Cambridge IGCSE Mathematics 0580 — 2007 Oct/Nov Paper 4 · Variant 1

1 Each year a school organises a concert. (a) (i) In 2004 the cost of organising the concert was $ 385. In 2005 the cost was 10% less than in 2004. Calculate the cost in 2005. [2] (ii) The cost of $ 385 in 2004 was 10% more than the cost in 2003. Calculate the cost in 2003. [2] (b) (i) In 2006 the number of tickets sold was 210. The ratio Number of adult tickets : Number of student tickets was 23 : 19. How many adult tickets were sold? [2] (ii) Adult tickets were $ 2.50 each and student tickets were $ 1.50 each. Calculate the total amount received from selling the tickets. [2] (iii) In 2006 the cost of organising the concert was $ 410. Calculate the percentage profit in 2006. [2] (c) In 2007, the number of tickets sold was again 210. Adult tickets were $ 2.60 each and student tickets were $ 1.40 each. The total amount received from selling the 210 tickets was $ 480. How many student tickets were sold? [4]

Mark scheme: 1 (a) (i) 385 × 0.9 oe M1 Implied by ans 346 or 347 ($) 346.5(0) cao A1 www2 (ii) 385 ÷ 1.1(0) oe M1 ($) 350 cao A1 www2 (b) (i) 23 M1 × 210 oe 23 +19 115 cao A1 www2 (ii) their (i) × .250 + (210 – their (i)) × 1.50 M1 (287.5 + 142.5) A1 www2 ($) 430 cao (iii) {their (ii) – 410} / 410 ( ×100 )oe M1 Dep on (ii) being greater than 410 A1 www2 (4.878 …) 4.88 After M0, SC1 for 104.9 or better or 4.9 ww (c) 2.6(210 – x) or 1.4(210 – x) seen M1 2.6(210 – x) + 1.4x = 480 M1 Allow 2.6x + 1.4 (210 – x ) = 480 546 – 480 = 2.6x – 1.4x or 2.6x – 1.4x = 480 – 294 M1 Dep on M2 55 cao A1 if trial and error, B4 or B0 if using simultaneous equations x + y = 210 M1 4.1 x + 6.2 y = 480 M1 variable eliminated by correct method M1d After 0 scored, SC2 for ans 155 [14]

More questions on Percentages

2 (a) Grade 1 2 3 4 5 6 7 Number of students 1 2 4 7 4 8 2 The table shows the grades gained by 28 students in a history test. (i) Write down the mode. [1] (ii) Find the median. [1] (iii) Calculate the mean. [3] (iv) Two students are chosen at random. Calculate the probability that they both gained grade 5. [2] (v) From all the students who gained grades 4 or 5 or 6 or 7, two are chosen at random. Calculate the probability that they both gained grade 5. [2] (vi) Students are chosen at random, one by one, from the original 28, until the student chosen has a grade 5. Calculate the probability that this is the third student chosen. [2] (b) Claude goes to school by bus. The probability that the bus is late is 0.1 . If the bus is late, the probability that Claude is late to school is 0.8 . If the bus is not late, the probability that Claude is late to school is 0.05 . (i) Calculate the probability that the bus is late and Claude is late to school. [1] (ii) Calculate the probability that Claude is late to school. [3] (iii) The school term lasts 56 days. How many days would Claude expect to be late? [1]

Mark scheme: 2 (a) (i) 6 B1 (ii) 4.5 B1 (iii) 1( × 1 + 2 × 2 + 4 × 3 + 7 × 4 + 4 × 5 + M1 Allow 1 slip 8 × 6 + 2 × 7 ) (127) ÷ 28 M1dep dep on 1st M1 4.54 A1 www 3 4.53571… (iv) 4 3 M1 Accept all probabilities as fracts/dec/% × -1 once for words or 2 sf, do not accept 28 27 ratios i.s. cancelling after correct answer. 1 A1 www2 e.g. ( 75612 , 0.0159 etc) o.e. 63 (v) 4 3 M1 × 21 20 1 A1 www2 e.g. ( 42012 , 0.0286 etc) o.e. 35 (vi) 24 23 4 M1 × × 28 27 26 92 A1 www2 e.g. ( 196562208 , 0.112) o.e. 819 (b) (i) 0.08 o.e. B1 (ii) 0.9 × 0.05 M1 their (b)(i) + 0.9 × 0.05 M1dep dep on 1st M1 0.125 o.e. A1 www3 (iii) 7 B1 ft their (ii) × 56 either correct to 3sf or better or r.o.t. [16] IGCSE – October/November 2007 0580/0581 04

More questions on Averages and measures of spread

3 y A NOT TO SCALE B C x 0 The diagram shows a sketch of y = x2 + 1 and y = 4 – x. (a) Write down the co-ordinates of (i) the point C, [1] (ii) the points of intersection of y = 4 – x with each axis. [2] (b) Write down the gradient of the line y = 4 – x. [1] (c) Write down the range of values of x for which the gradient of the graph of y = x2 + 1 is negative. [1] (d) The two graphs intersect at A and B. Show that the x co-ordinates of A and B satisfy the equation x2 + x – 3 = 0. [1] (e) Solve the equation x2 + x – 3 = 0, giving your answers correct to 2 decimal places. [4] (f) Find the co-ordinates of the mid-point of the straight line AB. [2]

Mark scheme: 3 (a) (i) (0, 1) B1 Accept w/out brackets/ commas, condone (ii) (4, 0) and (0, 4) B1B1 vectors, or states x = , y = (b) -1 cao B1 (c) (x) < 0 (allow ≤) B1 Any other variable < 0 B0 (d) x 2 + 1 = 4 − x o.e. B1 must be these 4 terms (e) M1 p +(-)√q where p = −1 and r = 2×1 r and q = 1² − 4(1)(-3) o.e. M1 q Allow second mark if in form p± r -2.30 , 1.30 cao www4 A1A1 If ww ans.correct but wrong acc - SC3 After A0, A0, SC1 for -2.3027756 and 1.3027756 rounded or truncated (f) (-0.5, 4.5 or 4.49) B1ft f.t (their –2.30 + their 1.30) ÷2 B1 ft ft (4 – their x co-ord dep on attempt at mid value of x from values in e) [12]

More questions on Graphs of functions

4 [The surface area of a sphere of radius r is 4πr2 and the volume is 4 πr 3 .] 3 (a) A solid metal sphere has a radius of 3.5 cm. One cubic centimetre of the metal has a mass of 5.6 grams. Calculate (i) the surface area of the sphere, [2] (ii) the volume of the sphere, [2] (iii) the mass of the sphere. [2] (b) NOT TO SCALE h 8 cm 16 cm 16 cm Diagram 1 Diagram 2 Diagram 1 shows a cylinder with a diameter of 16 cm. It contains water to a depth of 8 cm. Two spheres identical to the sphere in part (a) are placed in the water. This is shown in Diagram 2. Calculate h, the new depth of water in the cylinder. [4] (c) A different metal sphere has a mass of 1 kilogram. One cubic centimetre of this metal has a mass of 4.8 grams. Calculate the radius of this sphere. [3]

Mark scheme: 4 (a) (i) 4 π 5.3 2 = 153.86 to 153.96 or 154 M1A1 www2 (ii) 4 3 π 5.3 3 = 179.5 to 179. 62 or 180 M1A1 www2 (iii) their (ii)× 5.6 M1 1005 to 1006 or 1008or 1010 (g) A1ft their (ii)× 5.6 correct to 3sf or better (allow in kg) (b) π 82 × 8 (1608-1609) M1 Alt π 8 2 d = 2 × their (ii) M1 2 M1dep 2 π 8 h = 2×their (ii) + π 82 × 8 (2×their (a)(ii)) ÷ (π 8 ) M1dep (2×their (ii) + π 82 × 8 ) ÷( π 8 2 ) M1dep add 8 M1dep 9.78 to 9.79 (cm) A1 www4 (c) 1000 (or 1) ÷4.8 ÷ 43 π M1 49.7….. (or 0.0497) 3 3 M1dep Dep on previous M1 ans (or 10 × ans ) 3.67 to 3.68 (cm) A1 www3 figs 368 or ans 3.7 gets M2 [13] IGCSE – October/November 2007 0580/0581 04

More questions on Area and perimeter

5 S NOT TO SCALE D C 8 cm P 6 cm N R 7 cm M A B Q The diagram above shows the net of a pyramid. The base ABCD is a rectangle 8 cm by 6 cm. All the sloping edges of the pyramid are of length 7 cm. M is the mid-point of AB and N is the mid-point of BC. (a) Calculate the length of (i) QM, [2] (ii) RN. [1] (b) Calculate the surface area of the pyramid. [2] (c) P 7 cm D G C NOT TO H N 6 cm SCALE X M A B 8 cm The net is made into a pyramid, with P, Q, R and S meeting at P. The mid-point of CD is G and the mid-point of DA is H. The diagonals of the rectangle ABCD meet at X. (i) Show that the height, PX, of the pyramid is 4.90 cm, correct to 2 decimal places. [2] (ii) Calculate angle PNX. [2] (iii) Calculate angle HPN. [2] (iv) Calculate the angle between the edge PA and the base ABCD. [3] (v) Write down the vertices of a triangle which is a plane of symmetry of the pyramid. [1]

Mark scheme: 5 (a) (i) 2 2 M1A1 www2 5.74456… 7 − 4 = 5.74 (cm) (ii) 6.32 (cm) B1 6.32455….. (b) 1 1 M1 2× 2 × 8×’5.74’+2× 2 × 6×’6.32’ + 8×6 131.8 to 132(cm2) A1ft www2 ft 48 +8×their (a)(i) + 6×their (a)(ii) (c) (i) ((PX)² ) = (their (a)(i))² - 3² M1 or their a(ii)² - 4² or 7² - (3² + 4²) E1 24 soi or 4.898..… seen (ii) their ( c )(i ) M1 Alt correct trig methods involving their (a)(ii) Tan(PNX) = o.e. M1 for correct explicit statement 4 50.7 to 50.84 oe A1 www2 for a trig ratio (iii) (HPN) 180 – 2 × their (ii) M1 78.3 to 79 A1 www2 Alt – cos rule method – M1 at explicit stage (iv) their ( c )(i ) M2 M1 for recognition of angle PAX or PAC oe tan = o.e. 5 44.4 to 44.43° A1 Alt trig methods with PA = 7 used www3 44.4153086 (v) PHN or PGM o.e. (letters) B1 B0 if extras [15]

More questions on Pythagoras’ theorem

6 18 m D C NOT TO SCALE 30 m 80º A B 26 m The diagram shows the plan of a garden. The garden is a trapezium with AB = 26 metres, DC = 18 metres and angle DAB = 80°. A straight path from B to D has a length of 30 metres. (a) (i) Using a scale of 1 : 200, draw an accurate plan of the garden. [3] (ii) Measure and write down the size of angle ADB and the size of angle DCB. [2] (iii) A second path is such that all points on it are equidistant from AB and from AD. Using a straight edge and compasses only, construct this path on your plan. [2] (iv) A third path is such that all points on it are equidistant from A and from D. Using a straight edge and compasses only, construct this path on your plan. [2] (v) In the garden, vegetables are grown in the region which is nearer to AB than to AD and nearer to A than to D. Shade this region on your plan. [1] (b) Use trigonometry, showing all your working, to calculate (i) angle ADB, [3] (ii) the length of BC, [4] (iii) the area of the garden. [3]

Mark scheme: 6 (a) (i) AB=13 cm and BD=15 cm (± 2 mm) B1 Angle A = 80° (± 2°) B1 A,B,C,D correct within 4 mm B1 Dep. on B2 (ii) Angle ADB correct (57-61°) (± 2°) B1ft Either in working or written on diagram B1ft Angle DCB correct (101-105°) (± 2°) (iii) Acc. bisector of angle A with arcs B2ft B1 for accurate without/wrong arcs (at least 5 cm long) (± 2°)(± 2 mm) (iv) Acc. perp. bisector of AD with at least 1 B2ft B1 for accurate without/wrong arcs pair of arcs (± 2°)(± 2 mm) (at least 5 cm B1 for each if accurate with arcs but short long) (v) ‘Correct’ area shaded below their perp. B1 Dep. on at least B1 in (iii) and B1 in (iv) bisector and below their angle bisector (b) (i) sin D sin 80 M1 No M marks in (b) for measuring + using = lengths from diagram e.g. AD = 20 m 26 30 but allow 13, 15, 9 used for 26, 30, 18 in b 26 sin 80 M1dep dep on 1st M (sin D = ) 30 58.57 to 58.6° A1 www3 (ii) Angle BDC = 41.4 B1 ft Ft 100 – their 58.6 (BC2 =)182 + 302 – 2 × 18× 30 cos'41 '4. M1 Allow 41 or 42 for angle BDC square root of correct collection M1dep Dep on 1st M (413.88… ) 20.3 to 20.35 (m) cao A1 www4 (iii) 5.0 × 26 × 30 sin'41 '4.+ M2 M1 for correct area of one triangle oe (257.9 or 178.6). Must see calc for 5.0 × 18 × 30 sin'41 '4. trapezium height if used (30sin ‘41.4’) Allow 41 or 42 for angle BDC 436 to 437 (m2) cao A1 www3 [20] IGCSE – October/November 2007 0580/0581 04

More questions on Scale drawings

7 Answer the whole of this question on a sheet of graph paper. (a) Draw x and y axes from 0 to 12 using a scale of 1 cm to 1 unit on each axis. [1] (b) Draw and label triangle T with vertices (8, 6), (6, 10) and (10, 12). [1] (c) Triangle T is reflected in the line y = x. (i) Draw the image of triangle T. Label this image P. [2] (ii) Write down the matrix which represents this reflection. [2] 1 0 (d) A transformation is represented by the matrix 2 1 0 2 (i) Draw the image of triangle T under this transformation. Label this image Q. [2] (ii) Describe fully this single transformation. [3] (e) Triangle T is stretched with the y-axis invariant and a stretch factor of 1 . 2 Draw the image of triangle T. Label this image R. [2]

Mark scheme: 7 (a) Correct axes S1 must fit on paper 2mm acc throughout Ignore labels on triangles throughout (b) Correct triangle drawn (T) T1 vertices at (8, 6), (6, 10) and (10, 12) (c) (i) Correct reflection in y = x drawn (P) P2ft ft their T, P1 for two correct vertices drawn (6, 8), (10, 6), (12, 10) or line y = x correctly drawn (within 2mm of (12,12) if extended) (ii) B2 B1 for a correct column  0 1     1 0  (d) (i) Correct enlargement, scale factor 0.5, Q2ft (4, 3), (3, 5), (5, 6) centre (0,0) drawn (Q) Q1 for any enlargement s.f. ½ or 2 correct vertices drawn SC1 for 3 points within 5 mm if rays method used or for correct enlargement but of P (ii) Enlargement only B1 (scale factor) 0.5 B1 indep (centre) (0, 0) o.e. B1 indep (e) Correct stretch drawn (R) R2ft R1 for two correct vertices ft (4, 6), (3, 10), (5, 12) [13]

More questions on Transformations

38 f(x) = 2x – 1, g(x) = +1 , h(x) = 2x. x (a) Find the value of fg(6). [1] (b) Write, as a single fraction, gf(x) in terms of x. [3] (c) Find g−1(x). [3] (d) Find hh(3). [2] (e) Find x when h(x) = g −24 [2] 7 Question 9 is printed on the next page.

Mark scheme: 8 (a) 2 B1 (b) 3 M1 + 1 2 x − 1 3 + 2 x − 1 M1 Dep on 1st M1 2 x − 1 2 + 2x A1 www3 o.e. final ans 2x − 1 (c) 3 3 y = + 1 x = + 1 x y 3 M1 3 y − 1 = or xy = 3 + x Alt x − 1 = x y x ( y − )1 = 3 M1dep Dep on 1st M1 y ( x − )1 = 3 3 A1 3 o.e. final answer www3 o.e x − 1 x − 1 3 If answer is x = allow M2 x − 1 (d) 256 B2 8 B1 for 23 = 8 or 2 seen (e) x 3 M1 M for r.h.s. followed by attempt at 2 = + 1 24 recognising 2x = ………………… − 7 -3 A1 After M0, SC1 for 1/8 o.e seen www2 [11] IGCSE – October/November 2007 0580/0581 04

More questions on Functions

9 The table shows some terms of several sequences. Term 1 2 3 4 8 Sequence P 7 5 3 1 p Sequence Q 1 8 27 64 q 1 2 3 4 Sequence R r 2 3 4 5 Sequence S 4 9 16 25 s Sequence T 1 3 9 27 t Sequence U 3 6 7 −2 u (a) Find the values of p, q, r, s, t and u. [6] (b) Find the nth term of sequence (i) P, [1] (ii) Q, [1] (iii) R, [1] (iv) S, [1] (v) T, [1] (vi) U. [1] (c) Which term in sequence P is equal to −777? [2] (d) Which term in sequence T is equal to 177 147? [2]

Mark scheme: 9 (a) -7, 512, 8 , 81, 2187, -2106 B6 B1 each. Allow in any order ignore letters 9 (b) (i) (P) 9 – 2n B1 Accept correct expressions in any form e.g. 7 – 2(n – 1) (ii) (Q) n3 B1 If ‘n =’ withhold the first mark earned (iii) n B1 (R) n + 1 (iv) (S) (n + 1)2 B1 (v) n − 1 B1 (T) 3 (vi) (U) (n + 1)2 - 3 n − 1 B1ft their (iv)-their (v) dep on both algebraic expressions (c) their(b)(i) = – 777 M1 393 cao A1 www2 (d) 12 B2 SC1 for 11 or n - 1 = 11 or 312 3,11 seen [16]

More questions on Sequences