Cambridge IGCSE Mathematics 0580 — 2022 Feb/March Paper 2 · Variant 2

1 Measure the marked angle. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 40° 1

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2 Work out 5 # 62 . Give your answer correct to 2 decimal places. ................................................. [2]

Mark scheme: 2 80.50 cao 2 B1 for 80.498… or 80.5 or correctly rounding their more accurate decimal to 2 dp

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3 A journey starts at 21 15 one day and ends at 04 33 the next day. Calculate the time taken, in hours and minutes. .................... h .................... min [1]

Mark scheme: 3 7 [h] 18 [min] 1

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4 NOT TO SCALE 5 cm 4 cm 7 cm Calculate the total surface area of this cuboid. ......................................... cm2 [3]

Mark scheme: 4 166 3 M2 for [2 ×] (7×4 + 4×5 + 5×7) or M1 for 7×4 or 4×5 or 5×7

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5 (a) Write down the gradient of the line y = 5x + 7 . ................................................. [1] (b) Find the coordinates of the point where the line y = 5x + 7 crosses the y-axis. ( ................... , ................... ) [1]

Mark scheme: 5(a) 5 1 5(b) (0, 7) 1

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6 NOT TO 9.5 cm 8 cm SCALE 12 cm Using a ruler and compasses only, construct this triangle. Leave in your construction arcs. The side of length 12 cm has been drawn for you. [2]

Mark scheme: 6 correct triangle with arcs 2 B1 for correct triangle with incorrect or no arcs or for two correct arcs. or a triangle with arcs but one side not in range

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7 n – 2 – 1 0 1 Write down the inequality, in terms of n, shown by the number line. ................................................. [1]

Mark scheme: 7 n > − 1 oe 1

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8 y 6 5 4 3 A 2 1 B – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 9 10 x – 1 – 2 – 3 – 4 – 5 (a) On the grid, draw the image of (i) triangle A after a reflection in the y-axis, [1] - 3 (ii) triangle A after a translation by the vector [2] e- 4o. (b) Describe fully the single transformation that maps triangle A onto triangle B. ..................................................................................................................................................... ..................................................................................................................................................... [3]

Mark scheme: 8(a)(i) triangle at (− 1, 1) (− 4, 2) (− 3, 5) 1 8(a)(ii) triangle at (− 2, −3) (1, − 2) (0, 1) 2  −3   k  B1 for translation by   or by    k   −4  8(b) enlargement 3 B1 for each 1 [sf] 2 [centre] (9, − 1)

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9 Factorise completely. 12a 3 - 21a ................................................. [2]

Mark scheme: 9 2 2 3 2 3a 4 a − 7 final answer B1 for 3 4 a − 7 a or a 12 a − 21 ( ) ( ) ( ) or for 3a seen then spoilt ( 4 a 2 − 7 )

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10 (a) The nth term of a sequence is n 2 + 7 . Find the first three terms of this sequence. .................. , .................. , .................. [2] (b) These are the first four terms of a different sequence. 15 7 − 1 -9 Find the nth term of this sequence. ................................................. [2]

Mark scheme: 10(a) 8 11 16 2 B1 for two correct 10(b) 23 – 8n oe final answer 2 B1 for j – 8n or 23 – kn k ≠ 0 or 23 – 8n seen then spoilt

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11 As the temperature increases, people eat more ice cream. What type of correlation does this statement describe? ................................................. [1]

Mark scheme: 11 Positive 1

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12 (a) Sanjay invests $700 in an account paying simple interest at a rate of 2.5% per year. Calculate the value of his investment at the end of 6 years. $ ................................................ [3] (b) Meera invests $700 in an account paying compound interest at a rate of r % per year. At the end of 17 years the value of her investment is $1030.35 . Find the value of r. r = ................................................ [3]

Mark scheme: 12(a) 805 3 B2 for 105 700 × 2.5 × 6 or M2 for + 700 oe 100 700 × 2.5 [× 6 ] or M1 for oe 100 12(b) 2.3[0…] 3 1030.35 M2 for 17 oe 700 or M1 for 1030.35 = 700 ( k )17 oe for any k

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13 (a) Simplify h 2 # h 5 . ................................................. [1] 7 -3 (b) Simplify . b x l ................................................. [1] (c) a 8 ' a p = a 2 Find the value of p. p = ................................................ [1]

Mark scheme: 13(a) h 7 final answer 1 13(b) x 3 1 final answer 343 13(c) 6 1

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14 Calculate the circumference of a circle with radius 4.7 cm. ............................................ cm [2]

Mark scheme: 14 29.5 or 29.53… 2 M1 for 2 × π× 4.7 oe

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1 1115 Without using a calculator, work out 2 # . 3 14 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 15 7 M1 k 11 oe improper fraction or × where k > 3 3 3 14 5 A2 77 11 35 16 cao A1 for 42 or 6 or 142

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16 y 6 A 5 4 3 2 1 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 x – 1 – 2 – 3 B – 4 – 5 – 6 – 7 – 8 A is the point (- 6, 5) and B is the point (- 2, - 3). (a) Find the equation of the straight line, l, that passes through point A and point B. Give your answer in the form y = mx + c . y = ................................................ [2] (b) Find the equation of the line that is perpendicular to l and passes through the origin. ................................................. [2]

Mark scheme: 16(a) [ y = ] − 2 x − 7 final answer 2 B1 for − 2x + c or kx – 7, k ≠ 0 final answer 16(b) 1 2 1 y = x [ ±0 ] final answer FT −their gradientin(a) 2 B1 for y = kx [ ± 0 ] oe, k ≠ 0 1 or y = their x + c oe for any c 2 1 or their x [ ±0 ] oe 2

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17 O R X NOT TO 4 cm SCALE P Q 11 cm The diagram shows a rectangle OPQR with length 11 cm and width 4 cm. OQ is a diagonal and OPX is a sector of a circle, centre O. Calculate the percentage of the rectangle that is shaded. ............................................. % [5]

Mark scheme: 17 77.8 or 77.77 to 77.80 5 B4 for answer 22.2[%] or 22.20[%] to 22.23[%] OR − 111 − 1 4 M1 for tan oe or tan oe 4 11 their acuteangle 2 M2 for 4 × 11 − × π × 4 oe 360 their acuteangle 2 or M1 for π × 4 oe 360 their shaded area M1 for [×100] oe 4 ×1 1 their sector area or ×100 oe 4 ×1 1

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18 Mrs Kohli buys a jacket, 2 shirts and a hat. The jacket costs $x. The shirts each cost $24 less than the jacket and the hat costs $16 less than the jacket. Mrs Kohli spends exactly $100. Write down an equation in terms of x. Solve this equation to find the cost of the jacket. $ ................................................ [3]

Mark scheme: 18 A correct equation leading to 3 M2 for 4 x = 164 41 or M1 for x + 2 ( x − 24 ) + x − 16 = 100 oe or M1 for correctly simplifying their equation to the form kx = c provided at least one part correct from [ 2 ]( x − 24 ) oe or x − 16 or B1 for answer 41 without an equation in x shown

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19 y is inversely proportional to the square root of (x +4). When x = 5, y = 2. Find y when x = 77. y = ................................................ [3]

Mark scheme: 19 2 3 k oe M1 for y = 3 x + 4 theirk M1 for y = 77 + 4

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20 Solve the simultaneous equations. You must show all your working. 3x + y = 11 x 2 - 2 y = 18 x = ................................ y = ................................ x = ................................ y = ................................ [5]

Mark scheme: 20 x 2 + 6 x − 40 [=0] M2 M1 for correct method to eliminate one 2 variable e.g. or y − 40 y − 41 [=0] x 2 − 2 (11 − 3 x ) = 18 (11 − y ) 2 or − 2 y = 18 32 ( x − 4 )( x + 10 ) [=0] M1 or for correct factors for their quadratic equation or ( y − 41)( y + 1) [=0] or for correct use of quadratic formula for their quadratic equation or for correctly completing the square for their quadratic equation x = 4, y = − 1 B2 B1 for x = 4, x = − 10 x = −10, y = 41 or for y = − 1, y = 41 or for a correct pair of x and y values If B0 scored and at least 1 method mark scored SC1 for correct substitution shown of both of their x values or their y values into 3x + y = 11 or x2 – 2y = 18

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21 M H G E F NOT TO 14.5 cm SCALE D C 9 cm A B 18.6 cm The diagram shows an open rectangular box ABCDEFGH. AB = 18.6 cm, BC = 9 cm and CG = 14.5 cm. A straight stick AGM rests against A and G and extends outside the box to M. (a) Calculate the angle between the stick and the base of the box. ................................................. [4] (b) AM = 30 cm. Show that GM = 4.8 cm, correct to 1 decimal place. [3]

Mark scheme: 21(a) 35.1 or 35.05 to 35.06 4 14.5 M3 for tan = oe 18.6 2 + 9 2 or M2 for  AC 2 =  1 8.6 2 + 9 2 oe or better   or  AG 2 =  1 8.6 2 + 9 2 + 14.5 2   or M1 for recognising the angle GAC 21(b) M2 M1 for AG 2 = 18.6 2 + 9 2 + 14.5 2 oe or better 30 − 18.6 2 + 9 2 + 14.5 2 14.5 14.5 30 – or sin ( their(a) ) = sin ( their(a) ) AG or 18.6 2 + 9 2 or cos ( their(a) ) = 2 2 AG 18.6 + 9 30 – cos ( their(a) ) 4.75 to 4.78… A1

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22 R Q NOT TO a SCALE O P b The diagram shows a trapezium OPQR. O is the origin, OR = a and OP = b . 3 RQ = OP 5 (a) Find PQ in terms of a and b in its simplest form. PQ = ................................................ [2] (b) When PQ and OR are extended, they intersect at W. Find the position vector of W. ................................................. [2]

Mark scheme: 22(a) 2 2 3 a − b oe simplified M1 for −+b a + b or a correct route 5 5 22(b) 5 2 5  a oe B1 for ka where k >1 or OR 2 2

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