Finance x4w11

FINANCE 1.

At what interest rate, compounded annually, would you need to invest $100 in order to have $125 in 2 years? Working: Answer: ……………………………………………….. (Total 4 marks)

2.

Tony invested CHF 500 in a bank 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 Value at end of of year 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? (b) Complete the table for the fifth and sixth year of investment. Working: Answer: (a) ………………………………………….. (Total 4 marks)

3.

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; (2)

(ii)

the equivalent annual simple interest rate. (2)

Option B: Friendly Credit A 25% deposit, followed by 12 equal monthly payments of $ 287.50. (b)

(i)

How much is to be paid as a deposit under this option? (1)

(ii)

Find the cost of the loan under Friendly Credit . (2)

(c)

Give a reason why Angela might choose (i) (ii)

Option A Option B (2)

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. (4)

(e)

How many months will it take for Angela to completely pay off the $4000 loan? (2) (Total 15 marks)

4.

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 at the end of each year; (b) he had not removed the interest from his at the end of each year. Working: Answers: (a) ………………………………………….. (b) ………………………………………….. (Total 4 marks)

5.

Miranti deposits $1000 into an investment that pays 5% interest per annum. (a)

What will be the value of the investment after 5 years if the interest is reinvested? (2)

(b)

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 that pays 5% interest per annum. Interest is calculated annually and reinvested. (c)

How much would be in Brenda's after 5 years? (4) (Total 10 marks)

6.

John invests X USD in a bank. The bank's stated rate of interest is 6% per annum, compounded monthly. (a) (b)

Write down, in 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. Working: Answers: (a) ………………………………………….. (b) ………………………………………….. (Total 4 marks)

7.

(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.

(b)

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. Working: Answers:

(a) ………………………………………….. (b) .................................................................. (Total 4 marks)

8.

The table below gives the monthly repayments for a loan of 1000 Australian dollars (AUD). The interest rates are 18% per annum and 18.5% per annum, respectively. Monthly repayments on AUD 1000 Number of years of the loan

18% p.a.

18.5% p.a.

1

94.75

95.07

2

51.45

51.72

3

37.15

37.43

4

30.11

30.39

5

26.14

26.26

From the table, (a)

find the monthly repayment on a loan of AUD 1000 at 18.5% per annum taken over 1 year; (b) calculate the total amount to be repaid on a loan of AUD 8000 taken over 5 years, if the interest rate is 18% per annum. Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 4 marks)

9.

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) (ii)

p; q. (4)

(b)

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? (2)

(ii)

Assuming the bank charges 1.5% commission, (a)

how much in GBP does Joe pay in commission? (1)

(b)

how much in GBP does Joe actually receive for his USD 1500? (1)

(c)

Joe decides to invest GBP 700 of his money in a savings which pays interest at

5%, compounded annually. (i)

How much interest will the GBP 700 earn after 4 years? (2)

(ii)

For how many years must Joe invest his GBP 700 in order to earn at least GBP 200 in interest? (2)

(d)

After 4 years Joe has a total of GBP 900 in his savings on an investment at 5% interest compounded annually. How much did he invest? Give your answer to the nearest one GBP. (2) (Total 14 marks)

10.

The rate of inflation from the beginning of 1995 has been 4.5% per year. (a) (b)

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.) Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 4 marks)

11.

The following is a currency conversion table: FFR

USD

French Francs (FFR)

1

p

US Dollars (USD)

6.289

1

Japanese Yen (JPY)

0.057

0.009

British Pounds (GBP)

9.901

1.585

JPY

q

GBP

0.101 111.111 0.631

1

0.006 166.667 1

For example, from the table 1 USD = 0.631 GBP. Use the table to answer the following questions. (a)

Find the values of p and q. (2)

(b)

Mireille wants to change money at a bank in London. (i) (ii)

How many French Francs (FFR) will she have to change to receive 140 British Pounds (GBP)? The bank charges a 2.4% commission on all transactions. If she makes this transaction, how many British Pounds will Mireille actually receive from the bank? (4)

(c)

Jean invested 5000 FFR in Paris at 8% simple interest per annum. Paul invested 800 GBP in London at 6% simple interest per annum. (i) How much interest in FFR did Jean earn after 4 years? (ii) How much interest in US Dollars did Paul earn after 4 years?

(iii) (iv)

Who had earned more interest after 4 years? Explain your reasoning in part (c) (iii). (7) (Total 13 marks)

12.

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)

13.

The following is part of a display on the notice board of a bank in the United Kingdom. It shows the exchange rate between one British pound (GBP) and other currencies. EXCHANGE RATES Bank buys foreign currency

Bank sells foreign currency

Denmark (KR)

11.38

10.78

Finland (MKK)

7.00

6.60

10.05

9.45

(FFR) (DM)

2.854

Greece (DR)

292

2.798 266

NO COMMISSION CHARGED Geraldine eats a meal in a restaurant while on holiday in Greece. The meal costs 4256 drachma (DR). (a)

Use the bank-selling price to calculate the cost of the meal in British pounds.

The Williams family go to . Before leaving, they change GBP 600 into German marks. (b)

Calculate the number of German marks they receive for GBP 600, giving your answer correct to two decimal places.

They spend DM 824 in , and on returning to the United Kingdom, they change their remaining German marks into British pounds. (c) Calculate the number of British pounds they receive, correct to two decimal places. Working: Answers: (a) ………………………………………….. (b) .................................................................. (c) …………………………………….......... (Total 4 marks)

14.

The table below shows the deposits, in Australian dollars (AUD), made by Vicki in an investment 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 at the end of each month. Month Deposit (AUD)

(a)

January

600

February

1300

March

230

April

710

Show that the amount of money in Vicki's at the end of February is AUD

1918.78. (3)

(b)

Calculate the amount of Australian dollars in Vicki's at the end of April. (2)

Vicki makes no withdrawals or deposits after 1st April 1999. (c)

How much money is in Vicki's at the end of December 1999? (2)

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 AUD 3300? (2) (Total 9 marks)

15.

The exchange rate from US dollars (USD) to French francs (FFR) is given by 1 USD = 7.5 FFR. Give the answers to the following correct to two decimal places. (a) (b)

Convert 115 US dollars to French francs. Roger receives 600 Australian dollars (AUD) for 2430 FFR. Calculate the value of the US dollar in Australian dollars. Working: Answers: (a) …..….……………………… (b) ……………………………... (Total 8 marks)

16.

Keisha had 10 000 USD to invest. She invested m USD at the Midland Bank, which gave her 8% annual interest. She invested f USD at the First National Bank, which gave 6% annual interest. She received a total of 640 USD in interest at the end of the year. (a) Write two equations that represent this information. (b) Find the amount of money Keisha invested at each bank. Working: Answers: (a) ………………………………………….. (b) …………………………………….......... (Total 8 marks)

17.

Frederick had to change British pounds (GBP) into Swiss francs (CHF) in a bank. The exchange rate is 1 GBP = 2.5 CHF. There is also a bank charge of 3 GBP for each transaction. (a) (b)

How many Swiss francs would Frederick buy with 133 GBP? Let s be the number of Swiss francs received in exchange for b GBP. Express s in of b. (c) Frederick received 430 CHF. How many British pounds did he exchange? Working: Answers: (a) ………………………………………….. (b) .................................................................. (c) …………………………………….......... (Total 8 marks)

18.

th

On Vera's 18 birthday she was given an allowance from her parents. She was given the

following choices. Choice A $100 every month of the year. Choice B A fixed amount of $1100 at the beginning of the year, to be invested at an interest rate of 12% per annum, compounded monthly. Choice C $75 the first month and an increase of $5 every month thereafter. Choice D $80 the first month and an increase of 5% every month. (a)

Assuming that Vera does not spend any of her allowance during the year, calculate, for each of the choices, how much money she would have at the end of the year. (8)

(b)

Which of the choices do you think that Vera should choose? Give a reason for your answer. (2)

(c)

th

On her 19 birthday Vera invests $1200 in a bank that pays interest at r% per annum compounded annually. Vera would like to buy a scooter costing $1452 on her st 21 birthday. What rate will the bank have to offer her to enable her to buy the scooter? (4) (Total 14 marks)

19.

Zog from the planet Mars wants to change some Martian Dollars (MD) into US Dollars (USD). The exchange rate is 1 MD = 0.412 USD. The bank charges 2 % commission. (a)

How many US Dollars will Zog receive if she pays 3500 MD?

Zog meets Zania from Venus where the currency is Venusian Rupees (VR). They want to exchange money and avoid bank charges. The exchange rate is 1 MD = 1.63 VR. (b)

How many Martian Dollars, to the nearest dollar, will Zania receive if she gives Zog 2100 VR? Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 8 marks)

20.

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? Working: Answers: (i) ………………………………………….. (ii) .................................................................. (iii) …………………………………….......... (Total 8 marks)

21.

David invests 6000 Australian dollars (AUD) in a bank offering 6% interest compounded annually. (a) (b)

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.

Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 8 marks)

22.

A Swiss bank shows currency conversion rates in a table. Part of the table is shown below, which gives the exchange rate between British pounds (GBP), US dollars (USD) and Swiss francs (CHF). Buy

Sell

GBP

2.3400

2.4700

USD

1.6900

1.7700

This means that the bank will sell its British pounds to a client at an exchange rate of 1 GBP = 2.4700 CHF. (a)

What will be the selling price for 1 USD?

Andrew is going to travel from Europe to the USA. He plans to exchange 1000 CHF into dollars. The bank sells him the dollars and charges 2% commission. (b) How many dollars will he receive? Give your answer to the nearest dollar. Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 8 marks)

23.

Kurt wants to invest 2000 Euros in a savings 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 which offers interest at a rate of 3.6% per annum, compounding monthly. (b)

Giving your answer to the nearest Euro, calculate the value of Kurt's investment after 18 years if he follows Inge's advice. Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 8 marks)

24.

In 2000 Herman ed a tennis club. The fees were £ 1200 a year. Each year the fees increase by 3 %. (a) (b)

Calculate, to the nearest £ 1, the fees in 2002. Calculate the total fees for Herman who ed the tennis club in 2000 and remained a member for five years. Working: Answers: (a) ………………………………………….. (b) ..................................................................

(Total 8 marks)

25.

Bobby is spending a year travelling from America to and Britain. Consider the following exchange rates. 1 US dollar (USD) = 0.983 Euros 1 British Pound (GBP) = 1.59 Euros (a)

Bobby changes 500 USD into Euros. (i) Calculate how many Euros he receives. He spends 328 Euros in and changes the remainder into GBP. (ii) Calculate how many GBP he receives.

While in Britain Bobby decides to put this money in a bank that pays 6 % simple interest per annum, and he gets a part-time job to cover his expenses. Bobby remains in Britain for six months. (b) Calculate how much interest he receives for the six months. Working: Answers: (a) (i) ……………………………………... (ii) ........................................................... (b) …………………………………….......... (Total 8 marks)

26.

A bank in Canada offers the following exchange rate between Canadian dollars (CAD) and Euros (EUR). The bank sells 1 CAD for 1.5485 EUR and buys 1 CAD for 1.5162 EUR. A customer wishes to exchange 800 Canadian dollars for Euros. (a) (b)

Find how many Euros the customer will receive. The customer has to cancel his trip and changes his money back later when the rates are “sells 1 CAD = 1.5546 EUR, buys 1 CAD = 1.5284 EUR”. Use the “we sell” information to find how many Canadian dollars he receives. (c) How many Canadian dollars has he lost on the transaction? Working: Answers: (a) ………………………………………….. (b) .................................................................. (c) …………………………………….......... (Total 6 marks)

27.

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) (b)

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)

28.

William invests $1200 for 5 years at a rate of 3.75 % compounded annually. (a) (b)

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)

29.

(i)

Celia has $ 20 000 to invest. There are two different options that 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)

Copy and complete the table below giving the values of the investments to the nearest dollar for the first 4 years. (3)

(b)

Year

0

1

Option 1

20 000

20 700

Option 2

20 000

20 800

2

3

4

Calculate the values of each investment at the end of 15 years. (4)

(c)

If Option 1 is chosen find the total number of complete years before the values of the investment is first greater than $ 25 000. (2)

(d)

If Option 2 is chosen calculate the percentage increase in the investment for the final year. (2)

(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)

30.

A family in Malaysia received a gift of AUD $ 4000 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. (b)

Calculate the amount of interest earned from this investment. Give your answer to the nearest Ringgit. Working: Answers: (a) ………………………………………….. (b) .................................................................. (Total 8 marks)

MARKSCHEME r    1  100  1. A = C 

n

r   1  100  125 = 100 

2

(M1)



 1

1.25 = 

r   100 

2

(M1)

r 1.11803398 – 1 = 100 r = 11.8 % (3 s.f.)

(M1) (A1) (C4) [4]

2.

(a)

Compound interest 8 % per year

(A1 (A1)

(b) Year

Value at beginning of year CHF 500 CHF 540 CHF 583.20 CHF 629.86 CHF 680.25 CHF 734.67

1st 2nd 3rd 4th 5th 6th

Value at end of year CHF 540 CHF 583.20 CHF 629.86 CHF 680.25 CHF 734.67 CHF 793.44

(A1) (A1) [4]

3.

(a)

(b)

(c)

(d)

(e)

12

(i)

Cost of loan = 4000(1.01) – 4000 = $ 507.30 Note: Accept $ 4507.30 and then $ 4450 for part (b)(i) 507.30(100) 4000 (ii) Equivalent S.I. Rate = = 12.662503... = 12.7 % (3 s.f.) (i) Deposit = 25% of $4000 = $1000 (ii) Cost of the loan = 287.50(12) – 3000 (or equivalent) = $ 450.00 (i) Option A. Because she doesn’t need a deposit (or equivalent appropriate explanation). (ii) Option B. Because it is cheaper by $(507.30 – 450.00) = $57.30 (or equivalent appropriate explanation). x + 19y = 1540 x + 29y = 2140  x = 400 y = 60 400 + (n – 1)60 = 4000  n = 61 Therefore, 61 months.

(M1) (A1)

(M1) (A1)

4

(A1) (M1) (A1)

3

(R1) (R1) (M1) (M1) (A1) (A1) (M1)

2

(A1)

2

4

[15]

4.

(a) (b)

10000  5  28 100 I= = 14 000 CHF 28 I = 10 000(1.05) – 10 000 = 29 201.29 CHF (or 29 200 CHF to 3 s.f.)

(M1) (A1) (M1) (A1) [4]



5.

(a)

r   1  100  A = C 

 1

n

5   100 

= 1000  = $ 1276.28

5

(M1) (A1)

2



(b)

 1

2000 = 1000  n 2 = 1.05 n 10 20 15 14 n = 15 years

5   100 

n

(M1) (M1) n

1.05 10 1.05 = 1.6 20 1.05 = 2.7 15 1.05 = 2.07 14 1.05 = 1.98

(M1) (A1)

4

End of year balance $ 1050 $ 2152.50 $ 3310.125 $ 4525.63125 $ 5801.91 (A1)

4

(c) Year Deposit 1 $ 1000 2 $ 1000 3 $ 1000 4 $ 1000 5 $ 1000 After 5 years $ 5801.91

Start year balance $ 1000 $ 2050 $ 3152.50 $ 4310.125 $ 5525.63125

(M3) [10]

6.

12

(a)

X(1.005)

(b)

X(1.005) = X  Note: Award (M1) for equating follow through from (a). R = 100(1.0617) – 100 (or equivalent) Note: Award (M1) for isolating r correctly. Rate = 6.17 %



(A1)

12

 1

r   100 

(M1) (M1) (A1) [4]

7.

(a)

(b)

3

Price today = USD 180 000 × 1.03 = USD 196690.86...... = USD 196691 or US$ 196691 (to the nearest dollar) Note: Accept equivalent method 3 119102 = 100 000 × r 3 1.19102 = r r = 1.060001187 Rate = 6.00 % per annum (3 s.f.) (accept 6 %)

(M1) (A1) (C2)

(M1) (A1) (C2) [4]

8.

(a) (b)

Monthly repayment = AUD 95.07 Monthly repayment on AUD 8000 taken over 5 years at 18 % per annum = 8 × 26.14 = AUD 209.12 Note: Award (M1) for multiplying by 8 Total repaid = 209.12 × 5 × 12 = AUD 12547.20 Notes: Award (M1) for multiplying by (5 × 12) or 60. No accuracy specified, but the answer is exact.

(A1) (C1) (M1)

(M1) (A1) (C3) [4]

9.

Notes: If no method is shown, award (M1)(A1) if and only if answer is correct, otherwise award zero marks. However, award (M1) if correct method is shown; even if final answer is wrong. 1 (a) (i) p = 0.64 = 1.56 (2 d.p.) (M1)(A1) 1 (ii) q = 0.19 = 5.26 (2 d.p.) (M1)(A1) 4 Notes: For parts (a)(i) and (a)(ii) accept and follow through with conversions routed via candidate’s home currency. For example:

(b)

(c)

(d)

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 (M1)(A1) Note: Accept (1500 ÷ 1.56 (or candidate’s p)) = GBP 961.54 (ii) (a) (0.015 × 960) = GBP 14.40 (A1) Note: Follow through from part (b)(i) above. (b) (960 – 14.40) = GBP 945.60 (A1) Note: Follow through from parts (b)(i) and (b)(ii)(a). 4 (i) 700(1.05) = GBP 850.85 (M1) Therefore interest = GBP 150.85 (A1) 5 (ii) 700(1.05) = 893.397... = 893.40(2 d.p.) (M1) 6 700(1.05) = 938.066... = 938.07 (2 d.p.) therefore after 6 years (A1) Note: Accept other correct methods. 4 C(1.05) = 900 900 4  C = (1.05) (M1) Notes: Award the (M1) at the point where C has been correctly isolated 900 Accept C = 1.22 = GBP 738 = GBP 740 (nearest GBP) (A1)

2 1 1

2

2

2 [14]



10.

(a)

(b)

 1

4.5   100 

3

A = 1.70  = $ 1.939982413 = $1.94 (3 s.f.) 4.5    1  100  40000 = C 

(M1) (A1) 2

(M1) 40000 1.045 2 = C $36629.19805 = C $ 37000 = C (to the nearest thousand dollars) (A1) Note: Accuracy is specified in the question, therefore this is not a paper accuracy penalty. [4]

11.

(a) (b)

(c)

p = 0.159 (A1) q = 17.5 (A1) (i) 140(9.901) (M1) = 1386.14 = 1390 FFR (3 s.f.) (A1) OR 140 0.101 = 1386.14 (M1) = 1390 FFR (3 s.f.) (A1) Also accept 1387 FFR (you must go up to get 140 GBP) (ii) 140(0.024) = 3.36 GBP Commission (M1) 140 – 3.36 = 136.64 or 137 (3 s.f.) (A1) Note: Commission can be calculated in FFR also for (Ml). (i) I = 5000(0.08)(4) (M1) = 1600 FFR (A1) (ii) I = 800 (0.06)(4) = 192 GBP (M1) 192(1.585) = 304.32 USD or 304 USD (3 s.f.) (A1)

2

2 2 2 2

(iii) (iv)

Paul Jean: 1600 FFR 1600 ÷ 6.289 = 254.40 or 254 (3 s.f.) Paul: 304.32 or 304 (3 s.f.) Note: Accept any same currency comparison.

(A1)

1

(R1) (R1)

2 [13]

12.

Takaya: 1000 +1000(0.063)(15) = 1945 JPY (M1)(A1) 15 Morimi: 900(1.063) = 2250 JPY (M1)(A1) Morimi had more (A1) Note: Award (M0)(A0) for computing interest only for Takaya. Award (M1) follow through, (A1) follow through for follow through with only interest for Morimi. Award (A1) for “Morimi had more” for comparing the two interests. Award (A1) 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 (A1) for “Morimi had more” with no work shown at all.

5

[5]

13.

(a)

(b) (c)

4256 ÷ 266 = 16 GBP (A1) Note: Penalize only once in parts (b) and (c) if the buying and selling price are consistently confused throughout the question. Allow (ft) marks. 600 × 2.798 = 1678.80 DM (A1) 1678.80 – 824 = 854.80 DM (M1) = (854.80 ÷ 2.854) GBP (M1) = 299.51 GBP (2 d.p.) (A1) Note: Penalize only once for answer not given to 2 d.p. [4]

14.

(a)

(b)

(c) (d)

Amount End of month + Interest January 600 604.50 February 1904.50 1918.78 end January: 600 × 1.0075 = 604.50 (M1) begin February: 604.50 + 1300 = 1904.50 (M1) end February: 1904.50 × 1.0075 = 1918.78 (M1)(AG) March amount = 1918.78 + 230 = 2148.78 end of March = 2148.78 × 1.0075 (M1) = 2164.90 April amount = 2164.90 + 710 = 2874.90 end of April = 2874.90 × 1.0075 = 2896.46 (A1) 8 2896.46 × 1.0075 (M1) = 3074.88 (A1) n 3074.88 × 1.035 = 3300 n = 1 3074.88 × 1.035 = 3182.50 2 n = 2 3074.88 × 1.035 = 3293.89 3 n = 3 3074.88 × 1.035 = 3409.17 (M1) Hence after 3 years. (A1) OR n 3074.88 × 1.035 = 3300  n = 3, that is, after 3 years. (G2) Note: Candidates may also use logarithms to solve this.

3

2 2

2 [9]

15.

(a)

115 × 7.5 = 862.5

(M1)(A1)

(b)

862.50 USD METHOD 1 1 FF 2430  600 x AUD

(A1) (C3)

1 FF = 0.247 AUD 1 USD = 7.5 FF = 7.5 × 0.247 AUD = 1.8518... 1 USD = 1.85 AUD (2 d.p.) METHOD 2 2430 FFR = 600 AUD 600 1 FFR = 2430 AUD (= 0.247 AUD) 603 1 USD = 7.5 FFR = 7.5 × 2430 = 1.8518... 1 USD = 1.85 AUD (2 d.p.)

(M1) (A1) (M1)(A1) (A1) (C5) (M1) (M1) (M1)(A1) (A1) (C5) [8]

16.

(a) (b)

0.08 m + 0.06 f = 640 m + f = 10 000 8 m + 6 f = 64 000 –8 m – 8 f = 80 000 –2 f = –16 000 f = 8000 or $ 8 000 at First National Bank m = 2000 or $2 000 at Midland Bank Note: Allow ft from part (a).

(A2) (C2) (A2) (C2) (M1) (M1) (C2) (A1) (A1) (C2) [8]

17.

(a) (b) (c)

S = 2.5(133-3) S = 325 CHF S = 2.5(b – 3) or 2.5(b – 3) 430 = 2.5(b – 3) 172 = b – 3 175 = b 175 GBP

(M1) (A1) (C2) (A2) (C2) (M2) (A2) (C4) [8]

18.

(a)

Choice A = 100 × 12 = $1200

(A1)

12

 1  12  1200  = $ 1239.51 Choice B = 1100  Choice C = 75 + 80 + … 12 {2  75  11  5}  $1230 = 2 2 Choice D = 80 + 80 × 1.05 + 80 × 1.05 + … 80(1.0512 – 1)  $1273.37 = (1.05 – 1)

(M1) (A1) (M1) (A1) (M1)

(M1) (A1) Note: Award method marks if candidate works out each 0amount. But the answer has to be accurate to receive the mark of (A1)

(b)

Option D because the total allowance is the highest

(c)

 1  r  100  = 1452 1200 

(C1) (R1)

8 2

2

 1  r  100  

2

1452 = 1200 = 1.21

 1  r  100  = 1.21 = 1.1  r 100 = 0.1 R = 10%

(A1) (A1) (A1)

(A1)

4

[14]

19.

(a)

(b)

3500 × 0.412 USD = 1442 USD 1442 × 0.98 = 1413.16 USD OR 3500 × 0.412 USD = 1442 USD 1442 – 1442 × 0.02 = 1413.16 USD (accept nearest $ 1413 or 3 s.f. $ 1410). 2100 1.63 = 1288.34 = 1288 MD to the nearest dollar.

(M1)(A1) (M2)(A1) (M1)(A1) (M2)(A1) (C5)

(M1)(A1) (A1) (C3) [8]

20.

(i) (ii) (iii)

5

40000 × (1.05) = 51051.26 51051.26 × 1% = $ 510.51 (accept $ 511). 51051.26 – 510.51 = 50540.75 50540.75 – 5 × 2500 = $ 38040.75 (accept $ 38 000).

(M1)(A1) (M1)(A1) (M1)(A1) (M1)(A1) (C8) [8]

21.

(a) (b)

10

A = 6000(1.06) = 10745 (AUD) 10745 – 5000 = 5745 5000 × 1.08 + 5745 × 1.06 = 11489.80 = 11490 (to the nearest AUD)

(M1)(A1) (A1) (C3) (A1) (M1)(M1) (A1) (A1) (C5) [8]

22.

(a) (b)

1.7700 CHF Method 1 2% of 1000 CHF = 20 Amount = 1000 – 20 = 980 CHF 980 = 1.7700 USD = 553.67 USD = 554 USD (to nearest dollar) Method 2 1.7700 CHF = 1 USD 1000 1000 CHF = 1.77 = 564.97175 564.97175 × 0.98 = 553.67 = 554 USD (to the nearest dollar) (ft from answer in (a))

(A2) (C2) (A1) (M1) (A1) (M1) (A1) (A1) (C6)

(M1) (A1) (M1)(A1) (A1) (A1) (C6) [8]

23.

(a)

(b)

I = 0.04 × 2000 × 18 = 1440 Euros Total amount = I + 2000 = 3440 Euros. 1812 0.036    1  12  2000  = 3819.72 = 3820 Euros, to nearest Euro.

(M1)(A1) (M1)(A1) (C4) (M1)(A1) (A1) (A1) (C4) [8]

24.

(a)

2

1200 × 1.03 = 1273 (M1)(A1)(A2) (C4) Note: Award (M1) for using correct formula, (A1) for substituting correct numbers, (A2) for answer ((A1) of which is for answer to the nearest pound). Award (M1)(A0)(A2)ft if the candidate has put 3 for the power. Award (M1)(A0)(A1)ft if the candidate has put 1 for the power.

(b)

OR 3 % of 1200 = 36 (M1) After 1 year fees = 1236 (A1) 3 % of 1236 = 37.08 (M1) fees after 2 years = 1236 + 37.08 = 1273.08 = 1273 (A1) Note: Award (M1)(A1)(M1)(A0) if candidate has repeated the process another time. 5 1200(1.03  1) (1.03  1) (M1)(A1) = 6370.96 (6370 to 3 s.f.) (A2) (C4) Note: Award (M1) for using correct formula, (A1) for substituting correct numbers, (A2) for answer. Candidates can also write out each term and add them up. Award (M1)(A1) for and (A2) for the correct answer. Award only (A1) if answer is incorrect due to premature rounding. [8]

25.

(a)

(b)

(i) (ii)

500 × 0.983 = 491.50 Euros (492 Euros) 491.50 – 328 = 163.50 Euros 163.50 1.59 = 102.83 GBP (103 GBP) 102.83  0.5  6 100 = £ 3.08

(M1)(A1) (A1) (M1)(A1) (C5) (M1)(A1) (A1) (C3) [8]

26.

(a) (b) (c)

800 × 1.5162 for multiplying by 1.5162 = 1212.96 EUR (accept 1213) 1212.96 / 1.5546 = 780.24 (accept 780) 19.76 CAD

(M1) (A1) (C2) (M1) (A2) (C3) (A1) (C1) [6]



27.

(a)

 1

600  OR 669

 1

600  n = 25.6 n = 26 OR 26 

(c)

4

2.75   100 

n

= 668.77 (accept 669)

(M1)(A1) (G2) (C2)



(b)

2.75   100 

 1

600 

= 1200

(M1) (A1) (A1) (G2) (C2)

r   100 

20

= 1200 r 1 + 100 = 1.03526 r = 3.53% OR 3.53 %

(M1)

(G2) (C2) [6]

28.

(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. Accept $ 1440 or $ 1443.

(M1) (M1) (A1) (G3)

(b)

For using answer in part (a) in an expression. 3 For multiplying by (1.0325) $ 1587.79 accept $ 1588 or $ 1590

(M1) (M1) (A1) (G3) [6]

29.

(i)

(ii)

(a) 0 1 2 3 Option 1 20000 20700 21425 22174 Option 2 20000 20800 21600 22400 Award (A2) for Option 1 all correct. Award (A1) for Option 2 all correct. 15 (b) A = 20000(1.035) (A1) Option 1 = 33507 (A1) A = 20000 + 15 × 800 (A1) Option 2 = 32000 (A1) OR Option 1 = 33507 (G2) Option 2 = 32000 (G2) (c) 7 years (from reading the table from GDC) (G2) 800 31200 × 100 (d) (M1) Award (M1) for candidate ‘s difference divided by their original times 100. = 2.56 % (A1) 7 Option 3: 20000 × (1.035) + 800 × 8 (M1)(A1) = $ 31845.59 (A1) 8 Option 4: (20000 + 7 × 800) × (1.035) (M1)(A1) = $33710.31 (A1) OR $ 31845.59 (G3) $ 33710.31 (G3) She should choose Option 4 (R1)

4 22950 23200

(A3)

4 2

2

7 [18]

30.

(a) (b)

$ 4000 / 0.4504 = 8880.99 Ringgit (8880 to 3 s.f.) Note: Allow 8881 (nearest Ringgit). 30 ( 1 8880.99 × .0043) – 8880.99

(M1)(A1) (C2)

(M1)(A1)(A1)(M1) (A1) Note: Award (M1) for compound calculation, (A1) for multiplier, (A1) for 30 seen and (M1) for deducting principle. Use of 1.0043 (no recurrence) is premature rounding: award at most (M1)(A0)(A1)(M1)(A0), but final (A1) ft can be awarded below for 1220. = 1230 Ringgit (A1) (C6) Note: Final (A1) is for rounding candidate’s answer to nearest Ringgit. This is awarded only if the interest is calculated. = 1230.09

[8]

3