Exponential growth and decay — Paper 2 · IGCSE Mathematics 0580

4 The population of the world grows exponentially at a rate of 1.1% per year. Find the number of years it takes for the population to grow from 7 billion to 7.31 billion. Give your answer correct to the nearest whole number. … years [2]

2 marks

Mark scheme:  1.1  4 4 nfww 2 M1 for [7.31 =] 7  1 +  oe  100  2

This question in 0580/22 Feb/March 2017

9 The value of a motorbike is $12 400. Each year, the value of the motorbike decreases exponentially by 15%. Calculate the value of the motorbike after 3 years. $ … [2]

2 marks

Mark scheme: 9 7615.15 2 3  15  M1 for 12 400 ×  1 −  oe  100 

This question in 0580/21 Oct/Nov 2017

21 400 360 320 280 Value of 240investment ($) 200 160 120 80 40 0 0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 Number of years When Heidi was born, her grandfather invested some money in an account that paid compound interest. The graph shows the exponential growth of this investment. (a) Use the graph to find (i) the original amount of money invested, $ … [1] (ii) the number of years it took for the original amount to double, … years [1] (iii) the value of the investment after 54 years. $ … [1] (b) This account earned compound interest at a rate of r % per year. Use your answers to part (a)(i) and part (a)(ii) to write down an equation in terms of r. You do not have to solve your equation. … [2]

5 marks

Mark scheme: 21(a)(i) 20 1 21(a)(ii) 14 1 FT part (i) providing 20 < part (i) ⩽40 21(a)(iii) 280 1 21(b) 14 2 FT 2 marks for  r  2[× 20] = [20] 1 + oe isw their(a)(ii)    r   100  2[their (a)(i)]=[their (a)(i)]  1 +   100  M1 for n(x)14 or n(x)their(a)(ii) oe seen isw

This question in 0580/21 Oct/Nov 2018

9 There are 30 000 lions in Africa. The number of lions in Africa decreases exponentially by 2% each year. Find the number of lions in Africa after 6 years. Give your answer correct to the nearest hundred. … [2]

2 marks

Mark scheme: 9 26 600 cao 2 6  2  M1 for 30 000 ×  1 −  oe  100 

This question in 0580/22 Oct/Nov 2018

12 The population of a town decreases exponentially at a rate of 1.7% per year. The population now is 250 000. Calculate the population at the end of 5 years. Give your answer correct to the nearest hundred. … [3]

3 marks

Mark scheme: 12 229 500 cao 3 B2 for 229 460… OR  1.7  5 M1 for 250 000 ×  1 −  oe  100  B1 for their more accurate answer correctly rounded to the nearest 100

This question in 0580/22 Feb/March 2020

10 A town has a population of 45 000. This population increases exponentially at a rate of 1.6% per year. Find the population of the town at the end of 5 years. Give your answer correct to the nearest hundred. … [3]

3 marks

Mark scheme: 10 48 700 cao 3 5  1.6  M1 for 45 000 ×  1 +  oe  100  A1 for 48 710 to 48 720 If A0 scored B1 for their more accurate value correctly rounded to the nearest 100

This question in 0580/21 Oct/Nov 2020

13 The population of one variety of butterfly is decreasing exponentially at a rate of 34% per year. At the end of 2014, the population was 125.9 million. Calculate the population at the end of 2019. … million [2]

2 marks

Mark scheme: 13 15.8 or 15.76 to 15.77 2 5  34  M1 for 125.9 ×  1 −  oe  100 

This question in 0580/22 Feb/March 2021

12 The profit a company makes decreases exponentially at a rate of 0.9% per year. In 2014, the profit was $9500. Calculate the profit in 2019. $ … [2]

2 marks

Mark scheme: 12 9080 or 9080.13 2 5  0.9  M1 for 9500 ×  1 −   100 

This question in 0580/21 May/June 2021

12 (a) Sanjay invests $700 in an account paying simple interest at a rate of 2.5% per year. Calculate the value of his investment at the end of 6 years. $ … [3] (b) Meera invests $700 in an account paying compound interest at a rate of r % per year. At the end of 17 years the value of her investment is $1030.35 . Find the value of r. r = … [3]

6 marks

Mark scheme: 12(a) 805 3 B2 for 105 700 × 2.5 × 6 or M2 for + 700 oe 100 700 × 2.5 [× 6 ] or M1 for oe 100 12(b) 2.3[0…] 3 1030.35 M2 for 17 oe 700 or M1 for 1030.35 = 700 ( k )17 oe for any k

This question in 0580/22 Feb/March 2022

14 Carlos invests $4540 at a rate of r % per year compound interest. At the end of 10 years he has earned $1328.54 in interest. Calculate the value of r. r = … [3]

3 marks

Mark scheme: 14 2.6[0] or 2.600… 3 1328.54  4540 M2 for 10 4540 or M1 for 4540  k 10 = 1328.54 + 4540 for any k If 0 scored SC1 for answer –11.6 or –11.56…

This question in 0580/22 May/June 2022

15 The number of trees in a forest is decreasing exponentially at a rate of 1.75% per year. Eleven years ago there were 980 trees. Calculate the number of trees in the forest now. Give your answer correct to the nearest integer. … [2]

2 marks

Mark scheme: 15 807 2 11  1.75  M1 for 980   1   oe or better  100 

This question in 0580/23 May/June 2023

13 At the end of 2021 there were 27 000 rhinos living in the wild. The number of rhinos is expected to decrease exponentially by 3% each year. Work out the number of rhinos expected to be living in the wild 4 years later, at the end of 2025. Give your answer correct to the nearest whole number. … [3]

3 marks

Mark scheme: 13 23 903 cao 3 B2 for answer 23900, 23902, 23902.9… or 23 903 seen then rounded OR  3  4 M1 for 27 000 × 1 − oe    100  B1 for their more accurate value seen and correctly rounded to the nearest whole number

This question in 0580/21 Oct/Nov 2023

9 The value of a car is $8000. Each year the value of the car decreases exponentially by 25%. Calculate the value of this car after 3 years. $ … [2]

2 marks

Mark scheme: 9 3375 2 3  25  M1 for 8000  1  oe    100 

This question in 0580/21 May/June 2024

22 f ( )x = 2x + 5 g ( )x = x - 4 h ( )x = 5 x (a) Find f ( 3 ) . … [1] (b) Find f -1 ( )x . f -1 ( )x = … [2] (c) Solve fg ( )x = 25 . x = … [3] (d) Find x when h -1 ( )x = 2 . x = … [2]

8 marks

Mark scheme: 22(a) 11 1 22(b) x − 5 2 M1 for correct first step y − 5 2 x = 2y + 5 or y – 5 = 2x or = x 2 y 5 or = x + 2 2 22(c) 14 3 M2 for x – 4 = (25 – 5) ÷ 2 oe or better or 2x – 8 = 25 – 5 oe or better or M1 for 2(x – 4) + 5 = 25 22(d) 25 2 M1 for h(2) or 52

This question in 0580/23 May/June 2025