Exponential growth and decay — Paper 4 · IGCSE Mathematics 0580

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

11 marks

This question in 0580/42 Feb/March 2017

1 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) Alex has $20 and Bobbie has $25. (i) Write down the ratio Alex’s money : Bobbie’s money in its simplest form. … : … [1] 1 (ii) Alex and Bobbie each spend of their money. 5 Find the ratio Alex’s remaining money : Bobbie’s remaining money in its simplest form. … : … [1] (iii) Alex and Bobbie then each spend $4. Find the new ratio Alex’s remaining money : Bobbie’s remaining money in its simplest form. … : … [2] (b) (i) The population of a town in the year 1990 was 15 600. The population is now 11 420. Calculate the percentage decrease in the population. … % [3] (ii) The population of 15 600 was 2.5% less than the population in the year 1980. Calculate the population in the year 1980. … [3] (c) Chris invests $200 at a rate of x% per year simple interest. At the end of 15 years the total interest received is $48. Find the value of x. x = … [2] (d) Dani invests $200 at a rate of y% per year compound interest. At the end of 10 years the value of her investment is $256. Calculate the value of y, correct to 1 decimal place. y = … [3]

15 marks

Mark scheme: Question Answer Marks Partial marks 1(a)(i) 4 : 5 1 1(a)(ii) 4 : 5 1 1(a)(iii) 3 : 4 2 B1 for 12 : 16 or answer 4 : 3 1(b)(i) 26.8 or 26.79… 3 15600 − 11420 11420 M2 for [× 100 ] or × 100 15600 15600 11420 or M1 for 15600 1(b)(ii) 16 000 nfww 3 100 M2 for 15600 × oe 100 − 2.5 or M1 for 15600 associated with 97.5[%] seen 1(c) 8 2 200 × x × 15 1.6 or M1 for = 48 oe 5 100 or M1 for figs 16 1(d) 5 3 B2 for 2.49[9…] or 102.4[99…] or 1.024[99…] 2.5 or cao nfww or 2.50 or 102.5 or 1.025 2 256 or M2 for 10 oe 200 or M1 for 256 = 200 ( x )10 seen

This question in 0580/42 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

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

2 (a) A school has 240 students. The ratio girls : boys = 25 : 23. (i) Show that the number of boys is 115. [1] (ii) One day, there are 15 girls absent and 15 boys absent. Find the ratio girls : boys in school on this day. Give your answer in its simplest form. … : … [2] (iii) Next year, the number of students will increase by 15%. Calculate the number of students next year. … [2] (iv) Since the school was opened, the number of students has increased by 60%. There are now 240 students. Calculate the number of students when the school was opened. … [3] (b) The population of a city is increasing exponentially at a rate of 2% each year. The population now is 256 000. Calculate the population after 30 years. Give your answer correct to the nearest thousand. … [3] (c) A bacteria population increases exponentially at a rate of r% each day. After 32 days, the population has increased by 309%. Find the value of r. r = … [3]

14 marks

Mark scheme: 2(a)(i) 240 M1 × 23 (23 + 25) 2(a)(ii) 11 : 10 2 M1 for 110 : 100 or better or SC1 for 10 : 11, following boys 100, girls 110 2(a)(iii) 276 2  15  M1 for 240 ×  1 +  oe  100  or B1 for 36 seen 2(a)(iv) 150 3 240 M2 for [× 100] oe 100 + 60 or M1 for evidence of 160[%] associated 240 2(b) 464 000 3 30  2  M1 for 256 000 ×  1 +  oe  100  A1 for 463 700 to 463 710 B1 for their more accurate answer seen and rounded to nearest 1000 2(c) 4.5[0] 3 M2 for [x =] 32 .409 oe 32 = .409 oe or M1 for ( x ) If 0 scored, SC2 for answer 3.6 or 3.59 or 3.588… or SC1 for 32 3.09 or 1.0358 to 1.036 seen

This question in 0580/43 Oct/Nov 2018

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

3 (a) Dina invests $600 for 5 years at a rate of 2% per year compound interest. Calculate the value of this investment at the end of the 5 years. $ … [2] (b) The value of a gold ring increases exponentially at a rate of 5% per year. The value is now $882. (i) Calculate the value of the ring 2 years ago. $ … [2] (ii) Find the number of complete years it takes for the ring’s value of $882 to increase to a value greater than $1100. … [2]

6 marks

Mark scheme: 3(a) 662.45 2 5  2  M1 for 600 × 1 + oe    100  3(b)(i) 800 2 2 5  M1 for x 1 + = 882 oe    100  or SC1 for answer 82 3(b)(ii) 5 nfww 2 n  5  M1 for trial with 882 × 1 + with n > 1    100 

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

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

8 (a) Factorise completely. 3a 2 b - ab 2 … [2] (b) Solve the inequality. 3x + 12 1 5x - 3 … [2] (c) Simplify. 3 3x 2 y 4 ` j … [2] (d) Solve. 2 6 = x 2 - x x = … [3] (e) Expand and simplify. ( x - 2)( x + 5)( 2x - 1) … [3] (f) Alan invests $200 at a rate of r% per year compound interest. After 2 years the value of his investment is $206.46 . (i) Show that r 2 + 200r - 323 = 0 . [3] (ii) Solve the equation r 2 + 200r - 323 = 0 to find the rate of interest. Show all your working and give your answer correct to 2 decimal places. r = … [3]

18 marks

Mark scheme: 8(a) ab(3a – b) final answer 2 B1 for a(3ab – b2) or b(3a2 – ab) or ab(3a – b) seen 8(b) x > 7.5 final answer 2 B1 for 12+3 < 5x – 3x oe 8(c) 27x6y12 2 B1 for two of 27, x6 and y12 correct 8(d) 1 3 M2 for 4 = 6x + 2x or better 0.5 or 2 or M1 for 2(2 – x) = 6x oe 8(e) 2x3 + 5x2 – 23x + 10 final answer 3 B2 for correct expansion of three brackets unsimplified B1 for correct expansion of two brackets with at least 3 terms correct 8(f)(i) 2 M1  r  200  1 +  = 206.46 oe  100  2r r 2 M1 1 + + oe 100 100 2 r2 + 200r – 323 = 0 A1 Correct solution reached with no errors or omissions seen If 0 scored, SC1 for 200( n ) 2 = 206.46 8(f)(ii) 2 B2 2 −200 + 200 − 4(1)( −323) B1 for 200 − 4(1)( − 323) or (r + 100)2 2 × 1 −200 + q 2 B1 for or r = 323 + 100 – 100 2 × 1 OR  206.46  B2 for 100  − 1     200  206.46 or B1 for 200 1.60 cao final answer B1

This question in 0580/41 Oct/Nov 2020

3 (a) Beth invests $2000 at a rate of 2% per year compound interest. (i) Calculate the value of this investment at the end of 5 years. $ … [2] (ii) Calculate the overall percentage increase in the value of Beth’s investment at the end of 5 years. … % [2] (iii) Calculate the minimum number of complete years it takes for the value of Beth’s investment to increase from $2000 to more than $2500. … [3] (b) The population of a village decreases exponentially at a rate of 4% each year. The population is now 255. Calculate the population 16 years ago. … [3]

10 marks

Mark scheme: 3(a)(i) 2210 or 2208 or 2208.2, or 2208.16… 2 5  2  M1 for 2000 ×  1 +  oe  100  3(a)(ii) 10.4 or 10.5 or 10.40 to 10.41 2 their(a)(i) − 2000 M1 for [×100] or 2000 their(a)(i)  2  5 ×100 or  1 +  – 1 or 2000  100   2  5  1 +  × 100 oe  100  3(a)(iii) 12 3 B2 for 11.3 or 11.26 to 11.27 OR  2 11 M2 for [2000 ×]  1 +  oe  100   2 12 or [2000 ×]  1 +  oe seen  100   2  n or M1 for [2000 ×]  1 +  oe, n > 5 oe  100   2  n or for 2000 ×  1 +  = or > or ⩾ 2500  100  oe 3(b) 490 cao 3 16  4  M2 for p ×  1 −  = 255 oe soi by  100  490.0...  4  n or M1 for p ×  1 −  = 255 oe,  100  n > 1 oe

This question in 0580/42 Oct/Nov 2020

4 (a) The exchange rate is 1 euro = $1.142 . (i) Johann changes $500 into euros. Calculate the number of euros Johann receives. Give your answer correct to the nearest euro. … euros [2] (ii) Johann buys a computer for $329. The same computer costs 275 euros. Calculate the difference in cost in dollars. $ … [2] 3 (b) Lucy spends of the money she has saved this month on a book that costs $5.25 . 8 Calculate how much money Lucy has saved this month. $ … [2] (c) Kamal invests $6130 at a rate of r% per year compound interest. The value of his investment at the end of 5 years is $6669. Calculate the value of r. r = … [3]

9 marks

Mark scheme: 4(a)(i) 438 cao 2 500 M1 for 1.142 4(a)(ii) 14.95 2 M1 for [329 –] 275 × 1.142 oe 4(b) 14 2 8 M1 for 5.25 × oe 3 4(c) 1.7[0] or 1.699… 3 6669 M2 for 5 6130 or M1 for 6669 = 6130 ( k ) 5

This question in 0580/41 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

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

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

5 (a) $500 is invested at a rate of 3% per year. Calculate the total interest earned at the end of 7 years when (i) simple interest is paid, $ … [2] (ii) compound interest is paid. $ … [3] (b) The value of a car decreases exponentially by 10% each year. The value now is $6269.40 . Calculate the value of the car 3 years ago. $ … [3]

8 marks

Mark scheme: 5(a)(i) 105 2 3 M1 for × 500 [× 7 ] 100 5(a)(ii) 115 or 114.9... 3 7  3  M2 for 500 ×  1 +  [ −500 ]  100   3  k or M1 for 500 ×  1 +  , k integer ⩾ 2  100  5(b) 8600 3 6269.4 M2 for oe  10  3  1 −   100   10  3 or M1 for C ×  1 −  = 6269.4 oe  100 

This question in 0580/43 Oct/Nov 2021

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

3 (a) The table shows the numbers of tigers reported to be living in the wild in the year 2014 in some countries. Country Number India 2226 Indonesia 371 Nepal 198 Bangladesh 106 (i) Using the table, (a) find the number of tigers in Nepal as a percentage of the number of tigers in Bangladesh, … % [1] (b) find the ratio tigers in Bangladesh : tigers in Indonesia : tigers in India, giving your answer in its simplest form. … : … : … [2] (ii) Five years later, the number of tigers reported in India was 2967. Find the percentage increase in the population of tigers in India. … % [2] (iii) The number of tigers in India in the year 2014 is approximately 30.48% greater than in the year 2010. Find the number of tigers in India in the year 2010. Give your answer correct to the nearest integer. … [3] (b) At the start of June, a hive has a population of 2000 bees. Three months after the start of June the hive has a population of 2662 bees. The population of this hive can be calculated using the formula P = abx , where P is the population of the hive x months after the start of June. By finding the value of a and the value of b, calculate the population of the hive 7 months after the start of June. Give your answer correct to the nearest integer. … [5]

13 marks

Mark scheme: 3(a)(i)(a) 42 1 187 or 186.7 to 186.8 or 186 53 3(a)(i)(b) 2 : 7 : 42 cao 2 B1 for 106 : 371 : 2226 or any equivalent ratio If 0 scored, SC1 for 2 : 7 : 42 in the wrong order 3(a)(ii) 33.3 or 33.28 to 33.29 2 2967  2226 M1 for [ 100] oe 2226 2967 or  100 [– 100] oe 2226 3(a)(iii) 1706 cao nfww 3 B2 for 1705 to 1706.0… or 1710  30.48  or M1 for 1  x = 2226 oe or    100  better If 0 or M1 scored, SC1 for rounding their decimal answer seen to nearest integer 3(b) 3897 5 B1 for a = 2000 2662 M2 for [b =] 3 2000 or M1 for 2662 = 2000b3  2662  7 M1 for their 2000   3     their 2000  or for their a  (their b)7 provided their a and their b are clearly identified in the working If 0 or M1 scored, SC1 for rounding their decimal answer seen to nearest integer.

This question in 0580/43 May/June 2022

4 (a) (i) Zak invests $500 at a rate of 2% per year simple interest. Calculate the value of Zak’s investment at the end of 5 years. $ … [3] (ii) Yasmin invests $500 at a rate of 1.8% per year compound interest. Calculate the value of Yasmin’s investment at the end of 5 years. $ … [2] (iii) Zak and Yasmin continue with these investments. How many more complete years is it before the value of Yasmin’s investment is greater than the value of Zak’s investment? … [3] (b) Xavier buys a car for $2500. The value of the car decreases exponentially at a rate of 10% each year. Calculate the value of Xavier’s car at the end of 5 years. Give your answer correct to the nearest dollar. $ … [3] (c) The number of a certain type of bacteria increases exponentially at a rate of r % each day. After 22 days, the number of this bacteria has doubled. Find the value of r. r = … [3]

14 marks

Mark scheme: 4(a)(i) 550 nfww 3 500 2 5 M2 for + 500 oe 100 500 2 5 or M1 for oe 100 4(a)(ii) 546.65 2 5  1.8  M1 for 500  1 + oe    100  4(a)(iii) 8 nfww 3 B2 for final answer 13 OR M2 for trials correctly comparing both investments to 7 and 8 more years or M1 for at least two trials correctly comparing both investments 4(b) 1476 cao 3 B2 for 1480 or 1476.2 ... OR  10 5 M1 for 2500  1 − oe    100  B1 for their more accurate answer seen correctly rounded to the nearest dollar. 4(c) 3.2[0] or 3.200 to 3.201 3 M2 for (...) = 22 2 oe isw or M1 for [N]  (...)22 = 2[N]

This question in 0580/41 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

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

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

9 (a) Janna and Kamal each invest $8000. At the end of 12 years, they each have $12 800. (i) Janna invests in an account that pays simple interest at a rate of r% per year. Calculate the value of r. r = … [3] (ii) Kamal invests in an account that pays compound interest at a rate of R% per year. Calculate the value of R. R = … [3] (b) The population of a city is growing exponentially at a rate of 1.8% per year. The population now is 260 000. Find the number of complete years from now when the population will first be more than 300 000. … years [3]

9 marks

Mark scheme: 9(a)(ii) 4[.0] or 3.99… 3 12800 M2 for 12 8000 or M1 for 12800 = 8000  k 12 for any k 9(b) 9 nfww 3 8  1.8  M2 for 260 000 ×  1 +  oe evaluated to 4  100  sf or better  1.8 9 or 260 000 ×  1 +  oe evaluated to 2 sf  100  or better  1.8  n or M1 for [300 000 = ] 260 000 ×  1 +   100  oe soi (Accept any inequality sign in [300 000 = ])

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

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

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

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

19 The mass of a radioactive substance decays exponentially at a rate of 10% per day. The initial mass of the substance is 20 g. Find the number of whole days it takes for the mass of the substance to first be less than 1 g. … days [3]

3 marks

Mark scheme: 19 29 3 B2 for 28.4[3…] OR M2 for 20 × 0.928 or 20× 0.929 evaluated to at least 1 dp or for 0.928 or 0.929 evaluated to at least 2 dp or M1 for at least two trials of [20 ×] 0.9n soi  10  n or for 1  20 1 − oe    100 

This question in 0580/41 Oct/Nov 2025

22 Alex invests $200 at a rate of r% per year compound interest. At the end of 25 years the value of this investment is $301.10 . Find the value of r. r = … [3]

3 marks

Mark scheme: 22 1.65 or 1.649[9…] 3 301.10 M2 for 25 oe 200 or M1 for 200  [ ]25 = 301.10

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

15 The population of a town is 54 000. The population is decreasing exponentially at a rate of 2% per year. (a) Calculate the decrease in the population at the end of 4 years. … [3] (b) Find the number of complete years it takes for the population of 54 000 to first fall below 44 000. … years [2]

5 marks

Mark scheme: 15(a) 4192 or 4193 3 B2 for 49 807 to 49 808  2  4 or M2 for 54 000 − 54 000  1 −  oe  100   2  4 or M1 for 54 000   1 −  oe  100  15(b) 11 nfww 2 n  2  M1 for 54 000   1 −  evaluated  100  with n = 10 or n = 11 or B1 for 10.1 or 10.13 to 10.14

This question in 0580/43 Oct/Nov 2025