Exponential growth and decay — Paper 4 · IGCSE Mathematics (9-1) 0980

1 (a) The price of a newspaper increased from $0.97 to $1.13 . Calculate the percentage increase. … % [3] (b) One day, the newspaper had 60 pages of news and advertisements. The ratio number of pages of news : number of pages of advertisements = 5 : 7. (i) Calculate the number of pages of advertisements. … [2] (ii) Write the number of pages of advertisements as a percentage of the number of pages of news. … % [1] (c) On holiday Maria paid 2.25 euros for the newspaper when the exchange rate was $1 = 0.9416 euros. At home Maria paid $1.13 for the newspaper. Calculate the difference in price. Give your answer in dollars, correct to the nearest cent. $ … [3] (d) The number of newspapers sold decreases exponentially by x% each year. Over a period of 21 years the number of newspapers sold decreases from 1 763 000 to 58 000. Calculate the value of x. x = … [3] (e) Every page of the newspaper is a rectangle measuring 43 cm by 28 cm, both correct to the nearest centimetre. Calculate the upper bound of the area of a page. … cm2 [2]

14 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 16.5 or 16.49... 3 M2 for 1.13 − 0.97[× 100] oe or 1.13 × 100 oe 0.97 0.97 1.13 or M1 for oe 0.97 1(b)(i) 35 2 M1 for 60 ÷ ( 5 + 7 ) 1(b)(ii) 140 1 1(c) $1.26 final answer 3 B2 for 1.259... or 1.26 but not as final answer or M1 for 2.25 ÷ 0.9416 If 0 scored, SC1 for 1.13 × 0.9416 1(d) 15[.0…] 3 58000 M2 for 21 oe 1763000 or M1 for 58000 = 1763000 ( k ) 21 1(e) 1239.75 2 B1 for 43 + 0.5 or 28 + 0.5 oe seen

This question in 0980/42 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  oe M1 for x 1 +  = 882  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 0980/41 Oct/Nov 2019

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 0980/42 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 0980/41 Oct/Nov 2020

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 0980/41 Oct/Nov 2021

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 0980/41 Oct/Nov 2022

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 0980/42 May/June 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 0980/42 May/June 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 0980/42 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 0980/41 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 0980/41 Oct/Nov 2025