Cambridge IGCSE Mathematics 0580 — 2007 May/June Paper 4 · Variant 1

1 (a) The scale of a map is 1:20 000 000. On the map, the distance between Cairo and Addis Ababa is 12 cm. (i) Calculate the distance, in kilometres, between Cairo and Addis Ababa. [2] (ii) On the map the area of a desert region is 13 square centimetres. Calculate the actual area of this desert region, in square kilometres. [2] (b) (i) The actual distance between Cairo and Khartoum is 1580 km. On a different map this distance is represented by 31.6 cm. Calculate, in the form 1 : n, the scale of this map. [2] (ii) A plane flies the 1580 km from Cairo to Khartoum. It departs from Cairo at 11 55 and arrives in Khartoum at 14 03. Calculate the average speed of the plane, in kilometres per hour. [4]

Mark scheme: 1 (a) (i) 2 400 B2 SC1 for figures 24 (ii) 520 000 B2 SC1 for figures 52 (b) (i) 1 : 5 000 000 or n = 5 000 000 B2 SC1 for 5 000 000 seen in final answer or n = figs 5 oe in final answer (ii) Time = 2hrs 8 mins or 128 (mins) B1 = 2.13(33..) (hours) oe soi B1 Implies previous B1 Accept 128 60 1580 ÷ their time M1 soi is by correct answer 738 – 742 cso A1 www 4 (12.3 seen earns B1M1) [10]

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2 Answer the whole of this question on a sheet of graph paper. (a) Draw and label x and y axes from −6 to 6, using a scale of 1 cm to 1 unit. [1] (b) Draw triangle ABC with A (2,1), B (3,3) and C (5,1). [1] (c) Draw the reflection of triangle ABC in the line y = x. Label this A1B1C1. [2] (d) Rotate triangle A1B1C1 about (0,0) through 90° anti-clockwise. Label this A2B2C2. [2] (e) Describe fully the single transformation which maps triangle ABC onto triangle A2B2C2. [2]  1 0  (f) A transformation is represented by the matrix   .   −1 1  (i) Draw the image of triangle ABC under this transformation. Label this A3B3C3. [3]  1 0  (ii) Describe fully the single transformation represented by the matrix   . [2]   − 1 1  (iii) Find the matrix which represents the transformation that maps triangle A3B3C3 onto triangle ABC. [2]

Mark scheme: 2 (a) Axes to correct scale S1 Accept 2mm accuracy throughout (b) Correct triangle A(2,1)B(3,3)C(5,1) B1 Condone absence of labels (c) A1(1,2), C1(1,5), B1(3,3) B2 B1 for 2 correct points ft their ABC Condone absence of labels and sides but not incorrect suffices (d) A2(–2,1), C2(–5,1), B2(–3,3) B2 B1 for 2 correct points ft their A1B1C1 Condone absence of labels and sides but not incorrect suffices SC1 for rotation of their A1B1C1 90° clockwise about the origin If triangle ABC is rotated correctly treat as mis-read (e) Reflection B1 Indep (Only possible answer) y-axis oe cso B1 (f) (i) A3(2, –1), C3(5, –4), B3(3,0) B3 B2 for 2 correct points plotted Condone absence of labels and sides If B0, M1 for any set up of matrix multiplication seen for at least one point and A1 for correct result (If correct triangle A2B2C2 used treat as MR, and the co-ords are (–2, 3), (–5, 6), (–3, 6)) (ii) Shear, y-axis invariant oe B1,B1 Allow factor of either +1 or –1 if invariant line omitted, but dependent on shear or stretch (iii)  1 0  B2 B1 for the left hand column    1 1  [15] IGCSE – May/June 2007 0580 and 0581 04

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3 (a) A B 55° 40.3 cm NOT TO 26.826.826.8cmcmcm SCALE 92° X 20.120.120.1 cmcmcm C D A, B, C and D lie on a circle. AC and BD intersect at X. Angle ABX = 55° and angle AXB = 92°. BX = 26.8 cm, AX = 40.3 cm and XC = 20.1 cm. (i) Calculate the area of triangle AXB. You must show your working. [2] (ii) Calculate the length of AB. You must show your working. [3] (iii) Write down the size of angle ACD. Give a reason for your answer. [2] (iv) Find the size of angle BDC. [1] (v) Write down the geometrical word which completes the statement “Triangle AXB is to triangle DXC.” [1] (vi) Calculate the length of XD. You must show your working. [2] (b) P Q y cm cm +2) (y X NOT TO SCALE cm (y 1) + – (2y 1) cm R S In the diagram PQ is parallel to RS. PS and QR intersect at X. PX = y cm, QX = (y + 2) cm, RX = (2y – 1) cm and SX = (y + 1) cm. (i) Show that y2 – 4y – 2 = 0. [3] (ii) Solve the equation y2 – 4y – 2 = 0. Show all your working and give your answers correct to two decimal places. [4] (iii) Write down the length of RX. [1]

Mark scheme: 3 (a) (i) 0.5×40.3×26.8sin92 oe M1 Any other method must be complete (s = 58.13 – 58.15) 539.6 – 540 A1 ww scores zero 2 2 + 26 8. 2 (ii) AB 403. M1 ( AB ) = 403. = oe M1 sin 92 sin 55 − 2 × 40 3. × 26 8. cos 92 403. × sin 92 ( AB = ) M1 (AB = ) square root of above and a sin 55 correct combination M1 (dep) Accept if found in (i) ww scores zero 49.2 or 49.16 – 49.18 A1 (iii) 55 B1 Angles in the same segment oe B1dep (iv) 33 correct or ft B1 ft 88 – their 55, if answer is positive (v) Similar or enlarged B1 (vi) XD 201. M1 XD 201. = oe = 403. 26 8. sin their (iii ) sin their (iv ) A1 30.2(25) 30.2(309…) cao Any other method must be complete ww scores zero (b) (i) y y + 1 M1 May be implied by next line = oe Accept correct ratio statement y + 2 2 y − 1 y(2y – 1) = (y + 1)(y + 2) M1 May be implied by next line 2y2 – y = y2 + y + 2y + 2 Implies previous M2 y2 – 4y – 2 = 0 E1 Dep (no errors in any line) If M0, SC1 for y(2y – 1) – (y + 1)(y + 2) = 2y2 – y – y2 – y – 2y – 2 = y2 – 4y – 2 (ii) 4 ± 16 + 8 B1,B1 p + ( or − ) q If of form 2 r B1for 4 and 2, B1 for 42-4(1)(-2) q If of form p + ( or − ) r B1 for 42-4(1)(-2) but may recover the other B1 from answers -0.45, 4.45 cao B1,B1 SC1 for rounding or truncating to 1 dp or more – 0.44948…, 4.44948… ww scores max of 2 (iii) 7.9(0) or better 7.8989.. ft B1ft ft 2 × a positive root -1 [19] IGCSE – May/June 2007 0580 and 0581 04

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4 y 15 10 5 x –2 0 2 4 6 8 10 –5 –10 The diagram shows the accurate graph of y = f(x). (a) Use the graph to find (i) f(0), [1] (ii) f(8). [1] (b) Use the graph to solve (i) f(x) = 0, [2] (ii) f(x) = 5. [1] (c) k is an integer for which the equation f(x) = k has exactly two solutions. Use the graph to find the two values of k. [2] (d) Write down the range of values of x for which the graph of y = f(x) has a negative gradient. [2] (e) The equation f(x) + x – 1 = 0 can be solved by drawing a line on the grid. (i) Write down the equation of this line. [1] (ii) How many solutions are there for f(x) + x – 1 = 0? [1]

Mark scheme: 4 (a) (i) 3 B1 (ii) –4.25 to –4 B1 (b) (i) –1.6, 2.0, 8.6 to 8.63 B2 B1 for any one correct (ii) 9.2 B1 (c) –9, 3 B1,B1 –1 each extra incorrect value (d) 0<x<6, ( i.e.0 to 6 only) oe B2 Accept (0,6), [0,6], (0, 3) to (6, –9). SC1 for other inequality errors or answers using 0 and 6 as boundaries (e) (i) 1 – x oe B1 If re-arranged it must be correct equation with y or f(x) in it but exclude f(x) + x – 1 = 0 (ii) 3 B1 [11] IGCSE – May/June 2007 0580 and 0581 04

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5 B E 25 cm M O C 14 cm NOT TO SCALE D OBCD is a rhombus with sides of 25 cm. The length of the diagonal OC is 14 cm. (a) Show, by calculation, that the length of the diagonal BD is 48 cm. [3] (b) Calculate, correct to the nearest degree, (i) angle BCD, [2] (ii) angle OBC. [1] (c) DB = 2p and OC = 2q. Find, in terms of p and q, (i) OB, [1] (ii) OD. [1] (d) BE is parallel to OC and DCE is a straight line. Find, in its simplest form, OE in terms of p and q. [2] (e) M is the mid-point of CE. Find, in its simplest form, OM in terms of p and q. [2]  7  (f) O is the origin of a co-ordinate grid. OC lies along the x-axis and q =   .   0  (DB is vertical and DB = 48.) Write down as column vectors (i) p, [1] (ii) BC . [2] (g) Write down the value of │DE │. [1]

Mark scheme: 5 (a) Using a right-angled triangle with M1 25 and 7 seen is sufficient (or 50, 14) 25 and 7 252 - 72 oe (or 502 – 142) M1 Must be a correct numerical calculation oe includes trig methods, which can round to 24, then 48 for the E mark (BD) = 48 (or 24 × 2) E1 Dep on M2, correctly established (b) (i) −1  7  M1 If scale drawing seen then M0 cos   × 2 oe  25  147° cao A1 www 2 147.47…. score M1 only (ii) air 32 -34 or ft B1 ft 180 – their 147 (c) (i) q + p oe B1 (ii) q – p oe B1 (d) M1 any correct unsimplified expression OC + CE oe 2q + their (c) (i) e.g. their (q – p) + 2 × their (q + p) p + 3q cao A1 www 2 (e) 1 M1 any correct unsimplified expression OC + OB oe 2 2q + ½ their (c) (i) 0.5p + 2.5q cao A1 www 2 (f) Accept any reasonable notation in both parts (i)  0  B1    24  (ii) B1  7    B1  −24  (g) 50 B1 [16] IGCSE – May/June 2007 0580 and 0581 04

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6 Answer the whole of this question on a sheet of graph paper. Kristina asked 200 people how much water they drink in one day. The table shows her results. Amount of water (x litres) Number of people 0 < x 0.5 8 0.5 < x 1 27 1 < x 1.5 45 1.5 < x 2 50 2 < x 2.5 39 2.5 < x 3 21 3 < x 3.5 7 3.5 < x 4 3 (a) Write down the modal interval. [1] (b) Calculate an estimate of the mean. [4] (c) Make a cumulative frequency table for this data. [2] (d) Using a scale of 4 cm to 1 litre of water on the horizontal axis and 1 cm to 10 people on the vertical axis, draw the cumulative frequency graph. [5] (e) Use your cumulative frequency graph to find (i) the median, [1] (ii) the 40th percentile, [1] (iii) the number of people who drink at least 2.6 litres of water. [2] (f) A doctor recommends that a person drinks at least 1.8 litres of water each day. What percentage of these 200 people do not drink enough water? [2]

Mark scheme: 6 (a) 1.5 < x ≤ 2 B1 (b) (8×0.25 + 27×0.75 + 45×1.25 +…….. M1 For mid-values (allow two slips) ……….. 3×3.75) M1 For Σfx (allow two slips) dep on first M1, or mid-values ± 0.05 their 345.5 ÷ 200 M1 for ÷ 200 dep on second M1 1.7275, 1.727, 1.728 or 1.73 cso A1 www 4 (c) 8, 35, 80, 130, 169, 190, 197, 200 B2 If B0, allow M1 for clear attempt to add accumulatively (d) axes correct scale S1 Not reversed and must reach 200 vertically, even if not labelled 8 points plotted ft part (c) P3dep dep on at least M1 in (c) (0.5, 8), (1, 35), (1.5, 80), (2, 130), (2.5, 8 points from their values 169), (3, 190), (3.5, 197), For x-values (upper boundary values), (4, 200) points must touch grid line For y-values, even, must touch grid line, odd must be inside square. P2 for 6 or 7 points ft P1 for 4 or 5 points ft curve (or polygon) either correct or C1 Allow 1 mm tolerance through 8 points and correct shape Ignore any bars drawn if they do not compromise the points and graph (e) (i) 1.65-1.75 B1 (ii) 1.5 B1 (iii) 23 – 29 integers only B2 If B0 allow SC1 for non-integer in correct range, or 172 – 177 seen (may be written on graph) (f) 54 – 56.5 B2 SC1 for figures 108 – 113 or 87 – 92 Accept if written on graph www 2 [18] IGCSE – May/June 2007 0580 and 0581 04

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7 NOT TO SCALE 0.8 m 0.3 m 1.2 m The diagram shows water in a channel. This channel has a rectangular cross-section, 1.2 metres by 0.8 metres. (a) When the depth of water is 0.3 metres, the water flows along the channel at 3 metres/minute. Calculate the number of cubic metres which flow along the channel in one hour. [3] (b) When the depth of water in the channel increases to 0.8 metres, the water flows at 15 metres/minute. Calculate the percentage increase in the number of cubic metres which flow along the channel in one hour. [4] (c) The water comes from a cylindrical tank. When 2 cubic metres of water leave the tank, the level of water in the tank goes down by 1.3 millimetres. Calculate the radius of the tank, in metres, correct to one decimal place. [4] (d) When the channel is empty, its interior surface is repaired. This costs $0.12 per square metre. The total cost is $50.40. Calculate the length, in metres, of the channel. [4]

Mark scheme: 7 (a) 1.2 × 0.3 × 3 oe M1 (1.08) or 3 × 60 (180) × 60 oe M1dep × 1.2 × 0.3 (0.36) 64.8 cao A1 www 3 (b) 1.2 × 0.8 × 15 × 60 oe (= 864 seen) M1 Their (a) 8 3 × 5 oe seen Their 864 – their (a) M1ind or their 864 ÷ their (a) × 100 (1333.3..) ÷ their (a) × 100 M1dep subtract 100 (Dep on second M1) 1230 (%) or better (1233.3…) cao A1 www 4 (1330 or 1333.3…www M1M1M0) (c) πr2× figs13 = figs 2 oe M1 2 ÷ 0.0013 M1ind (implied by 1538.46…) 2 2 ( r ) = oe M1dep Dep on M2 (489.7..) π × .00013 22.1 or 22.12 – 22.14 cao A1 www 4 figs 221… imply first M1 (d) 0.8 + 1.2 + 0.8 = (2.8) M1 Accept 2.8 seen 50.40 = area × 0.12 oe M1ind Accept 420 seen Length × their perimeter = their area oe M1 150 cao A1 www 4 [15]

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8 A packet of sweets contains chocolates and toffees. (a) There are x chocolates which have a total mass of 105 grams. Write down, in terms of x, the mean mass of a chocolate. [1] (b) There are x + 4 toffees which have a total mass of 105 grams. Write down, in terms of x, the mean mass of a toffee. [1] (c) The difference between the two mean masses in parts (a) and (b) is 0.8 grams. Write down an equation in x and show that it simplifies to x2 + 4x – 525 = 0. [4] (d) (i) Factorise x2 + 4x – 525. [2] (ii) Write down the solutions of x2 + 4x – 525 = 0. [1] (e) Write down the total number of sweets in the packet. [1] (f) Find the mean mass of a sweet in the packet. [2]

Mark scheme: 8 (a) 105 B1 Do not allow x = , but allow other letter and condone presence of units x (b) 105 B1 Do not allow x = , but allow other letter and condone presence of units x + 4 (c) 105 105 M2 SC1 if ± signs between terms incorrect = 8.0 oe or SC1 for their (a) – their (b) = 0.8 oe x −x + 4 if (a) and (b) are fractions with linear denominators 105(x + 4) – 105x = 0.8x(x + 4) oe M1 Dep on M2 or SC1 and allow all over x(x + 4) at this stage Condone any sign error in any 0.8x2 + 3.2x – 420 = 0 oe expanding done first (this is taken into account in the E mark) Completed without any errors x2 + 4x – 525 = 0 E1 dep on M3 (d) (i) (x + 25)(x – 21) B2 B1 for (x – 25)(x + 21) (ii) -25, 21 B1 ft - allow 25 and -21 from above only (e) 46 B1 ft ft 2 × a positive root + 4 (f) 210 ÷ ( their (e)) M1 4.57 or better (4.565…) ft A1 ft www 2, but 4.6 ww scores zero [12] IGCSE – May/June 2007 0580 and 0581 04

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9 Diagram 1 Diagram 2 Diagram 3 The first three diagrams in a sequence are shown above. The diagrams are made up of dots and lines. Each line is one centimetre long. (a) Make a sketch of the next diagram in the sequence. [1] (b) The table below shows some information about the diagrams. Diagram 1 2 3 4 --------- n Area 1 4 9 16 --------- x Number of dots 4 9 16 p --------- y Number of one centimetre lines 4 12 24 q --------- z (i) Write down the values of p and q. [2] (ii) Write down each of x, y and z in terms of n. [4] (c) The total number of one centimetre lines in the first n diagrams is given by the expression 2 n3 + fn2 + gn. 3 (i) Use n = 1 in this expression to show that f + g =10 . [1] 3 32 (ii) Use n = 2 in this expression to show that 4f + 2g = . [2] 3 (iii) Find the values of f and g. [3] (iv) Find the total number of one centimetre lines in the first 10 diagrams. [1]

Mark scheme: 9 (a) Sketch of 4 by 4 diagram B1 (b) (i) 25, 40 B1,B1 (ii) n2 B1 (n + 1)2 oe B1 (n + 1)2 + n2 – 1 or 2n2 + 2n) or B2 Any one of these oe isw and if B0 2n(n + 1) oe allow SC1 for their (n + 1)2 + their (n2) – 1 or an expression containing 2n2, as the highest order term, soi (c) (i) 2 + f + g = 4 B1 3 (ii) 2 × 2 3 + f × 2 2 + g × 2 oe M1 ie for substituting 2 3 32 E1 No errors Allow 10, 2 3 10., 10.7, … 4 f + 2 g = 3 (iii) 2 f + 2 g = 20 M1 for correctly setting up for elimination 3 32 of one variable 4 f + 2 g = 3 ( f = ) (,2 g = ) 4 oe cao A1A1 www 3 accept 6 3 for 2 3 (iv) 880 cao B1 [14]

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