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Mathematics 0580/41 May/June 2019: question paper and mark scheme

Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0580_s19_qp_410580_s19_ms_410580/41/M/J/19

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Paper
Paper 4 (Extended)
Total marks
130
Time
2 hours 30 minutes
Code on the paper
0580/41/M/J/19
0580_s19_ms_41.pdf

0580/41 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 111 marks

Question

y 10 9 8 7 6 5 4 3 T 2 1 0 1 2 3 4 5 6 7 8 9 10 x -1 (a) (i) Translate shape T by the vector c m. 6 Label the image A. [2] (ii) Rotate shape T about the point (5, 3) through 180°. Label the image B. [2] (iii) Describe fully the single transformation that maps shape A onto shape B. …… [3] (b) (i) Reflect shape T in the line y = x. [2] (ii) Find the matrix that represents the transformation in part (b)(i). f p [2]

Mark scheme

1(a)(i) Image at (1, 7), (4, 7), (4, 9), (3, 9) 2  −1 k  B1 for translation by   or   k 6 1(a)(ii) Image at (5, 3), (6, 3), (8, 5), (5, 5) 2 B1 for 180° rotation with wrong centre 1(a)(iii) Rotation 3 B1 for rotation 180˚ B1 for 180° (4.5, 6) B1FT for centre from their (a)(i) OR Enlargement, B1 for enlargement [factor] – 1 B1 for – 1 (4.5, 6) B1FT for centre from their (a)(i) 1(b)(i) Image at (1, 2), (1, 5), (3, 5), (3, 4) 2 B1 for y = x drawn or for 3 correct points 1(b)(ii) 0 1 2 B1 for one correct row or one column   within a 2 by 2 matrix 1 0

Question 211 marks

Question

The table shows some values for y = x 3 + 3x 2 + 2 . x -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 y -4.1 5.1 6 5.4 4 2.6 2.9 12.1 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 + 3x 2 + 2 for - 3.5 G x G 1.5 . y 15 10 5 – 3.5 –3 – 2.5 –2 – 1.5 –1 – 0.5 0 0.5 1 1.5 x –5 [4] (c) Use your graph to solve the equation x 3 + 3x 2 + 2 = 0 for - 3.5 G x G 1.5 . x = …… [1] (d) By drawing a suitable straight line, solve the equation x 3 + 3x 2 + 2x + 2 = 0 for - 3.5 G x G 1.5 . x = …… [2] (e) For - 3.5 G x G 1.5 , the equation x 3 + 3x 2 + 2 = k has three solutions and k is an integer. Write down a possible value of k. k = …… [1]

Mark scheme

2(a) 2, 2, 6 3 B1 for each 2(b) Correct graph 4 B3FT for 10 or 11 correct plots or B2FT for 8 or 9 correct plots or B1FT for 6 or 7 correct plots 2(c) –3.3 to –3.1 1 FT their graph 2(d) y = –2x ruled M1 or B1 for y = −2 x stated –2.6 to –2.45 A1 2(e) 3 or 4 or 5 1 FT their graph Allow more than one correct value

Question 316 marks

Question

C North D 170 m 120 m NOT TO 150 m SCALE E 50 m A 100 m B The diagram shows a field ABCDE. (a) Calculate the perimeter of the field ABCDE. …… m [4] (b) Calculate angle ABD. Angle ABD = …… [4] (c) (i) Calculate angle CBD. Angle CBD = …… [2] (ii) The point C is due north of the point B. Find the bearing of D from B. …… [2] (d) Calculate the area of the field ABCDE. Give your answer in hectares. [1 hectare = 10 000 m2] …… hectares [4]

Mark scheme

3(a) 530 4 B3 for [DE] = 130 m and [DC] = 80 m or B2 for [DE] = 130 m or [DC] = 80 m or M1 for 502 + 1202 or 1702 – 1502 3(b) 52.9 or 52.89… 4 1002 + 1502 − 1202 M2 for 2 × 100 × 150 or M1 for 1202 = 1002 + 1502 – 2 × 100 × 150cos(…) 181 A1 for 0.603 or 0.6033…or 300 3(c)(i) 28.1 or 28.07… 2 15 M1 for cos = oe 17 3(c)(ii) 331.9 or 331.9… 2 FT 360 – their (c)(i) M1 for 360 – their (c)(i) oe 3(d) 1.5[0] or 1.498… nfww 4 1 M1 for × 50 × 120 oe 2 1 M1 for × 100 × 150sin(their (b)) oe 2 1 M1 for × 150 × theirCD oe 2 1 or × 150 × 170 × sin their (c)(i) 2 If 0 scored, SC1 for dividing their area by 10 000

Question 418 marks

Question

(a) The test scores of 14 students are shown below. 21 21 23 26 25 21 22 20 21 23 23 27 24 21 (i) Find the range, mode, median and mean of the test scores. Range = …… Mode = …… Median = …… Mean = …… [6] (ii) A student is chosen at random. Find the probability that this student has a test score of more than 24. …… [1] (b) Petra records the score in each test she takes. The mean of the first n scores is x. The mean of the first (n – 1) scores is (x + 1). Find the nth score in terms of n and x. Give your answer in its simplest form. …… [3] (c) During one year the midday temperatures, t°C, in Zedford were recorded. The table shows the results. Temperature (t°C) 0 1 t G 10 10 1 t G 15 15 1 t G 20 20 1 t G 25 25 1 t G 35 Number of days 50 85 100 120 10 (i) Calculate an estimate of the mean. …… °C [4] (ii) Complete the histogram to show the information in the table. 25 20 15 Frequency density 10 5 0 0 5 10 15 20 25 30 35 t Temperature (°C) [4]

Mark scheme

4(a)(i) range = 7 1 mode = 21 1 median = 22.5 2 M1 for evidence of middle value mean = 22.7 or 22.71… 2 M1 for use of Σx ÷ 14 4(a)(ii) 3 1 oe 14 4(b) x − n + 1 final answer 3 M2 for nx − (n − 1)( x + 1) or M1 for (n − 1)( x + 1) 4(c)(i) 16.6 or 16.60 to 16.61 nfww 4 M1 for 5, 12.5, 17.5, 22.5, 30 soi M1 for Σfx where x is in correct interval, including boundaries M1 dep on second M1 for Σfx 50 + 85 + 100 + 120 + 10 4(c)(ii) Correct histogram 4 B1 for each correct block If 0 scored, SC1 for 5, 20, 24, 1 seen

Question 59 marks

Question

NOT TO 1.2 m SCALE 3m The diagram shows the surface of a garden pond, made from a rectangle and two semicircles. The rectangle measures 3 m by 1.2 m. (a) Calculate the area of this surface. …… m2 [3] (b) The pond is a prism and the water in the pond has a depth of 20 cm. Calculate the number of litres of water in the pond. …… litres [3] (c) After a rainfall, the number of litres of water in the pond is 1007. Calculate the increase in the depth of water in the pond. Give your answer in centimetres. …… cm [3]

Mark scheme

5(a) 4.73 or 4.730 to 4.731... 3 M2 for 3 × 1.2 + π × 0.62 oe 1 or M1 for π × 0.62 or × π × 0.62 or 2 3 × 1.2 5(b) 946 or 946.0 to 946.2... 3 M2 for their (a) × 0.2 × 1000 oe or M1 for their (a) × 0.2 or 20 implied by figs 946[0] to 9462 5(c) 1.28 or 1.29 or 1.284 to 1.290 3 (1007 − their (b)) ÷ 1000 M2 for × 100 oe their (a) 1007 − their ( b ) or for × 20 oe their ( b ) 1007 − their ( b ) or M1 for figs or their ( a ) 1007 figs their ( a ) 1007 − their ( b ) or for or their ( b ) 1007 × 20 oe their ( b )

Question 67 marks

Question

 = {students in a school} F = {students who play football} B = {students who play baseball} There are 240 students in the school. • 120 students play football • 40 students play baseball • 90 students play football but not baseball. (a) Complete the Venn diagram to show this information. F B [2] (b) Find n ^F l + Blh . …… [1] (c) A student in the school is chosen at random. Find the probability that this student plays baseball but not football. …… [1] (d) Two students who play baseball are chosen at random. Find the probability that they both also play football. …… [3] 1

Mark scheme

6(a) 2 B1 for any one correct 90 30 10 110 6(b) 110 1 FT their 110 in Venn diagram 6(c) 10 1 their10 oe FT 240 240 6(d) 870 3 their 30 their 30 − 1 oe M2 for × 1560 40 39 p p −1 their 30 or M1 for × p < q or for q q −1 40 soi

Question 714 marks

Question

(a) s = ut + at 2 2 (i) Find s when t = 26.5, u = 104.3 and a = -2.2 . Give your answer in standard form, correct to 4 significant figures. s = …… [4] (ii) Rearrange the formula to write a in terms of u, t and s. a = …… [3] (b) NOT TO SCALE (x – 1) cm (x – 2) cm (2x + 3) cm (x + 1) cm The difference between the areas of the two rectangles is 62 cm2. (i) Show that x 2 + 2x - 63 = 0 . [3] (ii) Factorise x 2 + 2x - 63 . …… [2] (iii) Solve the equation x 2 + 2x - 63 = 0 to find the difference between the perimeters of the two rectangles. …… cm [2]

Mark scheme

7(a)(i) 1.991 × 103 4 B3 for 1991 or 1.99 × 103 or 1.991… × 103 or B2 for 1990 or 1991. … OR 1 M1 for 104.3 × 26.5 + × (−2.2) × 26.52 2 oe B1 for their seen value correctly rounded to 4 sf B1 for their seen value correctly converted into standard form 7(a)(ii) 2( s − ut ) 3 M1 for correct multiplication by 2 oe oe final answer M1 for correct rearrangement to isolate t2 term with a M1 for correct division by t2 for 3 marks e.g. cannot have a fraction in denominator nor ÷t 2 in numerator 7(b)(i) (2 x + 3)( x − 1) − ( x + 1)( x − 2) = 62 M1 2 x2 + 3x − 2 x − 3 oe B1 or x 2 + x − 2 x − 2 oe x2 + 2 x − 63 = 0 A1 Established with no errors or omissions 7(b)(ii) ( x + 9)( x − 7) 2 B1 for ( x + a)( x + b) where ab = – 63 or a+b=2 or for x( x − 7) + 9( x − 7) or for x( x + 9) − 7( x + 9) 7(b)(iii) 20 2 FT 2 × their positive root + 6 M1 for substituting their positive root into four lengths or for stating 2 x + 6

Question 811 marks

Question

(a) The price of a book increases from $2.50 to $2.65 . Calculate the percentage increase. …… % [3] (b) Scott invests $500 for 7 years at a rate of 1.5% per year simple interest. Calculate the value of his investment at the end of the 7 years. $ …… [3] (c) In a city the population is increasing exponentially at a rate of 1.6% per year. Find the overall percentage increase at the end of 20 years. …… % [2] (d) The population of a village is 6400. The population is decreasing exponentially at a rate of r% per year. After 22 years, the population will be 2607. Find the value of r. r = …… [3]

Mark scheme

8(a) 6 nfww 3 2.65 − 2.50 M2 for [×100] or for 2.50 2.65 × 100 2.50 2.65 or M1 for 2.50 8(b) 552.5[0] 3 B2 for 52.5[0] 1.5 or M2 for 500 × × 7 + 500 oe 100 1.5 or M1 for 500 × [× 7] oe 100 8(c) 37.4 or 37.36… 2  1.6  20 M1 for  1 +  oe soi 1.37…  100  8(d) 4[.00...] 3 2607 M2 for 22 6400 or M1 for 6400 × x22 = 2607 oe or better

Question 910 marks

Question

f (x) = 7x - 2 g (x) = x 2 + 1 h (x) = 3 x (a) Find gh(2). …… [2] (b) Find f – 1(x). f – 1(x) = …… [2] (c) gg (x) = ax 4 + bx 2 + c Find the values of a, b and c. a = …… b = …… c = …… [3] (d) Find x when hf(x) = 81. x = …… [3]

Mark scheme

9(a) 82 2 M1 for (3x)2+1 soi by (32)2+1 or g(9) isw 9(b) x+2 2 y 2 final answer M1 for y + 2 = 7x or = x − or 7 7 7 x = 7y – 2 9(c) [a =] 1, [b =] 2, [c =] 2 3 B2 for x4 + x2 + x2 + 1 + 1 or M1 for ( x 2 + 1) 2 + 1 9(d) 6 3 M2 for 7x – 2 = 4 oe or M1 for 3x = 81 soi f(x) = 4 7 or for 37 x −2 = 81 or better

Question 1012 marks

Question

The volume of each of the following solids is 1000 cm3. Calculate the value of x for each solid. (a) A cube with side length x cm. x = …… [1] (b) A sphere with radius x cm. 4 [The volume, V, of a sphere with radius r is V = rr 3. ] 3 x = …… [3] (c) NOT TO x 5 cm SCALE x cm A cone with radius x cm and slant height x 5cm. 1 [The volume, V, of a cone with radius r and height h is V = rr 2 h.] 3 x = …… [4] (d) x NOT TO cm 2 SCALE x cm 27x cm 2 A prism with a right-angled triangle as its cross-section. x = …… [4] Question 11 is printed on the next page.

Mark scheme

10(a) 10 1 10(b) 6.2[0] or 6.203 to 6.204 3 4 M2 for [x3 = ] 1000 ÷ π oe or better 3 4 3 or M1 for πx = 1000 3 10(c) 7.82 or 7.815 to 7.816 4 1 B3 for [ x3 =]1000 ÷ π ÷ 2 oe or better 3 ( ) − x soi by 4x or 2x 2 2 2 or M1 for x 5 1 M1dep for π × x 2 × theirh[= 1000] 3 10(d) 2 4 27 3x 6 or 6.67 or 6.666 to 6.667 B3 for [ x3 =]1000 ÷ oe or = 10 or 3 8 2 better 1 x 27 x or M2 for × x× × = 1000 oe 2 2 2 1 x or M1 for × x × 2 2 If 0 scored, SC2 for answer 5.29 or 5.291..

Question 1111 marks

Question

Brad travelled from his home in New York to Chamonix. • He left his home at 16 30 and travelled by taxi to the airport in New York. This journey took 55 minutes and had an average speed of 18 km/h. • He then travelled by plane to Geneva, departing from New York at 22 15. The flight path can be taken as an arc of a circle of radius 6400 km with a sector angle of 55.5°. The local time in Geneva is 6 hours ahead of the local time in New York. Brad arrived in Geneva at 11 25 the next day. • To complete his journey, Brad travelled by bus from Geneva to Chamonix. This journey started at 13 00 and took 1 hour 36 minutes. The average speed was 65 km/h. The local time in Chamonix is the same as the local time in Geneva. Find the overall average speed of Brad’s journey from his home in New York to Chamonix. Show all your working and give your answer in km/h. …… km/h [11]

Mark scheme

11 [Total time =]16 h 6 min or 16.1 h 2 B1 for 22 h 6 min or 22.1h or 966 mins If 0 scored, SC1 for 9 h 41 min [Distance to airport in New York =] 16.5 2 M1 for 18 × 55 [Arc length =] 3 55.5 6200 or 6199 to 6200. … M2 for × 2 × π × 6400 360 55.5 or M1 for or 2 × π × 2400 360 [Distance Geneva to Chamonix = ] 104 2 M1 for 65 × 1.6 or 65 × 96 oe 392 to 393 2 6316 to 6322.4 M1 for their 16.1 Must be correct value in numerator

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Where can I find the 0580/41 May/June 2019 mark scheme?

On this page, free: the mark scheme for IGCSE Mathematics 0580/41, May/June 2019 (0580_s19_ms_41), and the question paper (0580_s19_qp_41), to read or download, with each question's marking points written out underneath.

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Mathematics 0580/41 for May/June 2019 is out of 130 marks, with 2 hours 30 minutes to answer it.

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