Cambridge IGCSE Mathematics 0580 — 2013 Oct/Nov Paper 4 · Variant 1

1 David sells fruit at the market. For Examiner′s Use (a) In one week, David sells 120 kg of tomatoes and 80 kg of grapes. (i) Write 80 kg as a fraction of the total mass of tomatoes and grapes. Give your answer in its lowest terms. Answer(a)(i) ............................................... [1] (ii) Write down the ratio mass of tomatoes : mass of grapes. Give your answer in its simplest form. Answer(a)(ii) ...................... : ...................... [1] (b) (i) One day he sells 28 kg of oranges at $1.56 per kilogram. He also sells 35 kg of apples. The total he receives from selling the oranges and the apples is $86.38 . Calculate the price of 1 kilogram of apples. Answer(b)(i) $ ............................................... [2] (ii) The price of 1 kilogram of oranges is $1.56 . This is 20% more than the price two weeks ago. Calculate the price two weeks ago. Answer(b)(ii) $ ............................................... [3] (c) On another day, David received a total of $667 from all the fruit he sold. The cost of the fruit was $314.20 . 1 David worked for 10 2 hours on this day. Calculate David’s rate of profi t in dollars per hour. Answer(c) ................................ dollars/h [2] _____________________________________________________________________________________

Mark scheme: Qu Answers Mark Part Marks 1 2 (a) (i) cao 1 5 (ii) 3 : 2 cao 1 (b) (i) 1.22 2 M1 for 86.38 – 28 × 1.56 (ii) 1.3 [0] nfww 3 M2 for 1.56 ÷ 1.2 oe or M1 for 1.56 = 120% soi (c) 33.6[0] 2 M1 for (667 – 314.2) ÷ 10.5 oe

More questions on Ratio and proportion

2 Emily cycles along a path for 2 minutes. For Examiner′s She starts from rest and accelerates at a constant rate until she reaches a speed of 5 m/s after 40 seconds. Use She continues cycling at 5 m/s for 60 seconds. She then decelerates at a constant rate until she stops after a further 20 seconds. (a) On the grid, draw a speed-time graph to show Emily’s journey. 5 4 3 Speed (m/s) 2 1 0 10 20 30 40 50 60 70 80 90 100 110 120 Time (seconds) [2] (b) Find Emily’s acceleration. Answer(b) ....................................... m/s2 [1] (c) Calculate Emily’s average speed for the journey. Answer(c) ........................................ m/s [4] _____________________________________________________________________________________

Mark scheme: 2 (a) 3 correct lines on grid 2 Allow good freehand (0, 0) to (40, 5) SC1FT for 2 lines correct, FT from an incorrect (40, 5) to (100, 5) line (100, 5) to (120, 0) 5 (b) oe 1 40 (c) 3.75 4 M2 for 0.5 × 40 × 5 + 60 × 5 + 0.5 × 20 × 5 oe [450] or M1 for evidence of a relevant area = distance and M1dep their area (or distance) ÷ 120 IGCSE – October/November 2013 0580 41

More questions on Rates

3 For Examiner′s Use NOT TO SCALE 13 cm h 5 cm (a) The diagram shows a cone of radius 5 cm and slant height 13 cm. (i) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(a)(i) ........................................ cm2 [2] (ii) Calculate the perpendicular height, h, of the cone. Answer(a)(ii) h = ......................................... cm [3] (iii) Calculate the volume of the cone. 1 [The volume, V, of a cone with radius r and height h is V = 3 πr2h.] Answer(a)(iii) ........................................ cm3 [2] (iv) Write your answer to part (a)(iii) in cubic metres. Give your answer in standard form. Answer(a)(iv) .......................................... m3 [2] (b) For Examiner′s A Use O NOT TO SCALE 13 cm h O 5 cm B The cone is now cut along a slant height and it opens out to make the sector AOB of a circle. Calculate angle AOB. Answer(b) Angle AOB = ............................................... [4] _____________________________________________________________________________________

Mark scheme: 3 (a) (i) 204 or 204.2 to 204.23 2 M1 for π × 5× 13 implied by answer in range 204.1 to 204.3 (ii) 12 cao 3 M2 for 13 2 − 5 2 or states 5, 12, 13 triangle or M1 for 132 = 52 + h2 or better 1 2 (iii) 314 or 314.1 to 314.2 2 M1 for × π × 5 × their (a) (ii) implied by 3 answer in range 314 to 314.3 (iv) 3.14 × 10–4 or 3.141 to 2FT FT their (a) (iii) ÷ 1003 correctly evaluated and 3.142 × 10–4 given in standard form to 3 sig figs or better or M1 FT for their (a) (iii) ÷ 1003 or SC1 for conversion of their m3 into standard form only if negative power 10π (b) 138 or 138.3 to 138.5 4 M3 for × 360 oe or 26π π × 5 × 13 ortheir (a) (i) × 360 oe 2 π × 13 or M2 for a correct fraction without × 360 or M1 for π × 2× 13 oe [81.6 to 81.8] seen or π × 13 2 oe [530.6 to 531.2] seen 2 2

More questions on Area and perimeter

4 For Examiner′s D C Use 32° 70 m NOT TO SCALE 40° A B 55 m The diagram shows a school playground ABCD. ABCD is a trapezium. AB = 55 m, BD = 70 m, angle ABD = 40° and angle BCD = 32°. (a) Calculate AD. Answer(a) AD = ........................................... m [4] (b) Calculate BC. Answer(b) BC = ........................................... m [4] (c) (i) Calculate the area of the playground ABCD. For Examiner′s Use Answer(c)(i) .......................................... m2 [3] (ii) An accurate plan of the school playground is to be drawn to a scale of 1: 200 . Calculate the area of the school playground on the plan. Give your answer in cm2. Answer(c)(ii) ........................................ cm2 [2] (d) A fence, BD, divides the playground into two areas. Calculate the shortest distance from A to BD. Answer(d) ........................................... m [2] _____________________________________________________________________________________

Mark scheme: 4 (a) 45.[0] or 45.01 to 45.02 nfww 4 M2 for 552 + 702 – 2.55.70 cos 40 or M1 for correct implicit equation A1 for 2026. …. (b) 84.9 or 84.90 to 84.92 4 B1 for angle BDC = 40 soi 70 sin (their 40 ) M2 for sin 32 or M1 for correct implicit equation (c) (i) 4060 or 4063 to 4064 nfww 3 1 1 M2 for (55 × 70 sin 40 ) + 2 2 (70 × their (b ) sin (180 − their 40 − 32)) oe or M1 for correct method for one of the triangle areas (ii) 1020 or 1015 to 1016 2FT FT their (c) (i) ÷ 4 oe correctly evaluated or M1 their (c) (i) ÷ figs 4 oe distance (d) 35.4 or 35.35… nfww 2 M1 for sin 40 = or better 55 1 or for (55 × 70 sin 40) = (70 × distance) ÷ 2 2 or better IGCSE – October/November 2013 0580 41

More questions on Right-angled triangles

5 (a) For Examiner′s y Use 10 9 8 7 6 5 4 3 2 T U 1 x 0 1 2 3 4 5 6 7 8 9 10 (i) Draw the refl ection of triangle T in the line y = 5. [2] (ii) Draw the rotation of triangle T about the point (4, 2) through 180°. [2] (iii) Describe fully the single transformation that maps triangle T onto triangle U. Answer(a)(iii) .................................................................................................................... [3] (iv) Find the 2 × 2 matrix which represents the transformation in part (a)(iii). Answer(a)(iv) [2] f p (b) For Examiner′s P Q Use NOT TO SCALE p R O s S In the pentagon OPQRS, OP is parallel to RQ and OS is parallel to PQ. PQ = 2OS and OP = 2RQ. O is the origin, = p and = s. Find, in terms of p and s, in their simplest form, (i) the position vector of Q, Answer(b)(i) ............................................... [2] (ii) . Answer(b)(ii) = ............................................... [2] (c) Explain what your answers in part (b) tell you about the lines OQ and SR. Answer(c) .................................................................................................................................. [1] _____________________________________________________________________________________

Mark scheme: 5 (a) (i) Correct reflection to (4, 8) 2 SC1 for reflection in line x = 5 (2, 9) (4, 9) or reflection in y = k Ignore additional triangles (ii) Correct rotation to (4, 2), (4, 3) 2 SC1 for rotation 180˚ with incorrect centre (6, 3) Ignore additional triangles (iii) Shear, x-axis oe invariant, 3 B1 each (independent) [factor] 2  1 2  (iv)   2FT FT their shear factor 0 1   B1FT for one correct column or row in 2 by 2 matrix but not identity matrix  1 0  or SC1FT for   2 1   (b) (i) p + 2s final answer 2 M1 for recognising OQ as position vector soi 1 1 (ii) s + p final answer 2 B1 for s + kp or ks + p 2 2 or correct route (k ≠ 0) (c) parallel and OQ = 2SR oe 1

More questions on Transformations

6 (a) For Examiner′s y Use 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 The diagram shows the graph of y = f(x) for –3 Ğ x Ğ 3. (i) Find f(2). Answer(a)(i) ............................................... [1] (ii) Solve the equation f(x) = 0. Answer(a)(ii) x = ............................................... [1] (iii) Write down the value of the largest integer, k, for which the equation f(x) = k has 3 solutions. Answer(a)(iii) k = ............................................... [1] (iv) By drawing a suitable straight line, solve the equation f(x) = x. Answer(a)(iv) x = ..................... or x = ..................... or x = ..................... [3] (b) g(x) = 1 – 2x h(x) = x2 – 1 For Examiner′s Use (i) Find gh(3). Answer(b)(i) ............................................... [2] (ii) Find g–1(x). Answer(b)(ii) g–1(x) = ............................................... [2] (iii) Solve the equation h(x) = 3. Answer(b)(iii) x = .................... or x = .................... [3] (iv) Solve the equation g(3x) = 2x. Answer(b)(iv) x = ............................................... [3] _____________________________________________________________________________________

Mark scheme: 6 (a) (i) 1.4 to 1.6 1 (ii) 1.15 to 1.25 1 (iii) – 1 1 (iv) – 2.25 to – 2.1 3 B2 for 2 correct or B1 for one correct – 0.9 to – 0.75 or B1 for y = x drawn ruled to cut curve 3 times 2.2 to 2.35 (b) (i) – 15 2 B1 for [h(3) =] 8 seen or M1 for 1 – 2(x2 – 1) or better 1 − x 1 x (ii) or − oe final answer 2 M1 for2 x = 1 – y or x = 1–2y or better 2 2 2 (iii) – 2, 2 3 M1 for x2 – 1 = 3 or better B1 for one answer 1 3 M2 for 8x = 1 or 8x – 1 = 0 (iv) oe nfww 8 or M1 for 1 – 2(3x) [= 2x] IGCSE – October/November 2013 0580 41

More questions on Graphs of functions

7 120 students are asked to answer a question. For Examiner′s The time, t seconds, taken by each student to answer the question is measured. Use The frequency table shows the results. Time 0 < t Y 10 10 < t Y 20 20 < t Y 30 30 < t Y 40 40 < t Y 50 50 < t Y 60 Frequency 6 44 40 14 10 6 (a) Calculate an estimate of the mean time. Answer(a) ............................................ s [4] (b) (i) Complete the cumulative frequency table. Time t Y 10 t Y 20 t Y 30 t Y 40 t Y 50 t Y 60 Cumulative frequency 6 104 120 [2] (ii) On the grid below, draw a cumulative frequency diagram to show this information. 120 100 80 Cumulative 60 frequency 40 20 t 0 10 20 30 40 50 60 Time (seconds) [3] (iii) Use your cumulative frequency diagram to fi nd the median, the lower quartile and For Examiner′s the 60th percentile. Use Answer(b)(iii) Median ............................................ s Lower quartile ............................................ s 60th percentile ............................................ s [4] (c) The intervals for the times taken are changed. (i) Use the information in the frequency table on the opposite page to complete this new table. Time 0 < t Y 20 20 < t Y 30 30 < t Y 60 Frequency 40 [2] (ii) On the grid below, complete the histogram to show the information in the new table. One column has already been drawn for you. 4 3.5 3 2.5 Frequency density 2 1.5 1 0.5 t 0 10 20 30 40 50 60 Time (seconds) [3] _____________________________________________________________________________________

Mark scheme: 7 (a) 24.7 or 24.66 to 24.67 4 M1 for midpoints soi (condone 1 error or omission) (5, 15, 25, 35, 45, 55) and M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) and M1 (dependent on second M) for ∑fx ÷ 120 (b) (i) 50, 90, 114 2 B1 for 2 correct (ii) Correct curve 3 Ignore section to left of t = 10 or ruled polygon B1 for 6 correct horizontal plots and B1FT for 6 correct vertical plots If 0 scored SC1 for 5 out of 6 correct plots and B1FT for curve or polygon through at least 5 of their points dep on an increasing curve/polygon that reaches 120 vertically (iii) 21.5 to 23 B1 15 to 16.5 B1 24 to 26 4 B2 or B1 for 72 or 72.6 seen (c) (i) 50, 30 2 B1 each (ii) Correct histogram 3FT B1 for blocks of widths 0 – 20, 30 – 60 (no gaps) B1FT for block of height 2.5 or their 50 ÷ 20 and B1FT for block of height 1 or their 30 ÷ 30 IGCSE – October/November 2013 0580 41

More questions on Statistical charts and diagrams

8 (a) Solve the equation 8x2 – 11x – 11 = 0. For Examiner′s Show all your working and give your answers correct to 2 decimal places. Use Answer(a) x = ........................ or x = ........................ [4] (b) y varies directly as the square root of x. y = 18 when x = 9. Find y when x = 484. Answer(b) y = ........................... [3] (c) Sara spends $x on pens which cost $2.50 each. For Examiner′s She also spends $(x – 14.50) on pencils which cost $0.50 each. Use The total of the number of pens and the number of pencils is 19. Write down and solve an equation in x. Answer(c) x = ............................................... [6] _____________________________________________________________________________________

Mark scheme: 2  8 (a) (− 11) − 4(8 )(− 11) or better B1 Seen anywhere or for  x − 11  16  + − p q p q p = –(– 11), r = 2(8) or better B1 Must be in the form or r r 11  11  2 11 or B1 for +   + 8  16  16 – 0.67, 2.05 final answers B1B1 SC1 for – 0.7 or – 0.672 to – 0.671 and 2.0 or 2.046 to 2.047 or answers 0.67 and – 2.05 (b) 132 3 M1 for y = k x oe or x = ky oe A1 for k = 6 oe or better or for k = 0.1666 to 0.167 [k = 6 implies M1A1] oe x x − 14 5. (c) 20 with supporting algebraic working 6 B2 for + = 19 oe 5.2 5.0 x x − 14 5. or B1 for or 5.2 5. M1dep on B2 for first completed correct move to clear both fractions M1 for second completed correct move to collect terms in x to a single term M1 for third completed correct move to collect numeric term[s] leading to ax = b SC1 for 20 with no algebraic working

More questions on Percentages

9 For Examiner′s y Use L2 5 4 3 R 2 L1 1 x 0 1 2 3 4 5 6 7 8 9 10 L3 (a) Find the equations of the lines L1, L2 and L3. Answer(a) L1 ............................................... L2 ............................................... L3 ............................................... [5] (b) Write down the three inequalities that defi ne the shaded region, R. Answer(b) ................................................ ................................................ ................................................ [3] (c) A gardener buys x bushes and y trees. For Examiner′s The cost of a bush is $30 and the cost of a tree is $200. Use The shaded region R shows the only possible numbers of bushes and trees the gardener can buy. (i) Find the number of bushes and the number of trees when the total cost is $720. Answer(c)(i) ...................................... bushes ...................................... trees [2] (ii) Find the number of bushes and the number of trees which give the greatest possible total cost. Write down this greatest possible total cost. Answer(c)(ii) .......................................... bushes .......................................... trees Greatest possible total cost = $ ............................................... [3] _____________________________________________________________________________________

Mark scheme: 9 (a) y = 2 oe 1 y = 2x oe 2 M1 for y = kx, k ≠ 0 or gradient 2 soi 1 M1 for gradient – ½ soi or y = kx + 5oe y = – x + 5 oe 2 2 or x + 2y = k k ≠ 0 oe If L2 and L3 both correct but interchanged then SC3 (b) y ≥ 2 oe y ≤ 2x oe 1 y ≤ – x + 5 oe 3 B1 for each correct inequality, allow in any 2 order After 0 scored, SC1 for all inequalities reversed (c) (i) 4 [bushes], 3 [trees] 2 M1 for any correct trial using integer coordinates in region or 30x + 200y = 720 seen (ii) 2 [bushes], 4 [trees] 2 M1 for any correct trial using integer coordinates in region 860 1 IGCSE – October/November 2013 0580 41

More questions on Equations of linear graphs

10 (a) 1 = 1 For Examiner′s Use 1 + 2 = 3 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10 (i) Write down the next line of this pattern. Answer(a)(i) ...................................................................................................................... [1] n (ii) The sum of the fi rst n integers is (n + 1). k Show that k = 2. Answer(a)(ii) [2] (iii) Find the sum of the fi rst 60 integers. Answer(a)(iii) ............................................... [1] (iv) Find n when the sum of the fi rst n integers is 465. Answer(a)(iv) n = ............................................... [2] (n - 8)(n - 7) (v) 1 + 2 + 3 + 4 + ....... + x = 2 Write x in terms of n. Answer(a)(v) x = ............................................... [1] (b) 13 = 1 For Examiner′s Use 13 + 23 = 9 13 + 23 + 33 = 36 13 + 23 + 33 + 43 = 100 (i) Complete the statement. 13 + 23 + 33 + 43 + 53 = ...................... = (......................)2 [2] (ii) The sum of the fi rst n integers is n (n + 1). 2 Find an expression, in terms of n, for the sum of the fi rst n cubes. Answer(b)(ii) ............................................... [1] (iii) Find the sum of the fi rst 19 cubes. Answer(b)(iii) ............................................... [2] _____________________________________________________________________________________

Mark scheme: 10 (a) (i) 1 + 2 + 3 + 4 + 5 = 15 1 n (n + 1) (ii) Correct substitution equating to 2 M1 for using a value of n in k sum 2(2 + 1) 2 (2 + 1) e.g. = 3 and k = 2 stated e.g. = 3 k k with no errors seen or for a verification using k = 2 2 (2 + 1) e.g. = 3 2 (iii) 1830 1 n (n + 1) (iv) 30 2 M1 for = 465 or better 2 (v) n – 8 1 (b) (i) 225, 15 2 B1 either 2 n 2 (n + 1) (ii) oe 1 4 19 2 (19 + 1)2 (iii) 36100 2 M1 for oe or 1902 4

More questions on Types of number