Limits of accuracy — Paper 4 · IGCSE Mathematics 0580

2 The time taken for each of 90 cars to complete one lap of a race track is shown in the table. Time (t seconds) 70 1 t G 71 71 1 t G 72 72 1 t G 73 73 1 t G 74 74 1 t G 75 Frequency 17 24 21 18 10 (a) Write down the modal time interval. … 1 t G … [1] (b) Calculate an estimate of the mean time. … s [4] (c) (i) Complete the cumulative frequency table. Time (t seconds) t G 71 t G 72 t G 73 t G 74 t G 75 Cumulative frequency 17 [2] (ii) On the grid, draw a cumulative frequency diagram to show this information. 90 80 70 60 50 Cumulative frequency 40 30 20 10 0 t 70 71 72 73 74 75 Time (seconds) [3] (iii) Find the median time. … s [1] (iv) Find the inter-quartile range. … s [2] (d) One lap of the race track measures 3720 metres, correct to the nearest 10 metres. A car completed the lap in 75 seconds, correct to the nearest second. Calculate the upper bound for the average speed of this car. Give your answer in kilometres per hour. … km/h [4]

17 marks

Mark scheme: 2(a) 71 < t ⩽ 72 1 2(b) 72.3 or 72.27 to 72.28 nfww 4 M1 for midpoints soi (condone 1 error or omission) M1 for use of ∑fx with x in correct interval including both boundaries M1 (dep on 2nd M1) for ∑fx ÷ 90 2(c)(i) 41, 62, 80, 90 2 B1 for 2 correct values 2(c)(ii) Correct curve 3 B1FT their (c)(i) for 5 correct heights B1 for 5 points plotted at upper ends of intervals B1FT (dep on at least B1) for increasing curve or increasing polygon through 5 points If zero scored, SC1FT for 4 correct points plotted 2(c)(iii) 72.1 to 72.4 1 2(c)(iv) 1.9 to 2.2 2 M1 for UQ = 73.2 to 73.4 or LQ = 71.2 to 71.3 2(d) 180 cao nfww 4 B3 for 50 [m/s] nfww OR 3725 ÷ 1000 M3 for 74.5 ÷ 3600 OR M2 for 3725 ÷ 74.5 or M1 for 3725 or 74.5 seen or for (3715 to 3725) ÷ (74.5 to 75.5) M1 indep for multiply by 3.6 oe

This question in 0580/41 May/June 2017

1 (a) In 2016, a company sold 9600 cars, correct to the nearest hundred. (i) Write down the lower bound for the number of cars sold. … [1] (ii) The average profit on each car sold was $2430, correct to the nearest $10. Calculate the lower bound for the total profit. Write down the exact answer. $ … [2] (iii) Write your answer to part (a)(ii) correct to 4 significant figures. $ … [1] (iv) Write your answer to part (a)(iii) in standard form. $ … [1] (b) In April, the number of cars sold was 546. This was an increase of 5% on the number of cars sold in March. Calculate the number of cars sold in March. … [3] (c) The price of a new car grows exponentially by 3% per year. A new car has a price of $3000 in 2013. Find the price of a new car 4 years later. $ … [2]

10 marks

Mark scheme: Question Answer Marks Part marks 1(a)(i) 9550 1 1(a)(ii) 23 158 750 2FT FT their (a)(i) × 2425 correctly evaluated M1 for their lower bound × 2425 1(a)(iii) 23 160 000 1FT FT their (a)(ii) rounded to 4 sf 1(a)(iv) 2.316 × 107 1FT FT their (a)(iii) or their (a)(ii) rounded to 3sf or more and in standard form 1(b) 520 nfww 3 100 M2 for 546 × oe (100 + 5 ) or M1 for 105[%] associated with 546 oe 1(c) 3380 or 3376 to 3377 2 4  3  M1 for 3 000 ×  1 +  oe  100 

This question in 0580/43 May/June 2017

3 The graph shows information about the journey of a train between two stations. NOT TO SCALE 126 Speed (km / h) 0 09 00 09 04 09 48 09 55 Time of day (a) (i) Work out the acceleration of the train during the first 4 minutes of this journey. Give your answer in km/h2. … km/h2 [2] (ii) Calculate the distance, in kilometres, between the two stations. … km [4] (b) (i) Show that 126 km/h is the same speed as 35 m/s. [1] (ii) The train has a total length of 220 m. At 09 30, the train crossed a bridge of length 1400 m. Calculate the time, in seconds, that the train took to completely cross the bridge. … s [3] (c) On a different journey, the train took 73 minutes, correct to the nearest minute, to travel 215 km, correct to the nearest 5 km. Calculate the upper bound of the average speed of the train for this journey. Give your answer in km/h. … km/h [4]

14 marks

Mark scheme: 3(a)(i) 1890 2 M1 for 126 ÷ 4 [× 60] oe If zero scored, SC1 for answer 31.5 3(a)(ii) 103.95 4  44 55  M3 for 0.5 ×  +  × 126 oe  60 60  or SC3 for figs 10395 or figs 104 or M2 for two correct area methods or for a full method without minutes to hours conversion or M1 for one correct area with or without minutes to hours conversion 3(b)(i) 126 × 1000 ÷ (60 × 60) 1 3(b)(ii) 46.3 or 46.28 to 46.29 3 M2 for (1400 + 220) ÷ 35 oe or M1 for distance ÷ speed or 1400 + 220 3(c) 180 nfww 4 B3 for final answer 3 OR 217.5 M3 for × 60 oe 72.5 or M2 for 217.5 ÷ 72.5 oe 210 to 220 or × 60 72.5 217.5 or × 60 72 to 74 215 or M1 for 217.5 or 72.5 seen or × 60 73

This question in 0580/43 Oct/Nov 2017

10 (a) In 2017, the membership fee for a sports club was $79.50 . This was an increase of 6% on the fee in 2016. Calculate the fee in 2016. $ … [3] (b) On one day, the number of members using the exercise machines was 40, correct to the nearest 10. Each member used a machine for 30 minutes, correct to the nearest 5 minutes. Calculate the lower bound for the number of minutes the exercise machines were used on this day. … min [2] (c) On another day, the number of members using the exercise machines (E), the swimming pool (S) and the tennis courts (T) is shown on the Venn diagram. E S 33 20 5 4 7 8 16 T (i) Find the number of members using only the tennis courts. … [1] (ii) Find the number of members using the swimming pool. … [1] (iii) A member using the swimming pool is chosen at random. Find the probability that this member also uses the tennis courts and the exercise machines. … [2] (iv) Find n T + E , S ^ ^ hh. … [1]

10 marks

Mark scheme: 10(a) 75 3 M2 for 79.5 ÷ 1.06 oe or M1 for 79.5 associated with 106 [%] 10(b) 962.5 cao 2 B1 for 35 or 27.5 seen 10(c)(i) 16 1 10(c)(ii) 50 1 10(c)(iii) 4 2 FT their (c)(ii) for 1 or 2 marks oe 4 k 50 B1 for , k > 4 or , k < 50 k their 50 10(c)(iv) 19 1

This question in 0580/41 May/June 2018

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

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

1 (a) Campsite fees (per day) Tent … $15.00 Caravan … $25.00 The sign shows the fees charged at a campsite. Today there are 54 tents and 18 caravans on the site. Calculate the fees charged today. $ … [2] (b) In September the total income at the campsite was $37 054. This was a decrease of 4.5% on the total income in August. Calculate the total income in August. $ … [2] (c) The visitors to the campsite today are in the ratio men : women = 5 : 4 and women : children = 3 : 7. (i) Calculate the ratio men : women : children in its simplest form. … : … : … [2] (ii) Today there are 224 children at the campsite. Calculate the total number of men and women. … [3] (d) The space allowed for each tent is a rectangle measuring 8 m by 6 m, each correct to the nearest metre. Calculate the upper bound for the area of the space allowed for each tent. … m2 [2] (e) The value of the campsite has increased exponentially by 1.5% every year since it opened 30 years ago. Calculate the value of the campsite now as a percentage of its value 30 years ago. … % [2]

13 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 1260 2 M1 for 15 × 54 + 25 × 18 1(b) 38 800 2  4.5  M1 for 37054 ÷  1 −  oe  100  1(c)(i) 15 : 12 : 28 2 M1 for correct attempt to find a common multiple for the women oe 1(c)(ii) 216 3 M2 for 224 ÷ their 28 × their (15 + 12) or M1 for 224 ÷ their 28 1(d) 55.25 2 M1 for 8 + 0.5 or 6 + 0.5 seen 1(e) 156 or 156.3… 2 30  1.5  M1 for  1 +   100 

This question in 0580/43 May/June 2020

4 (a) A rectangle measures 8.5 cm by 10.7 cm, both correct to 1 decimal place. Calculate the upper bound of the perimeter of the rectangle. … cm [3] (b) B C D E 80° NOT TO SCALE 9 cm h 40° A 12 cm F ABDF is a parallelogram and BCDE is a straight line. AF = 12 cm, AB = 9 cm, angle CFD = 40° and angle FDE = 80°. (i) Calculate the height, h, of the parallelogram. h = … cm [2] (ii) Explain why triangle CDF is isosceles. … … [2] (iii) Calculate the area of the trapezium ABCF. … cm2 [3] (c) C B 12 cm NOT TO SCALE O 21° D A A, B, C and D are points on the circle, centre O. Angle ABD = 21° and CD = 12 cm. Calculate the area of the circle. … cm2 [5] (d) x° NOT TO 8 cm 9.5 cm SCALE The diagram shows a square with side length 8 cm and a sector of a circle with radius 9.5 cm and sector angle x°. The perimeter of the square is equal to the perimeter of the sector. Calculate the value of x. x = … [3]

18 marks

Mark scheme: 4(a) 38.6 3 M2 for [2 ×] (8.5 + 0.05 + 10.7 + 0.05) or M1 for 8.5 + 0.05 or 10.7 + 0.05 4(b)(i) 8.86 or 8.863… 2 h M1 for = sin 80 or better oe 9 4(b)(ii) ∠CDF = 100 leading to ∠DCF = 40 M1 Implied by 180-(100 + 40) = 40 Or or ∠EDF = 80 leading to ∠DCF = 40 80 – 40 ‘two equal angles’ A1 With no incorrect work seen 4(b)(iii) 66.5 or 66.45 to 66.47… 3 M2 for 0.5(3 + 12) × their (b)(i) or 12 × their (b)(i) – 0.5 × 9 × 9 × sin 100 oe or B1 for DC = 9 or BC = 3 4(c) 130 nfww or 129.6 to 129.8 5 B1 for ∠ACD = 21º or ∠CAD = 69º Method 1 12 M2 for cos 21 = oe AC or M1 for ∠ADC = 90 soi M1 for π(their AC/2)2 OR Method 2 12 r M2 for = oe sin138 sin 21 or M1 for ∠COD = 138 soi M1 for π (their r ) 2 OR Method 3 6 M2 for cos 21 = oe OC or M1 for ∠CXO = 90 soi where X is the point where the perpendicular from O meets the chord CD M1 for π ( their OC) 2 4(d) 78.4 or 78.37 to 78.41 3 M2 for x × 2 × π × 9.5 + 2 × 9.5 = 4 × 8 oe 360 x or M1 for × 2 × π × 9.5 360 After M0, SC1 for 9.5x + 19 = 32 oe

This question in 0580/41 Oct/Nov 2020

1 (a) A 2.5-litre tin of paint costs $13.50 . In a sale, the cost is reduced by 14%. (i) Work out the sale price of this tin of paint. $ … [2] (ii) Work out the cost of buying 42.5 litres of paint at this sale price. $ … [2] (b) Henri buys some paint in the ratio red paint : white paint : green paint = 2 : 8 : 5. (i) Find the percentage of this paint that is white. … % [1] (ii) Henri buys a total of 22.5 litres of paint. Find the number of litres of green paint he buys. … litres [2] (c) Maria paints a rectangular wall. The length of the wall is 20.5 m and the height is 2.4 m, both correct to 1 decimal place. One litre of paint covers an area of exactly 10 m2. Calculate the smallest number of 2.5-litre tins of paint she will need to be sure all the wall is painted. Show all your working. … [4]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 11.61 final answer 2  14  M1 for 13.5[0] ×  1 −  oe  100  or B1 for 1.89 1(a)(ii) 197.37 final answer 2 FT 17 × their (a)(i) exact or correct to nearest cent M1 for 42.5 ÷ 2.5 1(b)(i) 53.3 or 53.33… 1 1(b)(ii) 7.5 2 M1 for 22.5 ÷ (2 + 8 + 5) oe soi 1(c) 20.55 × 2.45 oe M2 M1 for 20.5 + 0.05 oe seen or 2.4 + 0.05 oe seen If 0 scored, SC1 here for 20.45 × 2.35 oe 3 nfww A2 M1 for their area ÷ 10 ÷ 2.5 oe

This question in 0580/42 May/June 2021

2 Bob, Chao and Mei take part in a run for charity. (a) Their times to complete the run are in the ratio Bob : Chao : Mei = 4 : 5 : 7. (i) Find Chao’s time as a percentage of Mei’s time. … % [1] (ii) Bob’s time for the run is 55 minutes 40 seconds. Find Mei’s time for the run. Give your answer in minutes and seconds. … min … s [3] (b) Chao collects $47.50 for charity. (i) Bob collects 28% more than Chao. Find the amount Bob collects. $ … [2] (ii) Chao collects 60% less than Mei. Find how much more money Mei collects than Chao. $ … [3] (c) When running, Chao has a stride length of 70 cm, correct to the nearest 5 cm. Chao runs a distance of 11.2 km, correct to the nearest 0.1 km. Work out the minimum number of strides that Chao could take to complete this distance. … [4] (d) In 2015, a charity raised a total of $1.6 million. After 2015, this amount increased exponentially by 2.4% each year for the next 5 years. Work out the amount raised by the charity in 2020. $ … million [2]

15 marks

Mark scheme: 2(a)(i) 71.4 or 71.42 to 71.43 1 2(a)(ii) 97 [min] 25 [s] 3 B2 for 13 min 55 sec seen or 97.4 or 97.41 to 97.42 seen or 5845 seen OR M2 for 55.66… ÷ 4 × 7 oe or 3340 ÷ 4 × 7 oe or for 7/4 × 55 + 7/4 × 40 oe or M1 for 55 min 40 sec ÷ 4 oe or M1 for total time ÷ 16 soi 2(b)(i) 60.8[0] 2  28  M1 for 47.5 ×  1 +  oe  100  or B1 for 13.3[0] 2(b)(ii) 71.25 3 B2 for 118.75  60  Or M2 for 47.50 ÷  1 −  – 47.50  100   60  or M1 for x ×  1 −  = 47.50 oe or  100  better 2(c) 15 380 4 M3 for (1 120 000 – 5000) ÷ (70 + 2.5) oe or B2 for answer figs 15 379 to figs 15 380 or M2 for (1 120 000 ± 5000) ÷ (70 ± 2.5) oe or M1 for one of figs 675, 725, 1115, 1125 seen 2(d) 1.8[0] or 1.801 to 1.802 [million] nfww 2 5  2.4  M1 for figs 16 ×  1 +  oe  100 

This question in 0580/41 Oct/Nov 2021

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]

13 marks

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

This question in 0580/42 Oct/Nov 2021

3 (a) 5 cm NOT TO SCALE 4 cm C D A B 10 cm The diagram shows a prism. The cross-section of the prism is a trapezium with CD parallel to AB and AC = BD. AB = 10 cm, CD = 4 cm and the height of the trapezium is 5 cm. The volume of the prism is 525 cm3. (i) The prism is made of iron. 1 cm3 of iron has a mass of 7.8 g. Calculate the mass of the prism. Give your answer in kilograms. … kg [2] (ii) Calculate the length of the prism. … cm [3] (iii) Calculate the total surface area of the prism. … cm2 [6] (iv) In a mathematically similar prism, the height of the trapezium is 10 cm. Calculate the volume of this prism. … cm3 [3] (b) A cuboid measures 10 cm by 4 cm by 6 cm. Each side is measured correct to the nearest centimetre. Complete the inequality for the volume, V, of this cuboid. … cm 3 G V 1 … cm3 [3]

17 marks

Mark scheme: 3(a)(i) 4.095 2 B1 for figs 4095 525 × 7.8 or M1 for 1000 3(a)(ii) 15 3 B2 for 35 OR 1 M2 for (10 + 4) × 5 × L = 525 oe 2 1 M1 for (10 + 4) × 5 oe 2 3(a)(iii) 455 or 454.9... 6 2 2 M3 for their [BD =] 3 + 5 × (their 15) [× 2] or B2 for 34 or 5.83 or 5.830 to 5.831 2  1  2 or M1 for 5 +  (10 − 4 )   2  and M1 for their 35 × 2 M1 for (their 15) × 10 and (their 15) × 4 3(a)(iv) 4200 3 3  10  M2 for 525 ×   oe  5   10  3  5  3 or M1 for   or   oe  5   10  3(b) 182.875 ... 307.125 final answer 3 B2 for either seen or M1 for 10 ± 0.5 or 6 ± 0.5 or 4 ± 0.5 oe

This question in 0580/43 Oct/Nov 2021

7 Two rectangular picture frames are mathematically similar. (a) The areas of the frames are 350 cm2 and 1134 cm2. The width of the smaller frame is 17.5 cm. Calculate the width of the larger frame. … cm [3] (b) A picture in the smaller frame has length 15 cm and width 10.5 cm, both correct to the nearest 5 mm. Calculate the upper bound for the area of this picture. … cm2 [2] (c) In a sale, the price of a large frame is reduced by 18%. Parthi pays $166.05 for 5 large frames in the sale. Calculate the original price of one large frame. $ … [2] (d) Parthi advertises a large frame for a price of $57 or 48.20 euros. The exchange rate is $1= 0.88 euros. Calculate the difference between these prices, in dollars and cents, correct to the nearest cent. $ … [3]

10 marks

Mark scheme: 7(a) 31.5 3 1134 M2 for 17.5 × oe 350 1134 350 or M1 for oe isw or oe isw 350 1134 1134  x  2 or for =   oe 350  17.5  7(b) 15 2 B1 for 15 + 0.25 or 10.5 + 0.25 or better seen 163.937 5 or 16316 final answer 7(c) 40.5[0] 2  18  166.05 M1 for x ×  1 −  = oe  100  [ 5 ] 7(d) $2.23 final answer 3 B2 for 2.227… or 2.23 seen OR 48.2 M2 for 57 – oe 0.88 48.2 or M1 for oe 0.88 If 0 scored SC1 for 57 × 0.88 oe seen

This question in 0580/42 Feb/March 2022

6 (a) At a festival, 380 people out of 500 people questioned say that they are camping. There are 55 300 people at the festival. Calculate an estimate of the total number of people camping at the festival. … [2] (b) 12 friends travel to the festival. 5 travel by car, 4 travel by bus and 3 travel by train. Two people are chosen at random from the 12 friends. Calculate the probability that they travel by different types of transport. … [4] (c) Arno buys a student ticket for $43.68 . This is a saving of 16% on the full price of a ticket. Calculate the full price of a ticket. $ … [2] (d) At a football match, there are 29 800 people, correct to the nearest 100. (i) At the end of the football match, the people leave at a rate of 400 people per minute, correct to the nearest 50 people. Calculate the lower bound for the number of minutes it takes for all the people to leave. … min [3] (ii) At a cricket match there are 27 500 people, correct to the nearest 100. Calculate the upper bound for the difference between the number of people at the football match and at the cricket match. … [2]

13 marks

Mark scheme: 6(a) 42 028 2 380 M1 for oe soi isw 500 6(b) 47 4 0.712[1…] oe 66  5 4   4 3   5 3  M3 for 2   2   2         12 11   12 11   12 11  oe  5 4 4 3 3 2  or 1 –      oe    12 11 12 11 12 11  or M2 for sum of 3 or more correct product pairs and no incorrect pairs 5 4 4 3 3 2 or for      and no other 12 11 12 11 12 11 pairs k j or M1 for  seen 12 11 94 If 0 scored SC1 for answer oe 144 6(c) 52 2 100  16 M1 for x   43.68 oe or better 100 6(d)(i) 70 or 70.16[5…] or 70.17 or 70.2 3 29750 to 29800 29750 to 29800 M2 for or or 400  25 400  24 29800  50 400to425 or B1 for 29 750 or 29 850 or 29 849 or 375 or 425 or 424 seen 6(d)(ii) 2399 2 B1 for 27 450 or 27 550 or 27 549 or 29 850 or or 2400 nfww 29 849 seen

This question in 0580/42 May/June 2022

2 (a) Write (i) 2994.99 correct to the nearest 10, … [1] (ii) 0.983 correct to 1 decimal place, … [1] (iii) 2090 correct to 2 significant figures. … [1] (b) Write down a prime number between 90 and 100. … [1] (c) Write 2 -6 as a fraction. … [1] (d) Write 0.007 01 in standard form. … [1] (e) Simplify 1.5 # 10 x + 1 .5 # 10 x - 1 giving your answer in standard form. … [2] (f) Write .037o as a fraction. You must show all your working. … [2]

10 marks

Mark scheme: 2(a)(i) 2990 cao 1 2(a)(ii) 1.0 cao 1 2(a)(iii) 2100 cao 1 2(b) 97 1 2(c) 1 1 final answer 64 2(d) 7.01[0]  10–3 1 2(e) 1.65  10x 2 M1 for final answer figs 165 or for 15  10 x −1 seen or for 0.15  10 x seen 2(f) 37.7... – 3.7... [= 34] oe M1 34 B1 oe fraction 90

This question in 0580/41 Oct/Nov 2022

10 (a) The lengths of the sides of a triangle are 11.4 cm, 14.8 cm and 15.7 cm, all correct to 1 decimal place. Calculate the upper bound of the perimeter of the triangle. … cm [2] (b) 15.6 cm NOT TO SCALE 150° The diagram shows a circle, radius 15.6 cm. The angle of the minor sector is 150°. Calculate the area of the minor sector. … cm2 [2] (c) r cm NOT TO x° SCALE The diagram shows a circle, radius r cm and minor sector angle x°. The perimeter of the major sector is three times the perimeter of the minor sector. 90 ( r - 2 ) Show that x = . r [4]

8 marks

Mark scheme: 10(a) 42.05 final answer 2 M1 for 11.4 + 0.05 oe or 14.8 + 0.05 oe or 15.7 + 0.05 oe 10(b) 319 or 318.5 to 318.6 2 150 2 M1 for   15.6 oe 360 10(c) 360 − x  x  M2  2πr + 2r = 3   2r + 2r  oe x 360  360  M1 for  2πr oe seen 360 or 360 − x  2πr oe seen 360 4 x M1 i.e. M mark for isolating and collecting terms in x  2π[r ] = 2π[r] – 4[r] oe 360 90 (− 2 ) A1 With no errors or omissions Leading to 

This question in 0580/42 Oct/Nov 2022

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]

14 marks

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

This question in 0580/43 Oct/Nov 2022

3 (a) The table shows information about the mass of each of 1000 eggs. Mass (m grams) 40 1 m G 50 50 1 m G 56 56 1 m G 64 64 1 m G 70 Frequency 126 520 154 200 (i) Calculate an estimate of the mean. … g [4] (ii) An egg is picked at random from the 1000 eggs. Find the probability that this egg has a mass greater than 56 g. Give your answer as a fraction in its simplest form. … [2] (b) One year, a farmer makes a profit of $24 730 selling eggs. Write this profit (i) correct to 2 significant figures $ … [1] (ii) in standard form. $ … [1] (c) On a farm, there are 500 hens, correct to the nearest 10. (i) In one year, the mean number of eggs laid per hen was 320 eggs, correct to the nearest 20. Calculate the upper bound for the total number of eggs all the hens lay in that year. … [3] (ii) Another farm has 800 hens, correct to the nearest 20. Calculate the lower bound for the difference between the number of hens on the two farms. … [2]

13 marks

Mark scheme: 3(a)(i) 55.87 4 M1 for midpoints soi M1 for use of where m is in the correct fm interval including boundaries M1 (dep on 2nd M1) for ÷1000 fm 3(a)(ii) 177 2 154  200 cao M1 for oe 500 1000 3(b)(i) 25000 1 3(b)(ii) 2.473  10 4 1 3(c)(i) 166 650 or 165816 nfww 3 M2 for (500 + 5) × ‘320 to 340’ or ‘500 to 510’ × (320 + 10) or M1 for 500  5 or 500  5 or 320 10 or 320 10 Alternative method M2 for 504 × ‘320 to 340’ or ‘500 to 510’ × 329 or M1 for 504 or 329 3(c)(ii) 285 or 286 nfww 2 M1 for 800 10

This question in 0580/41 May/June 2023

10 (a) H G F E NOT TO SCALE D C x cm A x cm B ABCDEFGH is a cuboid with a square base of side x cm. CG = 20 cm and AG = 28 cm . Calculate the value of x. x = … [4] (b) R Q N P NOT TO SCALE M L J K The diagram shows a different cuboid JKLMNPQR. MR = 30 cm correct to the nearest centimetre. KR = 37 cm correct to the nearest centimetre. Calculate the lower bound of the angle between KR and the base JKLM of the cuboid. … [4]

8 marks

Mark scheme: 10(a) 13.9 or 13.85 to 13.86 4 M3 for 2x2 = 282 – 202 or better or x  28 2  20 2 sin45 oe   or M2 for x2 + x2 + 202 = 282 oe x or sin45  28 2  20 2 ) or M1 for any correct Pythag in 2D or their AC × sin 45 oe dep on trig/Pythagoras attempt for AC 10(b) 51.9 or 51.87 to 51.88 4 29 to 30 30  0.5 M3 for sin = or oe 37  0.5 37 to 38 or M2 for correct trig statement for correct angle with values in range 29 to 31 and 36 to 38 or M1 for 30 + 0.5 or 30 – 0.5 or 37 + 0.5 or 37 – 0.5 seen or for identifying correct angle RKM

This question in 0580/42 May/June 2023

1 A grocer sells potatoes, mushrooms and carrots. (a) A customer buys 3 kg of mushrooms at $1.04 per kg and 4 kg of carrots at $1.28 per kg. Calculate the total cost. $ … [2] (b) In one week, the ratio of the masses of vegetables sold by the grocer is potatoes : mushrooms : carrots = 11 : 8 : 6. (i) Work out the mass of mushrooms sold as a percentage of the total mass. … % [2] (ii) The total mass of potatoes, mushrooms and carrots sold is 1500 kg. Find the mass of carrots the grocer sells this week. … kg [2] (iii) The profit the grocer makes selling 1 kg of carrots is $0.75 . Find the total profit the grocer makes selling carrots this week. $ … [1] (iv) On the last day of the week, the grocer reduces the price of 1 kg of potatoes by 8% to $1.15 . Calculate the original price of 1 kg of potatoes. $ … [2] (c) The grocer buys 620 kg of onions, correct to the nearest 20 kg. He packs them into bags each containing 5 kg of onions, correct to the nearest 1 kg. Calculate the upper bound for the number of bags of onions that he packs. … [3]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 8.24 cao 2 M1 for 3 1.04 + 4 1.28 1(b)(i) 32 2 8 M1 for  100  oe 11 + 8 + 6 1(b)(ii) 360 2 1500 M1 for  k where k = 1 , 11, 8 or 6 11 + 8 + 6 1(b)(iii) 270 1 FT 0.75 × their 360 1(b)(iv) 1.25 cao 2  8  M1 for x   1 −  = 1.15 oe or better  100  1(c) 140 nfww 3 620 to 640 620 + 10 M2 for or oe 5 − 0.5 4 to 5 or M1 for 620 +10 oe or 620 – 10 oe or 5 + 0.5 oe or 5 – 0.5 oe seen

This question in 0580/42 Feb/March 2024

1 (a) A fruit drink is made using 1.5 litres of apple juice and 450 millilitres of mango juice. Write the ratio apple juice : mango juice in its simplest form. … : … [2] (b) One litre of fruit drink is shared between three cups. The amount in the cups is in the ratio 9 : 6 : 10. Calculate the number of millilitres in each cup. … ml , … ml , … ml [3] (c) A shop buys bottles of the fruit drink for $3.20 each. It sells them at a profit of 15%. Calculate the selling price of each bottle of fruit drink. $ … [2] (d) The number of bottles of fruit drink sold has grown exponentially at a constant rate of 2.5% per year. 5 years ago, the shop sold 16 620 bottles. Calculate the number of bottles sold this year. … [2] (e) d cm NOT TO 23 cm SCALE 18.5 cm The bottles of juice are 18.5 cm tall, correct to the nearest millimetre. They are stored on shelves. The distance between the shelves is 23 cm, correct to the nearest centimetre. Calculate the lower bound for the distance, d cm, between the top of a bottle and the shelf above it. … cm [3]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 10 : 3 final answer 2 M1 for 1500 : 450 oe in ratio form If 0 scored SC1 for answer 3 : 10 1(b) 360 240 400 3 B2 for answer 0.36 0.24 0.4 or for answer two of 360 240 400 1000 or M1 for [ k ] where k = 1, 9, 9  6  10 6 or 10 If 0 scored, SC1 for answer with 3 values in ratio 9 : 6 : 10 in that order 1(c) 3.68 cao 2  15  M1 for  1    3.2 oe  100  or B1 for answer 0.48 1(d) 18 804[.0...] 2  2.5  5 1 for 16620   1   oe  100  1(e) 3.95 3 M2 for 22.5 – (18.5 to 18.6) or (22 to 23) −18.55 or M1 for 23 – 0.5 oe seen or 23 + 0.5 oe seen or 18.5– 0.05 oe seen or 18.5 + 0.05 oe seen

This question in 0580/42 May/June 2024

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

7 (a) C 60° North B NOT TO SCALE 85 m 129 m 39° A 72 m D The diagram shows a field, ABCD with B north of A. BD is a path across the field. AB = 85 m, AD = 72 m, BD = 129 m, angle BDC = 39° and angle BCD = 60°. (i) Show that angle CBD = 81°. [1] (ii) Calculate CD. … m [3] (iii) Show that angle ABD = 31.6°, correct to 1 decimal place. [4] (iv) Find the shortest distance from A to BD. … m [3] (v) Find the bearing of B from C. … [2] (vi) Trees are planted in the field. The number of trees planted is 1100 per hectare. Calculate the total number of trees planted in the field. [1 hectare = 10 000 m2] … [4] (b) A rectangle has an area of 9400 cm2, correct to the nearest 100 cm2. The length of the rectangle is 80 cm, correct to the nearest 10 cm. Calculate the upper bound of the width of the rectangle. … cm [3]

20 marks

Mark scheme: 7(a)(i) 180 – 60 – 39 [ = 81] 1 7(a)(ii) 147 or 147.1… 3 129sin(81) M2 for oe sin60 sin(81) sin60 or M1 for = oe CD 129 7(a)(iii) 85 2 + 129 2 − 72 2 M2 M1 for 72 2 = 852 + 129 2 −2 85  129cos ABD [cos = ] 2  85  129 31.58… A2 A1 for 0.851 to 0.852 9341 or or equivalent fraction 10965 7(a)(iv) 44.5 or 44.51 to 44.54 3 M2 for implicit correct method d e.g. = sin31.6 oe 85 or M1 for recognition that the line from A is perpendicular to BD 7(a)(v) 247 or 247.4… 2 M1 for 180 + (180 – 81 – 31.6) oe or for NBC = 180 – 81 – 31.6 oe or for NCB = 81 + 31.6 oe 7(a)(vi) 972 or 973 4 1 M1 for [ABD]  85  129sin31.6 oe 2 1 or  129  their 44.5 oe 2 1 M1 for [BCD ]  129  their147×sin39 oe 2 their total area M1 for  1100 10000 7(b) 126 nfww 3 9400 + 50 9400 to 9500 M2 for or 70 to 80 80 − 5 or M1 for 9350 or 9450 or 75 or 85 seen

This question in 0580/42 Oct/Nov 2024

20 A piece of metal has volume 1240 cm3, correct to the nearest 20 cm3. The mass of the piece of metal is 7800 g, correct to the nearest 100 g. Calculate the lower bound of the density of the metal. [Density = mass ÷ volume.] … /gcm3 [3]

3 marks

Mark scheme: 20 6.2 nfww 3 7800 − 50 7700 to 7800 M2 for or 1240 to 1260 1240 + 10 or M1 for 7850 or 7750 or 1250 or 1230 seen

This question in 0580/42 Feb/March 2025

12 The length of a rectangle is 16 cm, correct to the nearest centimetre. The width of the rectangle is 14 cm, correct to the nearest centimetre. Calculate the lower bound of the perimeter of the rectangle. … cm [2]

2 marks

Mark scheme: 12 58 nfww 2 M1 for 15.5 oe or 13.5 oe seen

This question in 0580/42 May/June 2025

20 A solid metal prism has a mass of 4810 g, correct to the nearest 10 g. The density of the metal is 7.7 g/cm3, correct to 1 decimal place. Calculate the lower bound for the volume of the prism. [Density = mass ' volume] … cm3 [3]

3 marks

Mark scheme: 20 620 nfww 3 4810 − 5 4800 to 4810 M2 for or oe 7.7 to 7.8 7.7 + 0.05 or M1 for 4815 or 4805 or 7.75 or 7.65 oe seen

This question in 0580/42 Oct/Nov 2025

25 Jenna has a length of wire measuring 68 cm, correct to the nearest cm. From this wire she cuts off two smaller pieces • a piece of length 4.7 cm, correct to the nearest mm • a piece of length 10.0 cm, correct to the nearest mm. Work out the lower bound and the upper bound for the length of the wire remaining. Lower bound = … cm Upper bound = … cm [3]

3 marks

Mark scheme: 25 [LB =] 52.7 3 B2 for answer [LB=] 52.7 or [UB=] 53.9 [UB =] 53.9 or M1 for 68 + 0.5 or 68 – 0.5 or 4.7 + 0.05 or 4.7 – 0.05 or 10[.0] + 0.05 or 10[.0] – 0.05. oe seen

This question in 0580/43 Oct/Nov 2025