + 4b = 4(a b ) 4(a b) = 4(a b)(a + b) 4(a b) = 4(a b)(a + b )) If a and b are both odd or both even, then (a-b) will be a multiple of and therefore, because of the coefficient of 4, the whole expression will be a multiple of 8. If one of a,b is odd and the other even, then (a+b-) will be a multiple of, and therefore employing similar logic to above, the whole expression will be a multiple of 8. Question 6 Upper bound of g is given by (4.505) (.5 ) (sin 9.5) So x = 0 ± 0 4 ( 400) x = 0 ± 000 x =.36 and x = 3.36 x =.36cm =.70 Lower bound of g is given by (4.495) (.35 ) (sin 30.5) = 9.79 Question 4 Area of ABC side AB is given by angle B is given by Now = (8)(5) sin 70 = 56.4cm AB = (8)(5) cos 70 AB = sin 70 AB = sin B 8 sin B = sin B = CX 5 CX = sin B 5 = 7.84cm g = = to significant figures given to significant figures, because that is the lowest number of significant figures in the inputed data Question 7 or p+q = p q = xy q = q q = y q = ( q ) = y p = p = x

9 8 xy = 3 x = 3 y Inserting into second equation ( ) 3 y = 3 y 64y = 3 y = and If y = If x = 64 x ( ) = 3 x = 64 q = q = p = 64 p = 6 Question 8 x mx = (x m) k k = x mx + m (x mx) k = x mx + m x + mx k = m Minimum value occurs when (x m) = 0 giving min. value = k = m Minimum value occurs when Question 9 x = m If the events were independent then the probability that both Betty and Colin will be late = = Since the actual probability is 0.0, the events are not independent Question 0 a = 50 b = 50 k = 360 λ = = 4