Exponential growth and decay — Paper 4 · IGCSE Mathematics - International 0607

4 (a) Marie has $260.50 and Luk has $208.40 . (i) Find, in its simplest form, the ratio Marie’s money : Luk’s money. Marie’s money : Luk’s money = … : … [2] (ii) Marie spends 16% of her money to buy a new coat. Calculate the cost of the coat. $ … [2] (iii) In a sale, the prices of all books are reduced by 10%. Luk buys a book for $11.25 . Calculate the original price of the book. $ … [3] (iv) Marie invests $200 at a rate of 2% per year simple interest. Calculate the total value of this investment at the end of 25 years. $ … [3] (v) Luk invests $190 at a rate of 2% per year compound interest. Calculate the value of this investment at the end of 25 years. $ … [3] (b) Fredrik invests $120 at a rate of 5.7% per year compound interest. Calculate the number of complete years it will take until the value of this investment is first greater than $300. … [3]

16 marks

Mark scheme: 4(a)(i) 5 : 4 2 B1 for any other correct ratio 4(a)(ii) 41.68 2 M1 for 0.16 × 260.5[0] oe 4(a)(iii) 12.5[0] 3 M2 for 11.25 ÷ 0.9 oe or M1 for recognising 11.25 as 90% 4(a)(iv) 300 nfww 3 200 × 2 × 25 M2 for + 200 oe 100 200 × 2 × 25 or M1 for oe (implied by 100 100 nfww) 4(a)(v) 311.72 3 M2 for 190 × 1.0225 oe or M1 for 190 × 1.02n oe where n > 1 4(b) 17 3 B2 for 16.5 or 16.52 to 16.53  300  log    120  or M2 for or appropriate sketch log1.057 or 120 × 1.057n = 300 and at least 2 trials which reach from 250 to 350 or M1 for 120 × 1.057n [ = 300]

This question in 0607/43 May/June 2017

5 (a) Carlos owns a vintage car. Each year the value of the car increases by 4% of its value at the start of the year. At the start of 2012 the value of the car was $17 500. Calculate the value of the car at the start of 2018. Give your answer correct to the nearest $100. $ … [4] (b) Alex invests $200 at a rate of r % per year compound interest. After 12 years, Alex has a total amount of $239.12 . Find the value of r. r = … [3]

7 marks

Mark scheme: 5(a) 22 100 4 B3 for 22 140 or 22 143. … seen and incorrectly or not rounded or M2 for 17 500 × 1.046 oe or M1 for 17 500 × 1.04k k > 1 oe 5(b) 1.5[0] or 1.499 to 1.5[00] 3 239.12 M2 for 12 200 or M1 for 200 × x12 = 239.12 or better

This question in 0607/42 Oct/Nov 2017

2 Alan, Brendan and Cieran work as gardeners. (a) The total amount of money they earn is shared in the ratio of the time each person works. One day Alan works for 2 hours 40 minutes, Brendan works for 5.5 hours and Cieran works for 200 minutes. They earn, in total, $379.50 . By changing all the times into minutes, find the amount of money each person earns. Alan $ … Brendan $ … Cieran $ … [5] (b) (i) Alan needs to buy some gardening tools. In shop A, the price of the tools is $70.20 . In shop B, the price of the tools is 5% less than in shop A. Find the price of the tools in shop B. $ … [2] (ii) The price of $70.20 is 8% higher than it was last year. Find the price last year. $ … [3] (c) (i) Brendan invests $450 for 5 years at a rate of 3.5% per year simple interest. Show that the total value of this investment after 5 years is $528.75 . [2] (ii) Cieran invests $450 for 5 years at a rate of x % compound interest. The value of Cieran’s investment after 5 years is $530.60 . Find the value of x. x = … [3]

15 marks

Mark scheme: 2(a) 88, 181.5, 110 5 B4 for any two correct or all three correct values seen OR M1 for converting times to same units e.g. 160 : 330 : 200 M2 for correct method to find any part their 160or330or200 e.g. × 379.5 oe their 690 or M1 for correct use of total 379.5 e.g. soi 0.55 their 690 A1 for any one value correct, correctly placed 2(b)(i) 66.69 2 M1 for 70.2 × 0.95 oe 2(b)(ii) 65[.00] cao 3 70.2 M2 for oe 1.08 or M1 for 70.2 = 108% soi 2(c)(i) 3.5 3.5 M1 450 × [ × 5] or 5 × [ × 450] or 100 100 better 3.5 A1 i.e. full and correct conclusion to 450 + 78.75 450 + 450 × 5 × [ = 528.75] 100 leading to 450 + 78.75 or better. 2(c)(ii) 3.35 or 3.350… 3 530.6 M2 for 5 450 or M1 for 450 × [ ]5 = 530.6 oe

This question in 0607/43 Oct/Nov 2017

2 Conrad, Delia and Eli share $8000 in the ratio Conrad : Delia : Eli = 5 : 7 : 8 . (a) Show that Eli receives $3200. [2] (b) Conrad buys a toy for $65. He sells it for $55. Calculate the percentage loss. … % [3] (c) Delia invests $2500 at a rate of 2.5% per year simple interest. Calculate the interest Delia has at the end of 8 years. $ … [2] (d) Eli invests $2400 at a rate of 2.4% per year compound interest. Calculate the interest Eli has at the end of 8 years. $ … [3] (e) Conrad buys a coat in a sale. The sale price is $79.80 after a reduction of 5%. Calculate the original price of the coat. $ … [3]

13 marks

Mark scheme: 2(a) 8000 M2 M1 for 8000 ÷ (5 + 7 + 8) × 8 [ = 3200] 5 + 7 + 8 3200 If 0 scored SC1 for × 20 = 8000 oe 8 2(b) 15.4 or 15.38… 3 65 − 55[× 100] or 55 × 100 or 1 − 55 M2 for 65 65 65 55 or M1 for 65 − 55 or 65 2(c) 500 2 2500 × 2.5 × 8 M1 for oe 100 2(d) 501.42 3 M2 for 2400 × 1.0248 oe (2901 or 2901.4[0] or 2901.42…) or M1 for 2400 × 1.024n oe where n > 1 2(e) 84 3  5  M2 for 79.80 ÷ 1 − oe    100  or M1 for recognising 79.80 is 95%

This question in 0607/41 May/June 2018

8 Every year the value of Xavier’s car decreases by 10%. The value is now $12 960. (a) Calculate the value of the car 2 years ago. $ … [2] (b) Calculate the number of complete years it will take for the value to decrease from $12 960 to less than $6480. … [3]

5 marks

Mark scheme: 8(a) 16 000 2 2  10  M1 for 12 960 ÷ 1 − oe    100  or B1 for 14400 8(b) 7 nfww 3 B2 for 6.58 or 6.578 to 6.579  6480  log    12960  or M2 for oe or appropriate log0.9 sketch or at least two trials with n > 3  10  n or M1 for 12960 × 1 − = 6480 oe    100  if 0 scored, SC1 for answer 9 nfww, coming from 16000

This question in 0607/41 May/June 2018

2 Flavia makes china cats. They each cost $22.60 to make. (a) Flavia sells some of them to Ari. She makes a profit of 35% on each cat. Calculate the price Ari pays for each cat. $ … [2] (b) Ari sells each cat for $43. Calculate Ari’s percentage profit. … % [3] (c) Jean buys 92 of Flavia’s cats. This is 15% more than the number Ari bought. Calculate the number of cats that Ari bought. … [3] (d) Jean bought the cats for $32 each. He sells some of the cats for $45 each. For the rest of the cats he reduces the price by 5% each day. Find the number of reductions he has made when the price first falls below $32. … [3]

11 marks

Mark scheme: 2(a) 30.51 2  35  M1 for 22.6 ×  1 +  oe  100  2(b) 40.9 or 40.93 to 40.94 3 43 − their 30.51 M2 for [× 100] oe their 30.51 43 or M1 for 43 – their 30.51 or their 30.51 2(c) 80 3  15  M2 for 92 ÷  1 +  oe  100  or M1 for 92 = 115% oe 2(d) 7 nfww 3 log( 3245 ) M2 for soi by 6.64 to 6.65 log0.95 or trials as far as n = 5 or M1 for 45 × 0.95n oe soi

This question in 0607/43 May/June 2018

5 The number of fish in a lake decreases by 4% each year. In January 2018 there are 30 000 fish in the lake. (a) Calculate the number of fish in the lake in (i) January 2019, … [2] (ii) January 2029, … [3] (iii) January 2017. … [3] (b) Find the last year in which there were at least 50 000 fish in the lake. … [4] (c) Philip runs a fishing business and he works 50 weeks every year. In 2018, he catches 800 kg of fish in each of these weeks. He sells all the fish he catches at a price of $3.50 for each kilogram. (i) Calculate the total amount he receives in 2018. $ … [3] (ii) For each of the 50 weeks, Philip’s business costs $2240 to run. Calculate his profit as a percentage of $2240. … % [3] (d) In 2019, Philip’s business costs 8% more to run than in 2018. The selling price of fish decreases by 10%. Find the amount of fish, in kilograms, Philip will need to catch each week to keep the percentage profit found in part (c)(ii) the same. … kg [4]

22 marks

Mark scheme: 5(a)(i) 28 800 2 100 − 4 M1 for 30000 × oe 100 5(a)(ii) 19 147 or 19 100 nfww 3 FT their 0.96, must be <1 and not 0.04 M2 for 30000 × (their 0.96)11 or 28800 × (their 0.96)10 or M1 for 30000 × ( their 0.96) k , k > 1 oe 5(a)(iii) 31 250 3 M2 for 30000 ÷their (0.96) or M1 for 30000 = their 0.96[ x ] 5(b) 2005 nfww 4 30000 M3 for n log(their 0.96) = log oe 50000 or M2 for (their 0.96) n = 0.6 oe or M1 for 50000 × (0.96) n = 30000 oe OR M3 for T and I with ‘12 and13’ seen or M2 for at least 3 correct trials or M1 for 50000 × (0.96) n = 30000 oe 5(c)(i) 140 000 3 M2 for 800 × 50 × 3.5 or M1 for multiplying any two 5(c)(ii) 25 3 their ( i ) − 2240 × M2 for 50[× 100] oe 2240 × 50 their ( i ) or × 100 oe 2240 × 50 800 × 3.5 − 2240 or [× 100 ] oe 2240 800 × 3.5 or × 100 2240 or M1 for their ( i ) − 2240 × 50 their ( i ) or 2240 × 50 or 800 × 3.5 − 2240 800 × 3.5 or 2240 5(d) 960 4 2240 × 1.08 × 1.25 M3 for oe 3.5 × 0.9 x× 3.5 × 0.9 − 2240 × 1.08 or for 2240 × 1.08 their ( c )( ii ) = oe 100 or B1 for 3.15 or 157.50 and B1 for 2419.2 or 120 960 or 3024

This question in 0607/41 Oct/Nov 2018

1 Adila has $10 000. (a) She uses some of the money to buy a car. The salesman reduces the price from $3800 to $3610. Calculate the percentage reduction. … % [3] (b) Adila invests the remaining $6390 at a rate of 3% per year compound interest. (i) Find the value of the investment at the end of 5 years. $ … [3] (ii) Find the least number of complete years after which the value of the investment is more than $9000. … [4]

10 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 5% 3 3800 − 3610 M2 for [× 100] oe 3800 3610 or × 100 3800 3610 or M1 for oe 3800 1(b)(i) 7410 or 7407 to 7408 3 3 5 M2 for 6390 × (1 + ) oe 100 3 k or M1 for 6390 × (1 + ) oe, k > 1 100 1(ii) 12 nfww 4  9000  M3 for nlog 1.03 = log   soi by 11.6  6390  or 11.58... oe or correct trials as far as 11 and 12 oe 9000 or M2 for 1.03n = 6390 or at least 3 correct trials with n ⩾ 5 or M1 for 6390 × 1.03n = 9000 soi.

This question in 0607/42 Oct/Nov 2018

1 In a sale, a shop reduces all its prices by 15%. (a) Calculate the sale price of a television originally costing $630. $ … [2] (b) The price of a fridge in the sale is $952. Calculate the original price. $ … [3] (c) After one week the shop reduces the price of the television in part (a) by a further 5% each week until it is sold. Calculate the number of weeks from the start of the sale until the television reaches half the original price. … [4]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 535.5[0] final answer 2  15  M1 for 630 ×  1 −  oe  100  1(b) $1120 3  15  M2 for 952 ÷  1 −  oe  100  or M1 for 85% associated with 952 1(c) 12 nfww 4  5   12 (630)  M3 for nlog  1 −  = log    100   their 535.50  oe soi by 10.3 or 10.4 or 10.34 to 10.36… or correct trials as far as 10 and 11 or suitable sketch(es) e.g. y = 535.5 × 0.95x and y = 315  5  n  12 (630)  or M2 for  1 −  =   oe  100   their 535.50  or at least 3 correct trials or final answer 11 nfww  5  n 1 or M1 for their 535.5 ×  1 −  = 2 (630)  100  soi oe

This question in 0607/41 May/June 2019

5 (a) Karl invests $200 at a rate of 1.5% per year simple interest. Calculate the value of Karl’s investment at the end of 8 years. $ … [3] (b) Lena invests $200 at a rate of 1.4% per year compound interest. Calculate the value of Lena’s investment at the end of 8 years. $ … [3] (c) The rates of interest remain the same as in part (a) and part (b). Find how many more complete years it will take for the value of Lena’s investment to be greater than the value of Karl’s investment. … [2]

8 marks

Mark scheme: 5(a) 224 3 200 × 1.5 × 8 M2 for 200 + oe 100 200 × 5.1 × 8 or M1 for oe implied by 24 100 5(b) 223.53 3 8 4.1  M2 for 200 × + 1  oe  100  4.1  k M1 for 200 × + 1  oe k integer > 1  100  If 0 scored, SC1 for 23.5 or 23.52 to 23.53 5(c) 3 nfww cao 2 M1 for trials with 1.5% and 1.4% beyond their 224 and their 223.53 respectively, implied by 11, or appropriate equation or graph sketch implied by 10.79..., 2.79...

This question in 0607/42 May/June 2019

7 (a) Sergio invests $2000 at a rate of 3% per year compound interest. (i) Find the value of his investment at the end of 5 years. $ … [3] (ii) After how many complete years is the value of his investment greater than $4000? … [3] (b) Anna invests $2000 at a rate of 0.24% per month compound interest. Find the value of her investment at the end of 5 years. $ … [3] (c) Calculate the monthly compound interest rate that is equal to a compound interest rate of 3% per year. … % [3]

12 marks

Mark scheme: 7(a)(i) 2318.55 3 5  3  M2 for 2000 ×  1 +   100  k  3  or M1 for 2000 ×  1 +  , k > 1  100  If 0 scored, SC1 for 318.5... or 319 or 320 7(a)(ii) 24 3 B2 for 23.4 or 23.44 to 23.45  4000  log    2000  or M2 for n = oe log1.03 or M1 for 2000 × 1.03 n = 4000 oe 7(b) 2309.37 3 60  0.24  M2 for 2000 ×  1 +   100  k  0.24  or M1 for 2000 ×  1 +  , k > 1  100  7(c) 0.247 or 0.2466… 3 3 M2 for 12 1 + implied by 1.00246[6..] 100 12 3 or M1 for x = 1 + oe 100

This question in 0607/43 May/June 2019

5 Each year the value of a motor bike decreases by 10% of its value at the start of the year. At the start of 2019, the value of the motor bike was $2025. (a) Find the value at the end of 4 years. Give your answer correct to the nearest dollar. $ … [4] (b) Find the value at the start of 2017. $ … [2] (c) Find the number of complete years it takes for the value of $2025 to decrease to a value less than $500. … [4]

10 marks

Mark scheme: 5(a) 1329 4 B3 for 1328.6 ... or 1330 or M2 for 2025 × 0.94 oe or M1 for 2025 × 0.9k , k > 1 oe 5(b) 2500 2 M1 for 2025 ÷ 0.92 oe 5(c) 14 4 B3 for 13.3 or 13.27 to 13.28 seen 500 or M3 for n log0.9 = log oe implied by or 2025 for correct trials reaching 13 and 14 or good sketch indicating value between 13 and 14 n 500 or M2 for 0.9 = oe or at least three 2025 correct trials with n > 4 or sketch that could lead to the solution or M1 for 2025 × 0.9 n = 500 oe or at least two correct trials with n > 4 If 0 scored, SC1 for answer 16 or for 15.3 or 15.27 to 15.28 seen

This question in 0607/41 Oct/Nov 2019

1 (a) Aisha invests $12 000 at a compound interest rate of 3.5% per year. Calculate the value of her investment at the end of 4 years. $ … [3] (b) 2 years ago, Byron invested $P at a compound interest rate of 3% per year. The value of his investment is now $10 078.55 . Calculate the value of P. P = … [3] (c) 5 years ago Cheng invested $Q at a simple interest rate of 4% per year. The value of his investment is now $20 400. Calculate the value of Q. Q = … [3]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 13 770.28 3 4  3.5  M2 for 12 000 ×  1 +  oe  100   3.5  k or M1 for 12 000 ×  1 +  , k > 1  100  oe 1(b) 9500 3 2  3  M2 for 10 078.55 ÷  1 +  oe  100   3  n or M1 for 10 078.55 ÷  1 +  oe  100  1(c) 17 000 3 Q× 4 × 5 M2 for Q + = 20 400 oe 100 Q× 4 × 5 or M1 for oe soi by 100 e.g. 0.2Q If 0 scored, SC1 for 16 800 or 16 760 to 16 770

This question in 0607/43 Oct/Nov 2019

7 (a) Louis invests $500 at a rate of 2.5% per year simple interest. Calculate the total amount of interest at the end of 8 years. $ … [2] (b) Martha invests $500 at a rate of 2.4% per year compound interest. Calculate the total amount of interest at the end of 8 years. $ … [4] (c) Naomi invests an amount of money at a rate of 2.1% per year compound interest. Find the number of complete years it takes for the value of Naomi’s investment to double. … [4] (d) Oscar invests an amount of money at a rate of r % per year compound interest. At the end of 31 years the value of Oscar’s investment is 2.5 times greater than the original amount of money. Find the value of r. r = … [3]

13 marks

This question in 0607/41 May/June 2020

6 Herman bought a motorbike on 1 January 2014. By 1 January 2015 the value of the motorbike had reduced by 16%. By 1 January 2016 the value of the motorbike had reduced by 12% of the value on 1 January 2015. The value of the motorbike on 1 January 2016 was $7392. (a) Find how much Herman paid for the motorbike. $ … [3] (b) From 2016, the value of the motorbike reduced by 8% each year. Calculate the number of complete years it will take for the value of the motorbike to decrease from $7392 to $5000. … [4]

7 marks

Mark scheme: 6(a) 10 000 3 7392 M2 for oe (1 − 0.16)(1 − 0.12) or M1 for ÷(1 − 0.16) or ÷ (1 − 0.12) oe or M1 for 88% is ‘equivalent’ to 7392 6(b) 5 4 5000 log 7392 M3 for [ k = ] oe log0.92 or correct trials as far as 4 and 5 k 5000 or M2 for 0.92 = oe 7392 or at least 3 correct trials or M1 for 7392 × 0.92 k = 5000 oe

This question in 0607/42 May/June 2020

3 (a) Riaz invests $5000 at a rate of 2.5% per year simple interest. (i) Calculate the value of the investment at the end of 4 years. $ … [3] (ii) Calculate the number of complete years it will take for the value of the investment to be $6500. … [2] (b) Yasmin invests $5000 at a rate of 2% per year compound interest. (i) Calculate the value of Yasmin’s investment at the end of 4 years. $ … [3] (ii) Calculate the number of complete years it will take for the value of Yasmin’s investment to first be worth more than $6500. … [4]

12 marks

Mark scheme: 3(a)(i) 5500 3 5000 × 2.5 × 4 M2 for 5000 + oe 100 5000 × 2.5 × 4 or M1 for oe 100 3(a)(i) 12 2 5000 × 2.5 × n M1 for = 6500 – 5000 100 oe 3(b)(i) 5412.16 3 4  2  M2 for 5000 ×  1 +   100   2  n or M1 for 5000 ×  1 +  , n > 1  100  3(b)(ii) 14 4  6500  log    5000  M3 for [n =] soi by 13.2 log2 or 13.24 to 13.25 or answer 13 or correct trials as far as 13 and 14  6500  or M2 for 1.02n =    5000  or at least 3 correct trials or suitable graph or M1 for 5000 × 1.02n = 6500 soi.

This question in 0607/43 May/June 2020

5 (a) Carla invests $600 at a rate of 1.8% per year compound interest. Calculate the value of Carla’s investment at the end of 7 years. $ … [3] (b) Dominic wants to invest his money so that it will double its value in 17 years. Find the lowest possible rate of compound interest per year that will give Dominic this result. Give your answer correct to 1 decimal place. … % [4] (c) Each year, the population of a village is decreasing at a rate of 4% of its value at the beginning of that year. The population is now 2120. Find the number of complete years since the population was last greater than 2700. … [4]

11 marks

Mark scheme: 5(a) 679.81 or 680 or 679.8... 3 7  1.8  M2 for 600  1 +   100   1.8  k or M1 for 600  1 +  , k > 1  100  5(b) 4.2 4 B3 for 4.16 or 4.161 to 4.162 or B2 for 17 2 oe or M1 for (P) × (...)17 = (2P) oe 5(c) 6 4 B3 for 5.92 or 5.924... OR  4   2120  M3 for n log  1 −  = log   oe  100   2700  or correct trials as far as 5 and 6 or good sketch indicating value between 5 and 6  4  n 2120 or M2 for  1 −  =  100  2700 or at least two trials with n > 2 or sketch that could lead to solution e.g. y = 0.96x  4  n or M1 for 2700  1 −  = 2120 oe  100  or at least 2 correct trials

This question in 0607/43 Oct/Nov 2020

2 (a) Write 260 512 correct to 3 significant figures. … [1] (b) Write 0.000 000 576 in standard form. … [1] (c) Calculate 27 2 - 6 # 31 0 .3 . Give your answer correct to 1 decimal place. … [2] (d) (i) Work out 37% of $820. $ … [2] (ii) Work out $36 as a percentage of $150. … % [1] (e) An amount of money is shared between Alan, Bjorn and Carlo in the ratio 3 : 7 : 5. Carlo receives $695. (i) Find the total amount of money shared. $ … [3] (ii) Carlo invests 40% of his $695 at a rate of 1.2% per year compound interest. Calculate the value of his investment at the end of 5 years. $ … [3] (f) Dana invests $2100 for 12 years at a rate of x% per year compound interest. At the end of the 12 years, the value of her investment is $2663.31 . Calculate the value of x. x = … [3]

16 marks

Mark scheme: 2(a) 261 000 1 2(b) 5.76 × 10−7 1 2(c) 26.7 2 B1 for 26.68 to 26.69 or answer 26.6 2(d)(i) 303.4[0] cao final answer 2 37 × 820 M1 for oe soi by 303 100 2(d)(ii) 24 1 2(e)(i) 2085 3 695 M1 for soi 5 M1 for (their 139) × (3 + 5 + 7) 2(e)(ii) 295.09 3 M2 for 0.4 × 695 × 1.0125 oe or M1 for 0.4 × 695 soi by 278 or A × 1.0125 2(f) 2[.00] or 1.998 to 2.001... 3 2663.31 M2 for 12 oe 2100 or M1 for 2100 × r12 = 2663.31 seen

This question in 0607/42 Feb/March 2021

6 Piero invests $5000 in Bank A and $5000 in Bank B. (a) Bank A pays simple interest at a rate of 6.5% each year. (i) Find the total amount Piero has in Bank A at the end of 4 years. $ … [3] (ii) Find the number of complete years it takes for the total amount that Piero has in Bank A to be greater than $10 000. … [3] (b) Bank B pays compound interest at a rate of 4% each year. (i) Find the total amount Piero has in Bank B at the end of 4 years. $ … [2] (ii) Find the number of complete years it takes for the total amount that Piero has in Bank B to be greater than $10 000. … [4] (c) By sketching suitable graphs, find the number of complete years it takes for the total amount that Piero has in Bank B to be greater than the total amount in Bank A. … [4]

16 marks

Mark scheme: 6(a)(i) 6300 3 M2 for 5000 + 5000 × 6.5 × 4 ÷ 100 oe or M1 for 5000 × 6.5 × 4 ÷ 100 oe implied by 1300 6(a)(ii) 16 3 B2 for 15.4 or 15.38... 5000 × 100 or M2 for oe 5000 × 6.5 5000 × 6.5 ×n or M1 for oe 100 6(b)(i) 5849.29 or 5850 2 4  4  M1 for 5000 ×  1 +  oe  100  6(b)(ii) 18 4 B3 for 17.7 or 17.67… as answer 10000  4  or M3 for log = n log  1 +  5000  100  oe or correct trials including 17 and 18 or good sketch indicating value between 17 and 18 10000  4  n or M2 for =  1 +  oe 5000  100  or at least 3 correct trials with n > 4 or sketch that could lead to solution  4  or M1 for 10000 = 5000 ×  1 +  oe  100  or at least 2 trials with n > 4 or suitable graph 6(c) Correct sketch M3 M2 for suitable graphs, e.g. y = 1.4x and y = 1 + 0.065x or M1 for one suitable graph, e.g. y = 1.04x or y = 1 + 0.0656x 24 B1

This question in 0607/41 May/June 2021

1 Ernst makes chairs. (a) The total cost of making a chair is $250. Total cost = cost of materials + $26 for each hour worked 1 Ernst works for 6 hours to make a chair. 2 Calculate the cost of the materials as a percentage of the total cost of $250. … % [3] (b) Ernst sells the chairs to a shop. The shop makes 24% profit when they sell a chair for $396.80 . Calculate the amount the shop pays Ernst for a chair. $ … [2] (c) In a sale the shop reduces the price, $396.80, of each chair by 3% each day until it is sold. Find the number of days until the price first goes below $200. … [4]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 32.4 3 250 − 6.5 × 26 M2 for [× 100 ] oe 250 6.5 × 26 or × 100 250 or M1 for 250 – 6.5 × 26 soi by 81 6.5 × 26 or 250 1(b) 320 2  24  M1 for ( ) ×  1 +  = 396.8 or  100  better 1(c) 23 4 B3 for 22.49… or 22.5 or 22 as answer or M3 for  3   200  n log  1 −  = log   oe  100   396.8  or correct trials as far as 22 and 23 or sketch indicating value between 22 and 23  3  n 200 or M2 for  1 −  = oe  100  396.8 or at least 3 correct trials or a sketch that could lead to solution e.g. y = 0.97x and y = 200  3  n or M1 for 396.8 ×  1 −  = 200  100  soi. or at least 2 correct trials

This question in 0607/42 May/June 2021

2 (a) Increase $55 by 250%. $ … [2] (b) (i) Beatrice invests $500 at a rate of 1.5% per year simple interest. Find the amount Beatrice has at the end of 12 years. $ … [3] (ii) Dan invests $500 at a rate of 1.5% per year compound interest. Find the difference between Dan’s amount and Beatrice’s amount at the end of 12 years. $ … [3] (c) Eva invests an amount of money at a rate of 2.1% per year compound interest. Find the number of complete years it takes for Eva’s investment to double in value. … [4] (d) Each year the value of Fred’s car reduces by 15% of its value at the start of that year. The value of the car is now $5158.65 . Find the value of Fred’s car 3 years ago. $ … [3]

15 marks

Mark scheme: 2(a) 192.5[0] 2 250 M1 for 55 × oe or better 100 2(b)(i) 590 3 500 × 1.5 × 12 M2 for 500 + oe 100 500 × 1.5 × 12 or M1 for 100 2(b)(ii) 7.81 3 B2 for 597.8 … or 598 seen OR  1.5 12 M2 for 500  1 +  − their (b)(i) oe  100   1.5 12 or M1 for 500  1 +  oe  100  2(c) 34 4 B3 for 33.4 or 33.35... OR  2.1  M3 for n log  1 +  = log2 oe  100  or for trials reaching 33 and 34 or good sketch indicating value between 33 and 34  2.1  n or M2 for  1 +  = 2 oe  100  or for at least 3 correct trials or for suitable graph  2.1  n or M1 for  1 +  oe soi by two trials  100  For M2 and M1 oe includes use of a sum of money 2(d) 8400 3 3  100 − 15  M2 for 5158.65 ÷   oe  100   100 − 15  n or M1 for 5158.65 ÷   , including n = 1  100 

This question in 0607/43 May/June 2021

2 (a) Find 12 kg as a percentage of 80 kg. … % [1] (b) Find 19% of $250. $ … [2] (c) Xavier invests $500 at a rate of 1.5% per year simple interest. At the end of y years, the value of Xavier’s investment is $612.50 . Find the value of y. y = … [3] (d) Each year the value of a car decreases by 12% of its value at the beginning of that year. The original value of the car is $20 000. (i) Calculate the value of the car at the end of 3 years. Give your answer correct to the nearest dollar. $ … [3] (ii) Find the number of complete years for the value of $20 000 to decrease until it is first below $1000. … [4] (e) Each year the value of another car decreases by r % of its value at the beginning of that year. At the end of 10 years, the value has decreased from $12 000 to $4673. Find the value of r. r = … [3]

16 marks

Mark scheme: 2(a) 15 1 2(b) 47.5[0] 2 19 M1 for × 250 oe 100 2(c) 15 3 500 × 1.5 × y M2 for 500 + = 612.50 oe 100 500 × 1.5 × y or M1 for oe 100 or for one year’s interest = 7.5[0] 2(d)(i) 13629 cao 3 B2 for 13630 or 13629. ...  12  3 or M1 for 20000 ×  1 −  oe  100  2(d)(ii) 24 nfww 4 B3 for 23.4 or 23.43... OR  12   1000  M3 y log  1 −  = log   oe  100   20000  or correct trials reaching 23 and 24 or good sketch indicating value between 23 and 24  12  y 1000 or M2 for  1 −  = oe  100  20000 or at least 3 correct trials or suitable graph with y > 1  12  y or M1 for 20000 ×  1 −  = 1000 oe soi by at  100  least 2 correct trials with n > 3 2(e) 9[.00] or 8.999 to 9.000... 3 4673 M2 for 10 12000 or M1 for 12 000 × ( … )10 = 4673

This question in 0607/42 Feb/March 2022

4 (a) $216 is shared in the ratio 5 : 1. Work out the larger share. $ … [2] (b) Luis shares some money between Ali, Betty and Clare in the ratio 3 : 4 : 6. Ali receives $171. Find the total amount of money Luis shared. $ … [2] (c) Farima invests $1400 in a savings account paying simple interest at a rate of 2.5% per year. Calculate the total amount in the account at the end of 3 years. $ … [3] (d) Emir invests $3000 at a rate of 2% per year compound interest. (i) Calculate the value of Emir’s investment at the end of 4 years. $ … [2] (ii) Find the number of complete years until Emir’s investment is first worth more than $4000. … [4]

13 marks

Mark scheme: 4(a) 180 2 216 M1 for [ 5] 5  1 4(b) 741 2 171 M1 for 3 4(c) 1505 3 2.5 M2 for 1400 + 3(1400 × ) oe 100 2.5 or M1 for 3(1400 × ) oe 100 4(d)(i) 3247.30 or 3250 2 4  2  M1 for 3000  1  oe    100  4(d)(ii) 15 4 B3 for 14.5 or 14.52 to 14.53 seen  4000  or M3 for n log(1.02)  log   oe  3000  or for correct trials reaching 14 and 15 or good sketch indicating value between 14 and 15 4000 or M2 for 1.02n = 3000 or at least three correct trials for n > 4 or suitable graph or M1 for 3000 × 1.02n = 4000 soi by at least 2 correct trials for n > 4

This question in 0607/41 May/June 2022

5 (a) Alenia, Bob and Cara share some money in the ratio 5 : 3 : 4 . Alenia’s share is $1240. (i) Show that Bob’s share is $744. [1] (ii) Cara spends $x from her share. The ratio of Bob’s money : Cara’s money is now 4 : 3 . Find the value of x. x = … [3] (b) A shop has a sale and all prices are reduced by 20%. (i) Bob buys a coat. The original price of the coat was $92. Work out the sale price of the coat. $ … [2] (ii) Cara buys a jacket in the sale for $132. Work out the original price of the jacket. $ … [2] (c) On 1 January 2022 Alenia buys a scooter for $1240. On 1 January 2023 the value of the scooter is reduced by 18%. On 1 January 2024 the value of the scooter is reduced by 12% of its 1 January 2023 value. (i) Calculate the value of the scooter on 1 January 2024. $ … [3] (ii) After 1 January 2024, the value of the scooter is reduced by 12% each year. Find the year in which the value of the scooter on 1 January will first be below $310. … [4]

15 marks

Mark scheme: 5(a)(i) 1240 1  3[  744] 5 5(a)(ii) 434 3 744  3 M2 for  x   992  oe 4 744  3 or M1 for oe (558) (186 × 3) or 4 3  1302 seen 7 or for 992 seen 5(b)(i) 73.6[0] 2 20  92 M1 for 92  oe 100 or B1 for 18.4 5(b)(ii) 165[.00] 2 100  20 M1 for x   132 oe or better 100 5(c)(i) 894.78 3 1240  (100  18)  (100  12) M2 for oe 100  100 1240  (100  18) or M1 for or 100 (their 1016.8)  (100  12) 100 5(c)(ii) 2033 nfww 4 B3 for 9.3 or 9.288 to 9.29… OR 8.3 or 8.29 as final value 310 or M3 for n log0.88  log1240  0.82 or 310 n log0.88  log their 894.78 or correct trials reaching 9 and 10 or 8 and 9 or good sketch indicating value between 9 and 10 or between 8 and 9 n 310 or M2 for 0.88  or 1240  0.82 n 310 0.88  their 894.78 or at least 3 correct trials or sketch that could lead to solution e.g. y = 0.88x and y = 0.3 or M1 for 1240  0.82  0.88n = 310 or their 894.78  0.88n = 310 or at least 2 trials or suitable graph e.g. y = 0.88x

This question in 0607/42 May/June 2022

1 (a) Anneka invests $2500 in an account paying compound interest at a rate of 1.6% per year. Find the amount in the account at the end of 3 years. $ … [2] (b) Bashir invests $2500 in an account paying simple interest at a rate of r% per year. At the end of 5 years the amount in the account is $2718.75 . Calculate the value of r. r = … [3] (c) Chanda invests $2500 in an account paying compound interest at a rate of 1.55% per year. Find the number of complete years until Chanda’s investment is first worth more than $4000. … [4]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 2621.93 2 3  1.6  M1 for 2500 × 1  oe    100  1(b) 1.75 3 r M2 for 2500 × × 5 = 2718.75 – 2500 oe 100 r 2718.75  2500 or M1 for 2500 × × 5 or 100 5 (43.75) or 8.75 seen 1(c) 31 4 B3 for 30.55 to 30.56 OR  1.55   4000  M3 for n log  1   = log   oe  100   2500  or correct trials as far as 30 and 31 or good sketch indicating value between 30 and 31  1.55  n 4000 or M2 for 1  = oe    100  2500 or at least 3 correct trials or sketch that could lead to solution 4000 e.g. y = 1.0155x and y = 2500  1.55  n or M1 for 2500 × 1  = 4000 soi.    100  or at least 2 correct trials

This question in 0607/43 May/June 2022

5 (a) Kris and Laila share $200 in the ratio 2 : 3. (i) Show that Kris receives $80. [1] (ii) Kris spends 30.8% of his $80 on a book. Calculate the cost of the book. $ … [2] (iii) Laila invests her $120 at a rate of 1.16% per year simple interest. Calculate the total amount Laila has at the end of 5 years. $ … [3] (b) On 1 January 2020, Sangita invests an amount of money at a rate of 2% per year compound interest. On 1 January 2023 the value of the investment is $5306.04 . (i) Calculate the amount Sangita invested on 1 January 2020. $ … [2] (ii) Calculate the value of the investment on 1 January 2025. $ … [2] (c) Tomas invests an amount of money at a rate of 1.4% per year compound interest. Find the number of complete years it takes for the value of his investment to increase by 50%. … [4]

14 marks

Mark scheme: 5(a)(i) 2 1 200 2 + 3 5(a)(ii) 24.64 cao 2 30.8 M1 for  80 oe 100 5(a)(iii) 126.96 cao 3 120  1.16  5 M2 for 120 + oe 100 120  1.16  5 or M1 for 100 5(b)(i) 5000 cao 2 3  2  M1 for X  1 + = 5306.04 oe    100  5(b)(ii) 5520.4[0] nfww 2 2  2  M1 for 5306.04  1 + oe    100   2  5 or for their 5000  1 +    100  5(c) 30 nfww 4 B3 for 29.9 or 29.16… OR  1.4   150  or M3 for n log  1 +  = log    100   100  oe or good sketch indicating value between 29 and 30 or correct trials reaching 20 and 21  1.4  n 150 or M2 for 1 + = oe    100  100 or suitable graph with n > 1 or at least 3 correct trials or M1 for  1.4  150    1 +  n =   oe soi  100  100 by at least 2 trials with n > 1

This question in 0607/43 Oct/Nov 2022

9 Henryk invests $5000 in Bank A and $5000 in Bank B. (a) Bank A pays compound interest at a rate of 3.5% each year. (i) Find the total amount Henryk has in Bank A at the end of 4 years. $ … [2] (ii) Calculate the number of complete years it takes for the value of Henryk’s investment of $5000 in Bank A to be first greater than $8000. … [4] (b) Bank B pays simple interest at a rate of 4% each year. (i) Find the total amount Henryk has in Bank B at the end of 4 years. $ … [3] (ii) Calculate the number of complete years it takes for the value of Henryk’s investment of $5000 in Bank B to be $8000. … [2] (c) At the end of x complete years, the total amount that Henryk has in Bank A is greater than the total amount he has in Bank B. Given that 5 1 x 1 10 , use a graphical method to find the value of x. x = … [4]

15 marks

Mark scheme: 9(a)(i) 5737.62 2 M1 for 5000 × 1.0354 oe 9(a)(ii) 14 4 B3 for 13.6 to 13.7 OR 8000 M3 for n log1.035 = log oe 5000 or good sketch indicating value between 13 and 14 or correct trials reaching 13 and 14 n 8000 or M2 for 1.035 = oe 5000 or exponential sketch or at least 3 correct trials with n > 4 or M1 for 5000 × 1.035n = 8000 oe or at least 2 correct trials If 0 scored, SC3 for answer 2 coming from use of 1.35 9(b)(i) 5800 3 5000 4 4 M2 for 5000 + oe 100 5000 4 4 or M1 for oe 100 9(b)(ii) 15 2 5000 4 n M1 for 5000 + = 8000 oe 100 9(c) 9 4 B3 for 8.556… or 8.56 OR M1 for 5000  1.035 n = 5000(1 + 0.04 n ) oe soi M1 for sketch of 1.035n M1 for sketch of 1 + 0.04n

This question in 0607/42 Feb/March 2023

4 (a) Alex invests $650 at a rate of 2% per year compound interest. (i) Calculate the value of this investment at the end of 10 years. $ … [2] (ii) Calculate the number of complete years it takes for the value of this investment of $650 to be first greater than $1000. … [4] (b) 2 years ago Chris invested $x at a rate of 3% per year compound interest. The value of this investment is now $607.90 correct to the nearest cent. Calculate the value of x. x = … [2] (c) Sam invested $200 at a rate of r % per year compound interest. At the end of 18 years, the value of this investment is $247.90 correct to the nearest cent. Find the value of r. r = … [3]

11 marks

Mark scheme: 4(a)(i) 792 or 792.35 2 10  2  M1 for 650   1    100  4(a)(ii) 22 4 B3 for 21.8 or 21.75... OR M3 for  2   1000  n log  1    log   oe  100   650  or sketch indicating value between 21 and 22 or correct trials as far as 21 and 22  2  n 1000 or M2 for  1   = oe  100  650 or graph which could lead to solution, 1000 e.g. y  1.02x and y = 650 or at least 3 correct trials with n > 10  2  n or M1 for 650  1   = 1000 oe  100  or at least 2 correct trials 4(b) 573 2 2  3  M1 for (...)   1   = 607.90 oe  100  4(c) 1.2[0] or 1.199 to 1.200 3 247.9 M2 for 18 200 or M1 for 200  […]18  247.90 oe or better

This question in 0607/41 May/June 2023

2 The population of a species of bird is estimated to be decreasing by 4% per year. At the end of 2020 the population was 4.32 million. (a) Find the population at the end of 2019. … million [2] (b) Calculate an estimate for the population at the end of 2025. … million [2] (c) Find the year in which the population is first expected to be below 2 million. … [4]

8 marks

Mark scheme: 2(a) 4.5 nfww 2 4  M1 for P  1    4.32 oe or better  100  2(b) 3.52 or 3.522... nfww 2 5  4  M1 for 4.32 1  oe    100  2(c) 2039 nfww 4 B3 for answer 18.9 or 18.86 to 18.87 or 19 nfww OR  4   2  M3 for n log  1   = log   oe  100   4.32  or good sketch indicating value between 18 and 19 or correct trials as far as 18 and 19  4  n 2 or M2 for 1  = oe    100  4.32 or sketch that could lead to solution or at least 3 correct trials  4  n or M1 for 4.32 × 1  = 2 soi.    100  or at least 2 correct trials

This question in 0607/42 May/June 2023

4 (a) $x is divided in the ratio 3 : 5. The larger share is $42. Find the value of x. x = … [2] (b) (i) Increase 124 by 16%. … [2] 2 (ii) The price of a coat is reduced by in a sale. 9 The new price of the coat is $73.50 . Find the original price of the coat. $ … [2] (c) Xiong invests $2000 in Bank A which pays simple interest at a rate of 3% each year. Find the total amount of interest Xiong receives at the end of 5 years. $ … [2] (d) Wendi invests $400 in Bank B which pays compound interest at a rate of 1.6% each year. Find the total amount of interest Wendi receives at the end of 3 years. $ … [3] (e) Pedro invests $1000 in Bank C for 18 years. Pedro also invests $1000 in Bank D for 18 years. Bank C pays simple interest at a rate of x % each year. Bank D pays compound interest at a rate of 0.7x % each year. At the end of 18 years Pedro has exactly the same amount of money in Bank C and Bank D. 18 18x 0 .7 x (i) Show that 1 + = 1 + . 100 e 100 o [2] (ii) Given that 5 1 x 1 7 , use a graphical method to find x. x 5 7 x = … [3]

16 marks

Mark scheme: 4(a) 67.2[0] 2 M1 for 42  5  k  where k is 8 or 3 or 1 4(b)(i) 143.84 cao 2 16 M1 for 124  (1  ) oe 100 or B1 for 19.84 or 143.84 seen 4(b)(ii) 94.5 [0] 2  2  M1 for  1   x  73.5 oe  9  4(c) 300 2 M1 for 2000  [0].03  5 oe 4(d) 19.51 3 3  1.6  M2 for 400  1    400 oe  100   1.6 3 or M1 for 400  1   oe  100  4(e)(i) 1000  x  18 M1 1000  oe 100 18 M1  0.7 x  1000  1   oe  100  4(e)(ii) 5.76 or 5.756... 3  18 x  M2 for sketch of [ y ]  1   and  100   0.7 x 18 [ y  ]  1   oe  100  with distinct curve and line with clear point of intersection  18 x  or M1 for sketch of [ y ]  1   or  100   0.7 x 18 [ y  ]  1   oe  100 

This question in 0607/43 May/June 2023

1 (a) Find $2.40 as a percentage of $1.60 . … % [1] (b) Calculate 7.2% of 2.5 g. … g [2] (c) Amir invests $400 at a rate of 1.8% per year compound interest. Calculate the value of this investment at the end of 6 years. $ … [2] (d) Each year the population of a small town increases by 4% of its value in the previous year. The population is now 29 640. (i) Calculate the population last year. … [2] (ii) Calculate the number of complete years it will take for the population of 29 640 to be first greater than 40 000. … years [4]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 150 1 1(b) [0].18 2 7.2 M1 for  2.5 oe 100 1(c) 445.19 or 445 or 445.19... 2 6  100 + 1.8  M1 for 400   oe  100  1(d)(i) 28 500 2  100 + 4  M1 for [...]    = 29640 oe  100  1(d)(ii) 8 nfww 4 B3 for 7.64 or 7.642 to 7.643 or  4   40000  M3 n log  1 +  = log   oe  100   29640  or for good sketch indicating value between 7 and 8 or for correct trials reaching 7 and 8 or  4  n 40000 M2 for 1 + = oe    100  29640 or suitable graph with n > 1 or at least 3 correct trials or  4  n M1 for 29640  1 + = 40000 oe    100  soi by at least 2 trials with n > 1

This question in 0607/41 Oct/Nov 2023

6 (a) Jade and Kim share $160. Jade receives $8 more than Kim. Find the ratio Jade’s money : Kim’s money. Give your answer in its simplest form. … : … [2] (b) Each year the height of a bush increases by x% of its height at the start of the year. It takes 6 years for the bush to grow from 1.2 m to 1.664 m. Find the value of x. x = … [3] (c) Work out, giving each answer in standard form. (i) ( 4. 5 # 10 85 ) # ( 3 # 10 36 ) … [2] (ii) ( 2 # 10 n ) + ( 2 # 10 n - 2 ) … [2]

9 marks

Mark scheme: 6(a) 21 : 19 cao 2 B1 for answer 84 : 76 or 1 1 M1 for (160) + 4 or (160) − 4 2 2 If 0 scored, SC1 for final answer of 19 : 21 6(b) 5.6[0] or 5.599 to 5.600 3 1.664 M2 for 6 oe 1.2 or M1 for 1.2 × [...]6 = 1.664 6(c)(i) 1.35 × 10122 cao 2 B1 for 13.5 × 10 121 oe seen 6(c)(ii) 2.02 × 10n cao 2 B1 for figs 202

This question in 0607/41 Oct/Nov 2023

5 A museum records the value of a picture every 5 years. The picture increases in value by 60% every 5 years. The value the museum recorded in 2020 was $20 000. (a) Calculate the value recorded in 2015. $ … [2] (b) Show that the value recorded in 2040 will be $131 072 . [1] (c) Calculate the year in which the value recorded will first be over $1 000 000. … [4]

7 marks

Mark scheme: 5(a) 12 500 2  60  M1 for P  1 +  = 20 000 oe  100  5(b) 4 1  60  20 000 × 1 + oe    100  [= 131 072] 5(c) 2065 nfww 4 B3 for 9 or 8.32 or 8.323... or for 45 or 41 to 42 OR  60   1000000  M3 for n log  1 +  = log    100   20000  oe or good sketch indicating value between 8 and 9 or correct trials as far as 8 and 9  60  n 1000000 or M2 for 1 + = oe    100  20000 or sketch that could lead to solution 1000000 e.g. y = 1.6x and y = 20000 or at least 3 correct trials or M1 for  60  n 20 000 × 1 + = 1 000 000 soi    100  or at least 2 correct trials

This question in 0607/43 Oct/Nov 2023

4 (a) The price of a coat is $84. The price is reduced by 12%. Find the new price of the coat. $ … [2] (b) The price of a table is reduced by 25%. The price is now $960. Find the original price of the table. $ … [2] (c) Samir invests $600 in a bank that pays compound interest at a rate of 5.1% each year. (i) Find the value of Samir’s investment after 4 complete years. $ … [2] (ii) Find the number of complete years for the value of Samir’s investment to be first worth more than $1000. … [4] (d) Amir and Bob work together and share their earnings in the ratio 3 : 5. (i) Find the amount Bob receives when their earnings are $120. $ … [2] (ii) They decide to change the ratio for all further earnings. Amir’s share of the earnings is increased by 20% of his original share. Bob’s share of the earnings is decreased by 20% of his original share. Show that the ratio of their earnings is now 9 : 10. [3]

15 marks

Mark scheme: 4(a) 73.92 2 (100  12) M1 for 84  soi 100 or B1 for 10.08 4(b) 1280 2 100  25 M1 for x   960 oe 100 4(c)(i) 732.09 2 M1 for 600  1.0514 oe 4(c)(ii) 11 4 B3 for 10.3 or 10.26 to 10.27 1000 or M3 for n log1.051  log 600 or good sketch indicating value between 10 and 11 or correct trials reaching 10 and 11 n 1000 or M2 for 1.051  oe 600 or sketch that could lead to solution e.g. y  1.051x , y  1.67 or at least 3 correct trials with n > 4 or M1 for 600  1.051n  1000 oe or suitable graph e.g. y  1.051x or at least 2 trials with n > 4 4(d)(i) 75 2 5 M1 for [120  ]5  3 oe 4(d)(ii) 3[ k ](1  0.2) : 5[ k ](1  0.2) M2 M1 for 5[ k ](1  0.2) or 3[ k ](1  0.2) or for their (d)(i)(1 – 0.2) Leading to 9:10 A1 No errors seen

This question in 0607/41 May/June 2024

6 Xavier started a new job in 2000. His annual pay increases each year by 2.5% of his pay in the previous year. (a) Calculate the number of complete years it took for Xavier’s annual pay to be 30% greater than his annual pay in 2000. … [4] (b) In 2024 Xavier’s annual pay is $25 215. Calculate the amount Xavier’s pay will increase from his annual pay in 2022 to his annual pay in 2027. Give your answer correct to the nearest dollar. $ … [4]

8 marks

Mark scheme: 6(a) 11 cao 4 B3 for 10.6 or 10.62 to 10.63 OR  2.5   130  or M3 for n log  1    log   oe  100   100  or good sketch indicating value between 10 and 11 or correct trials reaching 10 and 11  2.5  n 130 or M2 for  1    oe  100  100 or suitable graph with n > 1 or at least 3 correct trials  2.5  n 130 or M1 for [...]   1     [...] oe soi by  100  100 at least 2 trials with n > 1 6(b) 3154 cao 4 B3 for 3153.7 to 3153.8 OR 3 25215 M3 for 25215  1.025  oe 1.025 2 OR  2.5  2 M1 for X   1    25215  100   2.5  3 M1 for 25215   1    100   2.5  5 or for their 24000   1   oe provided their  100  2027 amount is greater than 25 215

This question in 0607/42 May/June 2024

2 (a) Ameera and Bertrand share some money in the ratio 4 : 5. Bertrand gets $3000. Calculate Ameera’s share. $ … [2] (b) Bertrand invests $3000 at a rate of r% per year simple interest. At the end of 10 years the value of the investment is $3840. Find the value of r. r = … [3] (c) Claudia invests $6000 at a rate of s% per year compound interest. At the end of 8 years the value of the investment is $7367.67 . Find the value of s. s = … [3] (d) Dieter invests $4000 at a rate of 1.8% per year compound interest. At the end of n complete years the value of the investment is more than $6000. Calculate the smallest value of n. n = … [4]

12 marks

Mark scheme: 2(a) 2400 2 3000 M1 for 5 2(b) 2.8 3 3000 r 10 M2 for = 840 oe 100 3000 r 10 or M1 for 100 or B1 for [1 year interest] = 84 2(c) 2.6[0] 3 7367.67 M2 for 8 oe 6000 or M1 for 6000  … 8  7367.67 2(d) 23 cao 4 B3 for 22.7 or 22.72 to 22.73 OR  1.8   6000  M3 n log  1    log   oe  100   4000  or good sketch indicating value between 22 and 23 or correct trials reaching 22 and 23  1.8  n 6000 or M2 for  1    oe  100  4000 or suitable graph with n > 1 or at least 3 correct trials  1.8  n or M1 for 4000   1    6000 oe soi by at least  100  2 correct trials with n > 1

This question in 0607/43 May/June 2024

4 (a) Alan, Beth and Imran share an amount of money in the ratio 3x : 2x : ( x + 1) where x is an integer. (i) Find the amount Beth receives when x = 4 and they share $400 in total. $ … [3] (ii) Find the amount that Alan receives when Beth receives $66. $ … [2] (iii) Find the value of x when Alan receives 2.5 times the amount Imran receives. x = … [2] (b) In a sale, a shop reduces the price of all furniture by 12%. (i) Find the sale price of a chair that has an original price of $90. $ … [2] (ii) Find the original price of a table that has a sale price of $440. $ … [2] (c) Kurt invests $X in a bank which pays simple interest at a rate of 4% each year. The total amount of money that Kurt has in the bank at the end of 6 years is $930. Show that X = 750 . [2] (d) Ivana invests $750 in a bank which pays compound interest at a rate of y % each year. The total amount of money that Ivana has in the bank at the end of 6 years is $921.94 . Find the value of y. y = … [3]

16 marks

Mark scheme: 4(a)(i) 128 3 8 M2 for  400  12 + 8 + 5 or M1 for 12, 8, 5 or 25 4(a)(ii) 99 2 3 M1 for  66 oe 2 4(a)(iii) 5 2 M1 for 3 x = 2.5( x + 1) oe 4(b)(i) 79.2 [0] 2 100 − 12 M1 for 90  oe 100 or B1 for 10.8 4(b)(ii) 500 2 100 − 12 M1 for x  = 440 oe 100 4(c) 24 X M1 930 = X + oe 100 930 A1 X = oe 1.24 4(d) 3.5 or 3.4999.. 3 921.94 M2 for 6 750 or M1 for 750  ( k ) 6 = 921.94 oe

This question in 0607/41 Oct/Nov 2024

2 (a) Work out 24% of $15.50 . $ … [2] (b) The price of a bookcase is $123. This price is increased by 7%. Calculate the new price. $ … [2] (c) An amount of money is shared between Ali, Kat and Lena in the ratio 5 : 3 : 4. Lena’s share is $76. Work out the total amount of money. $ … [3] (d) A library has 32 800 books. Each year the number of books in the library increases by 300. Calculate the number of years it takes until there are 40 000 books in the library. … [2] (e) A different library has 32 695 books at the end of 2024. Each year the number of books increases by 0.6% of the number of books in the library at the end of the previous year. (i) Calculate the number of books the library had at the end of 2023. … [2] (ii) Calculate the number of complete years from 2024 that it takes for the number of books to first be greater than 40 000. … [4]

15 marks

Mark scheme: 2(a) 3.72 2 24 M1 for  15.5 oe 100 2(b) 131.61 final answer 2 100 + 7 M1 for  123 oe 100 or B1 for 8.61 2(c) 228 3 76 M2 for  ( 5 + 3 + 4 ) oe 4 76 or M1 for 4 2(d) 24 2 M1 for 300 +x 32 800 = 40 000 oe 2(e)(i) 32500 2  100 + 0.6 M1 for [...]  = 32 695 oe  100  2(e)(ii) 34 nfww 4 B3 for 33.71… or 33.7 OR 0.6 40 000 M3 for n log(1 + ) = log oe 100 32 695 or a good sketch indicating value between 33 and 34 or correct trials reaching 33 and 34  0.6  n 40 000 or M2 for 1 + = oe    100  32 695 or suitable graph or at least three correct trials  0.6  n or M1 for 32 695 1 + = 40 000 oe soi    100  or at least 2 trials with n > 1

This question in 0607/43 Oct/Nov 2024

11 Vikram invests $450 at a rate of 3.1% per year compound interest. Calculate the number of complete years that it takes for the value of Vikram’s investment to first be greater than $900. … [4]

4 marks

Mark scheme: 11 23 4 B3 for 22.704… or 22.7 OR  3.1  900 M3 for n log 1 + = log oe    100  450 or a good sketch indicating value between 22 and 23 or correct trials reaching 22 and 23  3.1  n 900 or M2 for 1 + = oe    100  450 or suitable graph or at least three correct trials  3.1  n or M1 for 450 1 + = 900 oe soi    100  or at least 2 trials with n > 1

This question in 0607/42 Feb/March 2025

9 The value of a car depreciates exponentially by 15% each year. On 1 January 2023 its value was $20 400. (a) Find the value of the car on (i) 1 January 2022 $ … [2] (ii) 1 January 2025. $ … [2] (b) Find the year in which the value of the car on 1 January will first be less than $5000. … [4]

8 marks

Mark scheme: 9(a)(i) 24 000 2  15  M1 for P ×  1 −  = 20 400 oe or better  100  9(a)(ii) 14 739 2 2  15  M1 for 20 400 × 1 − oe    100  9(b) 2032 4 B3 for 8.65 or 8.651 to 8.652 (20 400) or 9.65 or 9.651 to 9.652 (24 000) or 6.65 or 6.651 to 6.652 (14 739) OR  15   5000  M3 for n log  1 −  = log   oe  100   P  where P = 20 400 or their 24 000 or their 14 739 or good sketch indicating value between 8 and 9 (20 400) or 9 and 10 (their 24 000) or 6 and 7 (their 14 739) or correct trials reaching 8 and 9 (20 400) or 9 and 10 (their 24 000) or 6 and 7 (their 14 739)  15  n 5000 or M2 for  1 −  = oe, same values  100  P of P or suitable graph with n > 1 or at least 3 correct trials  15  n or M1 for P   1 −  = 5000 oe, same  100  values of P soi by at least 2 correct trials with n > 1

This question in 0607/41 May/June 2025

12 Paula bought a house on 1 January 2023. On 1 January 2024 the value of the house increased by 10%. On 1 January 2025 the value of the house increased by 6% of its value on 1 January 2024. The value of the house on 1 January 2025 was $215 710. (a) Calculate the amount Paula paid for the house in 2023. $ … [3] (b) From 2025 the value of the house increases exponentially at the rate of 6% each year. The value of the house on 1 January 2025 was $215 710. Find the year in which the value of the house on 1 January will first be greater than $400 000. … [4]

7 marks

Mark scheme: 12(a) 185 000 3 215710 M2 for oe or better (1 + 0.1)(1 + 0.06) or M1 for A (1 + 0.1) = 215710 or A (1 + 0.06) = 215710 oe 12(b) 2036 4 B3 for 11 or 10.59 to 10.6 OR 400000 log 215710  400000  M3 for [ n = ] or log1.06   log1.06  215710  or suitable graph showing correct solution or correct trials as far as 10 and 11 n 400000 or M2 for 1.06 = oe 215710 or graph of y = 1.06 n and y = k or at least 3 correct trials or M1 for 215710  1.06 n = 400000 oe or graph of y = 1.06 n

This question in 0607/42 May/June 2025

18 (a) Bruce buys a new car on 1 January 2022. On 1 January 2023 the value of the car has decreased by 20%. On 1 January 2024 the value of the car has decreased by 15% of its value on 1 January 2023. Find the overall percentage decrease in the value of the car on 1 January 2024. … % [2] (b) Sangita buys a car with a value of $20 000. The value of the car decreases exponentially at a rate of 5% per year. Calculate the number of complete years it will take for the value of the car to decrease from $20 000 to $8000. … [4] (c) Sunil buys a car. The value of the car decreases exponentially at a rate of 6% per year. At the end of 9 years the value of the car is $8022. Calculate the original value of the car. $ … [2]

8 marks

Mark scheme: 18(a) 32 2  20  15  M1 for  1 −  1 −   100  100  18(b) 18 4 B3 for 17.9 or 17.86... OR  5   8000  M3 for n log  1 −  = log   oe  100   20000  or good sketch indicating value between 17 and 18 or correct trials reaching 17 and 18  5  n 8000 or M2 for 1 − = oe    100  20000 or suitable graph with n > 1 or at least 3 correct trials, n > 1  5  n or M1 for 20000  1 − = 8000 oe soi by    100  at least 2 trials with n > 1 18(c) 14 000 2 9  6  M1 for [...] ×  1 −  = 8022 oe  100 

This question in 0607/43 May/June 2025

6 (a) Jeanne invests $2000 at a rate of 3.5% per year simple interest. Calculate the interest earned at the end of 8 years. $ … [2] (b) Karim invests $2000 at a rate of 2.8% per year compound interest. Calculate the interest earned at the end of 10 years. $ … [3] (c) Lauren invests $2000 at a rate of 3.1% per year compound interest. At the end of n years, the value of the investment is $3066.56 correct to the nearest cent. Calculate the value of n. n = … [4]

9 marks

Mark scheme: 6(a) 560 2 2000  3.5  8 M1 for 100 6(b) 636 or 636.1 or 636.095 to 636.096 3 10  2.8  M2 for 2000 1 + − 2000 oe    100   2.8 10 or M1 for 2000  1 +  oe  100  6(c) 14 nfww 4 3066.56 M3 for n log1.031 = log oe 2000 or good sketch indicating value between 14 and 15 or correct trials reaching 14 n 3066.56 or M2 for 1.031 = oe 2000 or exponential sketch or at least 3 correct trials with n > 4 or M1 for 2000 × 1.031n = 3066.56 oe or at least 2 correct trials

This question in 0607/41 Oct/Nov 2025

15 Erik invests $x. (a) He receives compound interest at a rate of 7.5% each year. (i) At the end of 5 years, the value of Erik’s investment is $11 485. Show that x = 8000 correct to the nearest dollar. [3] (ii) Find the number of complete years it takes for the total value of his investment of $8000 to be first greater than $16 000. … [4] (b) The compound interest rate of 7.5% each year is equivalent to a compound interest rate of y % each month. Find the value of y. y = … [3]

10 marks

Mark scheme: 15(a)(i) 11485 M2 M1 for x  1.0755 = 11485 oe x = oe 1.0755 7999.9[…] A1 15(a)(ii) 10 nfww 4 B3 for answers which round to 9.58 OR log2 M3 for [ n = ]log1.075 oe or suitable graph showing correct soln or correct trials as far as 9 and 10 n 16000 or M2 for 1.075 = oe 8000 or graph of y = 1.075 n and y = k or at least 3 correct trials (n > 5) or M1 for 8000  1.075n = 16000 oe or graph of y = 1.075 n or at least 2 correct trials (n > 5) 15(b) 0.604 or 0.6044 to 0.6045 3 M2 for 121.075 oe or M1 for [...]12 = 1.075 oe

This question in 0607/42 Oct/Nov 2025