Cambridge IGCSE Mathematics 0580 — 2022 Oct/Nov Paper 4 · Variant 3

1 (a) Here are the ingredients needed to make a pasta bake to serve 12 people. 250 g butter 600 g pasta 460 g mushrooms 280 g cheese 800 ml milk (i) Find the mass of the cheese as a percentage of the mass of the mushrooms. ..............................................% [1] (ii) Find the mass of butter needed to make a pasta bake to serve 18 people. ............................................... g [2] (iii) Monica has 2.2 litres of milk and 1.5 kg of each other ingredient. Calculate the greatest number of people she can serve with pasta bake. ................................................. [3] (b) In 2019, a packet of pasta cost $2.40 . This was an increase of 25% of the cost of a packet in 2018. (i) Work out the cost in 2018. $ ................................................. [2] (ii) In 2020, the cost of a packet increased by 15% from the cost in 2019. Work out the total percentage increase in the cost of a packet from 2018 to 2020. ..............................................% [3] (c) width NOT TO SCALE Pasta is sold in packets with width 11.5 cm, correct to the nearest 0.5 cm. A shop places these packets in a single line on a shelf of length 2 m, correct to the nearest 0.1 m. Find the maximum number of these packets that will fit along this shelf. You must show all your working. ................................................. [3]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 60.9 or 60.86 to 60.87 1 1(a)(ii) 375 2 250 M1 for [ 18] oe 12 1(a)(iii) 30 nfww 3 M1 for figs2200 ÷ 800 [× 12]oe M1 for 1500 ÷ 600 [× 12] oe 1(b)(i) 1.92 2  25  M1 for k   1 +  = 2.4[0] oe or better  100  1(b)(ii) 3 3 43.75 or 43 4   25  15   100 oe M2 for   1 +  1 +  [ −1]     100  100    25  15  or  1 +  1 +   100 [–100]  100  100   15  2.40   1 +  or for  100  100 [– 100] oe their(b)(i)  15   25   15  or M1 for 2.40 ×  1 +  or  1 +    1 +  oe  100   100   100  1(c) 18 nfww 3 200 to 210 200 + 5 M2 for or oe 11.5 − 0.25 11 to 11.5 or M1 for 200 + 5, 200 – 5, 11.5 + 0.25 or 11.5 – 0.25

More questions on Ratio and proportion

2 (a) Simplify fully. (i) p 3 # p 11 ................................................. [1] 18 m 6 (ii) 2 3m ................................................. [2] 1 27x 9 y 27 - 3 (iii) e 64 o ................................................. [3] (b) A sequence has nth term 3n 2. Write down the first 3 terms of this sequence. .................... , .................... , .................... [2] (c) Find the nth term for each of these sequences. (i) 13, 16, 19, 22, 25, … ................................................. [2] (ii) 3, 17, 55, 129, 251, … ................................................. [2] (d) Solve. 3x - 22 = 23 4 x = ................................................. [3] (e) Use the quadratic formula to solve 3x 2 + 8x - 20 = 0 . Show all your working and give your answers correct to 2 decimal places. x = ................... , x = ................... [4]

Mark scheme: 2(a)(i) p14 final answer 1 2(a)(ii) 6m4 final answer 2 B1 for 6mk or km4 in final answer or correct answer seen and spoilt 2(a)(iii) 4 4 x −3 y −9 3 B2 for correct answer seen and spoilt or final answer or 2 correct elements in final answer 3 9 3x y 3 4 3 or B1 for one of or oe or x3 or y9 seen 3 4 2(b) 3, 12, 27 2 B1 for 12 or 27 2(c)(i) 3n + 10 oe final answer 2 B1 for 3n + k oe or jn + 10 oe (j ≠ 0) or for correct expression shown in working and then spoilt 2(c)(ii) 2n3 + 1 oe final answer 2 B1 for 3rd diff = 12 (both needed) or for cubic answer or for correct expression shown in working and then spoilt 2(d) 38 3 M2 for 3x = 4 × 23 + 22 or M1 for 3x – 22 = 4 × 23 3 x 22 or for = 23 + oe 4 4 2(e) 2 B2 2 −8 8 − 4(3)( −20) B1 for 8 − 4(3)( −20) oe 2  3 −+8 q −−8 q 2 or oe or oe or both −8 8 ( −20) 2  3 2  3 or  − 2  3 4  32 3 or better – 4.24, 1.57 final answers B2 B1 for each If B0, SC1 for answers – 4.2 or –4.23 or –4.240 to – 4.239 and 1.6 or 1.572 to 1.573 or – 4.24 and 1.57 seen in working or for –1.57 and 4.24 as final answer

More questions on Sequences

3 The height, h cm, of each of 100 plants is recorded. The table shows information about the heights of these plants. Height 10 1 h G 15 15 1 h G 25 25 1 h G 40 40 1 h G 60 60 1 h G 70 (h cm) Frequency 8 18 28 33 13 (a) Complete the histogram to show this information. The first two blocks have been drawn for you. 2.5 2 1.5 Frequency density 1 0.5 0 10 20 30 40 50 60 70 h Height (cm) [3] (b) Calculate an estimate of the mean height. ............................................ cm [4]

Mark scheme: 3(a) Correct histogram 3 B1 for each correct block 28 33 13 If 0 scored, SC1 for two of , , or 1.87 or 1.866 to 15 20 10 1.867, 1.65, 1.3 3(b) 38.65 4 M1 for 12.5, 20, 32.5, 50, 65 soi M1 for fx where x is in the correct interval including boundaries M1dep for fx ÷100

More questions on Histograms

4 y 6 5 4 A 3 2 T 1 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 (a) Draw the reflection of triangle T in the line y =- 2 . [2] 1 (b) Draw the enlargement of triangle T with scale factor and centre of enlargement ( - 5 , - 3 ). [2] 2 (c) Describe fully the single transformation that maps triangle T onto triangle A. ..................................................................................................................................................... ..................................................................................................................................................... [3]

Mark scheme: 4(a) Triangle drawn at (1, – 5), 2 B1 for reflection in any horizontal line (1, – 7), (5, – 5) If 0 scored, SC1 for reflection in x = – 2 4(b) Triangle drawn at (– 2, 0), 2 B1 for correct size and orientation but wrong position (– 2, – 1), (0, – 1) 4(c) Rotation 3 B1 for each 90 [anticlockwise] oe [centre] (– 1, 0)

More questions on Transformations

5 NOT TO 50 cm SCALE 40 cm 1.2 m 36 cm The diagram shows a water trough in the shape of a prism. The prism has a cross-section in the shape of an isosceles trapezium. The trough is completely filled with water. (a) Show that the volume of water in the trough is 206.4 litres. [3] (b) The water from the trough is emptied at a rate of 600 ml per second. Calculate the time taken, in minutes and seconds, for the trough to be emptied. .................... minutes .................... seconds [3] (c) All the water from the trough is emptied into a vertical cylindrical tank. The depth of the water in the tank is 84 cm. (i) Calculate the radius of the tank. ............................................ cm [3] (ii) The tank is 60% full. Calculate the height of the tank. ............................................ cm [2] (d) M NOT TO 50 cm SCALE 40 cm 1.2 m A 36 cm A steel rod AM is placed inside the empty water trough as shown in the diagram. A is a vertex at the base of the isosceles trapezium and M is the midpoint of the top edge on the opposite face. Calculate the length of the steel rod, AM. AM = ............................................ cm [4]

Mark scheme: 5(a)   (36 + 50)  40  120 oe M2  2  (36 + 50)  40 (0.36 + 0.5)  0.4 or M1 for oe or oe 2 2   (0.36 + 0.5)  0.4  1.2 oe  2  206400 ÷ 1000 = 206.4 A1 Must see an explicit conversion or 0.2064 × 1000 = 206.4 nfww 5(b) 5 [minutes] 44 seconds 3 B2 for 344 [seconds] oe 5.73…[mins] or M1 for figs206.4 ÷ figs 6 oe 5(c)(i) 28[.0] or 27.96 to 27.97 3 figs 2064 M2 for [r2=] ( figs84) or M1 for r 2  figs 84 = figs 2064 5(c)(ii) 140 cao 2 M1 for 0.6h = 84 oe ALT method 2 M1 for  ( their (c)(i) )  h = figs 206400  0.6 oe 5(d) 128 or 127.7 to 127.8 4 B3 for 40 2 + 120 2 + 18 2 oe OR B1 for horizontal length 18 soi M1 for any correct attempt at 2-dimensional Pythagoras’ 182 + 1202, 1202 + 402, 182 + 402

More questions on Pythagoras’ theorem and trigonometry

6 (a) P = 5k 2 - 7 (i) Find the value of P when k = 3 . P = ................................................. [2] (ii) Rearrange the formula to make k the subject. k = ................................................ [3] (b) (i) Solve. x - 3 G 5x + 7 ................................................. [2] (ii) Show your answer to part (b)(i) on the number line. x – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 [1] (c) The line y = 16 is drawn on the grid. y 40 30 20 10 0 10 20 30 40 x The region R satisfies the following inequalities. y H 16 x 2 2 2x + 3y H 72 y G 32 - x (i) By drawing three more lines and shading the region not required, find and label region R. [6] (ii) Find the integer coordinates (x, y) in the region R that give the maximum value of 2x + y . ( ...................... , ...................... ) [2]

Mark scheme: 6(a)(i) 38 2 M1 for 5 × 32 – 7 oe 6(a)(ii) P + 7 3 P 2 7  oe final answer M1 for P + 7 = 5k2 or = k − 5 5 5 M1 for k2 = ……. FT their first step M1 for square root to final answer Max M2 for incorrect answer 6(b)(i) x ⩾ – 2.5 final answer 2 M1 for –4x ⩽ 7 + 3 or better 6(b)(ii) 1 FT their inequality in (b)(i) –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 6(c)(i) x = 2 broken line B1 y = 32 – x solid line B1 2x + 3y = 72 solid line B2 B1 for line passing through (0, 24) or (36, 0) Correct region indicated cao B2 B1 for region satisfying 3 of the inequalities 1 1 R 1 1 6(c)(ii) (16, 16) 2 M1 for substitution into 2x + y for any integer point in their region

More questions on Equations

7 Regan is playing a game with these six number cards. – 3 – 2 2 3 5 7 (a) She takes two cards at random, without replacement, and multiplies the two numbers to give a score. Find the probability that (i) the score is 35 ................................................. [3] (ii) the score is a positive number. ................................................. [3] (b) Regan now takes three cards at random from the six cards, without replacement, and adds the three numbers to give a total. Find the probability that her total is 5. ................................................. [4]

Mark scheme: 7(a)(i) 1 3 1 1 oe M2 for 2   oe 15 6 5 1 1 or M1 for  oe 6 5 or list or indication of 2 correct pairs 1 If 0 scored, SC1 for answer oe 18 7(a)(ii) 7 3  4 3   1 1   1 1  oe M2 for    + 2    oe or 14    oe 15  6 5   6 5   6 5   2 4  or 1 – 2     6 5   4 3   1 1   2 4  or M1 for    or 2    oe or 2     6 5   6 5   6 5  or correct identification of 14 pairs 5 If 0 scored, SC1 for answer 9 7(b) 1 4  1 1 1   1 1 1  oe nfww M3 for 6     + 6     oe 10  6 5 4   6 5 4   1 1 1   1 1 1  or M2 for 6     oe or 2     oe  6 5 4   6 5 4   1 1 1  or M1 for k     where k is an integer and 1 ⩽ k ⩽ 12 but  6 5 4  not k = 2 or k = 6 or identifies –2, 2 and 5 or –3, 3 and 5 as the 3 cards needed 1 If 0 scored, SC1 for answer 18

More questions on Probability of combined events

8 A NOT TO SCALE 9.5 cm O 10 cm B 7.7 cm D C E A, B and C are points on the circle, centre O. DE is a tangent to the circle at C. AC = 10 cm , AB = 9.5 cm and BC = 7.7 cm . (a) Show that angle ABC = 70.2° , correct to 1 decimal place. [4] (b) Find (i) angle AOC Angle AOC = ................................................. [1] (ii) angle ACO Angle ACO = ................................................. [1] (iii) angle ACD. Angle ACD = ................................................. [1] (c) Calculate the radius, OC, of the circle. OC = ............................................ cm [3] (d) Calculate the area of triangle ABC as a percentage of the area of the circle. ..............................................% [4]

Mark scheme: 8(a) 9.52 + 7.7 2 − 10 2 M2 M1 for 102 = 9.52 + 7.72 – 2×9.5×7.7cosB oe or better [cos B = ] oe 2  9.5  7.7 70.206 to 70.207 or 70.21 to 70.22 A2 2477 A1 for oe or 0.339 or 0.3386…. 7315 8(b)(i) 140.4 1 8(b)(ii) 19.8 1 FT (180 – their (b)(i)) ÷ 2 8(b)(iii) 70.2 1 FT 90 – their (b)(ii) 8(c) 5.31 or 5.314 to 5.315 3 5 M2 for oe cos their(b)(ii) 5 or M1 for = cos(their (b)(ii)) oe r 8(d) 38.8 or 38.9 or 38.78 to 38.85 4 0.5  9.5  7.7  sin70.2 M3 for [ 100] their( (c)) 2 OR M1 for 0.5 × 9.5 × 7.7 × sin70.2 M1 for (their (c)2)

More questions on Circle theorems I

9 (a) Sketch the following graphs. On each sketch, indicate any intercepts with the axes. (i) 3x - 4y = 12 y O x [2] (ii) y = x 2 - 3x - 4 y O x [4] (iii) y = 6x y O x [2] dy 4 3(b) (i) Find the derivative, , of y = 5 + 8x - x . dx 3 ................................................. [2] 4 3 (ii) Find the gradient of y = 5 + 8 x - x at x =- 1. 3 ................................................. [2] 4 3 (iii) A tangent is drawn to the graph of y = 5 + 8x - x . 3 The gradient of the tangent is - 28 . Find the coordinates of the two possible points where this tangent meets the graph. ( ...................... , ...................... ) ( ...................... , ...................... ) [5]

Mark scheme: 9(a)(i) Correct sketch of 3x – 4y = 12 with 2 B1 for line with positive gradient y = –3 and x = 4 indicated on axes 4 –3 9(a)(ii) Correct sketch of y = x2 – 3x – 4 4 B3 for correct sketch with one value omitted or incorrect or for a with (0, – 4) indicated as y – intercept and x = – 1 and x = 4 poor sketch with all 3 intercepts correct. indicated as roots or B2 for roots x = – 1 and x = 4 soi with no extra roots -1 4 or for correct shape with y = – 4 indicated or B1 for correct shape -4 or for (x – 4) (x + 1) shown or for incorrect sketch with (0, – 4 ) indicated as y – intercept Minimum in fourth quadrant, not at x = 0 9(a)(iii) Correct sketch of y = 6x 2 B1 for increasing exponential graph seen on both sides of the with y-intercept indicated at (0, 1) y-axis. 1 9(b)(i) 8 – 4x2 [+ 0] 2 B1 for two terms correct and one extra incorrect term or for one of two terms correct or for correct answer seen and spoilt 9(b)(ii) 4 2 M1 for substitution of x = –1 into their (b)(i) 9(b)(iii) (3, –7) and (–3, 17) 5 B4 for (3, –7) or (–3, 17) or B3 for x = ± 3 or M2 for x2 = 9 or k(x – 3)(x + 3) = 0 oe or for correct method for solving their (b)(i) = – 28 or M1 for their (b)(i) = – 28

More questions on Differentiation

1 - 3 10 (a) a = b = e 2 o e 5o (i) On the grid, draw and label vector 2a. [1] (ii) On the grid, draw and label vector ( a - b ) . [2] (b) M C B NOT TO q SCALE N A O p OABC is a trapezium with OA parallel to CB. M is the midpoint of CB and N is the point on AB such that AN : NB = 1 : 2 . 3 O is the origin, OA = p , OC = q and CB = p . 4 (i) Find, in terms of p and/or q, in its simplest form (a) OB OB = ................................................. [1] (b) AB AB = ................................................. [2] (c) MN . MN = ................................................. [3] (ii) OA and MN are extended to meet at G. Find the position vector of G in terms of p. ................................................. [2]

Mark scheme: 10(a)(i) 2a drawn correctly with direction arrow 1 10(a)(ii) a − b drawn correctly with direction arrow 2  4  B1 for   seen or implied  −3  or M1 for correctly drawing their a – b with an arrow 10(b)(i)(a) 3 1 q + p final answer 4 10(b)(i)(b) 1 2 M1 for a correct route q – p final answer 4 10(b)(i)(c) 13 2 3 3 2 p – q final answer M2 for p – (their (b)(i)(b)) oe 24 3 8 3 3 1 or for – p – q + p + (their (b)(i)(b)) oe 8 3 or M1 for a correct route or for 2 [BN =] – (their (b)(i)(b)) 3 1 or [AN = ] (their (b)(i)(b)) 3 2 13 or final answer kp – q oe or p –kq oe 3 24 10(b)(ii) 19 2 3 p oe final answer M1 for AG = p ÷ 2 soi 16 8 or for answer kp oe

More questions on Vectors in two dimensions