Rates — Paper 4 · IGCSE Mathematics 0580

1 The Smith family paid $5635 for a holiday in India. The total cost was divided in the ratio travel : accommodation : entertainment = 10 : 17 : 8. (a) Calculate the percentage of the total cost spent on entertainment. … % [2] (b) Show that the amount spent on accommodation was $2737. [2] (c) The $5635 was the total amount Mr Smith received from an investment he made 5 years ago. Compound interest at a rate of 2.42% per year was paid on this investment. Calculate the amount he invested 5 years ago. $ … [3] (d) Mr Smith, his wife and their three children visit a theme park. The tickets cost 2500 Rupees for an adult and 1650 Rupees for a child. Calculate the total cost of the tickets. … Rupees [2] (e) One day the youngest child spent 130 Rupees on sweets. On this day the exchange rate was 1 Rupee = $0.0152 . Calculate the value of the sweets in dollars, correct to the nearest cent. $ … [2]

11 marks

This question in 0580/42 Feb/March 2017

1 (a) A library has a total of 10 494 fiction and non-fiction books. The ratio fiction books : non-fiction books = 13 : 5. Find the number of non-fiction books the library has. … [2] (b) The library has DVDs on crime, adventure and science fiction. The ratio crime : adventure : science fiction = 11 : 6 : 10. The library has 384 more science fiction DVDs than adventure DVDs. Calculate the number of crime DVDs the library has. … [2] (c) Every Monday, Sima travels by car to the library. The distance is 20 km and the journey takes 23 minutes. (i) Calculate the average speed for the journey in kilometres per hour. … km/h [2] (ii) One Monday, she is delayed and her average speed is reduced to 32 km/h. Calculate the percentage increase in the journey time. … % [5] (d) In Spain, the price of a book is 11.99 euros. In the USA, the price of the same book is $12.99 . The exchange rate is $1 = 0.9276 euros. Calculate the difference between these prices. Give your answer in dollars, correct to the nearest cent. $ … [3] (e) 7605 books were borrowed from the library in 2016. This was 22% less than in 2015. Calculate the number of books borrowed in 2015. … [3]

17 marks

Mark scheme: Question Answer Marks Partial marks 1(a) 2915 2 M1 for 10 494 ÷ (13 + 5) oe 1(b) 1056 2 M1 for 384 ÷ (10 – 6) oe 1(c)(i) 2 M1 for 20 ÷ 23 or 20 × 60 or 23 ÷ 60 isw 52.2 or 52.17… If zero scored, SC1 for answer 52.6 (from use of 0.38) 1(c)(ii) 63[.0] or 63.03 to 63.05… 5 their 52.17... − 32 M4 for × 100 oe 32 their 52.17... − 32 or M3 for oe or 32 their 52.17... × 100 oe 32 OR 5 B2 for [hours] oe or 37.5 [minutes] 8 or M1 for 20 ÷ 32 or better and their 37.5 − 23 M2 for × 100 oe 23 their 37.5 − 23 their 37.5 or M1 for or ×100 23 23 1(d) 0.06 final answer nfww 3 M1 for 11.99 ÷ 0.9276 or 12.99 × 0.9276 A1 for 12.93 or 12.925 to 12.926 1(e) 9750 3  22  M2 for 7605 ÷  1 −  oe  100  or M1 for (100 – 22)[%] correctly associated with 7605 seen

This question in 0580/41 Oct/Nov 2017

9 Luigi and Alfredo run in a 10 km race. Luigi’s average speed was x km/h. Alfredo’s average speed was 0.5 km/h slower than Luigi’s average speed. 10 (a) Luigi took hours to run the race. x Write down an expression, in terms of x, for the time that Alfredo took to run the race. … h [1] (b) Alfredo took 0.25 hours longer than Luigi to run the race. (i) Show that 2x 2 - x - 40 = 0 . [4] (ii) Use the quadratic formula to solve 2x 2 - x - 40 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (iii) Work out the time that Luigi took to run the 10 km race. Give your answer in hours and minutes, correct to the nearest minute. … h … min [3] Question 10 is printed on the next page.

12 marks

Mark scheme: 9(a) 10 1 20 oe final answer Accept x − 0.5 2 x − 1 9(b)(i) 10 10 M1 FT their (a) − = 0.25 oe x − 0.5 x 10x – 10(x – 0.5) = 0.25x (x – 0.5) M1 Clears algebraic denominators or collects as a oe single fraction FT their algebraic fractions dep on two fractions with algebraic denominators 10x – 10x + 5 = 0.25x2 – 0.125x or B1 Expands brackets better 2x2 – x – 40 = 0 A1 Dep on M1M1B1 and no errors seen 9(b)(ii) 2 B2 2 −−±1 ( − 1) − 4 × 2 × −40 B1 for ( −1) − 4(2)( −40) or better oe 2 × 2 −−+1 q −−−1 q or B1 for or or both 2 × 2 2 × 2 –4.23 and 4.73 final answers B1 B1 SC1 for –4.229… and 4.729… or for –4.23 and 4.73 seen in working or for –4.73 and 4.23 as final answer or for –4.2 or –4.22 and 4.7 or 4.72 as final answer 9(b)(iii) 2 [hours] 7 [minutes] 3 B2 for 2.11 or 2.114 to 2.115 or 126.8 to 126.9 or 127 or M1 for 10 ÷ their positive root from (b)(ii)

This question in 0580/43 Oct/Nov 2017

1 Amol and Priya deliver 645 parcels in the ratio Amol : Priya = 11 : 4. (a) Calculate the number of parcels Amol delivers. … [2] (b) Amol drives his truck at an average speed of 50 km/h. He leaves at 07 00 and arrives at 11 15. Calculate the distance he drives. … km [2] (c) Priya drives her van a distance of 54 km. She leaves at 10 55 and arrives at 12 38. Calculate her average speed. … km/h [3] (d) Priya has 50 identical parcels. Each parcel has a mass of 17 kg, correct to the nearest kilogram. Find the upper bound for the total mass of the 50 parcels. … kg [1] (e) 67 of the 645 parcels are damaged on the journey. Calculate the percentage of parcels that are damaged. … % [1] (f) (i) 29 parcels each have a value of $68. By writing each of these numbers correct to 1 significant figure, find an estimate for the total value of these 29 parcels. $ … [1] (ii) Without doing any calculation, complete this statement. The actual total value of these 29 parcels is less than the answer to part (f)(i) because … [1]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 473 2 M1 for 645 ÷ (11 + 4) 1(b) 212.5 2 M1 for 50 × 4.25 1(c) 31.5 or 31.45 to 31.46 3 43 M2 for 54 ÷ 160 oe or M1 for time =1h 43min or 103 [mins] or 54 ÷ their time 1(d) 875 1 1(e) 10.4 or 10.38 to 10.39 1 1(f)(i) 30 [×] 70 and 2100 1 1(f)(ii) both numbers rounded up oe 1

This question in 0580/42 Feb/March 2019

11 Brad travelled from his home in New York to Chamonix. • He left his home at 16 30 and travelled by taxi to the airport in New York. This journey took 55 minutes and had an average speed of 18 km/h. • He then travelled by plane to Geneva, departing from New York at 22 15. The flight path can be taken as an arc of a circle of radius 6400 km with a sector angle of 55.5°. The local time in Geneva is 6 hours ahead of the local time in New York. Brad arrived in Geneva at 11 25 the next day. • To complete his journey, Brad travelled by bus from Geneva to Chamonix. This journey started at 13 00 and took 1 hour 36 minutes. The average speed was 65 km/h. The local time in Chamonix is the same as the local time in Geneva. Find the overall average speed of Brad’s journey from his home in New York to Chamonix. Show all your working and give your answer in km/h. … km/h [11]

11 marks

Mark scheme: 11 [Total time =]16 h 6 min or 16.1 h 2 B1 for 22 h 6 min or 22.1h or 966 mins If 0 scored, SC1 for 9 h 41 min [Distance to airport in New York =] 16.5 2 M1 for 18 × 55 [Arc length =] 3 55.5 M2 for × 2 × π × 6400 6200 or 6199 to 6200. … 360 55.5 or M1 for or 2 × π × 2400 360 [Distance Geneva to Chamonix = ] 104 2 M1 for 65 × 1.6 or 65 × 96 oe 392 to 393 2 6316 to 6322.4 M1 for their 16.1 Must be correct value in numerator

This question in 0580/41 May/June 2019

1 Here is part of a train timetable for a journey from London to Marseille. All times given are in local time. The local time in Marseille is 1 hour ahead of the local time in London. London 07 19 Ashford 07 55 Lyon 13 00 Avignon 14 08 Marseille 14 46 (a) (i) Work out the total journey time from London to Marseille. Give your answer in hours and minutes. … h … min [2] (ii) The distance from London to Ashford is 90 km. The local time in London is the same as the local time in Ashford. Work out the average speed, in km/h, of the train between London and Ashford. … km/h [3] (iii) During the journey, the train takes 35 seconds to completely cross a bridge. The average speed of the train during this crossing is 90 km/h. The length of the train is 95 metres. Calculate the length, in metres, of this bridge. … m [4] (b) The fares for the train journey are shown in the table below. From London to Marseille Standard fare Premier fare Adult $84 $140 Child $60 $96 (i) For the standard fare, write the ratio adult fare : child fare in its simplest form. … : … [1] (ii) For an adult, find the percentage increase in the cost of the standard fare to the premier fare. … % [3] (iii) For one journey from London to Marseille, the ratio number of adults : number of children = 11 : 2. There were 220 adults in total on this journey. All of the children and 70% of the adults paid the standard fare. The remaining adults paid the premier fare. Calculate the total of the fares paid by the adults and the children. $ … [5] (c) There were 3.08 # 105 passengers that made this journey in 2018. This was a 12% decrease in the number of passengers that made this journey in 2017. Find the number of passengers that made this journey in 2017. Give your answer in standard form. … [3]

21 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 6h 27 mins 2 B1 for answer … h 27 mins 1(a)(ii) 150 km/h 3 90 M2 for × 60 36 90 or M1 for their time or B1 for 36 [mins] seen 1(a)(iii) 780 4  35  M3 for  90 ×  × 1000 – 95 oe  3600  or  35  M2 for  90 ×  × 1000 oe  3600  or B1 for figs 875 35 or M1 for 90 × seen 3600 1000 or for 90 × oe 3600 If 0 scored, SC1 for their distance (> 95) – 95 1(b)(i) 7 : 5 1 1(b)(ii) 66.7 or 66.66 to 66.67 3 140 − 84 M2 for [× 100] oe 84 140 or for × 100 oe 84 140 or M1 for oe 84 1(b)(iii) 24 576 5 M4 for complete method, 40 × 60 + 0.7 × 220 × 84 + 0.3 × 220 × 140 oe OR B1 for 40 [children] M1 for 0.7 × 220 × 84 oe M1 for 0.3 × 220 × 140 oe B1 for 2400 or 12936 or 9240 nfww 1(c) 3.5 × 105 nfww 3  100 − 12  M2 for 3.08 × 105 ÷   oe  100  or M1 for 3.08 [× 105] associated with (100–12)%

This question in 0580/43 May/June 2019

5 North NOT TO SCALE A 120 m 150 m B 180 m C The diagram shows a triangular field, ABC, on horizontal ground. (a) Olav runs from A to B at a constant speed of 4 m/s and then from B to C at a constant speed of 3 m/s. He then runs at a constant speed from C to A. His average speed for the whole journey is 3.6 m/s. Calculate his speed when he runs from C to A. … m/s [3] (b) Use the cosine rule to find angle BAC. Angle BAC = … [4] (c) The bearing of C from A is 210°. (i) Find the bearing of B from A. … [1] (ii) Find the bearing of A from B. … [2] (d) D is the point on AC that is nearest to B. Calculate the distance from D to A. … m [2]

12 marks

Mark scheme: 5(a) 4.29 or 4.285 to 4.286 3 150 M2 for 450 120 180 − − 6.3 4 3 or M1 for [time =] 120 ÷ 4 or 180 ÷ 3 or 150 + 180 + 120 450 ÷ 3.6 or 3.6 = total time 5(b) 82.8 or 82.81 to 82.82 using cosine 4 150 2 + 120 2 − 180 2 M2 for rule 2 × 150 × 120 or M1 for 180 2 = 120 2 + 150 2 − 2 × 120 × 150 cos(...) 4500 A1 for oe 36000 5(c)(i) 127.2 or 127.1 to 127.2 or 127 1 FT 210 – their (b) 5(c)(ii) 307.2 or 307.1 to 307.2 or 307 2 FT 180 + their(c)(i) M1 for 180 + their (c)(i) 5(d) 15 or 14.99 to 15.04 2 dist M1 for cos ( their ( b ) ) = oe 120

This question in 0580/41 Oct/Nov 2019

9 Car A and car B take part in a race around a circular track. One lap of the track measures 7.6 km. Car A takes 2 minutes and 40 seconds to complete each lap of the track. Car B takes 2 minutes and 25 seconds to complete each lap of the track. Both cars travel at a constant speed. (a) Calculate the speed of car A. Give your answer in kilometres per hour. … km/h [3] (b) Both cars start the race from the same position, S, at the same time. (i) Find the time taken when both car A and car B are next at position S at the same time. Give your answer in minutes and seconds. … min … s [4] (ii) Find the distance that car A has travelled at this time. … km [2]

9 marks

Mark scheme: 9(a) 171 or 171.0… 3 6.7 M2 for × 60 × 60 oe 160 7.6 7.6 7.6 or M1 for or or 160 2 2min 40sec 2 3 If 0 scored, SC1 for answer 189 or 188.6 to 188.7 9(b)(i) 77 [min] 20 [s] 4 32 M3 for × 29 oe 12 58 or B2 for 4640 or 1.29 or 1.288 to 1.289, 45 oe or 32 laps or 29 laps or M2 for 25× 5 × 29 oe or M1 for 2 m 40 sec ÷ (2 m 40 sec – 2 m 25 sec) soi for 2 m 25 sec ÷ (2 m 40 sec – 2 m 25 sec) soi or for an attempt to find LCM or 23 200 seen or correctly find prime factors of 145 or 160 7.6 7.6 7.6 or for or or oe, 145 5 2min 25sec 212 provided SC1 not earned in part (a) 9(b)(ii) 220.4 2 M1 for their (b)(i) ÷ 2min 40 sec [× 7.6] oe or their (a) × their (b)(i) ÷ 60 oe

This question in 0580/42 Oct/Nov 2019

1 (a) In a cycling club, the number of members are in the ratio males : females = 8 : 3. The club has 342 females. (i) Find the total number of members. … [2] (ii) Find the percentage of the total number of members that are female. … % [1] (b) The price of a bicycle is $1020. Club members receive a 15% discount on this price. Find how much a club member pays for this bicycle. $ … [2] (c) In 2019, the membership fee of the cycling club is $79.50 . This is 6% more than last year. Find the increase in the cost of the membership. $ … [3] (d) Asif cycles a distance of 105 km. On the first part of his journey he cycles 60 km in 2 hours 24 minutes. On the second part of his journey he cycles 45 km at 20 km/h. Find his average speed for the whole journey. … km/h [4] (e) Bryan invested $480 in an account 4 years ago. The account pays compound interest at a rate of 2.1% per year. Today, he uses some of the money in this account to buy a bicycle costing $430. Calculate how much money remains in his account. $ … [3] 1 2(f) The formula s = at is used to calculate the distance, s, travelled by a bicycle. 2 When a = 3 and t = 10 , each correct to the nearest integer, calculate the lower bound of the distance, s. … [2]

17 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1254 2 M1 for 342 ÷ 3 1(a)(ii) 27.3 or 27.27… 1 1(b) 867 2 15 M1 for 1020 × oe 100  15  or 1020 ×  1 −  oe  100  1(c) 4.5[0] 3 79.5 [ 0 M2 for ][× 6 ] oe 100 + 6 79.5 [ 0 ] or × 100 oe 100 + 6 or M1 for 79.5[0] associated with 106[%] 1(d) 22.6 or 22.58… nfww 4 45 M1 for or better 20 and 60 + 45 M2 for 45 their 2h 24min + their 20 45 or M1 for their + their 2h 24min 20 1(e) 91.6[0] to 91.61 3 4  2.1  M2 for 480 × 1 + − 430 oe    100   2.1  4 OR M1 for 480 × 1 + oe    100  A1 for 522, 521.6[0] to 521.61 1(f) 112.8125 2 B1 for 2.5 or 9.5 seen

This question in 0580/43 Oct/Nov 2019

7 North B 80 m NOT TO A SCALE 72° 115 m C The diagram shows the positions of three points A, B and C in a field. (a) Show that BC is 118.1 m, correct to 1 decimal place. [3] (b) Calculate angle ABC. Angle ABC = … [3] (c) The bearing of C from A is 147°. Find the bearing of (i) A from B, … [3] (ii) B from C. … [2] (d) Mitchell takes 35 seconds to run from A to C. Calculate his average running speed in kilometres per hour. … km/h [3] (e) Calculate the shortest distance from point B to AC. … m [3]

17 marks

Mark scheme: 7(a) [BC2 =] 802 + 1152 – 2 × 80 × M1 115 cos 72 oe 118.06… A2 A1 for 13939… 7(b) 67.8 or 67.9 or 67.83 to 67.88 3 115 × sin72 M2 for [sin B =] oe 118.1 115 118.1 or M1 for = oe sin B sin72 7(c)(i) 255 3 B1 for bearing of B from A is 75 soi M1 for 180 + 75 oe 7(c)(ii) [00]7.2 2 M1 for their (c)(i) – their (b) –180 7(d) 11.8 or 11.82 to 11.83 3 M1 for 115 ÷ 35 oe M1 for their speed in m/s × 60 × 60 ÷ 1000 7(e) 76.1 or 76.08 to 76.09 3 distance M2 for = sin72 oe 80 or M1 for distance required is perpendicular to AC soi

This question in 0580/41 May/June 2020

5 North D NOT TO SCALE 140° A 450 m 400 m B 350 m C The diagram shows a field ABCD. The bearing of B from A is 140°. C is due east of B and D is due north of C. AB = 400 m, BC = 350 m and CD = 450 m. (a) Find the bearing of D from B. … [2] (b) Calculate the distance from D to A. … m [6] (c) Jono runs around the field from A to B, B to C, C to D and D to A. He runs at a speed of 3 m/s. Calculate the total time Jono takes to run around the field. Give your answer in minutes and seconds, correct to the nearest second. … min … s [4]

12 marks

Mark scheme: 5(a) [0]38 or [0]37.9 or [0]37.87... 2 350 M1 for tan = oe 450 If 0 scored, SC1 for answer [0]52 or [0]52.1 or [0]52.12 to [0]52.13 5(b) 624 or 623.8 to 623.9 6 M2 for 450 – 400 sin 50 ... or M1 for sin 50 = 400 M2 for 350 + 400 cos 50 ... or M1 for cos 50 = 400 M1 for (their (450 – 400 sin 50))2 + (their (350 + 400 cos 50))2 5(c) 10 min 8 s 4 B3 for 10.1 or 10.13… or M2 for (400 + 350 + 450 + their DA) ÷ 3 [÷ 60] oe or M1 for any distance ÷ 3 M1 for rounding their minutes into minutes and seconds to nearest second if clearly seen

This question in 0580/42 May/June 2020

1 Karel travelled from London to Johannesburg and then from Johannesburg to Windhoek. (a) The flight from London to Johannesburg took 11 hours 10 minutes. The average speed was 813 km/h. Calculate the distance travelled from London to Johannesburg. Give your answer correct to the nearest 10 km. … km [3] (b) The total time for Karel’s journey from London to Windhoek was 15 hours 42 minutes. The total distance travelled from London to Windhoek was 10 260 km. (i) Calculate the average speed for this journey. … km/h [2] (ii) The cost of Karel’s journey from London to Windhoek was $470. (a) Calculate the distance travelled per dollar. … km per dollar [1] (b) Calculate the cost per 100 km of this journey. Give your answer correct to the nearest cent. $ … per 100 km [2] (c) Karel changed $300 into 3891 Namibian dollars. Complete the statement. $1 = … Namibian dollars [1]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 9080 cao 3 B2 for 9078 to 9081… or M1 for 813 × their 11h 10min 1(b)(i) 654 or 653.5… 2 M1 for 10260 ÷ 15 h 42 min oe 1(b)(ii)(a) 21.8 or 21.82 to 21.83 1 1(b)(ii)(b) 4.58 or 4.59 cao 2 M1 for 470 ÷ (10260 ÷ 100) oe or 100 ÷ their (b)(ii)(a) 1(c) 12.97 1

This question in 0580/42 Oct/Nov 2020

10 (a) A box is a cuboid with length 45 cm, width 30 cm and height 42 cm. The box is completely filled with 90.72 kg of sand. Calculate the density of this sand in kg/m3. [Density = mass ÷ volume] … kg/m3 [3] (b) A bag contains 15000 cm 3 of sand. Some of this sand is used to completely fill a hole in the shape of a cylinder. The hole is 30 cm deep and has radius 10 cm. Calculate the percentage of the sand from the bag that is used. … % [3] (c) Sand costs $98.90 per tonne. This cost includes a tax of 15%. Calculate the amount of tax paid per tonne of sand. $ … [3] (d) Raj buys some sand for 3540 rupees. Calculate the cost in dollars when the exchange rate is $1 = 70.8 rupees. $ … [2]

11 marks

Mark scheme: 10(a) 1600 3 B2 for answer figs 16 or M2 for 90.72 ÷ (figs45 × figs3 × figs42) or M1 for volume = figs 45 × figs 3 × figs 42 isw 10(b) 62.8 or 62.83 to 62.84 3 π× 10 2 × 30 M2 for × 100 15000 or M1 for π × 10 2 × 30 10(c) 12.9[0] 3 B2 for 86 OR 98.9 98.9 M2 for × 0.15 oe or 98.9 − oe 15 15 1 + 1 + 100 100  15  or M1 for  1 +  a = 98.9 oe isw  100  10(d) 50 2 M1 for 3540 ÷ 70.8

This question in 0580/42 Feb/March 2021

8 Darpan runs a distance of 12 km and then cycles a distance of 26 km. His running speed is x km / h and his cycling speed is 10 km / h faster than his running speed. He takes a total time of 2 hours 48 minutes. 12 (a) An expression for the time, in hours, Darpan takes to run the 12 km is . x Write an equation, in terms of x, for the total time he takes in hours. … [3] (b) Show that this equation simplifies to 7x 2 - 25x - 300 = 0 . [4] (c) Use the quadratic formula to solve 7x 2 - 25x - 300 = 0 . You must show all your working. x = … or x = … [4] (d) Calculate the number of minutes Darpan takes to run the 12 km. … min [2]

13 marks

Mark scheme: 8(a) 12 26 3 12 26 + = 2.8 oe isw B2 for + oe isw x x + 10 x x + 10 OR 26 B1 for seen x + 10 168 48 B1 for time = 2.8 or or 2 oe 60 60 8(b) 12 ( x + 10 ) + 26 x = 2.8 x ( x + 10 ) or M2 FT their time, provided 2 algebraic fractions one in x and other in ± x ± 10 better M1 for 12 ( x + 10 ) + 26 x seen or better 12 x + 120 + 26 x = 2.8 x 2 + 28 x M1 FT their equation dep on M2 2.8 x 2 − 10 x − 120 = 0 oe A1 or 30x + 300 + 65x = 7x2 + 70x or better 2 with no errors or omissions leading to 7 x − 25 x − 300 = 0 8(c) 2 B2 2 [ −− ]25 ± ( [ − ]25 ) − 4 × 7 × −300 B1 for ( [ − ]25 ) − 4(7)( −300) or better 2 × 7 [ −− ]25 + q [ −− ]25 − q oe or for or 2 × 7 2 × 7 − 5 and 8.57 or 8.571… B2 B1 for each or SC1 for final answers 5 and –8.57 8(d) 84 to 84.01… 2 720 FT to 3 sf or better their positive answer 12 M1 for [× 60 ] oe their positive answer

This question in 0580/42 Feb/March 2022

9 (a) O O NOT TO SCALE x° A B 2.4 cm AB The volume of a paper cone of radius 2.4 cm is 95.4 cm3. The paper is cut along the slant height from O to AB. The cone is opened to form a sector OAB of a circle with centre O. Calculate the sector angle x°. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … [6] (b) An empty fuel tank is filled using a cylindrical pipe with diameter 8 cm. Fuel flows along this pipe at a rate of 2 metres per second. It takes 24 minutes to fill the tank. Calculate the capacity of the tank. Give your answer in litres. … litres [4]

10 marks

Mark scheme: 9(a) 54[.0] or 53.99 to 54.03… 6 M2 for [h = ] 95.4 × 3 ÷ ( π × 2.42) oe 1 2 or M1 for 95.4 = × π × 2.4 × h 3 M2 for [slant ht , l = ] ( their h ) 2 + 2.4 2 or M1 for (their h)2 + 2.42 x M1 for × 2 × π × theirl = 2 × π × 2.4 oe 360 x 2 or × π × ( theirl ) = π × 2.4 × theirl 360 9(b) 14500 or 14470 to 14480 4 M3 for 200 × 60 × 24 × π × 42 [÷1000] or 2 × 60 ×24 ×π × 0.042 [×1000] or M2 for 200 × π × 42 or for 2 × π × 0.042 or M1 for π × 42 oe or π × 0.042 seen oe isw or 1000 cm3 = 1 litre soi or 1 m3 = 1000 litres soi or for 24 × 60 seen oe

This question in 0580/42 Feb/March 2022

1 (a) Find the lowest common multiple (LCM) of 30 and 75. … [2] (b) Share $608 in the ratio 4 : 5 : 7. $ … $ … $ … [3] 6 .39 # 10 4 (c) Work out 6 . 2 .45 # 10 Give your answer in standard form. … [2] (d) Write .027o o as a fraction. … [1] (e) A stone has volume 45 cm 3 and mass 126 g. Find the density of the stone, giving the units of your answer. [Density = mass ' volume] … … [2]

10 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 150 2 B1 for answer 150k or M1 for prime factors of 30 or 75 seen or a list of multiples of both 30 and 75 with at least 3 of each 30  75 or for oe 15 or for answer 2  3  52 1(b) 152 3 Accept in any order B2 for two correct answers 190 608 or M1 for  k oe where k =1, 4, 5, 7 4  5  7 266 1(c) 2.61  10–2 2.61  10 2 or 2 B1 for figs 2608 or 261 seen 2.608…  10–2 If 0 scored, SC1 for answer 2.6[0]  10–2 without more accurate value in standard form seen 1(d) 27 1 oe fraction 99 1(e) 2.8 1 g/cm3 or g cm–3 1

This question in 0580/42 May/June 2022

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]

15 marks

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

This question in 0580/43 Oct/Nov 2022

7 North C NOT TO SCALE 60 km 87 km 38° B A The diagram shows the straight roads between town A, town B and town C. AC = 60 km , CB = 87 km and B is due east of A. The bearing of C from A is 038°. (a) Show that angle ACB = 95.1° , correct to 1 decimal place. [5] (b) Without stopping, a car travels from town A to town C then to town B, before returning directly to town A. The total time taken for the journey is 3 hours 20 minutes. Calculate the average speed of the car for this journey. Give your answer in kilometres per hour. … km/h [6]

11 marks

Mark scheme: 7(a) Angle CAB = 52 B1 1 60sin their 52  M3 60sin their 52 180 – 52 – sin M2 for [sin[...]  ] oe    87  87 60 87 or M1 for  oe sin B sin their 52 95.08… A1 7(b) 77.1 or 77.08 to 77.11 6 B4 for dist travelled = 256.9 to 257[.0…] or B3 for [AB =] 109.9 to 110[.0…] or M3 for 60 + 87 + 60 2  87 2 – 2  60  87  cos 95.1 oe or M2 for 60 2  87 2 – 2  60  87  cos 95.1 oe or AB2 = 12093. … to 12097. … 87sin95.1 or oe sin their 52 or M1 for AB2 = 602 + 872 – 2 × 60 × 87 × cos 95.1 oe sin95.1 sin their 52 or  oe AB 87 20 M1 for their total distance ÷ 3 oe 60

This question in 0580/42 May/June 2023

7 (a) (i) A car travels 50 km at an average speed of 75 km/h. Find the time taken. Give your answer in minutes. … min [2] (ii) Another car travels 47 km, correct to the nearest kilometre. The average speed of this car is 75 km/h, correct to the nearest 5 km/h. Calculate the lower bound of the time taken. Give your answer in minutes. … min [3] (b) A train travels a total of 240 km. The train travels for t minutes at an average speed of 100 km/h. It then travels for ( t + 60 ) minutes at an average speed of 110 km/h. Find the average speed for the whole journey. … km/h [6]

11 marks

Mark scheme: 7(a)(i) 40 2 50 M1 for [ 60] oe 75 7(a)(ii) 36 nfww 3 47  0.5 46 to 47 M2 for [ 60] or [ 75to 80 75  2.5 60] or M1 for 47+0.5 or 47 – 0.5 or 75 + 2.5 or 75– 2.5 7(b) 107 or 107.2... 6 240 M5 for [speed = ]  60 oe 260  2  7  60  OR B5 for [total time = ] 134 or 134.2 to 134.3 or 2.24 or 2.238... or B4 for (t = ) 37.1 or 37.14... OR t t  60 M2 for  100 +  110 = 240 60 60 oe t t  60 or M1 for  100 or  110 oe 60 60 M1 for correct equation of form at = b from their equation containing two terms in t and involving the speeds. 240 M1 for [× 60] 2  theirt  60

This question in 0580/43 May/June 2024

1 (a) Anvi buys a new car. (i) The price of the car is $28 240. She is given a 7.5% discount. Calculate the amount she pays. $ … [2] (ii) The fuel tank in the new car has a capacity of 45 litres. This is 72% of the capacity of the fuel tank in her old car. Calculate the capacity of the fuel tank in her old car. … litres [2] (b) Aadi buys a new car costing $28 000. He pays for the car using a finance plan. The finance plan is • a deposit • 47 equal monthly payments of $330 • a final payment of $11 490. Using this finance plan, Aadi pays a total of $31 900 for the car. Calculate the deposit paid as a percentage of $28 000. … % [4] (c) A car travels 64 km and uses 2.5 litres of fuel. It then travels 128 km and uses 6 litres of fuel. Calculate the rate at which the car uses fuel during the whole journey. Give your answer in litres per 100 km. … litres per 100 km [2] (d) At the start of 2021 the value of a car was $46 500. At the end of 2021 the value of the car was 20% less. At the end of 2022 the value of the car was 15% less than its value at the end of 2021. Calculate the value of the car at the end of 2022. $ … [2]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 26 122 cao 2  7.5  M1 for 28 240   1 −  oe  100  or B1 for answer 2118 1(a)(ii) 62.5 2 72 M1 for C  = 45 oe or better 100 1(b) 17.5 4 31900 – 11490 – (47  330) M3 for [ 100] 28000 or M2 for 31 900 – 11 490 – (47  330) or M1 for 47  330 or for 31 900 – 11 490 1(c) 4.43 or 4.427… 2 2.5 + 6 M1 for [ 100] oe 64 + 128 1(d) 31 620 2  20    15   M1 for 46 500  1 −  1 −        100    100    15    20   or 46 500  1 −  1 −        100    100    20   15  or for  1 −    1 −   100   100 

This question in 0580/42 Oct/Nov 2024

5 (a) The cumulative frequency diagram shows information about the distance travelled by each of 80 motorists in a month. 80 60 Cumulative 40 frequency 20 0 0 400 800 1200 1600 2000 2400 Distance (km) (i) Use the cumulative frequency diagram to find an estimate for (a) the median … km [1] (b) the interquartile range … km [2] (ii) One of these motorists is picked at random. Find the probability that this motorist travels more than 1800 km. … [2] (b) The distance around a racing track is 5.104 km. The time taken by a car to complete one lap of the track is 1 min 18 s. Calculate the average speed of the car. Give your answer in km/h. … km/h [3] (c) The top speed, v km/h, of each of 160 cars is recorded. The histogram shows this information. 2.0 1.5 Frequency 1.0 density 0.5 0 v 100 140 180 220 260 300 Top speed (km / h) (i) Show that there are 8 cars with a top speed in the interval 120 1 v G 160 . [1] (ii) Calculate an estimate of the mean top speed. You must show all your working. … km/h [6]

15 marks

Mark scheme: 5(a)(i)(a) 1480 1 5(a)(i)(b) 440 2 M1 for [UQ =] 1600 soi or [LQ =] 1160 soi 5(a)(ii) 8 2 oe M1 for 72 or 8 written 80 5(b) 236 or 235.5 to 235.6 3 5.104 M2 for  60 oe 1.3 5.104 or M1 for their time 5(c)(i) (160 – 120)  0.2 [ = 8] 1 with no errors seen 5(c)(ii) 22, 36, 64, 30 seen B2 B1 for 2 or 3 correct frequencies or M1 for three of 1.1  (180 – 160), 1.8  (200 – 180), 1.6  (240 – 200) and 0.5  (300 – 240) oe (8 × 140 + their22 × 170 M3 + their36 × 190 + their64 × 220 M1 for midpoints soi + their30 × 270) ÷ 160 M1 for fx , x in interval or boundary of interval M1 dep on second M1 for fx ÷ 160 211.75 B1

This question in 0580/42 Oct/Nov 2024

5 (a) Naomi runs 100 m in 15 seconds. Calculate Naomi’s average speed in kilometres per hour. … km/h [2] (b) Olav runs for 45 minutes at a speed of 9.5 km/h. He then runs 8.1 km at a speed of 7.5 km/h. Calculate Olav’s average speed for the whole run. … km/h [3] (c) A train has length p metres. The train passes through a station of length q metres. The speed of the train is v kilometres per hour. Find an expression for the time the train takes to completely pass through the station. Give your answer in seconds, in terms of p, q and v. … s [3]

8 marks

Mark scheme: 5(a) 24 2 100or0.1 M1 for time or B1 for figs 24 5(b) 8.32 or 8.319 to 8.320 3 45 M1 for 9.5  oe 60 8.1 M1 for 7.5 5(c) 18( p + q ) 3 ( p + q ) oe final answer M1 for [k ] for some k  0 5v v 1000 M1 for v  oe soi 3600

This question in 0580/43 Oct/Nov 2024

24 Martha walks a distance of 10 km at a speed of x km/h. She then runs a distance of 5 km at a speed of ( x + 4 ) km/h. The total time taken for the whole journey is 3.5 hours. (a) Write down an expression in terms of x for the time Martha is walking. … h [1] (b) Show that 7x 2 - 2 x - 80 = 0 . [4] (c) Solve 7x 2 - 2 x - 80 = 0 , giving your answers correct to 2 decimal places. You must show all your working. x = … or x = … [3] (d) Calculate the difference between the time Martha is walking and the time she is running. Give your answer in hours and minutes correct to the nearest minute. … h … min [3]

11 marks

Mark scheme: 24(a) 10 1 x 24(b) their10 + 5 = 7 oe M1 x x + 4 2 20 x + 80 + 10 x = 7 x 2 + 28 x oe M2 Strict FT for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 and expanding all brackets Strict M1FT for correctly expressing their two algebraic fractions with two denominators in x and x + 4 as a single fraction or with a common denominator within a correct equation or for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 but not all brackets expanded Leading to 7 x 2 − 2 x − 80 = 0 A1 No errors or omissions 24(c) 2 B2 2 −−( 2 )  ([ − ]2) − 4 ( 7 )( −80 ) or B1 for ([ −]2) − 4 ( 7 )( −80 ) oe or for oe 2 ( 7 ) −−( 2) − p −−( 2) + p oe or for oe 2(7) 2(7)  2  2 or  x −   14  –3.24 and 3.53 B1 24(d) 2h 10min 3 B2 for 2.168 to 2.18 [h] or for 130.08 to 130.8 [min] or for 2hours 10.08 min to 2 hours 10.8 min OR 10 5 M2 for − their positive x their positive x + 4 or 10 5 M1 for or their positive x their positive x + 4

This question in 0580/41 Oct/Nov 2025