practice: simple & compound interest/depreciationkennethsokc.weebly.com/uploads/3/9/8/0/39809815/practice... · 2018-10-17 · practice: simple & compound interest/depreciation [145

practice: simple & compound interest/depreciation[145 marks]

1. [3 marks]

Jackson invested 12 000 Australian dollars (AUD) in a bank that offered simple interest at an annual interest rate of r %. The value ofJackson’s investment doubled after 20 years.

Maddison invests 15 000 AUD in a bank that offers compound interest at a nominal annual interest rate of 4.44 %,compounded quarterly.

Calculate the number of years that it will take for Maddison’s investment to triple in value.

Markscheme (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for a correctly substituted formula and correctly equated to 45000.

OR

(M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for a correctly substituted formula and correctly equated to 3.

n = 25 years (A1) (C3)

Notes: Award (A1)(M0)(A0) if 24.9 or 24.88 seen as a final answer, with no working seen. Award, at most, (A1)(M1)(A0) ifworking is seen and a final answer of 24.9 or 24.88 is given.

[3 marks]

45000 = 15000(1 + )4.44400

4n

3 = (1 + )4.44400

4n

2. [3 marks]

Veronica wants to make an investment and accumulate 25 000 EUR over a period of 18 years. She finds two investment options.

Option 2 offers a nominal annual interest rate of 4 %, compounded monthly.

Find the amount that Veronica has to invest with option 2 to have 25 000 EUR in her account after 18 years. Give your answer correct totwo decimal places.

Markscheme (M1)(A1)

Note: Award (M1) for substitution into a compound interest formula. Award (A1) for correct substitution and equation.

C =12183.39 (EUR) (A1) (C3)

Note: The final (A1) can only be given for seeing the correct figures.

[3 marks]

C = = 25000(1 + )0.0412

12×18

[1 mark]3a.

Cedric wants to buy an €8000 car. The car salesman offers him a loan repayment option of a 25 % deposit followed by 12 equalmonthly payments of €600 .

Write down the amount of the deposit.

Markscheme2000 (euros) (A1)

[1 mark]

[2 marks]3b. Calculate the total cost of the loan under this repayment scheme.

Markscheme (M1)

Note: Award (M1) for addition of two correct terms.

9200 (euros) (A1)(ft)(G2)

Note: Follow through from their part (a).

[2 marks]

2000 + 12 × 600

3c. [1 mark]Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repaythe loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.

The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The totalamount that Cedric’s mother receives after 24 months is €7100.

Write down a second equation involving x and y.

Markschemex + 23y = 7100 (A1)

[1 mark]

3d. [2 marks]Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repaythe loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.


Write down the value of x and the value of y.

Markschemex = 200, y = 300 (A1)(ft)(A1)(ft)(G2)

[2 marks]

3e. [3 marks]Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repaythe loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.


Calculate the number of months it will take Cedric’s mother to receive the €8000.

Markscheme (M1)

Note: Award (M1) for setting up the equation. Follow through from their x and y found in part (d).

n = 26 (A1)(ft)

26 + 1 = 27 (months) (A1)(ft)(G3)

Notes: Middle line n = 26 may be implied if correct answer given. The final (A1)(ft) is for adding 1 to their value of n (even if it isincorrect). Follow through from their part (d). If the final answer is not a positive integer award at most (M1)(A1)(ft)(A0). Award(G2) for final answer of 26.

OR

(M1)(A1)

Note: Award (M1) for division of difference by their value of y, (A1) for 24 seen.

27 (months) (A1)(ft)(G3)

Note: Follow through from their value of y.

[3 marks]

200 + n × 300 = 8000

+ 248000−7100300

3f. [5 marks]Cedric decides to buy a cheaper car for €6000 and invests the remaining €2000 at his bank. The bank offers two investmentoptions over three years.

Option A: Compound interest at an annual rate of 8 %.

Option B: Compound interest at a nominal annual rate of 7.5 % , compounded monthly.

Express each answer in part (f) to the nearest euro.

Calculate the value of his investment at the end of three years if he chooses

(i) Option A;

(ii) Option B.

Markscheme(i) (M1)

Note: Award (M1) for correct substitution in compound interest formula.

2519 (euros) (A1)(G2)

Note: If the answer is not given to the nearest euro award at most (M1)(A0).

(ii) (M1)(A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitutions.

2503 (euros) (A1)(G2)

Note: If the answer is not given to the nearest euro, award at most (M1)(A1)(A0), provided this has not been penalized in part (f)(i).

[5 marks]

2000(1 + )8100

3

2000(1 + )7.5100×12

3×12

4. [3 marks]

Jenny invested $20 000 in a bank account that paid 3.5 % annual simple interest. She withdrew her investment from the accountwhen its value was $31 200.

Ramón invests $18 000 in a bank account that pays 3.4 % nominal annual interest, compounded quarterly.

Find the minimum number of years that Ramón must invest the money for his investment to be worth $27 000.

Markscheme (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

(A1) (C3)

Note: Correct answer only. If 11.976… seen award (A2).

27000 = 18000[1 + ]3.4100×4

4n

(n =)12

5a. [3 marks]

Marcus has been given 500 Australian dollars (AUD) by his grandmother for his 18th birthday.

He plans to deposit it in a bank which offers a nominal annual interest rate of 6.0 %, compounded quarterly, for three years.

Calculate the total amount of interest Marcus would earn, in AUD, over the three years. Give your answer correct to twodecimal places.

Markscheme (M1)(A1)

Note: Award (M1) for substitution in correct formula (A1) for correct substitutions.

(A1) (C3)

Note: The answer must be given to 2 dp or the final (A1) is not awarded.

500 − 500(1 + )6100×4

4×3

= 97.81

5b. [3 marks]Marcus would earn the same amount of interest, compounded annually, for three years if he deposits the 500 AUD in asecond bank.

Calculate the interest rate the second bank offers.

Markscheme (M1)(A1)(ft)

Note: Award (M1) for substitution in correct formula, (A1)(ft) for their correct substitutions.

(6.13635...) (A1)(ft) (C3)

Note: Follow through from their answer to part (a).

97.8090... = 500 − 500(1 + )r

1003

= 6.14

[2 marks]6a.

Give your answers to parts (a) to (e) to the nearest dollar.On Hugh’s 18th birthday his parents gave him options of how he might receive his monthly allowance for the next two years.

Option A each month for two years

Option B in the first month, in the second month, in the third month, increasing by each month for two years

Option C in the first month and increasing by each month for two years

Option D Investing at a bank at the beginning of the first year, with an interest rate of per annum, compoundedmonthly.

Hugh does not spend any of his allowance during the two year period.

If Hugh chooses Option A, calculate the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(M1) Note: Award (M1) for correct product.

(A1)(G2)

[2 marks]

$60

$10 $15 $20 $5

$15 10%

$1500 6%

60 × 24

= 1440

6b. [5 marks]If Hugh chooses Option B, calculate

(i) the amount of money he will receive in the 17th month;

(ii) the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.(i) (M1)(A1) Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

(A1)(G2)

(ii) (M1)

OR (M1)

Note: Award (M1) for correct substitution in arithmetic series formula.

(A1)(ft)(G1)

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

[5 marks]

10 + (17 − 1)(5)

= 90

(2(10) + (24 − 1)(5))242

(10 + 125)242

= 1620

6c. [5 marks]If Hugh chooses Option C, calculate(i) the amount of money Hugh would receive in the 13th month;(ii) the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.(i) (M1)(A1) Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.

(A1)(G2)

Note: Award (M1)(A1)(A0) for .

Award (G1) for if workings are not shown.

(ii) (M1)

Note: Award (M1) for correct substitution in geometric series formula.

(A1)(ft)(G1)

Note: Follow through from part (c)(i).

[5 marks]

15(1.1)12

= 47

47.08

47.08

15( −1)1.124

1.1−1

= 1327

[3 marks]6d. If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.

Markscheme

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(M1)(A1)


OR

(A1)(M1) Note: Award (A1) for seen, (M1) for other correct entries.

OR

(A1)(M1)

Note: Award (A1) for seen, (M1) for other correct entries.

(A1)(G2)

[3 marks]

1500(1 + )6100(12)

12(2)

N = 2

I% = 6

PV = 1500

P/Y = 1

C/Y = 12

C/Y = 12

N = 24

I% = 6

PV = 1500

P/Y = 12

C/Y = 12

C/Y = 12

= 1691

6e. [1 mark]State which of the options, A, B, C or D, Hugh should choose to give him the greatest total value of his allowance at the end of

the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.Option D (A1)(ft) Note: Follow through from their parts (a), (b), (c) and (d). Award (A1)(ft) only if values for the four options are seen and only if their

answer is consistent with their parts (a), (b), (c) and (d).

[1 mark]

6f. [3 marks]Another bank guarantees Hugh an amount of after two years of investment if he invests $1500 at this bank. The interest

is compounded annually.

Calculate the interest rate per annum offered by the bank.

$1750

Markscheme (M1)(A1)

Note: Award (M1) for substituted compound interest formula equated to , (A1) for correct substitutions into formula.

OR

(A1)(M1)


(A1)(G2)

[3 marks]

1750 = 1500(1 + )r

1002

1750

N = 2

PV = 1500

FV = −1750

P/Y = 1

C/Y = 1

FV = 1750

= 8.01% (8.01234 … %, 0.0801)

[2 marks]7a.

Give all answers in this question correct to two decimal places.Arthur lives in London. On August 2008 Arthur paid euros ( ) for a new car from Germany. The price of the same car

in London was British pounds ( ).

The exchange rate on August 2008 was .

Calculate, in GBP, the price that Arthur paid for the car.

MarkschemeThe first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.

(M1) (A1)(G2)

[2 marks]

1st 37 500 EUR

34 075 GBP

1st 1 EUR = 0.7234GBP

37 500 × 0.7234

= 27 127.50

[1 mark]7b. Write down, in , the amount of money Arthur saved by buying the car in Germany.


(A1)(ft)(G1) Note: Follow through from part (a) irrespective of whether working is seen.

[1 mark]

GBP

6947.50

7c. [3 marks]Between August 2008 and August 2012 Arthur’s car depreciated at an annual rate of of its current value.

Calculate the value, in , of Arthur’s car on August 2009.

1st 1st 9%

GBP 1st


(A1)(M1) Note: Award (A1) for seen or equivalent, (M1) for their multiplied by OR

(A1)(M1) Note: Award (A1) for seen, and (M1) for .

(A1)(ft)(G2)

Note: Follow through from part (a). [3 marks]

27 127.50 × 0.91

0.91 27 127.50 0.91

27 127.50 − 0.09 × 27 127.50

0.09 × 27 127.50 27 127.50 − 0.09 × 27 127.50

= 24 686.03

7d. [3 marks]Between August 2008 and August 2012 Arthur’s car depreciated at an annual rate of of its current value.

Show that the value of Arthur’s car on August 2012 was , correct to the nearest .

1st 1st 9%

1st 18 600 GBP 100 GBP


(M1)(A1)(ft)

Notes: Award (M1) for substituted compound interest formula, (A1)(ft) for correct substitution.

Follow through from part (a).

OR

(M1)(A1)(ft) Notes: Award (M1) for substituted geometric sequence formula, (A1)(ft) for correct substitution.


OR (lists (i))

(M1)(A1)(ft) Notes: Award (M1) for at least the term correct (calculated from their ). Award (A1)(ft) for four correct terms (rounded

or unrounded).


Accept list containing the last three terms only ( may be implied).

OR (lists(ii))

(M1)(A1)(ft) Notes: Award (M1) for subtraction of four terms from .

Award (A1) for four correct terms (rounded or unrounded).


(A1)

(AG) Note: The final (A1) is not awarded unless both the unrounded and rounded answers are seen.

[3 marks]

27 127.50 × (1 − )9100

4

27 127.50 × (0.91)4

24 686.03, 22 464.28..., 20442.50..., 18602.67...

2nd (a) × 0.91

24 686.03

27 127.50 − (2441.47... + 2221.74... + 2021.79... + 1839.82…)

27 127.50

= 18 602.67

= 18 600

[1 mark]8a.

Ludmila takes a loan of 320 000 Brazilian Real (BRL) from a bank for two years at a nominal annual interest rate of 10%,

compounded half yearly.

Write down the number of times interest is added to the loan in the two years.

Markscheme4 (A1) (C1)[1 mark]

[3 marks]8b. Calculate the exact amount of money that Ludmila must repay at the end of the two years.

Markscheme (M1)(A1)


OR

(A1)(M1) Note: Award (A1) for seen, (M1) for correctly substituted values from the question into the finance application.

OR

(A1)(M1) Note: Award (A1) for seen, (M1) for correctly substituted values from the question into the finance application.

amount to repay (A1) (C3) Note: Award (C2) for final answer if not seen previously.

[3 marks]

320 000(1 + )102×100

2×2

N = 2

I% = 10

PV = −320000

P / Y = 1

C / Y = 2

C / Y = 2

N = 4

I% = 10

PV = −320000

P / Y = 2

C / Y = 2

C / Y = 2

= 388962

389000 388962

8c. [2 marks]Ludmila estimates that she will have to repay BRL at the end of the two years.

Calculate the percentage error in her estimate.

Markscheme (M1)

Note: Award (M1) for correctly substituted percentage error formula.

(A1)(ft) (C2)

Notes: Follow through from their answer to part (b).

[2 marks]

360 000

× 100∣∣360 000−388 962

388 962∣∣

= 7.45 (% ) (7.44597…)

[3 marks]9.

Astrid invests 1200 Euros for five years at a nominal annual interest rate of 7.2 %, compounded monthly.

Find the interest Astrid has earned during the five years of her investment. Give your answer correct to two decimal places.

Markscheme (M1)(A1)

I = 518.15 Euros (A1) (C3)

Notes: Award (M1) for substitution in the compound interest formula, (A1) for correct substitutions, (A1) for correct answer.

If final amount found is 1718.15 and working shown award (M1) (A1)(A0).

[3 marks]

I = 1200 − 1200(1 + )7.2600

5×12

10. [4 marks]Kunal borrows 200 000 Indian rupees (INR) from a money lender for 18 months at a nominal annual interest rate of ,compounded monthly.

Calculate the total amount that Kunal must repay at the end of the 18 months. Give your answer to the nearest rupee.

Markscheme (M1)(A1)


INR (A1)

INR (A1) (C4)

Note: Award final (A1) for their answer correct to the nearest rupee.

[4 marks]

15%

A = 200000(1 + )15100×12

1.5×12

= 250115.4788

= 250115

[2 marks]11a.

Give all answers in this question correct to two decimal places.

Part A

Estela lives in Brazil and wishes to exchange 4000 BRL (Brazil reals) for GBP (British pounds). The exchange rate is 1.00 BRL =0.3071 GBP. The bank charges 3 % commission on the amount in BRL.

Find, in BRL, the amount of money Estela has after commission.

Markscheme4000 × 0.97 = 3880.00 (3880) (M1)(A1)(G2)

Note: Award (M1) for multiplication of correct numbers.

OR

3 % of 4000 = 120 (A1)

4000 – 120 = 3880.00 (3880) (A1)(G2)

[2 marks]

[2 marks]11b. Find, in GBP, the amount of money Estela receives.

Markscheme3880 × 0.3071 = 1191.55 (M1)(A1)(ft)(G2)

Note: Award (M1) for multiplication of correct numbers. Follow through from their answer to part (a).

[2 marks]

11c. [2 marks]After her trip to the United Kingdom Estela has 400 GBP left. At the airport she changes this money back into BRL. Theexchange rate is now 1.00 BRL = 0.3125 GBP.

Find, in BRL, the amount of money that Estela should receive.

Markscheme (M1)

= 1280.00 (1280) (A1)(G2)

Note: Award (M1) for division of correct numbers.

[2 marks]

4000.3125

11d. [1 mark]Estela actually receives 1216.80 BRL after commission.

Find, in BRL, the commission charged to Estela.

Markscheme63.20 (A1)(ft)

Note: Follow through (their (c) –1216.80).

[1 mark]

[2 marks]11e. The commission rate is t % . Find the value of t.

Markscheme (M1)

t = 4.94 (A1)(ft)(G2)

Note: Follow through from their answers to parts (c) and (d).

[2 marks]

t = 63.20×1001280

[3 marks]11f.

Give all answers in this question correct to two decimal places.

Part B

Daniel invests $1000 in an account that offers a nominal annual interest rate of 3.5 % compounded half yearly.

Show that after three years Daniel will have $1109.70 in his account, correct to two decimal places.

Markscheme (M1)(A1)(A1)

= 1109.70 (AG)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for unrounded answer. If 1109.70 notseen award at most (M1)(A1)(A0).

OR

(M1)(A1)

A = 1109.7023... (A1)

= 1109.70 (AG)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for unrounded answer.

[3 marks]

A = 1000 = 1109.7023...(1 + )3.52×100

6

I = 1000 − 1000 = 109.7023(1 + )3.52×100

6

[1 mark]11g. Write down the interest Daniel receives after three years.

Markscheme109.70 (A1)

Note: No follow through here.

[1 mark]

[2 marks]12a.

Give all answers in this question to the nearest whole currency unit.

Ying and Ruby each have 5000 USD to invest.

Ying invests his 5000 USD in a bank account that pays a nominal annual interest rate of 4.2 % compounded yearly. Ruby investsher 5000 USD in an account that offers a fixed interest of 230 USD each year.

Find the amount of money that Ruby will have in the bank after 3 years.

Markscheme5000 + 3 × 230 = 5690 (M1)(A1)(G2)

Note: Accept alternative method.

[2 marks]

[3 marks]12b. Show that Ying will have 7545 USD in the bank at the end of 10 years.

Markscheme or equivalent (M1)(A1)

(A1)

(AG)

Note: Award (M1) for correct substituted compound interest formula, (A1) for correct substitutions, (A1) for unrounded answer seen.

If final line not seen award at most (M1)(A1)(A0).

[3 marks]

A = 5000(1 + )4.2100

10

= 7544.79…

= 7545 USD

[3 marks]12c. Find the number of complete years it will take for Ying’s investment to first exceed 6500 USD.

Markscheme5000(1.042) > 6500 (M1)(A1)

Notes: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

Follow through from their formula in part (b).

OR

List of values seen with at least 2 terms (M1)

Lists of values including at least the terms with n = 6 and n = 7 (A1)

Note: Follow through from their formula in part (b).

OR

Sketch showing 2 graphs, one exponential, the other a horizontal line (M1)

Point of intersection identified or vertical line (M1)

Note: Follow through from their formula in part (b).

n = 7 (A1)(ft)(G2)

[3 marks]

n

[3 marks]12d. Find the number of complete years it will take for Ying’s investment to exceed Ruby’s investment.

Markscheme5000(1.042) > 5000 + 230n (M1)(A1)

Note: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

OR

2 lists of values seen (at least 2 terms per list) (M1)

Lists of values including at least the terms with n = 5 and n = 6 (A1)

Note: One of the lists may be written under (c).

OR

Sketch showing 2 graphs of correct shape (M1)

Point of intersection identified or vertical line (M1)

n = 6 (A1)(ft)(G2)

Note: Follow through from their formulae used in parts (a) and (b).

[3 marks]

n

12e. [4 marks]Ruby moves from the USA to Italy. She transfers 6610 USD into an Italian bank which has an exchange rate of 1 USD =0.735 Euros. The bank charges 1.8 % commission.

Calculate the amount of money Ruby will invest in the Italian bank after commission.

Markscheme6610 × 0.735 (M1)

= 4858.35 (A1)

4858.35 × 0.982(= 4770.8997...) (M1)

= 4771 Euros (A1)(ft)(G3)

Note: Accept alternative method.

[4 marks]

12f. [5 marks]Ruby returns to the USA for a short holiday. She converts 800 Euros at a bank in Chicago and receives 1006.20 USD. Thebank advertises an exchange rate of 1 Euro = 1.29 USD.

Calculate the percentage commission Ruby is charged by the bank.

Markscheme800 × 1.29 (= 1032 USD) (M1)(A1)

Note: Award (M1) for multiplying by 1.29, (A1) for 1032. Award (G2) for 1032 if product not seen.

(1032 – 1006.20 = 25.8)

(A1)(M1)

Note: Award (A1) for 25.8 seen, (M1) for multiplying by .

OR

(M1)(A1)

OR

(M1)(A1)

(A1)(G3)

Notes: If working not shown award (G3) for 2.5.

Accept alternative method.

[5 marks]

25.8 × %1001032

1001032

= 0.9751006.201032

× 100 = 97.51006.201032

= 2.5 %

13. [3 marks]

Mr Tan invested 5000 Swiss Francs (CHF) in Bank A at an annual simple interest rate of r %, for four years. The total interest hereceived was 568 CHF.

Mr Black invested 5000 CHF in Bank B at a nominal annual interest rate of 3.6 %, compounded quarterly for four years.

Calculate the total interest he received at the end of the four years. Give your answer correct to two decimal places.

MarkschemeFinancial penalty (FP) applies in part (b).

(M1)(A1)

Note: Award (M1) for substitution into the compound interest formula, (A1) for correct values.

(FP) I = 770.70 CHF (A1) (C3)

[3 marks]

I = 5000(1.009 − 5000)16

14a. [3 marks]

Give all your numerical answers correct to two decimal places.

On 1 January 2005, Daniel invested at an annual simple interest rate in a Regular Saver account. On 1 January 2007,Daniel had in the account.

On 1 January 2005, Rebecca invested in a Supersaver account at a nominal annual rate of compoundedannually. Calculate the amount in the Supersaver account after two years.

30000 AUD31650 AUD

30000 AUD 2.5%

Markscheme (M1)(A1)

Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

(A1)(G2)

OR

(M1)(A1)

Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

(A1)(G2)

[3 marks]

Amount = 30000(1 + )2.5100

2

31518.75 AUD

I = 30000 − 30000(1 + )2.5100

2

31518.75 AUD

14b. [3 marks]On 1 January 2005, Rebecca invested in a Supersaver account at a nominal annual rate of compoundedannually.

Find the number of complete years since 1 January 2005 it would take for the amount in Rebecca’s account to exceed the amount inDaniel’s account.

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substitution in the correct formula for the two amounts, (A1) for correct substitution. Follow through from theirexpressions used in part (a) and/or part (b).

OR

2 lists of values seen (at least 2 terms per list) (M1)

lists of values including at least the terms with and (A1)(ft)

For

For

Note: Follow through from their expressions used in part (a) or/and (b).

OR

Sketch showing 2 graphs, one exponential and the other straight line (M1)

point of intersection identified (M1)

Note: Follow through from their expressions used in part (a) or/and (b).

(A1)(ft)(G2)

Note: Answer without working is awarded (G1).

Note: Accept comparison of interests instead of the total amounts in the two accounts.

[3 marks]

30000 AUD 2.5%

Rebecca's amount = 30000(1 + )2.5100

n

Daniel's amount = 30000 + 30000×2.75×n

100

n = 8 n = 9

n = 8 CI = 36552.09 SI = 36600

n = 9 CI = 37465.89 SI = 37425

n = 9

8.57

14c. [5 marks]On 1 January 2007, Daniel reinvested of the money from the Regular Saver account in an Extra Saver account at anominal annual rate of compounded quarterly.

(i) Calculate the amount of money reinvested by Daniel on the 1 January 2007.

(ii) Find the number of complete years it will take for the amount in Daniel’s Extra Saver account to exceed .

Markscheme(i) (M1)(A1)(G2)

Note: Award (M1) for correct use of percentages.

(ii) (M1)(M1)(ft)

Notes: Award (M1) for correct left-hand side of the inequality, (M1) for comparison to . Accept equation. Follow through fromtheir answer to part (d) (i).

OR

List of values from their seen (at least 2 terms) (M1)

Their correct values for ( ) and ( ) seen (A1)(ft)

Note: Follow through from their answer to (d) (i).

OR

Sketch showing 2 graphs – an exponential and a horizontal line (M1)

Point of intersection identified or vertical line drawn (M1)

Note: Follow through from their answer to (d) (i).

(A1)(ft)(G2)

Note: Award (G1) for answer with no working.

[5 marks]

80%3%

30000 AUD

0.80 × 31650 = 25320

25320 > 30000(1 + )34×100

4n

30000

25320(1 + )34×100

4n

n = 5 29401.18 n = 6 30293

n = 6

5.67

15. [4 marks]

An amount, C, of Australian Dollars (AUD) is invested for 5 years at 2.5 % yearly simple interest. The interest earned on thisinvestment is 446.25 AUD.

5000 AUD is invested at a nominal annual interest rate of 2.5 % compounded half yearly.

Calculate the length of time in years for the interest on this investment to exceed 446.25 AUD.

Markscheme (M1)(A1)

Notes: Award (M1) for substitution into compound interest formula. Award (A1) for correct values.

(A1)

n = 3.44

n = 3.5 (A1)

OR

5446.25 = 5000(1.0125) (A1)(M1)(A1)

Notes: Award (A1) for 5446.25 seen.

Award (M1) for substitution into compound interest formula.

Award (A1) for correct values.

n = 3.44 years

3.5 years required (A1) (C4)

Notes: For incorrect substitution into compound interest formula award at most (M1)(A0)(A1)(A0).

Award (A3) for 3.44 seen without working.

Allow solution by lists. In this case

Award (A1) for half year rate 1.25 % seen.

(A1) for 5446.25 seen.

(M1) for at least 2 correct uses of multiplication by 1.0125

5000 × 1.0125 = 5062.5 and 5062.5 × 1.0125 = 5125.78125

(A1) n = 3.5

If yearly rate used then award (A0)(A1)(M1)(A0)

[4 marks]

446.25 = 5000 − 5000(1 + )2.52(100)

2n

5446.25 = 5000(1 + )2.52(100)

2n

2n

16. [3 marks]

Inge borrows € 4500 for 2 years.

Bank 1 charges compound interest at a rate of 15 % per annum, compounded quarterly.

Calculate the total amount to be repaid at the end of the 2 years. Give your answer correct to two decimal places.

MarkschemeNote: Financial penalty (FP) applies in this part

(M1)(A1)

Note: Award (M1) for substitution into CI formula, (A1) for correct substitution.

(FP) A = € 6041.12 (€ not required) (A1) (C3)

[3 marks]

A = 4500(1 + )15400

4×2

[3 marks]17a.

Yun Bin invests in an account which pays a nominal annual interest rate of , compounded monthly.

Give all answers correct to two decimal places.

Find the value of the investment after 3 years.

Markscheme (M1)(A1)


OR

(M1)(A1)


OR

(M1)(A1)


(A1) (C3)

Note: The answer should be given correct to two decimal places or the final (A1) is not awarded.

5000 euros 6.25%

FV = 5000(1 + )6.251200

3×12

N = 3I% = 6.25PV = −5000P/Y = 1C/Y = 12

C/Y = 12

N = 36I% = 6.25PV = −5000P/Y = 12C/Y = 12

C/Y = 12

= 6028.22

[3 marks]17b. Find the difference in the final value of the investment if the interest was compounded quarterly at the same nominal rate.

Markscheme (M1)

Note: Award (M1) for correctly substituted compound interest formula.

OR

(M1)

Note: Award (M1) for all correct entries seen.

OR

(M1)

Note: Award (M1) for all correct entries seen.

(A1) (A1)(ft) (C3)

Notes: Accept . This answer should be given correct to two decimal places or the final (A1) is not awarded unless this hasalready been penalized in part (a). Follow through from part (a).Notes: Illustrating use of GDC notation acceptable in this case only. However on P2 an answer given with no working would receiveG2.

FV = 5000(1 + )6.25400

3×4

N = 3I% = 6.25PV = −5000P/Y = 1C/Y = 4

N = 12I% = 6.25PV = −5000P/Y = 4C/Y = 4

FV = 6022.41Difference = 5.80

5.81

[3 marks]18.

Ben inherits $6500. Ben invests his money in a bank that pays compound interest at a rate of 4.5% per annum.

Calculate the value of Ben’s investment at the end of 6 years. Give your answer correct to 2 decimal places.

Markscheme (M1)(A1)

(A1)

(M1)(A1)(A0) if interest only found (=$1964.69) (C3)

[3 marks]

Ben Amount = 6500(1 + )4.5100

6

= $8464.69

19. [3 marks]Eva invests at a nominal annual interest rate of compounded half-yearly.

Calculate the value of her investment after years, correct to the nearest dollar.

USD2000 8%

5

Markscheme (M1)(A1)

Note: (M1) for substitution into CI formula. (A1) for correct substitution.

(A1)

Note: Award the final A1 for rounding their answer correctly to the nearest Yuan.

OR

(M1)(A1)

Note: (M1) for substitution into CI formula. (A1) for correct substitution.

(A1) (C3)

Note: Award the final A1 for rounding their answer correctly to the nearest Yuan.

2000(1.04)10

2960

2000 − 2000(1 + )8200

10

2960

20. [3 marks]Charles invests in a bank that offers compound interest at a rate of per annum, compounded half-yearly.

Calculate the number of years that it takes for Charles’s money to double.

Markscheme (M1)(A1)

Note: (M1) for substituting values into a compound interest formula, (A1) for correct values with a variable for the power.

(A1) (C3)

Note: If used in formula instead of , can allow as long as final answer is halved to get .

[3 marks]

3000 USD 3.5%

6000 = 3000(1 + )3.5200

2n

n = 20 years

n 2n 20

21a. [3 marks]

Emma places in a bank account that pays a nominal interest rate of per annum, compounded quarterly.

Calculate the amount of money that Emma would have in her account after 15 years. Give your answer correct to the nearestEuro.

Markscheme (M1)(A1)

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

only (A1) (C3)

[3 marks]

€8000 5%

FV = 8000(1.0125)60

€16857

21b. [3 marks]After a period of time she decides to withdraw the money from this bank. There is in her account. Find the numberof months that Emma had left her money in the account.

€9058.17

Printed for Victoria Shanghai Academy

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme (M1)

Note: (M1) for equating compound interest formula to

correct answer only (A1)

So 30 months, (ft) on their (A1)(ft) (C3)

Note: Award (C2) for seen with no working.

[3 marks]

8000(1.0125 = 9058.17)n

9058.17

n = 10

n

2.5

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