Save a Tree Please Do Not Print Me

SAVE A TREE – PLEASE DO NOT PRINT ME

IB Math Studies – Chapter 13 – Financial Mathematics – Review Questions

1. 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)

2. 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)

3. The table shows part of a currency conversion chart. For example GBP 1 is equivalent to FFR8.33.

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.

(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)

(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 account 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)

4. 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.

(b) 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)

5. 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.

(b) 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.

(Total 6 marks)

6. Emma places €8000 in a bank account that pays a nominal interest rate of 5% per annum, compounded quarterly.

(a) Calculate the amount of money that Emma would have in her account after 15 years. Give your answer correct to the nearest Euro.

(3)

(b) After a period of time she decides to withdraw the money from this bank. There is €9058.17 in her account. Find the number of months that Emma had left her money in the account.

(3)

(Total 6 marks)

7. 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.

(b) Find the total amount in Philip’s account after these 6 years.

(Total 6 marks)

8. (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) Complete the table below giving the values of the investments to the nearest dollar for the first 4 years.

(3)

Year / 0 / 1 / 2 / 3 / 4
Option 1 / 20 000 / 20 700
Option 2 / 20 000 / 20 800

(b) 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 value 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)

9. The table below shows the monthly repayments per $10 000 borrowed for various nominal annual interest rates.

Loan term / Table of monthly repayments in $, per $10 000
(years) / Annual interest rate
7% / 8% / 9%
5 / 198.0112 / 202.7634 / 207.5836
10 / 116.1085 / 121.3276 / 126.6758
15 / 89.8828 / 95.5652 / 101.4267
20 / 77.5299 / 83.6440 / 89.9726
25 / 70.6779 / 77.1816 / 83.9196

Beryl borrows $150 000 to buy an apartment at an interest rate of 8%, to be repaid over 20 years.

(a) Calculate Beryl’s exact monthly repayment.

(b) Find the exact amount of interest paid for the loan over the 20 years.

(Total 6 marks)

IB Math Studies – Chapter 13 – Financial Mathematics – Review Questions Mark Scheme

1. (a) 1.7700 CHF (A2) (C2)

(b) Method 1
2% of 1000 CHF = 20 (A1)
Amount = 1000 – 20 (M1)
= 980 CHF (A1)
= USD (M1)
= 553.67 USD (A1)
= 554 USD (to nearest dollar) (A1) (C6)

Method 2
1.7700 CHF = 1 USD
1000 CHF = (M1)
= 564.97175 (A1)
564.97175 × 0.98 (M1)(A1)
= 553.67 (A1)
= 554 USD (to the nearest dollar) (ft from answer in (a)) (A1) (C6)

[8]

2. (a) 3500 × 0.412 USD = 1442 USD (M1)(A1)
1442 × 0.98 = 1413.16 USD (M2)(A1) (C5)

3500 × 0.412 USD = 1442 USD (M1)(A1)
1442 – 1442 × 0.02 = 1413.16 USD (M2)(A1) (C5)

(accept nearest $1413 or 3 s.f. $1410).

(b) = 1288.34 (M1)(A1)
= 1288 MD to the nearest dollar. (A1) (C3)

[8]

3. 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.

(a) (i) p = = 1.56 (2 d.p.) (M1)(A1)

(ii) q = = 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:
USD 1 = GBP 0.64
GBP 1 = FFR 8.33
\ USD 1 = FFR (0.64) (8.33)
Þ q = 5.33 instead of 5.26

(b) (i) GBP (1500 × 0.64) = GBP 960.00 (M1)(A1) 2

Note: Accept (1500 ÷ 1.56 (or candidate’s p)) = GBP 961.54

(ii) (a) (0.015 × 960) = GBP 14.40 (A1) 1

Note: Follow through from part (b)(i) above.

(b) (960 – 14.40) = GBP 945.60 (A1) 1

Note: Follow through from parts (b)(i) and (b)(ii)(a).

(ii) 700(1.05)5 = 893.397... = 893.40(2 d.p.) (M1)
700(1.05)6 = 938.066... = 938.07 (2 d.p.)
therefore after 6 years (A1) 2

Note: Accept other correct methods.

(d) C(1.05)4 = 900
Þ C = (M1)

Notes: Award the (M1) at the point where C has been correctly isolated
Accept C = = GBP 738

= GBP 740 (nearest GBP) (A1) 2

[14]

4. (a) A = 6000(1.06)10 (M1)(A1)
= 10745 (AUD) (A1) (C3)

(b) 10745 – 5000
= 5745 (A1)
5000 × 1.08 + 5745 × 1.06 (M1)(M1)
= 11489.80 (A1)
= 11490 (to the nearest AUD) (A1) (C5)

[8]

5. (a) 600 = 668.77 (accept 669) (M1)(A1)
OR
669 (G2) (C2)

(b) 600 = 1200 (M1)
n = 25.6
n = 26 (A1)
OR
26 (G2) (C2)

[6]

6. (a) FV = 8000 (1.0125)60 (M1)(A1)

Note: (M1) for substituting in compound interest
formula, (A1) for correct substitution

€16857 only (A1) (C3)

(b) 8000 (1.0125)n = 9058.17 (M1)

Note: (M1) for equating compound interest formula
to 9058.17

n =10 correct answer only (A1)

So 30 months, (ft) on their n (A1)(ft)

Note: Award (C2) for 2.5 seen with no working (C3)

[6]

7. (a) Swiss francs (M1)(A1) (C2)

Note: Award (M1) for formula with correct values.

(b) 20 000 (1.0125)24 – 20 000 (M1)(A1)

= 6947.02 Swiss francs (A1)

Principal = 26947.02 Swiss francs. (A1)(ft) (C4)

Note: (M1) for correct substituted formula, (A1) for correct values inserted. (A1) for 6947.02 and (A1) for adding back the 20 000. The last (A1) follows through from the previous answer.

20000 (1.0125)24 (M2)(A1)

= 26947.02 Swiss francs (A1)

Note: (M2) for correct substituted formula, (A1) for correct substitution, (A1) for correct answer.

(M2)(A1) (A1)

[6]

8. (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 (A1) for Option 2 all correct.

(b) A = 20000(1.035)15 (A1)
Option 1 = 33507 (A1)
A = 20000 + 15 × 800 (A1)
Option 2 = 32000 (A1)

Option 1 = 33507 (G2)
Option 2 = 32000 (G2) 4

(d) × 100 (M1)

Note: Award (M1) for candidate’s difference divided by their
original times 100.

= 2.56% (A1) 2

(ii) Option 3: 20000 × (1.035)7 + 800 × 8 (M1)(A1)
= $31845.59 (A1)
Option 4: (20000 + 7 × 800) × (1.035)8 (M1)(A1)
= $33710.31 (A1)

$31845.59 (G3)
$33710.31 (G3)

She should choose Option 4 (R1) 7

[18]

9. Financial accuracy penalty (FP) is applicable where indicated.

(FP) (a) 15 ´ 83.6440 = $1254.66 (A1)(M1)(A1) (C3)

Note: The first (A1) for 15 seen. (M1) for 83.6440 in a product.
(A1) for 1254.66.

(FP) (b) Total repayment is 24 ´ 1254.66 = $301118.40 (M1)(A1)(ft)

Note: (M1) for multiplying calculated value from (a) by 240.

Interest paid is 301118.40 – 150000 = $151118.40 (A1)(ft) (C3)

Note: The final (A1) is for 150000 correctly subtracted from their value.

[6]