Percentages — Paper 4 · IGCSE Mathematics 0580

1 An energy company charged these prices in 2013. Electricity price Gas price 23.15 cents per day 24.5 cents per day plus plus 13.5 cents for each unit used 5.5 cents for each unit used (a) (i) In 90 days, the Siddique family used 1885 units of electricity. Calculate the total cost, in dollars, of the electricity they used. $ … [2] (ii) In 90 days, the gas used by the Khan family cost $198.16 . Calculate the number of units of gas used. … units [3] (b) In 2013, the price for each unit of electricity was 13.5 cents. Over the next 3 years, this price increased exponentially at a rate of 8% per year. Calculate the price for each unit of electricity after 3 years. … cents [2] (c) Over these 3 years, the price for each unit of gas increased from 5.5 cents to 7.7 cents. (i) Calculate the percentage increase from 5.5 cents to 7.7 cents. … % [3] (ii) Over the 3 years, the 5.5 cents increased exponentially by the same percentage each year to 7.7 cents. Calculate the percentage increase each year. … % [3] (d) In 2015, the energy company divided its profits in the ratio shareholders : bonuses : development = 5 : 2 : 6 . In 2015, its profits were $390 million. Calculate the amount the company gave to shareholders. $ … million [2] (e) The share price of the company in June 2015 was $258.25 . This was an increase of 3.3% on the share price in May 2015. Calculate the share price in May 2015. $ … [3]

18 marks

Mark scheme: Question Answer Marks Part marks 1(a)(i) 275.31 2 M1 for 90 × 23.15 + 1885 × 13.5 oe 1(a)(ii) 3202 3 198.16 – 90 × 0.245 M2 for oe 0.055 M1 for 90 × 0.245 or 90 × 24.5 oe 1(b) 17.[0] or 17.00 to 17.01 2 3 § 8 · M1 for 13.5 × ¨ 1 + ¸ © 100 ¹ 1(c)(i) 40 3 7.7 − 5.5 7.7 M2 for [×100] oe or ×100 5.5 5.5 7.7 or M1 for oe 5.5 1(c)(ii) 11.9 or 11.86 to 11.87 3 7.7 M2 for 3 oe 5.5 or M1 for 5.5 × x3 = 7.7 oe 1(d) 150 [million] oe 2 M1 for 390 [million] ÷ ( 5 + 2 + 6) 1(e) 250 nfww 3 M2 for 258.25 ÷ ((100 + 3.3) ÷ 100) or M1 for 258.25 associated with 103.3[%]

This question in 0580/41 May/June 2017

1 (a) Annie and Dermot share $600 in the ratio 11 : 9. (i) Show that Annie receives $330. [1] (ii) Find the amount that Dermot receives. $ … [1] (b) (i) Annie invests $330 at a rate of 1.5% per year compound interest. Calculate the amount that Annie has after 8 years. Give your answer correct to the nearest dollar. $ … [3] (ii) Find the amount of interest that Annie has, after the 8 years, as a percentage of the $330. … % [2] (c) Dermot has $70 to spend. He spends $24.75 on a shirt. (i) Find $24.75 as a fraction of $70. Give your answer in its lowest terms. … [1] (ii) The $24.75 is the sale price after reducing the original price by 10%. Calculate the original price. $ … [3] (d) After one year, the value of Annie’s car had reduced by 20%. At the end of the second year, the value of Annie’s car had reduced by a further 15% of its value at the end of the first year. (i) Calculate the overall percentage reduction after the two years. … % [2] (ii) After three years the overall percentage reduction in the value of Annie’s car is 40.84%. Calculate the percentage reduction in the third year. … % [2]

15 marks

Mark scheme: Question Answer Marks Part marks 1(a)(i) 600 ÷ (11+ 9) × 11 [ =330] M1 Could be in separate steps with no errors seen 1(a)(ii) 270 1 1(b)(i) 372 cao nfww 3 B2 for answer 371.7…  1.5  8 or M1 for 330 × 1 + oe not spoiled    100  After zero scored, SC1 for answer 42 or 41.7… 1(b)(ii) 12.6 or 12.7 or 12.63 to 12.73 2 their (b)(i) − 330 their (b)(i) M1 for or × 100 soi by 112.7 330 330 or 113 After zero scored, SC1 for answer 12% 1(c)(i) 99 1 cao final answer 280 1(c)(ii) 27.5[0] 3 100 − 10 M2 for 24.75 ÷ oe 100 or M1 for recognising 24.75 as 90[%] oe 1(d)(i) 32 cao 2  20  15  M1 for  1 −  1 −  [x]oe  100  100  or for 0.15 × 0.8 [x] oe 1(d)(ii) 13 cao 2  20  15  M1 for  1 −  1 −  × x = 40.84 – 32 oe seen  100  100    their (d)(i)   or for their (d)(i) +  1 −    x = 40.84 oe   100  

This question in 0580/42 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

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

3 (a) The price of a house decreased from $82 500 to $77 500. Calculate the percentage decrease. … % [3] (b) Roland invests $12 000 in an account that pays compound interest at a rate of 2.2% per year. Calculate the value of his investment at the end of 6 years. Give your answer correct to the nearest dollar. $ … [3]

6 marks

Mark scheme: 3(a) 6.06 or 6.060 to 6.061 3 82500 − 77500[× 100] oe M2 for 82500 or M1 for 77500[× 100] soi 82500 3(b) 13 674 cao 3 6  2.2  M1 for 12000  1 +   100  A1 for 13673.7...

This question in 0580/41 May/June 2018

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 (a) Here is a list of ingredients to make 20 biscuits. 260 g of butter 500 g of sugar 650 g of flour 425 g of rice (i) Find the mass of rice as a percentage of the mass of sugar. … % [1] (ii) Find the mass of butter needed to make 35 of these biscuits. … g [2] (iii) Michel has 2 kg of each ingredient. Work out the greatest number of these biscuits that he can make. … [3] (b) A company makes these biscuits at a cost of $1.35 per packet. These biscuits are sold for $1.89 per packet. (i) Calculate the percentage profit the company makes on each packet. … % [3] (ii) The selling price of $1.89 has increased by 8% from last year. Calculate the selling price last year. $ … [3] (c) Over a period of 3 years, the company’s sales of biscuits increased from 15.6 million packets to 20.8 million packets. The sales increased exponentially by the same percentage each year. Calculate the percentage increase each year. … % [3] (d) The people who work for the company are in the following age groups. Group A Group B Group C Under 30 years 30 to 50 years Over 50 years The ratio of the number in group A to the number in group B is 7 : 10. The ratio of the number in group B to the number in group C is 4 : 3. (i) Find the ratio of the number in group A to the number in group C. Give your answer in its simplest form. … : … [3] (ii) There are 45 people in group C. Find the total number of people who work for the company. … [3]

21 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 85 1 1(a)(ii) 455 2 M1 for 260 ÷ 20 × 35 oe 1(a)(iii) 61 3 B2 for 61.5… seen or M1 for 2000 ÷ 650 soi x 20 or for = oe or other attempt at 2000 650 scaling up with 650 or for 650 ÷ 20 oe 1(b)(i) 40 3 1.89 − 1.35 M2 for [× 100] oe 1.35 1.89 or × 100 oe 1.35 or M1 for oe 1.89[× 100] soi 1.35 1(b)(ii) 1.75 nfww 3  100 + 8  M2 for 1.89 ÷   or better  100  or M1 for 1.89 associated with 108 [%] 1(c) 10.1 or 10.06… 3 20.8 M2 for 3 oe 15.6 or M1 for 15.6 × k 3 = 20.8 oe 1(d)(i) 14:15 3 B2 for correct unsimplified 3 term ratio A: B: C or correct unsimplified two term ratio A : C or M1 for attempt to find common multiple of 4 and 10 or other common value for B 4 10 or for 7 × oe or 3 × oe 10 4 1(d)(ii) 147 3 45 M2 for (14 + 20 [ +15] ) oe or 15 45 ÷ 3 × 4 + (45 ÷ 3 × 4) ÷ 10 × 7 [+ 45] or M1 for 45 ÷ 3 oe or 45 ÷ their (d)(i) value for C shown

This question in 0580/42 May/June 2018

1 (a) Rowena buys and sells clothes. (i) She buys a jacket for $40 and sells it for $45.40 . Calculate the percentage profit. … % [3] (ii) She sells a dress for $42.60 after making a profit of 20% on the cost price. Calculate the cost price. $ … [3] (b) Sara invests $500 for 15 years at a rate of 2% per year simple interest. Calculate the total interest Sara receives. $ … [2] (c) Tomas has two cars. (i) The value, today, of one car is $21 000. The value of this car decreases exponentially by 18% each year. Calculate the value of this car after 5 years. Give your answer correct to the nearest hundred dollars. $ … [3] (ii) The value, today, of the other car is $15 000. The value of this car increases exponentially by x % each year. After 12 years the value of the car will be $42 190. Calculate the value of x. x = … [3]

14 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 13.5 3 45.4[0] − 40 45.4[0] M2 for [× 100] or × 100 40 40 45.4[0] or M1 for [× 100] 40 1(a)(ii) 35.5[0] 3  20  M2 for 42.6[0] ÷ 1 + or better    100  or M1 for recognising 42.6[0] as 120[%] 1(b) 150 cao 2 500 × 2 × 15 M1 for oe 100 1(c)(i) 7800 cao 3 B2 for 7790 or 7785 to 7786  18  5 or M1 for 21000 × 1 − oe isw    100  If 0 or 1 scored, SC1 for their 7785… seen and rounded correctly to nearest 100 1(c)(ii) 9[.00…] 3 42190 M2 for 12 or better 15000  x 12 or M1 for 15000 1 + = [42190]    100 

This question in 0580/43 May/June 2018

1 Marianne sells photos. (a) The selling price of each photo is $6. (i) The selling price for each photo is made up of two parts, printing cost and profit. For each photo, the ratio printing cost : profit = 5 : 3. Calculate the profit she makes on each photo. $ … [2] (ii) Calculate her profit as a percentage of the selling price. … % [1] (iii) Calculate the selling price of a photo in euros (€) when the exchange rate is €1 = $1.091 . € … [2] (b) Marianne sells two sizes of photo. These photos are mathematically similar rectangles. The smaller photo has length 15 cm and width 12 cm. The larger photo has area 352.8 cm2. Calculate the length of the larger photo. … cm [3] (c) In a sale, Marianne buys a new camera for $483. This is a reduction of 8% on the original price. Calculate the original price of the camera. $ … [3]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 2.25 final answer 2 3 6 M1 for or oe 5 + 3 5 + 3 1(a)(ii) 37.5 1 (a)(i) FT their × 100 6 1(a)(iii) 5.5[0] or 5.499 to 5.500 2 M1 for 6 ÷ .1091 1(b) 21 3 3528. M2 for 15 × oe 15 × 12 or SC2 for answer 16.8 352 8. 15× 12 or M1 for or seen 15 × 12 352 8. or M1 for a correct implicit statement for the length 1(c) 525 3 483 M2 for [× 100] oe 100 − 8 or M1 for 483 associated with 92 [%]

This question in 0580/41 Oct/Nov 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

8 (a) The price of a book increases from $2.50 to $2.65 . Calculate the percentage increase. … % [3] (b) Scott invests $500 for 7 years at a rate of 1.5% per year simple interest. Calculate the value of his investment at the end of the 7 years. $ … [3] (c) In a city the population is increasing exponentially at a rate of 1.6% per year. Find the overall percentage increase at the end of 20 years. … % [2] (d) The population of a village is 6400. The population is decreasing exponentially at a rate of r% per year. After 22 years, the population will be 2607. Find the value of r. r = … [3]

11 marks

Mark scheme: 8(a) 6 nfww 3 M2 for 2.65 − 2.50[× 100] or for 2.50 2.65 × 100 2.50 2.65 or M1 for 2.50 8(b) 552.5[0] 3 B2 for 52.5[0] 1.5 or M2 for 500 × × 7 + 500 oe 100 1.5 or M1 for 500 × [× 7] oe 100 8(c) 37.4 or 37.36… 2 20  1.6  M1 for 1 + oe soi 1.37…    100  8(d) 4[.00...] 3 2607 M2 for 22 6400 or M1 for 6400 × x22 = 2607 oe or better

This question in 0580/41 May/June 2019

2 (a) Ali and Mo share a sum of money in the ratio Ali : Mo = 9 : 7. Ali receives $600 more than Mo. Calculate how much each receives. Ali $ … Mo $ … [3] (b) In a sale, Ali buys a television for $195.80 . The original price was $220. Calculate the percentage reduction on the original price. … % [3] (c) In the sale, Mo buys a jacket for $63. The original price was reduced by 25%. Calculate the original price of the jacket. $ … [3]

9 marks

Mark scheme: 2(a) [Ali] 2700 3 B2 for one correct or for correct values [Mo] 2100 reversed or M1 for 600 ÷ (9 – 7) or for any equation that would lead to an answer of 300, 2700 or 2100, or 4800 (for the total) 2(b) 11 3 220 − 1958. M2 for [× 100] or for 220 1958. [100 − ] × 100 220 195 8. or M1 for 220 – 195.8 or for or a 220 correct implicit equation for percentage 195.8 − 220 reduction or for 220 2(c) 84 3 63 M2 for oe 25 1 − 100 or M1 for associating 63 with (100 – 25)% or a correct implicit equation for the original price.

This question in 0580/41 Oct/Nov 2019

1 (a) Mohsin has 600 pear trees and 720 apple trees on his farm. (i) Write the ratio pear trees : apple trees in its simplest form. … : … [1] (ii) Each apple tree produces 16 boxes of apples each year. One box contains 18 kg of apples. Calculate the total mass of apples produced by the 720 trees in one year. Give your answer in standard form. … kg [3] (b) (i) One week, the total mass of pears picked was 18 540 kg. For this week, the ratio mass of apples : mass of pears = 13 : 9. Find the mass of apples picked that week. … kg [2] (ii) The apples cost Mohsin $0.85 per kilogram to produce. He sells them at a profit of 60%. Work out the selling price per kilogram of the apples. $ … [2] (c) Mohsin exports some of his pears to a shop in Belgium. The shop buys the pears at $1.50 per kilogram. The shop sells the pears for 2.30 euros per kilogram. The exchange rate is $1 = 0.92 euros. Calculate the percentage profit per kilogram made by the shop. … % [5] (d) Mohsin’s earnings increase exponentially at a rate of 8.7% each year. During 2018 he earned $195 600. During 2027, how much more does he earn than during 2018? $ … [3]

16 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 5 : 6 1 1(a)(ii) 2.0736[0] × 105 final answer 3 B2 for 207360 oe or M1 for 16 × 18 × 720 1(b)(i) 26780 2 M1 for 18540 ÷ 9 soi 1(b)(ii) 1.36 2 M1 for 0.85 × 1.6 oe or B1 for 0.51 or 51 1(c) 66.7 or 66.66 to 66.67 5 (2.3 − 1.5 × M4 for 0.92)[× 100] oe or 1.5 × 0.92 2.3 × 100 oe 1.5 × 0.92 OR Working in euros B2 for [€]1.38 or M1 for 1.5[0] × 0.92 M2dep on B2 or M1 for 2.3 − their 1.38[× 100] oe their 1.38 2.3 − their 1.38 or × 100 oe their 1.38 2.3 or M1 for 2.3 – their 1.38 or their 1.38 OR Working in dollars B2 for [$]2.50 or M1 for or 2.3[0] ÷ 0.92 M2dep on B2 or M1 for their 5.2 − 5.1 their 2.5 [×100] oe or × 100 5.1 1.5 their 2.5 or M1 for their 2.5 – 1.5 or 1.5 1(d) 219 000 3 B2 for 414000 or 414414[.3.…] rounded to 4 or 218814[.3.…] rounded to 4 sf or sf or more more 9  8.7  or M2 for 195600 ×  1 +  [– 195600]  100   8.7  k or M1 for 195600 ×  1 +  or better  100  (k >1 and an integer)

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

3 (a) Manjeet uses 220 litres of water each day. She reduces the amount of water she uses by 15%. Calculate the number of litres of water she now uses each day. … litres [2] (b) Manjeet has two mathematically similar bottles in her bathroom. The large bottle holds 1.35 litres and is 29.7 cm high. The small bottle holds 0.4 litres. Calculate the height of the small bottle. … cm [3] (c) Water from Manjeet’s shower flows at a rate of 12 litres per minute. The water from the shower flows into a tank that is a cuboid of length 90 cm and width 75 cm. Calculate the increase in the level of water in the tank when the shower is used for 7 minutes. … cm [3]

8 marks

Mark scheme: 3(a) 187 2  15  M1 for 220 ×  1 −  oe  100  or B1 for 33 seen 3(b) 19.8 3 0.4 M2 for 29.7 × 3 oe 1.35 0.4 1.35 or M1 for 3 or 3 oe seen 1.35 0.4 29.7 3 1.35 or for = oe x 3 0.4 3(c) 12.4 or 12.44… 3 M1 for 90 × 75 × h = 7 × figs 12 B1 for 1000 cm3 = 1 litre soi

This question in 0580/42 Feb/March 2020

1 (a) In 2018, Gretal earned $32 000. (i) She paid tax of 24% on these earnings. Work out the amount she paid in tax in 2018. $ … [2] (ii) In 2019, Gretal’s earnings increased by 7%. Work out her earnings in 2019. $ … [2] (b) Gretal invests $5000 at a rate of 2% per year compound interest. Calculate the value of her investment at the end of 3 years. $ … [2] (c) One month, Gretal spent a total of $360 on presents. She spent 15 of this total on presents for her parents. She spent 23 of the remaining money on presents for her friends. She spent the rest of the money on presents for her sisters. Calculate the percentage of the $360 that she spent on presents for her sisters. … % [4] (d) Arjun earned $36 515 in 2019. This was an increase of 9% on his earnings in 2018. Work out his earnings in 2018. $ … [2] (e) Arjun and Gretal each pay rent. In 2018, the ratio of the amount each paid in rent was Arjun : Gretal = 5 : 7. In 2019, the ratio of the amount each paid in rent was Arjun : Gretal = 9 : 13. Arjun paid the same amount of rent in both 2018 and 2019. Gretal paid $290 more rent in 2019 than she did in 2018. Work out the amount Arjun paid in rent in 2019. $ … [4]

16 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 7680 2 M1 for 0.24 × 32 000 oe 1(a)(ii) 34 240 2 100 + 7 M1 for 32 000 × oe 100 1(b) 5306.04 2 3  2  M1 for 5000 ×  1 +  oe  100  1(c) 26.7 or 26.66... to 26.67 4 96 B3 for 96 or oe 360 OR 1 M3 for (1 − ) × (1 −2 ) × 100 oe 5 3 1 or M2 for (1 − ) and (1 −2 ) oe 5 3 OR M1 for 360 ÷ 5 [× 4] oe M1 for their 288 ÷ 3 [× 2] 1(d) 33 500 2 100 + 9 M1 for 36 515 ÷ oe 100 1(e) 6525 4  65 63  M3 for  −  [ A ] = 290 oe  45 45   13 7  or M2 for  −  [ A ] = 290 oe  9 5  or M1 for correct attempt to convert to a common ratio value for Arjun 13 7 or for − oe 9 5

This question in 0580/41 May/June 2020

1 (a) (i) Divide $24 in the ratio 7 : 5. $ … , $ … [2] (ii) Write $24.60 as a fraction of $2870. Give your answer in its lowest terms. … [2] (iii) Write $1.92 as a percentage of $1.60 . … % [1] (b) In a sale the original prices are reduced by 15%. (i) Calculate the sale price of a book that has an original price of $12. $ … [2] (ii) Calculate the original price of a jacket that has a sale price of $38.25 . $ … [2] (c) (i) Dean invests $500 for 10 years at a rate of 1.7% per year simple interest. Calculate the total interest earned during the 10 years. $ … [2] (ii) Ollie invests $200 at a rate of 0.0035% per day compound interest. Calculate the value of Ollie’s investment at the end of 1 year. [1 year = 365 days.] $ … [2] (iii) Edna invests $500 at a rate of r % per year compound interest. At the end of 6 years, the value of Edna’s investment is $559.78 . Find the value of r. r = … [3]

16 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 14, 10 2 M1 for 24 ÷ (7 + 5) 1(a)(ii) 3 2 B1 for correct fraction not in lowest terms 350 1(a)(iii) 120 1 1(b)(i) 10.2[0] 2 15 M1 for × 12 oe or better 100 1(b)(ii) 45 2 38.25 M1 for oe 15 1 − 100 1(c)(i) 85 2 500 × 1.7 × 10 M1 for oe 100 1(c)(ii) 203 or 202.5 to 202.6 2 365  0.0035  M1 for 200 ×  1 +   100  1(c)(iii) 1.9 3 559.78 M2 for 6 500  r 6 or M1 for 500  1 +  = 559.78  100 

This question in 0580/42 May/June 2020

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

9 (a) There are 32 students in a class. 5 do not study any languages. 15 study German (G). 18 study Spanish (S). G S (i) Complete the Venn diagram to show this information. [2] (ii) A student is chosen at random. Find the probability that the student studies Spanish but not German. … [1] (iii) A student who studies German is chosen at random. Find the probability that this student also studies Spanish. … [1] (b) A bag contains 54 red marbles and some blue marbles. 36% of the marbles in the bag are red. Find the number of blue marbles in the bag. … [2] (c) Another bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its colour and replaces it in the bag. She then picks another bead at random. (i) Find the probability that she picks two red beads. … [2] (ii) Find the probability that she does not pick two red beads. … [1] (d) A box contains 15 red pencils, 8 yellow pencils and 2 green pencils. Two pencils are picked at random without replacement. Find the probability that at least one pencil is red. … [3]

12 marks

Mark scheme: 9(a)(i) 2 B1 for two correct values 5 Or 9 6 12 B1 5 outside and total in G = 15 and total in G S S = 18 9(a)(ii) 3 1 their 12 oe FT 8 32 9(a)(iii) 2 1 their 6 oe FT 5 15 9(b) 96 2 36 54 54 M1 for = oe or 36 = × 100 64 x ( 54 + b ) oe If 0 scored SC1 for answer 150 9(c)(i) 9 2 15 15 oe M1 for × oe 25 25 25 9(c)(ii) 16 1 FT 1 – their (c)(i) oe 25 9(d) 17 3 10 9 oe M2 for 1 − × oe 20 25 24 15 14 15 8 15 2 8 15 or for × + × + × + × 25 24 25 24 25 24 25 24 2 15 + × oe 25 24 or M1 for one correct relevant product

This question in 0580/41 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

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

1 (a) (i) Yasmin and Zak share an amount of money in the ratio 21 : 19. Yasmin receives $6 more than Zak. Calculate the total amount of money shared by Yasmin and Zak. $ … [2] (ii) In a sale, all prices are reduced by 15%. (a) Yasmin buys a blouse with an original price of $40. Calculate the sale price of the blouse. $ … [2] (b) Zak buys a shirt with a sale price of $29.75 . Calculate the original price of the shirt. $ … [2] (b) Xavier’s salary increases by 2% each year. In 2010, his salary was $40 100. (i) Calculate his salary in 2015. Give your answer correct to the nearest dollar. $ … [3] (ii) In which year is Xavier’s salary first greater than $47 500? … [3] (c) In January 2020, the population of a town was 5% more than its population in January 2018. In January 2021, the population of this town was 2% less than its population in January 2020. Calculate the overall percentage increase in the population from January 2018 to January 2021. … % [2]

14 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 120 2 M1 for 6 ÷ (21 – 19) oe soi 2 x or for = 6 40 1(a)(ii)(a) 34 2 15 M1 for 40 – × 40 oe or better 100 or B1 for 6 1(a)(ii)(b) 35 2  15  M1 for  1 −  × p = 29.75 or better  100  1(b)(i) 44 274 cao 3 B2 for 44273 to 44274 or 44270  2  5 or M1 for 40100 ×  1 +  oe  100  1(b)(ii) 2019 nfww 3 M2 for one correct trial of n = 8 or n = 9 either to find a salary or, if working with 1.02n and 47 500÷ 40 100 [= 1.1845], to find a value of 1.02n or B2 for final answer 9 or 4 nfww or M1 for  2  n their 44 274 ×  1 +  = 47 500 oe  100   2  n or 40 100 ×  1 +  = 47 500 oe  100  or for at least one trial giving a value greater than their 44 274 1(c) 2.9 [increase] 2  5  2  M1 for  1 + × 1 −  oe  100  100  implied by 1.029 or 102.9[%]

This question in 0580/43 May/June 2021

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]

14 marks

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 

This question in 0580/42 Oct/Nov 2021

1 A company employed 300 workers when it started and now employs 852 workers. (a) Calculate the percentage increase in the number of workers. … % [2] (b) Of the 852 workers, the ratio part-time workers : full-time workers = 5 : 7. Calculate the number of full-time workers. … [2] (c) The company makes 40 600 headphones in one year. Write this number (i) in words, … [1] (ii) in standard form. … [1] (d) In one month, the company sells 3 000 headphones. 3 Of these, 48% are exported, are sold to shops and the rest are sold online. 8 Calculate the number of headphones that are sold online. … [3] (e) One year, sales increased by 15%. The following year sales increased by 18%. Calculate the overall percentage increase in sales. … % [3]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 184 2 852 − 300 M1 for [× 100 ] oe 300 852 or for × 100 [ −100 ] oe 300 1(b) 497 2 852 M1 for × k oe where k = 1, 5 or 7 5 + 7 1(c)(i) Forty thousand six hundred 1 1(c)(ii) 4.06 × 10 4 1 1(d) 435 3  48 3  M2 for 3000 ×  1 − −  oe  100 8  or B2 for 2565, or 1440 and 1125 or 1875 and 1440 or 1560 and 1125 48 3  48 3  or M1 for 1 − − or 3000 ×  +  oe 100 8  100 8  or B1 for 1440 or 1125 or 1560 or 1875 If 0 scored SC1 for answer 975 1(e) 35.7 3 100 + 15 100 + 18 M2 for × [–1] oe or better 100 100 100 + 15 100 + 18 or M1 for k × × oe 100 100

This question in 0580/42 Feb/March 2022

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

2 (a) Alex, Bobbie and Chris share strawberries in the ratio Alex : Bobbie : Chris = 3 : 2 : 2. Chris receives 12 strawberries. Calculate the total number of strawberries shared. … [2] (b) In a sale, a shop reduces all prices by 12%. (i) Dina buys a book which has an original price of $6.50 . Calculate how much Dina pays for the book. $ … [2] (ii) Elu pays $11 for a toy. Calculate the original price of the toy. $ … [2] (c) Feri invests some money. The rate of interest for the first year is 2.5%. At the end of the second year the overall percentage increase of Feri’s investment is 6.6%. Find the rate of interest for the second year. … % [2] (d) A radioactive substance decays at an exponential rate of 2% per day. The initial mass is 80 g. (i) Find the mass at the end of 5 days. … g [2] (ii) Find how many more whole days, after day 5, it takes for the mass to reduce to less than 67 g. … [3]

13 marks

Mark scheme: 2(a) 42 2 M1 for 12 ÷ 2 or better 2(b)(i) 5.72 2 M1 for 100 12 6.50 100   oe or B1 for 0.88 oe 2(b)(ii) 12.5[0] 2 M1 for 100 12 11 100    x or better oe Question Answer Marks Partial Marks 2(c) 4 2 M1 for 100 2.5 100 6.6 [...] 100 100     oe 2(d)(i) 72.3 or 72.31... 2 M1 for 5 100 2 80 100        oe 2(d)(ii) 4 nfww 3 B2 for answer 9 nfww or M2 for correct trials with values giving either side of 67 or M1 for 100 2 80 100        n = 67 or  100 2 67 100          k their i or an evaluated trial with n ⩾ 6 or k ⩾ 1

This question in 0580/41 May/June 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

1 (a) Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2] (ii) a solid hemisphere with radius 6 cm. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm3 [2] (b) NOT TO SCALE 14 cm 6 cm The cylinder and hemisphere in part (a) are joined to form the solid in the diagram. The solid is made of steel and 1 cm 3 of steel has a mass of 7.85 g. (i) Show that 1 cm 3 of steel has a mass of 0.007 85 kg. [1] (ii) Calculate the total mass of the solid. … kg [2] (c) 2000 cm 3of iron is melted down and some of it is used to make 50 spheres with radius 2 cm. (i) Calculate the percentage of iron that is left over. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3] (ii) The iron left over is then made into a cube. Calculate the length of an edge of the cube. … cm [1] (d) A solid cone has radius 3R cm and slant height 9R cm. A solid cylinder has radius x cm and height 7x cm. The total surface area of the cone is equal to the total surface area of the cylinder. Given that R = kx , find the value of k. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] k = … [4]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1580 or 1583 to 1584 2 M1 for π  6 2  14 1(a)(ii) 452 or 452.3 to 452.4... 2 3 M1 for 1 4  π  6  2 3 1(b)(i) 7.85 ÷ 1000 [= 0.00785] M1 1(b)(ii) 16[.0] or 15.95 to 15.99 2 FT {their (a)(i) + their (a)(ii)}  0.00785 evaluated to 3 sig fig or better M1 for (their (a)(i) + their (a)(ii)) × 0.00785 1(c)(i) 16.2 or 16.21 to 16.23 3 4 3 2000 − 50  π 2 3 M2 for 100  2000 4 3 50  π 2 3 or for  100 2000 4 3 50  π 2 3 or M1 for 2000 1(c)(ii) 6.87 or 6.870 to 6.872 1  4 3  FT 3 2000 − their 50  π 2    3  evaluated to 3sf or better 1(d) 2 4 M1 for [π](3 R ) 2 + [π]3 R  9 R oe oe 3 M1 for 2[π]x 2 + 2[π]x  7 x oe M1 for their area of cone = their area of cylinder seen

This question in 0580/41 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

1 (a) (i) Alain and Beatrice share $750 in the ratio Alain : Beatrice = 8 : 7. Show that Alain receives $400. [1] (ii) (a) Alain spends $150. Write $150 as a percentage of $400. … % [1] (b) He invests the remaining $250 at a rate of 2% per year simple interest. Calculate the amount Alain has at the end of 5 years. $ … [3] (iii) Beatrice invests her $350 at a rate of 0.25% per month compound interest. Calculate the amount Beatrice has at the end of 5 years. Give your answer correct to the nearest dollar. $ … [3] (b) Carl, Dina and Eva share 100 oranges. The ratio Carl’s oranges : Dina’s oranges = 3 : 5. The ratio Carl’s oranges : Eva’s oranges = 2 : 3. Find the number of oranges Carl receives. … [2] (c) Fred buys a house. At the end of the first year, the value of the house increases by 5%. At the end of the second year, the value of the house increases by 3% of its value at the end of the first year. The value of Fred’s house at the end of the second year is $60 564. Calculate how much Fred paid for the house. $ … [3] (d) Gabrielle invests $500 at a rate of r % per year compound interest. At the end of 8 years the value of Gabrielle’s investment is $609.20 . Find the value of r. r = … [3]

16 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 750 M1  8 [= 400] 8 + 7 1(a)(ii)(a) 37.5 1 1(a)(ii)(b) 275 3 250 2 5 M2 for 250 + oe 100 250 2 5 or M1 for oe 100 1(a)(iii) 407[.00] cao nfww 3 B2 for 406.5 to 406.7  0.25  60 or M1 for 350  1 + oe isw    100  If 0 scored SC1 for answer 354 or answer 406 1(b) 24 2 M1 for [C : D =] 6 : 10 oe and [C : E =] 6 : 9 oe 6 or for [ 100] oe 6 + 10 + 9 1(c) 56 000 nfww 3  3   5  M2 for 60564  1 +  1 + oe      100   100   3  5  or M1 for [ x ] 1 +  1 +     100  100   3   5  or for 60564  1 + oe or 60564  1 +      100   100  If 0 scored, SC1 for answer 65499 to 65500 1(d) 2.5[0] or 2.499... 3 609.20 M2 for 8 oe 500 or M1 for 500  (...)8 = 609.2[0] oe

This question in 0580/42 Feb/March 2023

3 12 cm NOT TO SCALE 3 cm The diagram shows a cylinder containing water. There is a solid metal sphere touching the base of the cylinder. Half of the sphere is in the water. The radius of the cylinder is 12 cm and the radius of the sphere is 3 cm. (a) The sphere is removed from the cylinder and the level of the water decreases by h cm. Show that h = 0.125 . 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 [3] (b) The water in the cylinder is poured into another cylinder of radius R cm. The depth of the water in this cylinder is 18 cm. Calculate the value of R. R = … [3] (c) The sphere is melted down and some of the metal is used to make 30 cubes with edge length 1.5 cm. Calculate the percentage of metal not used. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3]

9 marks

Mark scheme: 3(a) 1 4 3 M3  π 3 2 3 [ h = ] oe π  12 2 leading to 0.125 or 2 1 4 3  12 −3  π 3 2 3 3 – oe π  12 2 2 1 4 3 M2 for π  12  h =  π 3 oe 2 3 leading to 0.125 or for π ×122 × 3 = π×122 × x + ⅔×π×33 oe 1 4 3   π  3 2 3 h or for = oe π  12 2  3 3 2 1 4 3 or M1 for π  12  h or  π 3 oe 2 3 or  12 2  3 3(b) 4.8[0] or 4.795 to 4.796 3 M2 for π  12 2  (3 − 0.125) = π  R 2  18 oe or π ×122 × 3 – ⅔×π×33 = π  R 2  18 or B1 for 3 – 0.125 or for 414 oe 3(c) 10.5 or 10.47 to 10.49 3 4 3 3  π  3 − 30  1.5 3 3 30  1.5 M2 for or  100 oe 4 3 4 3  π  3 π 3 3 3 4 3 3 30  1.53 or M1 for π 3 − 30  1.5 or oe 3 4 3  π  3 3

This question in 0580/42 Feb/March 2023

2 (a) Anil changes $830 into euros when the exchange rate is 1 euro = $1.16 . He spends 500 euros. He then changes the remaining money back into dollars at the same exchange rate. Work out how much, in dollars, Anil receives. $ … [3] (b) In 2021, Anil earns $37 000. (i) He spends $12 400 on bills in 2021. Calculate the percentage of his earnings he spends on bills. … % [2] (ii) His earnings of $37 000 increase by 3.2% in 2022. Calculate his earnings in 2022. $ … [2] (c) Anil invests $3500 in an account that pays a rate of 2.4% per year compound interest. (i) Calculate the total interest earned at the end of 5 years. $ … [3] (ii) Find the number of complete years before Anil has at least $5000 in this account. … years [3]

13 marks

Mark scheme: 2(a) 249.98 to 250[.0…] 3 M2 for 830 – 500 × 1.16 or M1 for 500 × 1.16 OR M1 for 830 ÷ 1.16 M1 for (their 715.5… – 500 ) × 1.16 2(b)(i) 33.5 or 33.51… 2 12400 M1 for [ 100] oe 37000 If 0 scored, SC1 for answer 66.5 or 66.48 to 66.49 2(b)(ii) 38 184 cao 2  3.2  M1 for 37 000   1   oe  100  or B1 for 1184 2(c)(i) 441 or 440.6 3 B2 for answer 3941 or 3940.6 or 3940.64 or 440.64 to 440.65 to 3940.65  2.4  5 or M2 for 3500 ×  1   – 3500  100   2.4  5 or M1 for 3500 ×  1   oe isw  100  2(c)(ii) 16 3 B2 for 15[.0] nfww to 15.1  2.4 15 or M2 for 3500 ×  1   oe seen  100   2.4 16 or 3500 ×  1   oe seen  100  or M1 for  2.4  n (3500 or their 3941) ×  1    100  associated with 5000 oe

This question in 0580/42 May/June 2023

1 (a) Tomas sells a computer, a bike and a phone. The amounts he receives are in the ratio computer : bike : phone = 14 : 17 : 9. (i) Calculate the amount he receives for the phone as a percentage of the total. … % [2] (ii) The total amount he receives is $560. Calculate how much he receives for the bike. $ … [2] (iii) Tomas originally bought the bike for $195. He wanted to make a profit of at least 25% when he sold it. Does Tomas make a profit of at least 25%? You must show all your working to support your decision. [3] (b) Ulla invests $725 for 6 years in an account paying simple interest at a rate of 1.3% per year. Calculate the total interest earned at the end of 6 years. $ … [2] (c) In a sale, all prices are reduced by 24%. Victor pays $36.86 for a pair of shoes in the sale. Calculate the original price of the shoes. $ … [2]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 22.5 2 9 M1 for [ 100] 14  17  9 1(a)(ii) 238 2 FT their 14 + 17 + 9 = N seen in (a)(i) 560 M1 for  k, their 14  17  9  where k = 1, 9, 14 or 17 1(a)(iii) METHOD 1 M2 25 M1 for  195 100 1.25  195 oe 243[.75] and No oe A1 Strict FT yes if their (a)(ii) > 243.75 If M0 scored, then SC1 for 243.75 and a correct conclusion. METHOD 2 (M2) their 238 M1 for = 1.22... oe 195 their 238 – 1 = 0.22… oe 195 22[%] (or better) and No oe (A1) Strict FT yes if their (a)(ii) gives answer > 25 If M0 scored, then SC1 for 22.05 and a correct conclusion. METHOD 3 (M2) M1 for 0.25  195 195  0.25 = 48.75 oe and their 238 – 195 = 43 43 and 48.75 and NO (A1) Strict FT yes if their (a)(ii) gives profit > 48.75 If M0 scored, then SC1 for 43 and 48.75 and a correct conclusion. METHOD 4 (M2)  25  M1 for x   1   = their 238  100  their 238  100 125 190.4 and NO (A1) Strict FT yes if their (a)(ii) gives answer > 195 If M0 scored then SC1 for 190.4 and a correct conclusion. 1(b) 56.55 2 725  1.3[ 6] M1 for oe 100 1(c) 48.5[0] 2  24  M1 for x   1   = 36.86 oe  100 

This question in 0580/43 May/June 2023

3 (a) The table shows information about some of the planets in the solar system. Planet Diameter (km) Average distance from the Sun (km) Earth 12 800 1.496 # 108 Mars 6 800 2.279 # 108 Jupiter 143 000 7.786 # 108 Saturn 120 500 1.434 # 109 Neptune 49 500 4.495 # 109 (i) The average distance of Mars from the Sun is 2.279 # 108 km . Write this distance as an ordinary number. … km [1] (ii) The planet Uranus has a diameter that is 35.8% of the diameter of Jupiter. Calculate the diameter of Uranus. … km [2] (iii) The ratio diameter of Neptune : diameter of Saturn can be written in the form 1 : n. Find the value of n. n = … [1] (iv) Find the average distance of Neptune from the Sun as a percentage of the average distance of the Earth from the Sun. … % [2] (v) Distances within the solar system are also measured in astronomical units (AU). The average distance of Jupiter from the Sun is 5.20 AU. Calculate the average distance of Mars from the Sun in astronomical units. … AU [2] (vi) The diameter of Mars is 39.2% greater than the diameter of Mercury. Calculate the diameter of Mercury. … km [2] (b) One light year is the distance that light travels in a year of 365.25 days. The speed of light is .29979 # 105 kilometres per second. (i) Show that one light year is 9.461 # 1012 km , correct to 4 significant figures. [2] (ii) The distance from the Andromeda Galaxy to Earth is 2.40 # 1019 km . Calculate the time taken for light to travel from this galaxy to Earth. Give your answer in millions of years. … million years [2]

14 marks

Mark scheme: 3(a)(i) 227 900 000 1 3(a)(ii) 51 200 or 51 190 or 51 194 2 35.8 M1 for  143 000 100 After 0 scored SC1 for answer figs 512 or figs 5119 or figs 51194 3(a)(iii) 2.43 or 2.434… 1 3(a)(iv) 3000 or 3004 to 3005 2 4.495  10 9 M1 for [ 100] oe 1.496  108 After 0 scored SC1 for answer figs 3 or figs 3004…. or figs 3005 3(a)(v) 1.52 or 1.522… 2 B1 for 1AU = 1.5[0]  108 or 1.497…  108 [km] or 1km = 6.68  10−9 or 6.678 10−9  AU OR 5.2  2.279[108 ] M1 for oe 7.786[108 ] After 0 scored SC1 for answer figs 152 or figs 1522…… 3(a)(vi) 4890 or 4885… 2  39.2  M1 for d   1 +  = 6800 oe  100  3(b)(i) 2.9979  105  602  24  M1 365.25 After M0 SC1 for 2.9979  105  31557600 oe = 9.4606…  1012 A1 3(b)(ii) 2.54 or 2.536 to 2.537 2 2.4  1019 M1 for 12 oe 9.461  10

This question in 0580/41 Oct/Nov 2023

3 (a) The value of Priya’s car decreases by 10% every year. The value today is $7695. (i) Calculate the value of the car after one year. $ … [2] (ii) Calculate the value of the car one year ago. $ … [2] (b) Ali invests $600 at a rate of 2% per year simple interest. Calculate the value of Ali’s investment at the end of 5 years. $ … [3] (c) Sara invests $500 at a rate of r % per year compound interest. At the end of 12 years, the value of Sara’s investment is $601.35, correct to the nearest cent. Find the value of r. r = … [3] (d) The mass of a radioactive substance decreases exponentially at a rate of 3% each day. (i) Find the overall percentage decrease at the end of 10 days. … % [2] (ii) Find the number of whole days it takes until the mass of this substance is one half of its original amount. … [3]

15 marks

Mark scheme: 3(a)(i) 6925.5[0] cao 2 100 − 10 M1 for 7695 × oe 100 or B1 for answer 769.5 3(a)(ii) 8550 2 100 − 10 M1 for X  = 7695 oe 100 3(b) 660 3 B2 for 60 600 2 5 or M2 for 600 + oe 100 600  2[5] or M1 for oe 100 3(c) 1.55 or 1.549 to 1.550 3 601.35 M2 for 12 500 or M1 for 500  (...)12 = 601.35 3(d)(i) 26.3 or 26.25 to 26.26 2 10  100 − 3  M1 for [k]   oe  100  3(d)(ii) 23 3 M2 for a correct trial evaluated with n =22 or n = 23 or M1 for [k] (0.97)n < 0.5[k] oe soi or for [k](0.97)n = 0.5[k] oe soi, implied by one correct trial n > 10 or for [k](0.97)23 oe seen If 0 scored SC1 for answer 22

This question in 0580/42 Oct/Nov 2023

2 The table shows the number of each type of bird seen in a garden on Monday. Type of bird Frequency Pie chart sector angle Goldfinch 8 96° Jay 6 Starling 11 Robin 5 (a) Find the percentage of the birds that are Starlings. … % [2] (b) (i) In the table, complete the column for the pie chart sector angle. [2] (ii) Complete the pie chart to show the information in the table. Goldfinch [2] (c) On Tuesday, the number of Goldfinches seen in the garden increased by 262.5%. Calculate the number of Goldfinches seen on Tuesday. … [2] (d) One of the most common birds in the world is the Red-Billed Quelea which lives in Sub-Saharan Africa. There are approximately 1500 million of these birds in this area. (i) Write 1500 million in standard form. … [1] (ii) The land area of Sub-Saharan Africa is approximately 21.2 million square kilometres. Work out the average number of these birds per square kilometre. … birds/km2 [2]

11 marks

Mark scheme: 2(a) 2 2 11 36.7 or 36.66 to 36.67 or 36 M1 for [ 100] oe 3 8 + 6 + 11 + 5 2(b)(i) 72, 132 and 60 2 M1 for 360 ÷ (8 + 6 + 11 + 5) oe or 96 ÷ 8 2(b)(ii) Correct pie chart drawn 2 For 2 marks, strict FT their angles for correct pie chart only if angles add up to 360. B1FT for one correct sector 2(c) 29 2  262.5  M1 for 8   1 +  oe  100  or B1 for 21 2(d)(i) 1.5  109 1 2(d)(ii) 70.8 or 70.75… 2 M1 for 1500 [million] ÷ 21.2 [million]

This question in 0580/43 Oct/Nov 2023

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

1 (a) In 2023 a football club had 50 adult members and 70 child members. The membership fee for an adult was $40 and the membership fee for a child was $15. (i) Calculate the total of the membership fees received by the club in 2023. $ … [2] (ii) The cost of running the club in 2023 was $2780. Calculate $2780 as a percentage of the total of the membership fees received by the club. … % [1] (iii) In 2023 there were 120 members. This was a decrease by 4% of the number of members in 2022. Calculate the number of members in 2022. … [2] (iv) In 2024 the total number of members increased from the 120 members in 2023. The number of adult members and the number of child members each increased by the same number. The ratio number of adult members : number of child members changed to 14 : 19. (a) Find the total number of members in 2024. … [2] (b) Calculate the percentage increase in the total number of members from 2023 to 2024. … % [2] (b) The population of a village is 2500. The population is decreasing exponentially at a rate of 3% per year. (i) Calculate the population at the end of 3 years. … [2] (ii) Find the number of complete years it takes for the population to first fall below 2000. … years [2]

13 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 3050 2 M1 for 50 × 40 + 70 × 15 or better 1(a)(ii) 91.1 or 91.14 to 91.15 1 2780 FT  100 their 3050 1(a)(iii) 125 nfww 2 100  4 M1 for [...] × =120 oe 100 1(a)(iv)(a) 132 2 B1 for increase of 6 in adult or junior or M1 for 56 : 76 or for multiples of 33 seen 33, 66, 99, 132, … or 50 + x : 70 + x = 14 : 19 oe 19  14 or (70 – 50)  oe 19  14 14 or 50 + x = (120 + 2x)  oe 19  14 1(a)(iv)(b) 10 2 their (a)  120 FT  100 120 dep on their (a) > 120 their (a)  120 M1 for  100  or 120 their (a)  100 [ 100] 120 1(b)(i) 2280 or 2281 to 2282 nfww 2 3  3  M1 for 2500  1  oe    100  1(b)(ii) 8 2 n  3  M1 for 2500  1  or 0.97n    100  evaluated with n > 3

This question in 0580/43 May/June 2024

4 (a) Enzo, Rashid and Blessy each swim as many lengths of a swimming pool as they can in 15 minutes. The results are shown in the table. Name Number of lengths Enzo 11.25 Rashid 18.75 Blessy 20 (i) Find the number of lengths Enzo swims as a percentage of the total number of lengths all three people swim. … % [2] (ii) Write the ratio of the number of lengths each person swims in the form Enzo : Rashid : Blessy. Give your answer in its simplest form. … : … : … [2] (iii) Each length of the pool is 25 m. (a) Work out Blessy’s average swimming speed for the 15 minutes. Give your answer in metres per second. … m/s [3] (b) Rashid continues to swim at the same rate. Calculate the time it takes Rashid to swim a total distance of 5 km. Give your answer in hours and minutes. … h … min [4] (iv) Blessy swims for one hour. The number of lengths she swims decreases by 5% every 15 minutes. Calculate the number of lengths she swims in the final 15 minutes. … [3] (b) Another swimmer, Adam, swims 450 m, correct to the nearest 25 metres. This takes 10 minutes, correct to the nearest minute. Calculate the minimum distance Adam swims in one hour at this rate. … m [3]

17 marks

Mark scheme: 4(a)(i) 22.5 2 11.25 M1 for 100  oe 11.25 + 18.75 + 20 4(a)(ii) 9 : 15 : 16 2 M1 for 1125 : 1850 : 2000 or better 4(a)(iii)(a) 5 3 or 0.556 or 0.5555 to 0.5556 9 20  25 M2 for oe 15[ 60] or M1 for 20 × 25 or for their distance ÷ (15 [× 60]) oe 4(a)(iii)(b) 2 h 40 mins 4 Approach 1 8 B3 for [h]oe or 160 [mins] or 9600[s] 3 Or M3 for 5000 ÷ (18.75 × 25 × 4)[h] oe or 5000 ÷ (18.75 × 25 ÷ 15)[mins] oe or 5000 ÷ ((18.75 × 25 × 4) ÷ (60 × 60))[secs] oe Or M2 for (18.75 × 25 × 4)[m/h] oe or (18.75 × 25 ÷ 15)[m/min] oe or (18.75 × 25 × 4) ÷ (60 × 60))[m/sec] oe Or B1 for 200 or 1 km =1000m soi After 0 scored SC1 for time Figs 267 or figs 2666 to 2667 or figs 16 or figs 96 Approach 2 B3 for 160 [mins] Or M3 for 15 × 5000 ÷ (18.75 × 25) [mins] oe Or M2 for 5000 ÷ (18.75 × 25) oe Or B1 for 200 or 1 km =1000m soi After 0 scored SC1 for time figs 16 4(a)(iv) 17.1 or 17.14 to 17.15 3 3  100 − 5  M2 for 20 ×   oe  100   100 − 5  k or M1 for 20 ×   where k is 2, or  100  4  100 − 5  3 or for 20 ×   oe seen and spoiled  100  4(b) 2500 3 425to450 450 − 12.5 M2 for or or 10 + 0.5 10 to 11 425 to 450 450 − 12.5 or 630 600 to 660 or M1 for 10.5 or 9.5 or 437.5 or 462.5 or 630[s] or 570[s]

This question in 0580/41 Oct/Nov 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

1 Dinari sells fruit and vegetables. (a) One day the mass of fruit and vegetables he sells is in the ratio fruit : vegetables = 9 : 8. He sells 48 kg of vegetables. Find the mass of the fruit he sells. … kg [2] (b) On another day he receives $280 for the fruit and vegetables he sells. The $280 is in the ratio fruit : vegetables = ( c + 3) : ( c - 1) . Find the amount he receives from selling the fruit. $ … [3] (c) In one week Dinari buys fruit and vegetables for $1620. He sells the fruit and vegetables for $1750. Calculate his percentage profit. … % [2] (d) In another week Dinari sells fruit and vegetables for $1738. He makes a profit of 10%. Calculate the amount he paid for the fruit and vegetables in that week. $ … [2]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 54 2 48 M1 for 8 1(b) 142 3 B2 for 2c = 278 or better or M1 for c + 3 + c – 1 = 280 OR B2 for 2x = 276 or better or M1 for c + 3 – (c – 1) = 4 1(c) 8.02 or 8.024 to 8.025 2 1750 − 1620 M1 for [100] 1620 1750 or for 100[ −100] 1620 1(d) 1580 2 + 10 M1 for 100...  =1738 oe or better 100

This question in 0580/43 Oct/Nov 2024

11 (a) Midhil invests $1500 at a rate of 4.2% per year compound interest. Calculate the value of the investment at the end of 5 years. $ … [2] (b) Hitanshi invests some money at a rate of x% per year compound interest. At the end of 11 years the value of the investment has doubled. Calculate the value of x. x = … [3]

5 marks

Mark scheme: 11(a) 1842.59 2 5  4.2  M1 for 1500 ×  1 +  oe  100  11(b) 6.5[0…] 3 1 M2 for 11 2 or 2 11 or M1 for [k×]y11 = 2[×k] oe

This question in 0580/42 Feb/March 2025

11 In a sale, the original price of a shirt is reduced by 15%. The sale price of the shirt is $23.63 . Find the original price of the shirt. $ … [2]

2 marks

Mark scheme: 11 27.8[0] 2  M1 for X ×  100 − 15 = 23.63 oe  100 

This question in 0580/42 May/June 2025

17 (a) Alex invests $400 at a rate of 2.3% per year simple interest. Find the total amount Alex has at the end of 5 years. $ … [3] (b) Virat has $100 to spend. In February he spends $x . In March he spends 10% more than he spends in February. In April he spends 10% more than he spends in March. At the end of April, Virat has $33.80 remaining. Find the value of x. x = … [3] (c) Bobbie invests $500 in an account that pays compound interest each year. At the end of 17 years, the value of Bobbie’s investment is $700.13 . Find the value of Bobbie’s investment at the end of 20 years. $ … [4]

10 marks

Mark scheme: 17(a) 446 3 B2 for answer 46 400  2.3  5 or M2 for 400 + oe 100 400  2.3 5  or M1 for oe 100 17(b) 20 nfww 3 M2 for x + 1.1x + 1.12x = [100 –] 33.80 oe  10  2 oe seen or M1 for  1 +  x  100  or for one correctly evaluated trial 17(c) 742.97 to 742.99 4 B3 for 1.02[0…] or interest rate = 2[.0…][%] OR  700.13  20 M3 for 500  17  oe    500   700.13  3 or for 700.13×  17  oe    500  700.13 or M2 for 17 oe 500 OR M1 for 500(…)17 = 700.13 oe M1 dep on previous M1 for  their r  20  their r 3 500  1 + or 700.13  1 +      100   100 

This question in 0580/42 May/June 2025

14 A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm. 256 (a) Show that the volume of the sphere is rcm 3. 3 [1] (b) Calculate the percentage of the cube that is not occupied by the sphere. … % [3] (c) The density of the metal of the sphere is 7.86 g/cm3. Calculate the mass of the sphere. Give your answer in kilograms. [Density = mass ' volume] … kg [2] (d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder. … cm2 [4]

10 marks

Mark scheme: 14(a) 4 3 256 1 π 4 [= π ] 3 3 14(b) 47.6 nfww or 3 50 B2 for 52.4 or 52.35 to 52.37 or π nfww 47.63 to 47.64… nfww 3 OR 3 256 8 − π 3 M2 for 3  100 oe 8 3 256 256 8 − π π 3 3 or M1 for 3 [ 100] oe or 3  100 8 8 oe 14(c) 2.11 or 2.107… 2 256 M1 for π  7.86 3 14(d) 233 or 233.3 to 233.4 4 2 256 M1 for π  3.1  h = π 3 M2dep for 2  π  3.12 + 2  π  3.1  their h or M1dep for 2 π 3.1  theirh or M1 for 2π  3.12

This question in 0580/41 Oct/Nov 2025

14 (a) Carlos invests $24 000 at a rate of 3.2% per year compound interest. Calculate the value of his investment at the end of 4 years. $ … [2] (b) Carlos buys a painting for $x. He sells the painting for $40 870. He makes a profit of 34%. Calculate the value of his profit. $ … [3] (c) Carlos also buys a car with a value of $32 500. The value of the car decreases exponentially by 23% each year. Find a formula for the value, $V, of the car at the end of n years. … [3]

8 marks

Mark scheme: 14(a) 27 223 2 4  3.2  M1 for 24 000 ×  1 +   100  14(b) 10 370 3 B2 for 30 500 40 870 or M2 for  34 oe 134  34  or M1 for  1 +  x = 40 870 oe  100  14(c) V = 32 500 × 0.77n final 3 B2 for answer 32 500 × 0.77n answer or for correct explicit formula seen in working, may be unsimplified or M1 for 32 500 × (1 – 0.23)n oe seen or for correct implicit formula seen, may be unsimplified or answer of form V = 32 500 × kn or answer of form V = p × (0.77 oe)n

This question in 0580/42 Oct/Nov 2025

21 A dressmaker takes 75 hours to make 31 dresses. In week 1, she takes a total of 12 hours 30 minutes to make the first 4 dresses. In week 2, she makes the remaining 27 dresses at a constant hourly rate. Work out the percentage increase in her hourly rate of making dresses from week 1 to week 2. … % [4]

4 marks

Mark scheme: 21 35 4 27  ( 75 − 12.5 ) M3 for  100  −100  4  12.5 oe  27  ( 75 − 12.5 )  or  − 1  100  oe 4  12.5   27  ( 75 − 12.5 ) − ( 4  12.5 ) or  100  4  12.5 oe  12.5 62.5  or    × 100[– 100]  4 27  OR M1 for 4 ÷ 12.5 oe M1 for 27 ÷ (75 –12.5) oe

This question in 0580/43 Oct/Nov 2025