Cambridge IGCSE Mathematics 0580 — 2018 May/June Paper 4 · Variant 3

1 (a) Rowena buys and sells clothes. (i) She buys a jacket for $40 and sells it for $45.40 . Calculate the percentage profit. ............................................ % [3] (ii) She sells a dress for $42.60 after making a profit of 20% on the cost price. Calculate the cost price. $ ............................................... [3] (b) Sara invests $500 for 15 years at a rate of 2% per year simple interest. Calculate the total interest Sara receives. $ ............................................... [2] (c) Tomas has two cars. (i) The value, today, of one car is $21 000. The value of this car decreases exponentially by 18% each year. Calculate the value of this car after 5 years. Give your answer correct to the nearest hundred dollars. $ ............................................... [3] (ii) The value, today, of the other car is $15 000. The value of this car increases exponentially by x % each year. After 12 years the value of the car will be $42 190. Calculate the value of x. x = ............................................... [3]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 13.5 3 45.4[0] − 40 45.4[0] M2 for [× 100] or × 100 40 40 45.4[0] or M1 for [× 100] 40 1(a)(ii) 35.5[0] 3  20  M2 for 42.6[0] ÷ 1 + or better    100  or M1 for recognising 42.6[0] as 120[%] 1(b) 150 cao 2 500 × 2 × 15 M1 for oe 100 1(c)(i) 7800 cao 3 B2 for 7790 or 7785 to 7786  18  5 or M1 for 21000 × 1 − oe isw    100  If 0 or 1 scored, SC1 for their 7785… seen and rounded correctly to nearest 100 1(c)(ii) 9[.00…] 3 42190 M2 for 12 or better 15000  x 12 or M1 for 15000 1 + = [42190]    100 

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2 (a) (i) y = 2x Complete the table. x 0 1 2 3 4 y 2 4 8 [2] (ii) y = 14 - x 2 Complete the table. x 0 1 2 3 4 y 13 10 5 [2] (b) On the grid, draw the graphs of y = 2x and y = 14 - x 2 for 0 G x G 4 . y 16 14 12 10 8 6 4 2 0 x 1 2 3 4 −2 [6] (c) Use your graphs to solve the equations. (i) 2 x = 12 x = ............................................... [1] (ii) 2 x = 14 - x 2 x = ............................................... [1] (d) (i) On the grid, draw the line from the point (4, 2) that has a gradient of - 4 . [1] (ii) Complete the statement. This straight line is a .................................. to the graph of y = 14 - x 2 at the point ( .......... , .......... ). [2]

Mark scheme: 2(a)(i) 1, ….., ….., …., 16 2 B1 for each 2(a)(ii) 14, ….., ….., …., – 2 2 B1 for each 2(b) Fully correct smooth curves 6 B3 for correct curve of y = 2 x or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points B3 for correct curve of y = 14 − x 2 or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 2(c)(i) 3.5 to 3.7 1 2(c)(ii) 2.65 to 2.8 1 2(d)(i) Correct line 1 Ruled, through (4, 2) and gradient −4 2(d)(ii) Tangent 2 B1 for each (2, 10)

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3 (a) The scatter diagram shows the physics mark and the chemistry mark for each of 12 students. 7 6 5 4 Chemistry mark 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 Physics mark (i) What type of correlation is shown in the scatter diagram? ................................................ [1] (ii) On the scatter diagram, draw a line of best fit. [1] (iii) Find an estimate of the chemistry mark for another student who has a physics mark of 4. ................................................ [1] (b) A teacher records the number of days each of the 24 students in her class are absent. The frequency table shows the results. Number of days 0 1 2 3 4 5 Frequency 10 8 3 2 0 1 Find the mode, the median and the mean. Mode = ............................................... Median = ............................................... Mean = ............................................... [5] (c) Three sizes of eggs are sold in a shop. The table shows the number of eggs of each size sold in one day. Size Small Medium Large Mass (m grams) 46 1 m G 52 52 1 m G 62 62 1 m G 80 Number of eggs sold 78 180 162 (i) Calculate an estimate of the mean mass. ............................................. g [4] (ii) On the grid, draw a histogram to show the information in the table. 20 18 16 14 12 Frequency 10 density 8 6 4 2 0 m 40 50 60 70 80 Mass (grams) [4]

Mark scheme: 3(a)(i) Positive 1 Ignore strong, weak, etc. 3(a)(ii) Correct ruled line 1 3(a)(iii) 2 1 3(b) [mode = ] 0 5 B1 [median = ] 1 B1 [mean = ] 1.04 or 1.041 to 1.042 B3 or M2 for ([10 × 0] + 8 × 1 + 3 × 2 + 2 × 3 + [0 × 4] + 1 × 5) ÷ 24 oe or M1 for [10 × 0] + 8 × 1 + 3 × 2 + 2 × 3 + [0 × 4] + 1 × 5 oe 3(c)(i) 60.9 or 60.91... nfww 4 M1 for 49, 57, 71 correct M1 for use of Σfx with x in the correct interval including both boundaries M1 (dep on 2nd M1) for their (78 × 49 + 180 × 57 + 162 × 71) ÷ (78 + 180 + 162) 3(c)(ii) Correct histogram 4 B1 for correct widths in correct position B1 height 13 B1 height 18 B1 height 9 If 0 scored B1 for 13, 18 and 9 seen

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4 (a) The diagram shows two sets of cards. Set A 1 1 2 2 2 Set B 0 1 1 1 2 (i) Jojo chooses two cards at random from Set A without replacement. Find the probability that the two cards have the same number. ................................................ [3] (ii) Jojo replaces the two cards. Kylie then chooses one card at random from Set A and one card at random from Set B. Find the probability that the two cards have the same number. ................................................ [3] (iii) Who is the most likely to choose two cards that have the same number? Show all your working. ................................................ [1] (b) Set C 4 4 5 5 5 Lena chooses three cards at random from Set C without replacement. Find the probability that the third card chosen is numbered 4. ................................................ [3]

Mark scheme: 4(a)(i) 8 3 2 1 3 2 oe M2 for × + × 20 5 4 5 4 or M1 for one of these products OR M1 for probability tree identifying all 20 outcomes with the correct 8 identified OR M1 for completed possibility space / 2-way table identifying the 8 possible outcomes out of 20, oe 13 SC1 for with replacement 25 4(a)(ii) 9 3 2 3 3 1 oe M2 for × + × oe 25 5 5 5 5 or M1 for one of these products OR M1 for probability tree identifying all 25 outcomes with the correct 9 identified OR M1 for completed possibility space / 2-way table identifying the 9 possible outcomes out of 25, oe 4(a)(iii) 40 36 1 1FT their (i) and (ii) dep on being in range 0 to 1 Jojo and e.g. > 100 100 4(b) 24 3 2 3 1 3 2 1 3 2 2 oe M2 for × × + × × + × × oe 60 5 4 3 5 4 3 5 4 3 or M1 for any one correct product OR M1 for 4, 5, 4 and 5, 4, 4 and 5, 5, 4 clearly identified on a tree or in a list

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5 (a) At a football match, the price of an adult ticket is $x and the price of a child ticket is $ x - 2.50 ^ h. There are 18 500 adults and 2400 children attending the football match. The total amount paid for the tickets is $320 040. Find the price of an adult ticket. $ ................................................ [4] (b) (i) Factorise y 2 + 5y - 84 . ................................................ [2] (ii) NOT TO y cm SCALE (y + 5) cm The area of the rectangle is 84 cm2. Find the perimeter. .......................................... cm [3] (c) In a shop, the price of a monthly magazine is $m and the price of a weekly magazine is $ m - 0.75 ^ h. One day, the shop receives • $168 from selling monthly magazines • $207 from selling weekly magazines. The total number of these magazines sold during this day is 100. (i) Show that 50m 2 - 225m + 63 = 0 . [3] (ii) Find the price of a monthly magazine. Show all your working. $ ............................................... [3]

Mark scheme: 5(a) 15.6[0] 4 B3 for 20 900x = 326 040 or better or M2 for 18 500x + 2400(x – 2.5[0]) = 320 040 or M1 for 18 500x or 2400(x – 2.5[0]) 5(b)(i) ( y + 12)( y − 7) final answer 2 B1 for ( y + a )( y + b ) where ab = – 84 or a + b = 5 or y ( y + 12 ) − 7 ( y + 12 ) or y(y – 7) + 12(y – 7) 5(b)(ii) 38 cao 3 B2 for y = 7 or M1 for y(y + 5) = 84 oe 5(c)(i) 168(m – 0.75) + 207m =100m(m – 0.75) M2 May be all over common denominator oe 168 207 M1 for or used m m − 0.75 OR 126 207 = 100m – 168 – 75 + m at least one interim line A1 No errors or omissions leading to 50m2 – 225m + 63 = 0 5(c)(ii) (10 m − 3)(5 m − 21) B2 M1 for (10m + a)(5m + b) where ab = 63 or 5a + 10b = –225 or 10m(5m – 21) – 3(5m – 21) or 5m(10m – 3) – 21(10m – 3) OR OR −−( 225) ± ( −225) 2 − 4(50)(63) M1 for ( −225) 2 − 4(50)(63) or for p = –(–225), m = oe 2(50) p + q p − q r = 2(50) if in form or r r OR OR 225  225  2 63  225  2 m = ±   − oe M1 for  m −  oe 100  100  50  100  4.2[0] cao B1

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6 (a) E 6 cm D NOT TO SCALE A 12 cm C 6 cm B In the pentagon ABCDE, angle ACB = angle AED = 90°. Triangle ACD is equilateral with side length 12 cm. DE = BC = 6 cm. (i) Calculate angle BAE. Angle BAE = ............................................... [4] (ii) Calculate AB. AB = ......................................... cm [2] (iii) Calculate AE. AE = ......................................... cm [3] (iv) Calculate the area of the pentagon. ......................................... cm2 [4] (b) S R P Q 5 cm NOT TO SCALE D C 4 cm A 8 cm B The diagram shows a cuboid. AB = 8 cm, BC = 4 cm and CR = 5 cm. (i) Write down the number of planes of symmetry of this cuboid. ................................................ [1] (ii) Calculate the angle between the diagonal AR and the plane BCRQ. ................................................ [4]

Mark scheme: 6(a)(i) 116.6 or 116.56 to 116.57 4 6 M1 for sin[ EAD ] = oe 12 6 M1 for tan[ BAC ] = oe 12 B1 for [angle DAC] = 60 6(a)(ii) 13.4 or 13.41 to 13.42 2 M1 for 12 2 + 6 2 6(a)(iii) 10.4 or 10.39… 3 2 2 M2 for 12 − 6 or M1 for AE 2 + 6 2 = 12 2 6(a)(iv) 130 or 129.5… to 129.6 4 M1 for 0.5 × 6 ×theirAE oe M1 for 0.5 × 12 × 12 × sin60 oe M1 for 0.5 × 6 × 12 oe 6(b)(i) 3 1 6(b)(ii) 51.3 or 51.30 to 51.34… 4 8 8 M3 for tan = or sin = oe 4 2 + 5 2 4 2 + 5 2 + 8 2 or M2 for 4 2 + 5 2 or 4 2 + 5 2 + 8 2 or M1 for angle ARB clearly indicated

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7 (a) NOT TO SCALE 1.5 cm Water flows through a cylindrical pipe at a speed of 8 cm/s. The radius of the circular cross-section is 1.5 cm and the pipe is always completely full of water. Calculate the amount of water that flows through the pipe in 1 hour. Give your answer in litres. ....................................... litres [4] (b) NOT TO SCALE 12 cm y cm x cm 6 cm The diagram shows three solids. The base radius of the cone is 6 cm and the slant height is 12 cm. The radius of the sphere is x cm and the radius of the hemisphere is y cm. The total surface area of each solid is the same. (i) Show that the total surface area of the cone is 108r cm2. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] [2] (ii) Find the value of x and the value of y. [The surface area, A, of a sphere with radius r is A = 4 r r 2 .] x = ............................................... y = ............................................... [4]

Mark scheme: 7(a) 204 or 203.5 to 203.6… nfww 4 M2 for π × 1.5 2 × 8 × 60 × 60 or M1 for π × 1.52 M1 for dividing their volume by 1000 If 0 scored SC1 for an answer figs 204 or figs 2035 to 2036 without working 7(b)(i) π × 6 × 12 + π × 62 = 108π M2 M1 for π × 6 × 12 7(b)(ii) [x = ] 5.2[0] or 5.196… 4 B2 or M1 for 4πx 2 = 108π seen [y = ] 6 B2 or M1 for ½(4πy2) + πy2 or better seen

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2 1 4 8 (a) M = N = 1 2 P = c4 3m ^ h f 1 p (i) For the following calculations, put a tick (ü) if it is possible or put a cross (û) if it is not possible. There is no need to carry out any of the calculations. Calculation ü or û N + P NP M2 N2 MN NM [4] 1 (ii) Work out + P . f 2 p ................................................ [1] (iii) Work out PN. ................................................ [2] (iv) Work out M -1 . ................................................ [2] 0 - 1 (b) Describe fully the single transformation represented by the matrix f1 0p. ............................................................................................................................................................. ............................................................................................................................................................. [3]

Mark scheme: 8(a)(i) × 4 B3 for 5 correct 9 B2 for 4 correct 9 B1 for 3 correct × × 9 8(a)(ii) 5 1 Fraction line and/or missing brackets scores 0  3 8(a)(iii)  4 8  2 B1 for 2 or 3 correct elements (dep on 2 × 2 matrix)    1 2  8(a)(iv) 1  3 −1  2  3 −1    oe isw B1 for k   or determinant = 2 soi 2  −4 2   −4 2  8(b) Rotation 3 B1 for each Origin oe 90 [anticlockwise] oe

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19 (a) Find the equation of the straight line that is perpendicular to the line y = x + 1 and passes through 2 the point (1, 3). ................................................ [3] (b) y 8 7 6 5 4 R 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 (i) Find the three inequalities that define the region R. ................................................ ................................................ ................................................ [4] (ii) Find the point (x, y), with integer co-ordinates, inside the region R such that 3x + 5y = 35 . ( .................... , ....................) [2]

Mark scheme: 9(a) y = −2 x + 5 oe 3 B2 for –2x + 5 or 1 M1 for gradient = −÷1 or better 2 M1 for substituting (1, 3) into y = (their m)x + c oe If 0 scored SC1 for (1, 3) satisfying their wrong 1 equation (c ≠ 0) with gradient ≠ 2 9(b)(i) x . 2 oe 1 SC3 for x > 2 and y < 5 and y > x 2 y - 5 oe OR B1 for x ⩾ 2 1 B1 for y ⩽ 5 y . x oe 4 1 2 B2 for y ⩾ x 2 or M1 for y ⩾ kx (k > 0) OR SC2 for all three boundary lines identified but with incorrect sign(s) If 0 scored SC1 for one or two correct boundary lines with incorrect sign(s) 9(b)(ii) (5, 4) 2 M1 for one trial of an integer point inside region or for 3 x + 5 y = 35 drawn

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10 (a) f (x) = 2x - 3 g (x) = x 2 + 1 (i) Find gg(2). ................................................ [2] (ii) Find g (x + 2) , giving your answer in its simplest form. ................................................ [2] (iii) Find x when f ()x = 7 . x = ............................................... [2] (iv) Find f -1 ()x . f -1 ()x = ............................................... [2] (b) h (x) = x x , x 2 0 (i) Calculate h(0.3). Give your answer correct to 2 decimal places. ................................................ [2] (ii) Find x when h(x) = 256. x = ............................................... [1]

Mark scheme: 10(a)(i) 26 2 2 2 M1 for g(5) or for x + 1 + 1 ( ) 10(a)(ii) x 2 + 4 x + 5 2 M1 for ( x + 2 ) 2 + 1 10(a)(iii) 5 2 M1 for 2 x − 3 = 7 10(a)(iv) x + 3 2 y 3 oe M1 for x = 2 y − 3 or y + 3 = 2 x or = x − oe 2 2 2 10(b)(i) [0].70 cao 2 B1 for [0].696 to [0].697 10(b)(ii) 4 cao 1

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