Cambridge IGCSE Mathematics

Cambridge IGCSE Mathematics 004 Model Answers Note the instructions ask you to give answers to 3 sig figs, where appropriate. (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 1

http://www.maccer.co.uk Paper 1-0580/01-0581/01, May/June 004 Question 1 4 3 5 Question 8 Question 9 4xy 6xz = x(y 3z) Question Question 3 Question 4 Question 5 Actual length Question 6 Question 7 = 64 5 = 39 447 ( 395) 447 + 395 = 84m 75% = 75 100 = 3 4 0.07 = 7 100 49 31 = 18 5 = 450cm = 4.5m 1 4 1 = 9 4 1 = 18 4 = 9 141.5cm height < 14.5cm 00 = 190.476... = 190.48 euros to nearest cent 1.05 (from the given exchange information, when you hand over euros are you going to get a larger or smaller no. of dollars? - that will tell you whether to multiply by 1.05 or divide by 1.05) Question 10 0 (a line of symmetry acts like a mirror. If you have a proposal for a line of symmetry, cover up one side of the line, imagine that the line is a mirror and imagine whether this mirror will re-create the figure you are examining) Question 11 Sinced OP Q is isceles Question 1 Question 13 P OQ = 180 ( 35) = 110 ( ) x 1 = 1 8 x = 3 7 y = 1 y = 0 00 40 5 5.58... = 5.6 to sig figs (a useful technique here is being practised here - if you have a feel for the numbers and therefore an idea of the general magnitude of the final answer, you can avoid the common error of being out by a factor of 10 (or 100 etc))

http://www.maccer.co.uk 3 Question 14 To work out cost per one milliliter Size A Size B So size B is cheapest 130 800 = 0.165 30 1500 = 0.15 3 Question 19 Area = π(30) = 87.43... = 830cm to 3 sig figs (instructions at the front of the paper request you to state an answer like this to 3 sig figs) Volume = π(30) 80cm Question 15 sin 3 = CB 5 = π(30) 80 1000 litres = 6.19... = 6 litres to 3 sig figs Question 16 Question 17 CB = 5 sin 3 CB =.649... =.65 to 3 sig figs y = a + bc y = ( 3) + ()(8) = 3 + 16 = 13 y = a + bc a + bc = y bc = y a c = y a b Construct bar chart - have bars of equal width and with height corresponding to figures in the table. Could have every square representing 5 vertically. Question 18 Money brought in = 50 0.5 = $1.50 Question 0 Question 1 4x 5 = 31 4x = 36 x = 9 4(y 5) = 36 4y 0 = 36 4y = 56 y = 14 00 : 15 Monday From, aircraft leaves at 00:15 Dubai time. So journey lasts from Average Speed 00 : 15 07 : 45 = 7h 30m = 560 7.5 = 749. 3 = 749km/h to 3 sig figs Profit As a fraction As a percentage = 1.50 8 = $4.50 4.5 8 4.5 8 100 = 56.5%

http://www.maccer.co.uk 4 Paper - 0580/0-0581/0, May/June 004 Question 1 takes Question 3 : 0(18th) to 0 : 40(19th) 3h 0m 10.87... = 10.9 to 3 sig figs Question 9 Interest = 7.5 3000 = $5 100 After two years the interest will be 5 = $450 (could al calculate the interest as 0.075 3000) Question 10 80 000 km Question 3 Question 4 0.5 3, 0.5, 3 p1 3 4 p8 = 6 1 p0 0.5 Question 11 (1) 8.0 10 4 km 1 x + y = 5 (1) x y = 6 () x + y = 10 (3) Question 5 = 1 p0 Add () and (3) x = 16 x = 8 x 4 8 = and from () 8 y = 6 x 3 = 8 y = 8 6 = x = 4 y = 1 Question 6 Question 7 1 3 8 1 + 3 8 6375 P < 6385 = 4 8 3 8 4 8 + 3 8 = 7 8 1 8 = 7 (If you are confident with the procedure, you could get rid of all the denominators of 8 on the second line instead of the third) Question 8 4 4 Question 1 The sector for N Ireland has a central angle of 1, 6 corresponds to 1 million England 50 Scotland 5 Wales 3 N Ireland Question 13 c = kd + e kd + e = c kd = c e d = c e k c e d = k

http://www.maccer.co.uk 5 Question 14 Placing compasses (with a width of 4cm) on P, draw an arc inside the garden Area = 1 4 π(4) = 4π = 1.56.. = 1.6 to 3 sig figs (top line is one quarter of the area of a circle) Question 15 Since the triangles are similar, the ratio of the areas will equal the square of the scale factor So scale factor is 3 Thus Question 16 using BC = P Q 3 99 11 = 9 P Q = 3 BC = 1 3 = 4cm OB = a + c CA = a c OM = c + 1 (a c) = c + 1 a 1 c Question 17 Join BD. With compasses on B and D in turn, draw arcs above and below BD such that they intersect. Use straight edge to join these intersections. With compasses on D, draw arcs intersecting DC and DA. With compasses on each of these intersects in turn, construct arcs that themselves intersect with each other. Join D to this latter intersection. Use bisectors and choose area that is both to the right of the bisector of BD and to the right of bisector of angle ADC. Question 18 BR is side between Bruges and Rotterdam. Using Pythagoras BR = 78 + 83 BR = 113.89.. = 114km to 3 sig figs Bearing is identical; to angle at Rotterdam (alternate angles), tan B = 83 78 B(bearing) Question 19 1 83 = tan 78 = 46.778... = 47 to nearest degree ( ) x + 1 gf(x) = + 1 = x + 1 + 1 = x + gf(9) = 9 + = 11 now = 1 (a + c) BM = OM OB = 1 (a + c) (a + c) = 1 a + 1 c a c = 1 a 1 c = 1 (a + c) given already in, i.e. gf(x) = x + y = x + 1 x + 1 = y x = y 1 x = y 1 This g 1 (y), but by just swopping the parameter (which we can do at will because the parameter is only a dummy parameter - we can use any letter to represent it) g 1 (x) = x 1

http://www.maccer.co.uk 6 Question 0 Question 1 Acceleration 1x 3y 3(4x y ) Completing the square Sp p = 3, q = 1 (x 3) = (x 3)(x 3) = x 3x 3x + 9 = x 6x + 9 x 6x + 10 (x 3) 9 + 10 = (x 3) + 1 = 18 10 = 1.8m/s (ie the gradient of the line between 0-10 secs) Distance travelled in the first 30 secs equals the area under the graph to 30 secs = 1 (10)(18) + (0 18) = 90 + 360 = 450m Between 30 and 45 secs, the distance travelled So total distance travelled So average speed Question = 1 (15)(18) = 135m = 450 + 135 = 585m = 585 45 = 13m/s BA (the no of columns of B is not equal to no. of rows of A) BC = [ 6 5 4 [ 4 6 5 = [ 38 0 0 38 B 1 = 1 38 [ 4 6 5 ( since BB 1 = I)

http://www.maccer.co.uk 7 Paper 3-0580/03-0581/03, May/June 004 Question 1 data in order is Median (iii) Mean 51 1, 34, 35, 39, 48, 50, 51, 51, 65, 75 = 1 (48 + 50) = 49 = 460 10 = 46 There are 90 students, 1 student corresponds to an angle of 4 So required angles are A : 0 B: 60 C: 160 D: 80 E: 40 Draw pie chart using these angles above (iii) Question Area of roof Area of one wall Area of 3 walls No of pieces of wood (d) Volume (e) Posts : 4.80 Wood : 9.00 Roof : 14.40 Nails :.10 Total 30.30 40 90 = 4 9 0 90 + 10 90 = 30 90 = 1 3 = 3 3 = 9m = 3 = 6m = 6 3 = 18m 3 0. = 0.6m = 18 0.6 = 30 = 0.1 0.1 = 0.0m Question 3 Completing the table y = 8 x x = 1 y = 8 ( 1) = 8 1 = 7 x = 0 y = 8 x = y = 8 = 8 4 = 4 x = 3 y = 8 3 = 8 9 = 1 Draw graph with data from - parabola, symmetric about the y-axis with a maximum at y = 8 Answers in the region of (d) Completing table ±.8 y = x + 5 x = 3 y = ( 3) + 5 = 6 + 5 = 1 x = 0 y = 5 (e) Draw line for data above. A straight line, intersecting the y-axis at y = 5 and with gradient (f) (g) (i.e. m in y = mx + c) Question 4 x = 3, x = 1 s = 360 (80 + 50 + 110) = 10 3t = 360 (75) 3t = 10 t = 70 x + y = 180 50 x + y = 130 x + y = 360 60 = 100 (iii) Eqn from less the eqn from produces Into the eqn from y = 30 x + 30 = 100 x = 70

http://www.maccer.co.uk 8 Question 5 00m 0.km The angle between a tangent to a circle and the corresponding radius line is a right angle P C = 7.8 + 0. = 8km (iii) Using Pythagoras on P SC 7.8 + P S = 8 P S = 8 7.8 P S = 3.16km (iii) Using alternate angles al 540 5 = 108 x = 70 y = 180 160 = 0 x + y = 70 + 0 = 90 tan BAC = 10 100 = 1. BAC = 50.1944.. = 50. to 1 dp (iv) Area Question 6 translation given by = 1 (7.8)(3.16) = 1.34 = 1km to sig figs [ 10 Rotation about origin counter-clockwise by 90 (iii) Bearing (iv) Question 8 = 70 + 50. = 10 to 3 sig figs 180 + 10 = 300 to 3 sig figs (making use of alternate angles) Print-outs of the exam paper might not be exactly the same size as the original exam paper, distances could be a bit distorted Treat the y-axis as a mirror to draw the reflection of E. All you need to do is make the x-coordinates positive, to produce a figure with the following coordinates 6cm 10km (5, 3), (8, 3), (6, 1), (6, ), (5, ), (8, ), (7, 1), (7, ) (iii) (when the center of enlargement is the origin, we just need to multiply all the coordinates of the vertices by the scale factor) Question 7 F maps to a triangle with vertices Pentagon Sum of interior angles (, 4), (4, 4), (, 8) = (n ) 180 = 3 180 = 540 75km Draw a line 1cm parallel to road, on both sides of road. This is best done using a set square which is placed with one edge along the road, another edge being placed on a straight edge (ruler), allowing it to slide along parallel to the road for a distance of 1cm. With compasses on A and B in turn, construct two arcs from each point that intersect above and below the points. Join these intersections up. (d) Draw a circle of 4cms centered on C (e) The airport is at the intersection of the bisector of AB and a line 1cm from the road. The circle from (d) allows to choose which point, i.e. the one below the road.

http://www.maccer.co.uk 9 Question 9 1 General formula is n + if n = 9 (iii) as above n + = 0 n + General formula is 4n where n is the no. of the diagram for diagram 5 4 5 = 0 dots General formula here is n for diagram 5 5 = 5 squares no of dots for n = 1 144 1 n = 1 = 4 1 = 48 (iii) No of squares for n = 10 4n = 40 n = 10 = 10 = 100

http://www.maccer.co.uk 10 Paper 4-0580/04-0581/04, May/June 004 Question 1 Fatima pays 60 10 + (6 10) 100 = 7 + 60 = $13 13 100 = 110% 10 (i.e. state as a fraction and then multiply by 100) $159.10 represents 86%, original price Frame Size (d) Fatima rides = 159.10 86 The total distance Question 100 = $185 156 5 = 48cm 169 11 36 = 19.8cm 0 = 36 3 = 414km f(x) = x x 3 p = ( 3) ( 3) 3 = 9 + 3 3 = 9 q = 1 1 3 = 3 r = 4 4 3 = 9 Draw graph using given data. Parabola with minimum between (0, 3) and (1, 3) (iii) Estimate a line representing gradient at x = 1. Extend this line and calculate its gradient -expect answer in region of 3 u = 6 ( 1)3 3 g(x) = 6 x3 3 ( = 6 1 ) = 6 1 3 3 v = 6 Draw graph using given data x x 3 = 6 x3 3 3x 3x 9 = 18 x 3 x 3 + 3x 3x 7 = 0 Need the points where the two lines intersect. Question 3 (iii) (iv) Something of the order of (.5, 0.6) 36cm (the nearest whole number corresponding to 18/183) 6 46cm (0cm) 33.5cm number corresponding to 145 80 85 below 5cm, (365 85) above 5cm q = 15 from graph. p = 365 (17 + 41 + 6 + 98 + 85 + 15) = 365 318 = 47 Depth mid pt No. days m.p. x days 0 < d 10 5 17 85 10 < d 0 15 41 615 0 < d 30 5 6 1550 30 < d 40 35 98 3430 40 < d 50 45 85 385 50 < d 60 55 47 585 60 < d 70 65 15 975 Total = 13065 365 = 35.8.. to 3 sig figs No of squares occupied by second column = 0 8 = 160 First column has the same width as the second column, height = 8 58 =.9cm 160 Third column is 1.5 times as wide, its height = 3 8 147 = 4.9cm 160

http://www.maccer.co.uk 11 Question 4 Using Cosine Rule AC = 11.1 + 9.5 (11.1)(9.5) cos 70 = 141.379.. AC = 11.88... = 11.9 to 3 sig figs In cyclic quadrilaterals, opposite angles sum to 360, ADC = 180 ABC (d) Using Sine Rule ACD = 180 (110 + 37) = 33 11.9 sin 110 = AD sin 33 11.9 sin 33 AD = sin 110 = 6.89715... = 6.90 to 3 sig figs 70 (angle subtended by AC at circumference) To find CE Question 5 Time taken Area of ACE EAC = 1 (180 70) = 55 CE sin 55 = 11.9 sin 70 11.9 sin 55 CE = sin 70 = 1 (11.9)(CE) sin 55 = 1 11.9 sin 55 sin 70 = 50.560... = 50.6cm to 3 sig figs = 10 x 10 x 10 x + 1 = 1 (x + 1)(10) x(10) = x(x + 1)( 1 ) 10x + 10 10x = 1 (x + x) 0x + 0 0x = x + x 0 = x + x x + x 0 = 0 or x + x 0 = 0 (x + 5)(x 4) = 0 x + 5 = 0 x = 5 x 4 = 0 x = 4 (d) Only physical lution from is x = 4, time to waterfall = 10 4 =.5h Question 6 Volume of cone Volume of hemisphere Total Mass = 1 3 π(7) (13) = 3 π(7)3 = 1 3 7 π(13 + 7) = 49π 3 (7) = 49π(9) = 441π = 1385.443... = 1385.4cm to 1 dp = 1385.443... 0.94 = 130.3158...g = 1.3kg to 1 dp Length of sloping edge l is given by Surface Area Area of hemisphere l = 13 + 7 = 169 + 49 = 18 l = 18 = π(7) 18 = 34.6954... = 35cm to 3 sig figs Cost of gold plate per cm = = πr = π(7) = 98π 411.58 34.6954... + 98π = 0.65064... = $0.65 to nearest whole cent

http://www.maccer.co.uk 1 Question 7 Complete Venn Diagram - 3 are outside the circles and 8 in the intersection, leaving 1 studying Physics only and 7 studying Chemistry only. 8 Question 8 Anti-Clockwise (or positive) rotation of 90 about the origin translation described by [ 5 (iii) (as already mentioned in 1 30 (iii) Reflection about the line y = x (iv) Clockwise (or negative) rotation of 180 about (1, 1) (v) Enlargement by scale factor of about origin (iv) (vi) Shear in the y-direction, x unchanged 1 0 = 3 5 Applying the transformation to the vertices of A. [ [ [ 0 1 1 3 1 1 1 = 1 0 1 1 1 1 3 = P (BB) = 3 9 4 10 = 1 90 = 15 P (W W ) + P (W B) + P (BW ) ( 6 9 6 ) ( 6 + 10 9 4 ) ( 3 + 10 9 6 ) 10 = 36 90 + 4 90 + 18 90 = 78 90 = 13 15 and the resultant coordinates belong to B. B onto D is a reflection about the x-axis. The required matrix is therefore [ 1 0 0 1 Question 9 [ 1 0 1 1 (iii) = P (W W W W ) ( 6 9 6 10 5 8 5 ) 9 ie 15x + 5y 000 3x + 5y 400 = 5 36 (iv) P (BBBB) = ( 3 9 4 10 8 3 ) 9 x y y 35 = 1 90 (d) set up coordinates Probability not all same color = 1 5 36 1 90 = 1 5 180 = 1 7 180 = 153 180 = 17 0 draw three lines mentioned, i.e. in each case, replace the inequality sign by an equal sign and draw the resultant straight line. The area then satisfying each inequality is to one side of this straight line - which side to choose can be determined by just choosing one point and testing this point to see whether it satisfies the inequality or not. You are asked to shade the unwanted area, which is the area outside a triangle which vertices (35, 35), (74, 35), (50, 50)

http://www.maccer.co.uk 13 (e) If 70 pencils are bought 3x + 5y 400 10 + 5y 400 5y 190 y 38 So largest possible number of pens is 38 (f) The largest possible profit will occur at a point represented by a vertex of the triangle mentioned in (d) The expression for profit (P) P = 5x + 7y using the data at each of the three vertices in turn gives P = 35(5) + 35(7) = 40 P = 74(5) + 35(7) = 615 P = 50(5) + 50(7) = 600 So the greatest possible profit is 615 cents = $6.15