Compound Interest Practice Problems (99-07)

IB Math Studies 1: Algebra & Numbers: Compound Interest Alei - Desert Academy 2013-14 Compound Interest Practice Problems (99-07) 1. Two brothers Adam and Ben each inherit $6500. Adam invests his money in a bond that pays simple interest at a rate of 5 per annum. Ben invests his money in a bank that pays compound interest at a rate of 4.5 per annum. (a) Calculate the value of Adam s investment at the end of 6 years. Calculate the value of Ben s investment at the end of 6 years. Give your answer correct to 2 decimal places. (3) (Total 6 marks) 2. Alex invests 3600 euros in an account that pays a nominal rate of 5.4 interest per year, compounding monthly. The interest is added to the account at the end of each month. (a) Calculate the number of whole months it will take for Alex s investment to double. (i) Calculate the value of Alex s investment after nine years. (ii) Find the rate of simple interest per year that would give the same value for the investment after nine years. (6) (Total 10 marks) 3. Andrew invests 20 000 Swiss francs in a bank that offers a 2 simple interest per year for 8 years. (a) Find the interest he has after these 8 years. Philip invests 20 000 Swiss francs for 6 years in a bank at a nominal rate of 5 interest compounded quarterly. Find the total amount in Philip s account after these 6 years. (Total 6 marks) 4. Bob invests 600 EUR in a bank that offers a rate of 2.75% compounded annually. The interest is added on at the end of each year. (a) Calculate how much money Bob has in the bank after 4 years. Calculate the number of years it will take for the investment to double. Ann invests 600 EUR in another bank that offers interest compounded annually. Her investment doubles in 20 years. (c) Find the rate that the bank is offering. (Total 6 marks) 5. William invests $1200 for 5 years at a rate of 3.75% compounded annually. (a) Calculate the amount of money he has in total at the end of the 5 years. The interest rate then drops to 3.25%. If he continues to leave his money in the bank find how much it will be worth after a further 3 years. (Total 6 marks) (3) (4) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 1 of 5

IB Math Studies 1: Algebra & Numbers: Compound Interest Alei - Desert Academy 2013-14 6. (i) Celia has $20,000 to invest. There are two different options from which she can choose. Option 1: The investment grows at a rate of 3.5% compound interest each year. Option 2: The total value of the investment increases by $800 each year. The money is to be invested for 15 years. (a) Complete the table below giving the values of the investments to the nearest dollar for the first 4 years. (c) Year 0 1 2 3 4 Option 1 20 000 20 700 Option 2 20 000 20 800 Calculate the values of each investment at the end of 15 years. If Option 1 is chosen find the total number of complete years before the value of the investment is first greater than $25,000. (d) If Option 2 is chosen calculate the percentage increase in the investment for the final year. (ii) Two more Options are available to Celia. After 7 years she can change the investment conditions. Option 3: If Celia has chosen Option 1 she can change and then receives $800 each year until the end of the 15 years. Option 4: If Celia has chosen Option 2 she can change and then receive 3.5% interest compounded annually. If Celia wishes to receive the maximum amount of money at the end of the 15 years which option should she choose? (7) (Total 18 marks) 7. A family in Malaysia received a gift of $4000 AUD from a cousin living in Australia. The money was converted to Malaysian Ringgit. One Ringgit can be exchanged for 0.4504 AUD. (a) Calculate the amount of Ringgit received. The money was invested for 2 years and 6 months at 5.2% p.a. compounding monthly. Calculate the amount of interest earned from this investment. Give your answer to the nearest Ringgit. (Total 8 marks) 8. Ali, Bob and Connie each have 3 000 USD (US dollar) to invest. Ali invests his 3 000 USD in a firm that offers simple interest at 4.5 per annum. The interest is added at the end of each year. Bob invests his 3 000 USD in a bank that offers interest compounded annually at a rate of 4 per annum. The interest is added at the end of each year. Connie invests her 3 000 USD in another bank that offers interest compounded half-yearly at a rate of 3.8 per annum. The interest is added at the end of each half year. (a) Calculate how much money Ali and Bob have at the beginning of year 7. Show that Connie has 3 760.20 USD at the beginning of year 7. (c) Calculate how many years it will take for Bob to have 6 000 USD in the bank. At the beginning of year 7, Connie moves to England. She transfers her money into a Bank there at an exchange rate of 1 USD = 0.711 GBP (British pounds). The bank charges 2 commission. (d) (i) Calculate, in USD, the commission that the bank charges. (ii) Calculate the amount of money, in GBP, that Connie transfers to the bank in England. (3) (4) (6) (3) (3) (5) (Total 17 marks) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 2 of 5

IB Math Studies 1: Algebra & Numbers: Compound Interest Alei - Desert Academy 2013-14 9. Kurt wants to invest 2000 Euros in a savings account for his new grandson. (a) Calculate the value of Kurt s investment based on a simple interest rate of 4% per annum, after 18 years. Inge tells Kurt about a better account which offers interest at a rate of 3.6% per annum, compounding monthly. Giving your answer to the nearest Euro, calculate the value of Kurt s investment after 18 years if he follows Inge s advice. (Total 8 marks) 10. In 2000 Herman joined a tennis club. The fees were 1200 a year. Each year the fees increase by 3%. (a) Calculate, to the nearest 1, the fees in 2002. Calculate the total fees for Herman who joined the tennis club in 2000 and remained a member for five years. (Total 8 marks) 11. Mario has spent $40000 to buy some land. The land increases in value by 5% each year. (i) What is the value of the land after the end of five years? At the end of five years, Mario sells the land. He pays 1% tax on the sale and spends the rest of the money on a car. The car loses value at a rate of $2500 every year. (ii) How much tax does Mario pay? (iii) How much is the car worth five years after Mario buys it? (Total 8 marks) 12. David invests 6000 Australian dollars (AUD) in a bank offering 6% interest compounded annually. (a) Calculate the amount of money he has after 10 years. David then withdraws 5000 AUD to invest in another bank offering 8% interest compounded annually. Calculate the total amount he will have in both banks at the end of one more year. Give your answer correct to the nearest Australian dollar. (Total 8 marks) 13. The rate of inflation from the beginning of 1995 has been 4.5% per year. (a) A loaf of bread cost $1.70 on January 1, 1996. What did it cost on January 1, 1999? A car cost $40 000 on January 1, 1999. What did it cost on January 1, 1997? (Give your answer to the nearest thousand dollars.) (Total 4 marks) 14. Takaya invested 1000 JPY at 6.3% simple interest for 15 years. Morimi invested 900 JPY at 6.3% interest compounded annually for 15 years. Who had more money at the end of the 15th year? Justify your answer clearly. (Total 5 marks) 15. The table below shows the deposits, in Australian dollars (AUD), made by Vicki in an investment account on the first day of each month for the first four months in 1999. The interest rate is 0.75% per month compounded monthly. The interest is added to the account at the end of each month. Month Deposit (AUD) January 600 February 1300 March 230 April 710 (a) Show that the amount of money in Vicki s account at the end of February is 1918.78 AUD. (3) Calculate the amount of Australian dollars in Vicki s account at the end of April. Vicki makes no withdrawals or deposits after 1st April 1999. (c) How much money is in Vicki s account at the end of December 1999? From 1st January 2000 the bank applies a new interest rate of 3.5% per annum compounded annually. (d) In how many full years after December 1999 will Vicki s investment first exceed 3300 AUD? (Total 9 marks) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 3 of 5

IB Math Studies 1: Algebra & Numbers: Compound Interest Alei - Desert Academy 2013-14 16. (a) In city Y, house prices have increased by 3% each year for the last three years. If a house cost USD 180 000 three years ago, calculate, to the nearest dollar, its value today. In city Z, a house worth USD 100 000 three years ago is now valued at USD 119 102. Calculate the yearly percentage increase in the value of this house. (Total 4 marks) 17. The table shows part of a currency conversion chart. For example GBP 1 is equivalent to FFR 8.33. GBP USD FFR GBP 1 p 8.33 USD 0.64 1 q FFR 0.12 0.19 l For all calculations in this question give your answers correct to two decimal places. (a) Calculate the value of (i) p; (ii) q. Joe has USD 1500 to exchange at a bank. (i) Assuming no commission is charged, how much in GBP will Joe receive from the bank? (ii) Assuming the bank charges 1.5% commission, (a) how much in GBP does Joe pay in commission? how much in GBP does Joe actually receive for his USD 1500? (c) Joe decides to invest GBP 700 of his money in a savings account which pays interest at 5%, compounded annually. (i) How much interest will the GBP 700 earn after 4 years? (ii) For how many years must Joe invest his GBP 700 in order to earn at least GBP 200 in interest? (d) After 4 years Joe has a total of GBP 900 in his savings account on an investment at 5% interest compounded annually. How much did he invest? Give your answer to the nearest one GBP. (Total 14 marks) 18. Hassan invested 10 000 CHF at the end of 1971. The interest rate was 5% per annum. How much interest in total would Hassan have earned at the end of the year 1999 if (a) he had removed the interest from his account at the end of each year; he had not removed the interest from his account at the end of each year. 19. Miranti deposits $1000 into an investment account that pays 5% interest per annum. (a) What will be the value of the investment after 5 years if the interest is reinvested? (4) (1) (1) (Total 4 marks) How many years would it take Miranti s investment of $1000 to double in value? (4) At the beginning of each year Brenda deposits $1000 into an investment account that pays 5% interest per annum. Interest is calculated annually and reinvested. (c) How much would be in Brenda s account after 5 years? (4) (Total 10 marks) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 4 of 5

IB Math Studies 1: Algebra & Numbers: Compound Interest Alei - Desert Academy 2013-14 20. John invests X USD in a bank. The bank s stated rate of interest is 6% per annum, compounded monthly. (a) Write down, in terms of X, an expression for the value of John s investment after one year. What rate of interest, when compounded annually (instead of monthly) will give the same value of John s investment as in part (a)? Give your answer correct to three significant figures. (Total 4 marks) 21. At what interest rate, compounded annually, would you need to invest $100 in order to have $125 in 2 years? (Total 4 marks) 22. Tony invested CHF 500 in a bank account at a constant rate of interest. The bank calculates his balance at the end of each year, rounded to two decimal places, as shown in the table below. Year Value at beginning of year Value at end of year 1st CHF 500 CHF 540 2nd CHF 540 CHF 583.20 3rd CHF 583.20 CHF 629.86 4th CHF 629.86 CHF 680.25 5th CHF 680.25 6th (a) What is the rate and type of interest? Complete the table for the fifth and sixth year of investment. 23. Angela needs $4000 to pay for a car. She was given two options by the car seller. Option A: Outright Loan A loan of $4000 at a rate of 12% per annum compounded monthly. (a) Find (i) the cost of this loan for one year; (ii) the equivalent annual simple interest rate. Option B: Friendly Credit Terms A 25% deposit, followed by 12 equal monthly payments of $287.50. (i) How much is to be paid as a deposit under this option? (c) (ii) Find the cost of the loan under Friendly Credit Terms. Give a reason why Angela might choose (i) Option A (ii) Option B To help Angela, her employer agrees to give her an interest free loan of $4000 to buy the car. The employer is to recover the money by making the following deductions from Angela s salary: $x in the first month, $y every subsequent month. The total deductions after 20 months is $1540 and after 30 months it is $2140. (d) Find x and y. (e) How many months will it take for Angela to completely pay off the $4000 loan? (Total 4 marks) (1) (4) (Total 15 marks) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 5 of 5

Compound Interest Practice Problems (99-07): MarkScheme 1. Financial penalty (FP) is applicable in question part only. Crn (a) Adam I 100 6500 5 6 = 100 Adam has 1950 + 6500 = $8450 6 (C3) 4.5 FP Ben Amount = 6500 1 = $8464.69 (C3) Note: (A0) if interest only found (=$1964.69) 12n 5.4 2. (a) 3600 1 7200 1200 Note: for either CI formula with substitutions. for correct substitutions. 12n = 154.37 (154.4) Allow n = 12.86 (12.9) (G2) = 155 months (ft)(g3) Notes: Award final for a wrong answer rounded correctly to whole months. Note that 156 is premature rounding and receives (A0). For answer of 154 with no working award (G2) If 0.45% or equivalent is seen and correct answer is given with no further working, award (G4) or with answers of 154 or 154.4 (G3). If 0.45% or equivalent is seen with no further working and incorrect answer award. 108 (i) 5.4 At end of ninth year 3600 1 1200 Note: for substituted formula for correct substitutions. = 5846.50 euros (5847 or 5850) (G2) Notes: If 108 seen and correct answer given with no further working, award (G3). If 108 seen with incorrect answer award. If annual compounds used in parts (a) and, follow though for answer of 5779.20 euros. If Simple Interest is used in parts (a) and, can award final in part (a) and, at most, (ft)(a0) in part for answer of 5349.60. (ii) I = 5846.50 3600 = 2246.50 (ft) 3600 i% 9 = 2246.50 Rate i% = 6.93% (0.0693) (ft)(g3) Notes: Allow 6.94% (0.0694) as follow through from 5850 Answer of 18.0% receives (A0)(ft) If Simple Interest has been used in part (i), award only for answer of 5.4%. 3. (a) 20000 2 8 3200 Swiss francs 100 Note: Award for formula with correct values. (C2) 20 000 (1.0125) 24 20 000 = 6947.02 Swiss francs Principal = 26947.02 Swiss francs. (ft) (C4) Note: for correct substituted formula, for correct values C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 1 of 8 [6] [10]

inserted. for 6947.02 and for adding back the 20 000. The last follows through from the previous answer. 20000 (1.0125) 24 (M2) = 26947.02 Swiss francs Note: (M2) for correct substituted formula, for correct substitution, for correct answer. (M2) 4 4. (a) 2.75 600 1 = 668.77 (accept 669) 669 (G2) (C2) n 2.75 600 1 = 1200 n = 25.6 n = 26 26 (G2) (C2) 20 (c) r 600 1 = 1200 r 1 + = 1.03526 100 r = 3.53% 3.53% (G2) (C2) 5. (a) For attempting to find 5 years by compound interest formula or any alternative method. For using (1.0375) $1442.52 accept 3 s.f. (G3) Note: Accept $1440 or $1443. For using answer in part (a) in an expression. For multiplying by (1.0325) 3 $1587.79 accept $1588 or $1590 (G3) 6. (i) (a) 0 1 2 3 4 Option 1 20000 20700 21425 22174 22950 Option 2 20000 20800 21600 22400 23200 (A3) 3 Notes: Award (A2) for Option 1 all correct. Award for Option 2 all correct. A = 20000(1.035) 15 Option 1 = 33507 A = 20000 + 15 800 Option 2 = 32000 Option 1 = 33507 (G2) Option 2 = 32000 (G2) 4 (c) 7 years (from reading the table from GDC) (G2) 2 [6] [6] [6] C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 2 of 8

800 (d) 100 31200 Note: Award for candidate s difference divided by their original times 100. = 2.56% 2 (ii) Option 3: 20000 (1.035) 7 + 800 8 = $31845.59 Option 4: (20000 + 7 800) (1.035) 8 = $33710.31 $31845.59 (G3) $33710.31 (G3) She should choose Option 4 (R1) 7 7. (a) $4000 / 0.4504 = 8880.99 Ringgit (8880 to 3 s.f.) (C2) Note: Allow 8881 (nearest Ringgit). 30 8880.99 ( 1.0043) 8880.99 = 1230.09 Note: Award for compound calculation, for multiplier, for 30 seen and for deducting principle. Use of 1.0043 (no recurrence) is premature rounding: award at most (A0)(A0), but final ft can be awarded below for 1220. = 1230 Ringgit (C6) Note: Final is for rounding candidate s answer to nearest Ringgit. This is awarded only if the interest is calculated. (3000 4.5 6) 8. (a) Ali : 3000 $ 3810 100 Note: Award for correct formula, for correct numbers in formula, for correct answer. 6 4 Bob : 3000 1 $3 795.96( $ 3800) or (G3) 6 100 Note: Award for correct formula, for correct numbers in formula, for correct answer. 12 3.8 Connie : 3000 1 200 $3 760.204(482) Note: Award for correct formula, for correct numbers in formula for full correct answer. = $3760.20 (AG) 3 (c) n 3000(1.04) 6000 n 1.04 2 n = 18 (by trial and error or solver or by using logarithms) or (G3) 3 (accept 17.7) (d) (i) 2% of 3760.20 = 75.20 USD 2 (ii) 3760.20 75.20 = 3685 3685 0.711 = 2620.035 GBP (accept 2620.04, 2620.03, 2620) 3 9. (a) I = 0.04 2000 18 = 1440 Euros Total amount = I + 2000 = 3440 Euros. (C4) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 3 of 8 [18] [8] [17]

18 12 0.036 2000 1 12 = 3819.72 = 3820 Euros, to nearest Euro. (C4) 10. (a) 1200 1.03 2 = 1273 (A2) (C4) Note: Award for using correct formula, for substituting correct numbers, (A2) for answer ( of which is for answer to the nearest pound). Award (A0)(A2) ft if the candidate has put 3 for the power. Award (A0) ft if the candidate has put 1 for the power. 3% of 1200 = 36 After 1 year fees = 1236 3% of 1236 = 37.08 fees after 2 years = 1236 + 37.08 = 1273.08 = 1273 Note: Award (A0) if candidate has repeated the process another time. 5 1200(1.03 1) (1.03 1) = 6370.96 (6370 to 3 s.f.) (A2) (C4) Note: Award for using correct formula, for substituting correct numbers, (A2) for answer. Candidates can also write out each term and add them up. Award for terms and (A2) for the correct answer. Award only if answer is incorrect due to premature rounding. 11. (i) 40000 (1.05) 5 = 51051.26 (ii) 51051.26 1% = $510.51 (accept $511). (iii) 51051.26 510.51 = 50540.75 50540.75 5 2500 = $38040.75 (accept $38 000). (C8) 12. (a) A = 6000(1.06) 10 = 10745 (AUD) (C3) 10745 5000 = 5745 5000 1.08 + 5745 1.06 = 11489.80 = 11490 (to the nearest AUD) (C5) 3 13. (a) 4.5 A = 1.70 1 = $1.939982413 = $1.94 (3 s.f.) 2 4.5 40000 = C 1 40000 2 = C 1.045 $36629.19805 = C $37000 = C (to the nearest thousand dollars) C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 4 of 8 [8] [8] [8] [8]

Note: Accuracy is specified in the question, therefore this is not a paper accuracy penalty. 14. Takaya: 1000 +1000(0.063)(15) = 1945 JPY Morimi: 900(1.063) 15 = 2250 JPY Morimi had more 5 Note: Award (M0)(A0) for computing interest only for Takaya. Award follow through, follow through for follow through with only interest for Morimi. Award for Morimi had more for comparing the two interests. Award for Morimi had more for any reasoning that shows understanding of difference between compound and simple interest. Award (A0) for Morimi had more after computing interest for Takaya, but interest + principal for Morimi. Award for Morimi had more with no work shown at all. 15. (a) Amount End of month + Interest January 600 604.50 February 1904.50 1918.78 end January: 600 1.0075 = 604.50 begin February: 604.50 + 1300 = 1904.50 end February: 1904.50 1.0075 = 1918.78 (AG) 3 March amount = 1918.78 + 230 = 2148.78 end of March = 2148.78 1.0075 = 2164.90 April amount = 2164.90 + 710 = 2874.90 end of April = 2874.90 1.0075 = 2896.46 2 (c) 2896.46 1.0075 8 = 3074.88 2 (d) 3074.88 1.035 n = 3300 n = 1 3074.88 1.035 = 3182.50 n = 2 3074.88 1.035 2 = 3293.89 n = 3 3074.88 1.035 3 = 3409.17 Hence after 3 years. 3074.88 1.035 n = 3300 n = 3, that is, after 3 years. (G2) 2 Note: Candidates may also use logarithms to solve this. 16. (a) Price today = USD 180 000 1.03 3 = USD 196690.86... = USD 196691 or US$196691 (to the nearest dollar) (C2) Note: Accept equivalent method 119102 = 100 000 r 3 1.19102 = r 3 r = 1.060001187 Rate = 6.00% per annum (3 s.f.) (accept 6%) (C2) 17. Notes: If no method is shown, award if and only if answer is correct, otherwise award zero marks. However, award if correct method is shown; even if final answer is wrong. C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 5 of 8 [4] [5] [9] [4]

(a) (i) 1 p = = 1.56 (2 d.p.) 0.64 (ii) 1 q = = 5.26 (2 d.p.) 0.19 4 Notes: For parts (a)(i) and (a)(ii) accept and follow through with conversions routed via candidate s home currency. For example: USD 1 = GBP 0.64 GBP 1 = FFR 8.33 USD 1 = FFR (0.64) (8.33) q = 5.33 instead of 5.26 (i) GBP (1500 0.64) = GBP 960.00 2 Note: Accept (1500 1.56 (or candidate s p)) = GBP 961.54 (ii) (a) (0.015 960) = GBP 14.40 1 Note: Follow through from part (i) above. (960 14.40) = GBP 945.60 1 Note: Follow through from parts (i) and (ii)(a). (c) (i) 700(1.05) 4 = GBP 850.85 Therefore interest = GBP 150.85 2 (ii) 700(1.05) 5 = 893.397... = 893.40(2 d.p.) 700(1.05) 6 = 938.066... = 938.07 (2 d.p.) therefore after 6 years 2 Note: Accept other correct methods. (d) C(1.05) 4 = 900 900 C = 4 (1.05) Notes: Award the at the point where C has been correctly isolated 900 Accept C = = GBP 738 1.22 = GBP 740 (nearest GBP) 2 18. (a) 10000 5 28 I = 100 = 14 000 CHF I = 10 000(1.05) 28 10 000 = 29 201.29 CHF (or 29 200 CHF to 3 s.f.) r 19. (a) A = C 1 100 n 5 5 = 1000 1 = $1276.28 2 n 5 2000 = 1000 1 2 = 1.05 n n 1.05 n 10 1.05 10 = 1.6 20 1.05 20 = 2.7 15 1.05 15 = 2.07 C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 6 of 8 [14] [4]

(c) 14 1.05 14 = 1.98 n = 15 years 4 Year Deposit Start year balance End of year balance 1 $1000 $1000 $1050 2 $1000 $2050 $2152.50 3 $1000 $3152.50 $3310.125 4 $1000 $4310.125 $4525.63125 5 $1000 $5525.63125 $5801.91 (M3) After 5 years $5801.91 4 20. (a) X(1.005) 12 X(1.005) 12 r = X 1 100 Note: Award for equating follow through from (a). r = 100(1.0617) 100 (or equivalent) Note: Award for isolating r correctly. Rate = 6.17% r 21. A = C 1 100 r 125 = 100 1 n 2 2 C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 7 of 8 r 1.25 = 1 r 1.11803398 1 = 100 r = 11.8% (3 s.f.) (C4) 22. (a) Compound interest 8% per year Year Value at beginning Value at end of year of year 1st CHF 500 CHF 540 2nd CHF 540 CHF 583.20 3rd CHF 583.20 CHF 629.86 4th CHF 629.86 CHF 680.25 5th CHF 680.25 CHF 734.67 6th CHF 734.67 CHF 793.44 23. (a) (i) Cost of loan = 4000(1.01) 12 4000 = $507.30 Note: Accept $4507.30 and then $4450 for part (i) (ii) 507.30(100) Equivalent S.I. Rate = 4000 = 12.662503... = 12.7% (3 s.f.) 4 (i) Deposit = 25% of $4000 = $1000 [10] [4] [4] [4]

(ii) Cost of the loan = 287.50(12) 3000 (or equivalent) = $450.00 3 (c) (i) Option A. Because she doesn t need a deposit (or equivalent appropriate explanation). (R1) (ii) Option B. Because it is cheaper by $(507.30 450.00) = $57.30 (or equivalent appropriate explanation). (R1) 2 (d) x + 19y = 1540 x + 29y = 2140 x = 400 y = 60 4 (e) 400 + (n 1)60 = 4000 n = 61 Therefore, 61 months. 2 [15] C:\Users\Bob\Documents\Dropbox\Desert\Studies\1NumAlg\LP_MS1NumAlg13-14.doc Page 8 of 8