Cambridge IGCSE Mathematics 0580 — 2021 Oct/Nov Paper 4 · Variant 2

1 (a) Malena has 450 fruit trees. The fruit trees are in the ratio apple : pear : plum = 8 : 7 : 3. (i) Show that Malena has 200 apple trees. [2] (ii) Find the number of plum trees. ................................................. [1] (iii) Malena wants to increase the number of pear trees by 32%. Calculate the number of extra pear trees she needs. ................................................. [2] (iv) Each apple tree produces 48.5 kg of apples. The apples have an average mass of 165 g each. Calculate the total number of apples produced by the 200 trees. Give your answer correct to the nearest 1000 apples. ................................................. [3] (b) Malena’s land is valued at three million and seventy-five thousand dollars. (i) Write this number in figures. ................................................. [1] (ii) Write your answer to part (b)(i) in standard form. ................................................. [1] (c) In 2020, each plum tree produced 37.7 kg of plums. This was 16% more than in 2019. Calculate the mass of plums produced by each plum tree in 2019. ............................................. kg [2] (d) Malena invests $1800 at a rate of 2.1% per year compound interest. Calculate the value of her investment at the end of 15 years. $ ................................................. [2]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 450 2 450 × 8 oe M1 for 8 + 7 + 3 8 + 7 + 3 1(a)(ii) 75 1 1(a)(iii) 56 2 32 M1 for × (450 – 200 – their 75) oe 100 32 450 or × × 7 oe 100 8 + 7 + 3 If 0 scored, SC1 for answer 231 1(a)(iv) 59 000 nfww 3 B2 for 58 600 to 58 800 or B1 for 293 to 294 figs485 × 200 or M1 for oe 165 If 0 scored, SC1 for their more accurate answer seen and rounded to the nearest 1000 1(b)(i) 3 075 000 1 1(b)(ii) 3.075 × 106 1 FT their (b)(i) 1(c) 32.5 2  16  M1 for x ×  1 +  = 37.7 or better  100  1(d) 2460 or 2458. … 2 15  2.1  M1 for 1800 1 +  oe  100 

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2 (a) The cumulative frequency diagram shows information about the floor area, a m 2, of each of 80 houses. 80 70 60 50 Cumulative 40frequency 30 20 10 0 a 40 60 80 100 120 140 160 180 200 Floor area (m2) Use the diagram to find an estimate of (i) the median, ........................................... m2 [1] (ii) the lower quartile, ........................................... m2 [1] (iii) the interquartile range, ........................................... m2 [1] (iv) the number of houses with a floor area greater than 120 m2. ................................................. [2] (b) The information about the 80 floor areas is shown in this frequency table. Floor area 2 40 1 a G 60 60 1 a G 80 8 0 1 a G 100 100 1 a G 130 130 1 a G 160 160 1 a G 200 ( a m ) Frequency 14 17 18 15 9 7 (i) Calculate an estimate of the mean floor area. ........................................... m2 [4] (ii) Complete the histogram to show the information in the frequency table. 1.0 Frequency 0.5density 0 a 40 60 80 100 120 140 160 180 200 Floor area (m2) [4] (iii) Two of the houses are picked at random. Find the probability that one of the houses has a floor area greater than 130 m 2 and the other has a floor area 60 m 2 or less. ................................................. [3]

Mark scheme: 2(a)(i) 90 1 2(a)(ii) 68 1 2(a)(iii) 52 1 FT 120 – their (a)(ii) 2(a)(iv) 20 2 B1 for 60 in working or as answer 2(b)(i) 97.5 4 M1 for mid-points soi (50, 70, 90, 115, 145, 180) M1 for use of Σfm with m in correct interval including both boundaries M1 for (dep on 2nd M1) for Σfm ÷ 80 2(b)(ii) Bars with heights 0.9, 0.5, 0.3, 0.175 4 B1 for each correct bar and with correct widths If 0 scored, SC1 for 3 or 4 correct frequency densities 2(b)(iii) 28 3 16 14 oe M2 for [2 ×] × oe 395 80 79 16 16 14 or M1 for or oe or oe or 80 79 80 14 oe seen 79 7 If 0 scored, SC1 for answer oe 100

More questions on Cumulative frequency diagrams

3 (a) C 38.6 m 56.5 m D B 94° 78.4 m NOT TO SCALE 46.1 m 64° E A ABCDE is a pentagon. (i) Calculate AD and show that it rounds to 94.5 m, correct to 1 decimal place. [2] (ii) Calculate angle BAC. Angle BAC = ................................................ [3] (iii) Calculate the largest angle in triangle CAD. ................................................. [4] (b) Q L 34.3 cm P NOT TO 21.5 cm SCALE 111° R N M 27.6 cm Triangle PQR has the same area as triangle LMN. Calculate the shortest distance from R to the line PQ. ............................................ cm [3]

Mark scheme: 3(a)(i) AD M1 = tan 64 oe or better 46.1 94.51 to 94.52 A1 3(a)(ii) 46[.0] or 45.96… nfww 3 sin94 M2 for 56.5 × oe 78.4 56.5 78.4 or M1 for = oe sin BAC sin94 3(a)(iii) 102.3 or 102.4 or 102.34 to 102.38 4 38.6 2 + 78.4 2 − 94.5 2 M2 for [cosC = ] 2 × 38.6 × 78.4 or M1 for 94.5 2 = 38.6 2 + 78.4 2 − 2 × 38.6 × 78.4 × cos C and A1 for –0.214 or –0.2144 to –0.2137 If 0 scored, SC2 for [CAD =] 23.5 or 23.51 to 23.52 or for [CDA =] 54.1 or 54.14… 3(b) 16.2 or 16.15… 3 1 1 M2 for × 21.5 × 27.6sin111 = × 34.3 × d 2 2 oe 1 or M1 for × 21.5 × 27.6sin111 seen or 2 1 × 34.3 × d oe soi 2

More questions on Non-right-angled triangles

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

Mark scheme: 4(a)(i) Image at (–5, 6) (–5, 8) (–6, 7) 2  − 4  k B1 for translation by   or   k  5 4(a)(ii) Image at (3, 1) (3, 3) (4, 2) 2 B1 for reflection in y = 1 or x = k 4(a)(iii) Image at (3, 4) (3, 8) (1, 6) 2 B1 for enlargement, sf 2, in wrong position 4(b) Rotation 3 B1 for each 90º [anticlockwise] oe (–3, 0)

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5 The table shows some values for y = x 3 - 3x 2 + 3 . x - 1 - .05 0 0.5 1 1.5 2 2.5 3 y 2.125 3 2.375 1 - 1 - .0125 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 - 3x 2 + 3 for - 1 G x G 3 . y 3.5 3 2.5 2 1.5 1 0.5 – 1 – 0.5 0 0.5 1 1.5 2 2.5 3 x – 0.5 – 1 – 1.5 [4] (c) By drawing a suitable straight line on the grid, solve the equation x 3 - 3x 2 + x + 1 = 0 . x = .................... or x = .................... or x = .................... [4]

Mark scheme: 5(a) –1, –0.375, 3 3 B1 for each 5(b) Correct graph 4 B3FT for 8 or 9 correct points or B2FT for 6 or 7 correct points or B1FT for 4 or 5 correct points 5(c) y = 2 – x ruled correctly 4 B2 for y = 2 – x ruled AND or B1 for [y = ] 2 – x soi –0.45 to –0.35 or y = k – x ruled 1 or y = kx + 2 ruled, but not y = 2 2.35 to 2.45 B2 for all three values or B1 for any two values

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6 (a) Solve. (i) 4 ( 2x - 3) = 24 x = ................................................ [3] (ii) 6x + 14 2 6 ................................................. [2] (b) Rearrange the formula V = 2x 3 - 3y 3 to make y the subject. y = ................................................ [3] 2 (c) Show that 2n - 5 - 13 is a multiple of 4 for all integer values of n. ` j [3] 2 2(d) The expression 5 + 12x - 2x can be written in the form q - 2 x + p . ` j (i) Find the value of p and the value of q. p = ................ , q = ................ [3] (ii) Write down the coordinates of the maximum point of the curve y = 5 + 12x - 2x 2 . (................ , ................) [1] (e) The energy of a moving object is directly proportional to the square of its speed. The speed of the object is increased by 30%. Calculate the percentage increase in the energy of the object. ............................................. % [2]

Mark scheme: 6(a)(i) 1 9 3 M1 for 8x – 12 = 24 or 2x – 3 = 6 4.5, 4 or M1 for reaching ax = b correctly FT their 2 2 first step 6(a)(ii) 4 2 14 x > − or x > –11 final answer M1 for 6x > 6 – 14 or x + > 1 3 3 6 6(b) 3 3 M1 for isolating term in y 2 x − V [y =] 3 oe final answer M1 for division by 3 or FT their first step 3 M1 for cube root or FT their previous step to the final answer 6(c) 4n2 – 20n + 12 M2 B1 for 4n2 – 10n – 10n + 25 4(n2 – 5n + 3) A1 with no errors seen or e.g. 4, [–]20 and 12 are all multiples of 4 or correct explanation linked to divides each term or each coefficient by 4 expression 6(d)(i) p = –3 and q = 23 3 B2 for 23 – 2(x –3)2 OR M1 for [q] – 2x2 – 4px – 2p2 or –2(x – 3)2 seen B1 for either p = –3 or q = 23 or FT q = 5 + 2(their p)2 6(d)(ii) (3, 23) 1 FT their (d)(i) 6(e) 69 2 M1 for figs 132 oe

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7 (a) The diagram shows a container for storing grain. The container is made from a hemisphere, a cylinder and a cone, each with radius 2 m. 2 m The height of the cylinder is 5.2 m and the height of the cone is h m. 5.2 m NOT TO SCALE 2 m h m (i) Calculate the volume of the hemisphere. Give your answer as a multiple of r. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 ........................................... m3 [2] 88 r (ii) The total volume of the container is m3 . 3 Calculate the value of h. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 h = ................................................ [4] (iii) The container is full of grain. Grain is removed from the container at a rate of 35 000 kg per hour. 1m3 of grain has a mass of 620 kg. Calculate the time taken to empty the container. Give your answer in hours and minutes. .................... h .................... min [3] (b) O NOT TO r cm 140° B SCALE A A and B are points on a circle, centre O, radius r cm. The area of the shaded segment is 65cm2. Calculate the value of r. r = ................................................ [4]

Mark scheme: 7(a)(i) 16π 2 1 4 3 or 51 π final answer M1 for × π × 2 oe 3 3 2 3 7(a)(ii) 2.4[0] 4 B3 for answer in range 2.396… to 2.40… OR 16π M3 for their + π × 22 × 5.2 + 3 1 2 88π π× 2 × h = oe 3 3 88π 16π or M2 for – their – π × 22 × 5.2 3 3 oe or M1 for π × 22 × 5.2 oe 1 2 or π× 2 × h oe soi 3 7(a)(iii) 1 hour 38 min or 1 hour 37.8 min to 1 3 B2 for 1.63[2…] or 98 [mins] or 97.8 to hour 37.9… min 97.9… ] 88π × 620 3 or M1 for [× 60] oe 35000 7(b) 8.5[0] or 8.496 to 8.497 4 65 M3 for [r= ] oe 140 1 π − sin140 360 2 140 1 or M2 for π × r2 – r2 × sin140 [=65] 360 2 oe or M1 for either area expression seen

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8 (a) Kaito runs along a 12 km path at an average speed of x km/h. (i) Write down an expression, in terms of x, for the number of hours he takes. ........................................ hours [1] (ii) Yuki takes 1.5 hours longer to walk along the same path as Kaito. She walks at an average speed of ( x - 4 ) km/h. Write down an equation, in terms of x, and show that it simplifies to x 2 - 4x - 32 = 0 . [4] (iii) Solve by factorisation. x 2 - 4x - 32 = 0 x = .................... or x = .................... [3] (iv) Find the number of hours it takes Yuki to walk along the 12 km path. ........................................ hours [2] (b) A bus travels 440 km, correct to the nearest 10 km. The time taken to complete the journey is 6 hours, correct to the nearest half hour. Calculate the lower bound of the speed of the bus. ........................................ km/h [3]

Mark scheme: 8(a)(i) 12 1 or 12 ÷ x final answer x 8(a)(ii) 12 12 M1 Accept 3 or more term equivalents – their = 1.5oe x − 4 x 12x – 12(x – 4) = 1.5x(x – 4) M1 Correctly clearing fractions, or correctly or collecting into a ‘single fraction’ 12 x − 12( x − 4) FT their expression dep on two fractions both [= 1.5] x ( x − 4) with algebraic denominators 12x – 12x + 48 = 1.5x2 – 6x M1 Correctly multiplying their two sets of brackets FT their expression dep on two fractions both with algebraic denominators or first M1 given [1.5x2 – 6x – 48 = 0 ] A1 One further step either 3 term equation or division throughout by 1.5 leading to solution x2 – 4x – 32=0 With no errors or omissions seen, dep on M3 8(a)(iii) (x + 4)(x – 8) M2 M1 for (x + a)(x + b) where ab = –32 or a + b = –4 or for x(x + 4) – 8(x + 4) or x(x – 8) + 4(x – 8) –4 and 8 B1 8(a)(iv) 3 2 12 FT their 8 − 4 12 12 M1 for or + 1.5 oe their 8 − 4 their 8 12 or for answer their 8 8(b) 69.6 3 430 to 440 440 − 5 M2 for or oe 6 + 0.25 6 to 6.5 or M1 for 440 + 5 oe or 440 – 5 oe or 6 + 0.25 oe or 6 – 0.25 oe seen

More questions on Graphs in practical situations

- 2 9 (a) F is the point (5, - 2 ) and FG = e 3o. Find (i) the coordinates of point G, (.................... , ....................) [1] (ii) 5 FG , [1] f p (iii) FG . ................................................. [2] (b) Q A B a P NOT TO SCALE O c C OABC is a parallelogram. P is a point on AC and Q is the midpoint of AB. OA = a and O C = c . (i) Find, in terms of a and/or c (a) AQ, AQ = ................................................. [1] (b) O Q . O Q = ................................................. [1] 2 1 (ii) OP = a + c 3 3 (a) Show that O, P and Q lie on a straight line. [2] (b) Write down the ratio OP : OQ. Give your answer in the form 1 : n. 1 : ................................................ [1]

Mark scheme: 9(a)(i) (3, 1) 1 9(a)(ii)  −10  1    15  9(a)(iii) 3.61 or 3.605 to 3.606 2 M1 for (–2)2 + 32 oe 9(b)(i)(a) 1 1 c 2 9(b)(i)(b) 1 1 FT a + their (b)(i)(a) a + c oe 2 9(b)(ii)(a)  1 2  O P = + c) oe B1 for O P or PQ factorised 3(2a  1 and OQ = (2a + c) oe or for correct multiplicative statement on 2 relationship without factorised vectors   OR 2 e.g. OQ = 1.5 O,P OQ = O,P 2 1 O=P (a + c) 3  3 2 2 PQ = O,P OR  2 1  1 1 1.5 a +  c = a + 1 c PQ = + c)  3 3  2 3(a 2 and correct comment e.g. have the same base vector or that they are multiples of one another and they share a common point OR    e.g. OQ = 1.5 O,P 2 PQ = O P 9(b)(ii)(b) 1.5 oe 1

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10 (a) Find the coordinates of the turning points of the graph of y = x 3 - 12x + 6 . You must show all your working. (........... , ...........) and (........... , ...........) [5] (b) Determine whether each turning point is a maximum or a minimum. Show how you decide. [3]

Mark scheme: 10(a) (2, –10) and (–2, 22) 5 B2 for 3x2 – 12 isw or B1 for 3x2 + k or px2 – 12 (p ≠ 0) or for 3x2 – 12 + 6 isw M1 for setting their derivative = 0 dy or = 0 dx B1 for x = ± 2 or for one correct coordinate pair 10(b) (2, –10) minimum with correct reason 3 B2 for 1 correct with correct reasoning or sketch or B2FT for correct evaluation with correct 2nd derivative for both of their different x (–2, 22) maximum with correct reason values or sketch or M1 for showing [2nd derivative =] 6x or gradients for one value on either side of one correct stationary point or for reasonable sketch of cubic

More questions on Differentiation

11 (a) R Z NOT TO 20.4 cm 15.3 cm 13.6 cm SCALE X Y P Q Triangle PQR is mathematically similar to triangle XYZ. (i) Find YZ. YZ = ........................................... cm [2] (ii) The area of triangle XYZ is 63.6 cm 2. Calculate the area of triangle PQR. ......................................... cm2 [3] (b) Two containers are mathematically similar. The larger container has a capacity of 64.8 litres and a surface area of 0.792 m 2. The smaller container has a capacity of 37.5 litres. Calculate the surface area of the smaller container. ........................................... m2 [3]

Mark scheme: 11(a)(i) 10.2 2 YZ 15.3 M1 for = oe or better 13.6 20.4 11(a)(ii) 143.1 3 2  20.4  M2 for   × 63.6 oe  13.6   20.4  2  13.6  2 or M1 for   or   oe  13.6   20.4  Alt method M2 for ½ × 20.4 × 15.3 × sin R where −1 63.6  R is sin    0.5 ×13.6 × their (a)(i)  −1 63.6  or M1 for R = sin    0.5 ×13.6 × their (a)(i)  11(b) 0.55 3 2  37.5  M2 for [ratio of areas] =  3  or    64.8  2  64.8   3  oe    37.5  37.5 or M1 for [ratio of lengths] = 3 or 64.8 64.8 3 oe 37.5  0.792 3  64.8  2 or for   =   oe  x   37.5 

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