Money — Paper 4 · IGCSE Mathematics - International 0607

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

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

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

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 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

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

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

3 Alana, Bev and Cara work as decorators. (a) The total amount of money they earn is shared in the ratio of the time each person works. During one week Alana works for 16 hours 40 minutes, Bev works for 30 hours and Cara works for 1200 minutes. They earn a total of $1680.80 . Change all the times into minutes and find the amount of money each person earns. Alana $ … Bev $ … Cara $ … [4] (b) (i) Alana pays a weekly rent of $255 for her apartment. The price of this weekly rent is 2% higher than one year ago. Find the price of her weekly rent one year ago. $ … [2] (ii) Alana can pay one full year’s rent in advance. One year is 52 weeks. She will receive a discount of 3% on each weekly rent of $255. Calculate the cost of paying the full year’s rent in advance. $ … [2] (c) One week Bev earns $x. 1 2 She spends of these earnings on rent and on food. 4 9 1 She spends of the remaining money on clothes and saves the rest. 3 She saves $152. Find the value of x. x = … [3] (d) Cara invests $500 for 5 years at a rate of y % per year simple interest. The value of Cara’s investment at the end of 5 years is $530.75 . Find the value of y. y = … [3]

14 marks

Mark scheme: 3(a) 420.2[0] 4 B3 for one correct answer 756.36 or M2 for 1000 + 1800 + 1200 oe 504.24 or B1 for 1000 or 1800 seen 3(b)(i) 250 nfww 2  2  M1 for x   1 +  = 255 oe  100  3(b)(ii) 12862.2[0] 2 3 M1 for 255  52 − (255  52) oe 100 or B1 for 247.35 or 397.8 3(c) 432[.00] 3 19 35 B2 for oe or oe or 235 or 234.5... 54 54  1 2  or M1 for  1 − −  oe or better  4 9  3(d) 1.23 3 500  y  5 M2 for 500 + = 530.75 oe or better 100 500  y  5 or M1 for seen 100 or B1 for 6.15 If 0 scored, SC1 for 1.2[00] to 1.201 from compound interest used

This question in 0607/41 Oct/Nov 2022

3 (a) Sachin earns $63 000 per year before paying tax. He pays tax on his earnings at a rate of 16%. Calculate the amount Sachin has after paying tax. $ … [2] (b) Britte has $60 480 per year after paying tax at a rate of 16%. Calculate the amount that Britte earns before paying tax. $ … [2] (c) (i) Sachin opens a savings account with $1500 on 1 January. The account pays 1.8% per year simple interest. Show that the amount in Sachin’s account at the end of 3 years is $1581. [2] (ii) Britte also opens a savings account on 1 January. The account pays 2% per year compound interest. Britte pays $500 into her account on 1 January every year. Find who has the greater amount in their account at the end of 3 years. Give the difference correct to the nearest cent. … by $ … [4]

10 marks

Mark scheme: 3(a) 52 920 2 16 M1 for 63 000 × or better 100 3(b) 72 000 2  16  M1 for A   1 −  = 60 480 oe or  100  better 3(c)(i) 1.8 2 1.8 1500 + 1500 × 3 × oe M1 for 1500 3 100 100 3(c)(ii) Sachin 20.2[0] 4 B3 for 1560.8... or 20.2 or 20.19 to 20.20  2 3 or M2 for 500 1 + +    100   2  2  2  500  1 +  + 500  1 +  oe  100   100   2 3 or M1 for 500 1 + or    100   2  2 500 1 + oe    100 

This question in 0607/42 Oct/Nov 2022

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