Cambridge IGCSE Mathematics 0580 — 2022 May/June Paper 2 · Variant 3

1 The probability of picking a red sweet from a bag is 0.05 . Find the probability of not picking a red sweet. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 0.95 oe 1

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m k 32 Work out the value of when m = 4 and k = 7 . 3 ................................................. [2]

Mark scheme: 2 792 or 792.1… 2 4  7 3 M1 for oe 3 or B1 for 1372

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3 T NOT TO S SCALE R 73° 75° P 129° Q PQRS is a quadrilateral. RST is a straight line. Find angle PST. Angle PST = ................................................ [2]

Mark scheme: 3 97 2 M1 for 360 – (73 + 129 + 75)

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4 These are the masses, in kg, of 12 parcels. 0.3 0.4 1.2 0.8 1.1 2.1 1.7 1.8 1.2 2.3 0.7 1.1 (a) Complete the stem-and-leaf diagram for the 12 parcels. 0 3 4 1 2 Key: 0 q3 represents 0.3 kg [2] (b) Find the median. ............................................ kg [1]

Mark scheme: 4(a) 2 B1 for two rows correct or for fully correct 0 (3 4) 7 8 unordered stem-and-leaf diagram or for a correct diagram with one error or omission 1 1 1 2 2 7 8 2 1 3 4(b) 1.15 1

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5 The nth term of a sequence is n 2 - 1. Find the first three terms of this sequence. ....................... , ....................... , ....................... [2]

Mark scheme: 5 0, 3, 8 2 B1 for 2 correct terms in correct position or SC1 for −1, 0, 3

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6 Simplify. (a) y 3 ' y 5 ................................................. [1] (b) 7x 0 ................................................. [1]

Mark scheme: 6(a) –2 1 1 y or final answer y 2 6(b) 7 1

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7 The scatter diagram shows the number of visitors and the total amount spent, in thousands of dollars, at a zoo on each of eight days. 45 40 35 30 Total 25 amount spent 20 ($1000) 15 10 5 0150 200 250 300 350 400 450 500 550 600 Number of visitors (a) On one of the eight days there are 410 visitors. Find the total amount spent by visitors during this day. $ ................................................ [1] (b) Information for the ninth day is shown in the table. Number of visitors 175 Total amount spent ($1000) 9 Plot this information on the scatter diagram. [1] (c) Draw a line of best fit on the scatter diagram. [1] (d) On the tenth day the total amount spent is $22 000. Estimate the number of visitors on this day. ................................................. [1]

Mark scheme: 7(a) 27 000 1 7(b) Point plotted at (175, 9) 1 7(c) Correct single ruled line of best fit 1 7(d) 300 to 350 1 FT their straight line of best fit provided positive gradient

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2 58 Without using a calculator, work out ' . 9 6 You must show all your working and give your answer as a fraction in its simplest form. ................................................. [2]

Mark scheme: 8 2 6 4 15 M1  or  oe 9 5 18 18 4 A1 cao 15

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9 Change 300 m / min to km / h. ....................................... km / h [2]

Mark scheme: 9 18 2 300  60 M1 for oe 1000 or B1 for figs 18 in their answer

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10 A B 25 33 12 3 Find n A + B l . ` j ................................................. [1]

Mark scheme: 10 40 1

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11 ABC, DEF and GHK are triangles with all vertices on the circumference of a circle. D A K 8° 85° NOT TO 6° SCALE 85° H G B 94° 82° C F E From the list, draw a ring around the line that is a diameter of the circle. AB AC DE DF GH GK [1]

Mark scheme: 11 DE 1

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12 f is a common factor of 14 and 28. m is a common multiple of 10 and 25. p is a prime number. f Work out the largest possible value of . mp ................................................. [4]

Mark scheme: 12 0.14 oe nfww 4 14 M3 for with at least 2 out of 3 50  2 values correct and for the one incorrect value: f must be 1, 2 or 7 m must be a multiple of 50 p must be prime OR B1 for f = 14 B1 for m = 50 B1 for p = 2 If 0 scored SC1 for a correct multiple for m, factor for f or prime for p

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13 Factorise completely. (a) 18px - 27p ................................................. [2] (b) mt - n - m + nt ................................................. [2]

Mark scheme: 13(a) 9p(2x – 3) final answer 2 B1 for 9(2px – 3p) or p(18x – 27) or 3p(6x – 9) or 9p(2x – 3) seen and spoilt 13(b) (m + n)(t – 1) final answer 2 B1 for m(t – 1) + n(t – 1) or t(m + n) – [1](m + n) or correct answer seen and spoilt

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14 Find the nth term of this sequence. 8, 17, 32, 53, 80, … ................................................. [2]

Mark scheme: 14 3n2 + 5 oe final answer 2 M1 for correctly finding second differences or an answer that is a quadratic sequence

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15 Solve. 12x - 3 H 4x + 13 ................................................. [2]

Mark scheme: 15 x ⩾ 2 final answer 2 M1 for 12x – 4x ≥ 13 + 3 oe

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16 Abdul draws this speed–time graph for a journey. The graph has four sections A, B, C and D. 12 B 10 C 8 Speed A 6 (m/s) 4 D 2 0 0 10 20 30 40 50 60 70 80 90 100 110 Time (s) Complete these statements about the speed–time graph. Section ....................... cannot be correct. Section ....................... shows constant speed. Section ....................... shows deceleration. Section A shows acceleration of .................................. m / s2. The distance travelled in the first 30 seconds of the journey is ....................... m. [4]

Mark scheme: 16 D B1 B C 1 B1 or 0.333… 3 150 B2 1 or M1 for  30  10 2

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17 A NOT TO SCALE B D C In triangle ABC, AC = AB. D is the point on BC such that AD is perpendicular to BC. Complete the following statements to show that triangle ACD and triangle ABD are congruent. AD is perpendicular to BC so that Angle ............... = Angle ............... = ............... ° AC = AB is given information. Side ...................... is common to both triangles. Triangle ACD is congruent to triangle ABD because of the congruency criterion ...................... [3]

Mark scheme: 17 ADC and ADB and 90 3 B1 for each correct line AD RHS

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18 The bearing of B from A is x°. The bearing of A from B is y°. x : y = 2 : 7 Calculate the value of y. North North NOT TO SCALE B y° x° A y = ................................................ [3]

Mark scheme: 18 252 3 M2 for 180 ÷ (7 – 2) oe OR M1 for 180 – x + y = 360 oe M1 for correct use of ratio

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1 7x - 2 3 - 10x19 f( )x = kx2 g ( x) = h ( x) = j( )x = x 5 14 (a) f(- 5)k = 675 Find the value of k. k = ................................................ [2] (b) Find gh ( x) . ................................................. [1] (c) Find h -1 ( x) + j( x) . Give your answer in its simplest form. ................................................. [4]

Mark scheme: 19(a) 3 2 M1 for k(–5k)2 = 675 or better 19(b) 5 1 final answer 7 x  2 19(c) 1 4 7 or 0.5 B3 for answer 2 14 OR 5 x  2 B2 for 7 or M1 for correct first step for h –1(x) 7 y  2 e.g. x = 5 y  7 x  2 5 2 7 x y +  5 5 2  5 x  2  3  10 x M1FT for  oe with 14 14 common denominator

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20 k cm B C NOT TO SCALE E A M D The diagram shows a square ABCD with side length k cm. MDE is a sector of a circle, centre D. E lies on the diagonal, BD, of the square. M is the midpoint of AD. Find the percentage of the square that is shaded. ............................................. % [4]

Mark scheme: 20 90.2 or 90.18… 4 B3 for 9.82[%] OR  2 45  k  2  2  k M3 for 100   k        360  2     oe 45  k  2 2  k oe or M2 for 100      360  2  2 45  k  2 or k     360  2  or 100× (k2 – mπk2 ) ÷ k2 c  k  2 or M1 for    oe 360  2  or for (k2 – mπk2 ) ÷ k2 or for 100× (k2 – mk2 ) ÷ k2

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21 Neha has a piece of ribbon of length 23 cm, correct to the nearest cm. From this ribbon she cuts off a piece with length 87 mm, correct to the nearest mm. Work out the lower bound and the upper bound for the length of the remaining ribbon. Give your answer in centimetres. Lower bound = ........................................... cm Upper bound = ............................................cm [3]

Mark scheme: 21 13.75 3 B2 for one correct answer or both correct 14.85 answers seen in working then rounded to 3sf or both correct but reversed or M1 for 2 correct seen from 23 + 0.5, 23 – 0.5, 8.7 + 0.05 or 8.7 – 0.05 or better

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22 Simplify. 5 x - x 2 25 - x 2 ................................................. [3]

Mark scheme: 22 x 3 B1 for x(5 – x) final answer nfww B1 for (5 – x)(5 + x) or 5  x

More questions on Algebraic fractions

23 Solve the equation 3 sinx + 3 = 1 for 0° G x G 360° . x = .......................... or x = .......................... [3]

Mark scheme: 23 221.8 or 221.81... 3 B2 for one correct and 2 or M1 for sin x = – oe 318.2 or 318.18 to 318.19 3 If 0 scored, SC1 for two reflex angles with a sum of 540 or two non-reflex angles with a sum of 180

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24 y is inversely proportional to the cube of ( x - 1) . y = 9.45 when x = 3 . Find y when x = 4 . y = .................................................. [3]

Mark scheme: 24 2.8 3 k M1 for y   x  1 3 k M1 for y their  4 –1 3 OR M2 for y(4 – 1)3 = 9.45(3 – 1)3

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1 25 m - 4 = 27m -1 Find the value of m. m = ................................................ [3]

Mark scheme: 25 81 3 3 M2 for m 4  27 or better 1 27 or M1 for or better 1 m m 4 1  1 or m 4 = 27 1 If 0 scored SC1 for answer 81

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