Question 3 [Maximum mark: 19]
Give your answers in parts (a), (d)(i), (e) and (f) to the nearest dollar.
Daisy invested 37 000 Australian dollars (AUD) in a fixed deposit account with an annual interest rate of 6.4 % compounded quarterly.
a. Calculate the value of Daisy’s investment after 2 years. [3]
After m months, the amount of money in the fixed deposit account has appreciated to more than 50 000 AUD.
b. Find the minimum value of m , where m∈N. . [4]
Daisy is saving to purchase a new apartment. The price of the apartment is 200 000 AUD.
Daisy makes an initial payment of 25 % and takes out a loan to pay the rest.
c. Write down the amount of the loan. [1]
The loan is for 10 years, compounded monthly, with equal monthly payments of 1700 AUD
made by Daisy at the end of each month.
d. For this loan, find
(i)the amount of interest paid by Daisy.
(ii)the annual interest rate of the loan. [5]
After 5 years of paying off this loan, Daisy decides to pay the remainder in one final payment.
e. Find the amount of Daisy’s final payment. [3]
f. Find how much money Daisy saved by making one final payment after 5 years. [3]
Answer/Explanation
(a) EITHER
\(N=2\) or \( N=8\)
\(PV= -37000\) or \(PV=-37000\)
\(I%=6.4\) or \( I% = 6.4 \)
\(P/Y = 1\) or \(P/Y= 4 \)
\(C/Y = 4\) OR \( C/Y = 4\)
OR
\(FV = 37000\times (1+\frac{6.4}{100\times 4})^{4\times 2}\)
\(= 42010AUD\)
(b) EITHER
\(PV=-37000 \) or \( PV=-37000\)
\(FV = 50000\) or \( FV = 50000\)
\(I%=6.4\) or \(I%=6.4\)
\( P/Y = 1\) or \(P/Y = 4\)
\(C/Y = 4\) or \(C/Y = 4\)
OR
\(50000< 37000\times (1+\frac{6.4}{100\times 4})^{4\times n}\)
OR \( 50000< 37000\times (1+\frac{6.4}{100\times 4})^{ n}\) THEN
\(N = 4.74 (year ) (4.74230…)\) OR \(N =18.9692… m = 57 months\) .
(c) 150000 AUD
(d) (i) \(1201700-1\left [ \frac{9.39}{sin100} =\frac{4.5}{sin(60-\Theta)}=\frac{7.5}{sin(\Theta +20)}\right ]5000= 54000\) AUD
(ii)
\(N=120\) , \(PV= -150000\) , \(PMT = 1700\) , \(FV = 0\) , \(P/Y = 12 \) , \(C/Y= 12\)
r = 6.46(%)(6.45779…)
(e) \(N= 60\) , \(L= 6.6=46(6.45779..)\) , \(PV = – 150000\) , \(PMT = 1700\) , \( P/Y = 12\) , \(V/Y= 12 \), \(FV = 86973\) AUD
(f) 204000 – (60 × 1700 + 86973) OR \(204000 – 188973 \rightarrow 15027\) AUD
Question
Give all answers in this question correct to the nearest dollar.
Clara wants to buy some land. She can choose between two different payment options. Both options require her to pay for the land in 20 monthly installments.
Option 1: The first installment is \(\$ 2500\). Each installment is \(\$ 200\) more than the one before.
Option 2: The first installment is \(\$ 2000\). Each installment is \(8\% \) more than the one before.
If Clara chooses option 1,
(i) write down the values of the second and third installments;
(ii) calculate the value of the final installment;
(iii) show that the total amount that Clara would pay for the land is \(\$ 88000\).[7]
If Clara chooses option 2,
(i) find the value of the second installment;
(ii) show that the value of the fifth installment is \(\$ 2721\).[4]
The price of the land is \(\$ 80000\). In option 1 her total repayments are \(\$ 88000\) over the 20 months. Find the annual rate of simple interest which gives this total.[4]
Clara knows that the total amount she would pay for the land is not the same for both options. She wants to spend the least amount of money. Find how much she will save by choosing the cheaper option.[4]
Answer/Explanation
Markscheme
(i) Second installment \( = \$ 2700\) (A1)
Third installment \( = \$ 2900\) (A1)
(ii) Final installment \( = 2500 + 200 \times 19\) (M1)(A1)
Note: (M1) for substituting in correct formula or listing, (A1) for correct substitutions.
\( = \$ 6300\) (A1)(G2)
(iii) Total amount \( = \frac{{20}}{2}(2500 + 6300)\)
OR
\( \frac{{20}}{2}(5000 + 19 \times 200)\) (M1)(A1)
Note: (M1) for substituting in correct formula or listing, (A1) for correct substitution.
\( = \$ 88000\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[7 marks]
(i) Second installment \(2000 \times 1.08 = \$ 2160\) (M1)(A1)(G2)
Note: (M1) for multiplying by \(1.08\) or equivalent, (A1) for correct answer.
(ii) Fifth installment \( = 2000 \times {1.08^4} = 2720.98 = \$ 2721\) (M1)(A1)(AG)
Notes: (M1) for correct formula used with numbers from the problem. (A1) for correct substitution. The \(2720.9 \ldots \) must be seen for the (A1) mark to be awarded. Accept list of 5 correct values. If values are rounded prematurely award (M1)(A0)(AG).[4 marks]
Interest is \( = \$ 8000\) (A1)
\(80000 \times \frac{r}{{100}} \times \frac{{20}}{{12}} = 8000\) (M1)(A1)
Note: (M1) for attempting to substitute in simple interest formula, (A1) for correct substitution.
Simple Interest Rate \( = 6\% \) (A1)(G3)
Note: Award (G3) for answer of \(6\% \) with no working present if interest is also seen award (A1) for interest and (G2) for correct answer.[4 marks]
Financial accuracy penalty (FP) is applicable where indicated in the left hand column.
(FP) Total amount for option 2 \( = 2000\frac{{(1 – {{1.08}^{20}})}}{{(1 – 1.08)}}\) (M1)(A1)
Note: (M1) for substituting in correct formula, (A1) for correct substitution.
\( = \$ 91523.93\) (\( = \$ 91524\)) (A1)
\(91523.93 – 88000 = \$ 3523.93 = \$ 3524\) to the nearest dollar (A1)(ft)(G3)
Note: Award (G3) for an answer of \(\$ 3524\) with no working. The difference follows through from the sum, if reasonable. Award a maximum of (M1)(A0)(A0)(A1)(ft) if candidate has treated option 2 as an arithmetic sequence and has followed through into their common difference. Award a maximum of (M1)(A1)(A0)(ft)(A0) if candidate has consistently used \(0.08\) in (b) and (d).[4 marks]
Question
Daniel wants to invest \(\$ 25\,000\) for a total of three years. There are two investment options.
Option One pays compound interest at a nominal annual rate of interest of 5 %, compounded annually.
Option Two pays compound interest at a nominal annual rate of interest of 4.8 %, compounded monthly.
An arithmetic sequence is defined as
un = 135 + 7n, n = 1, 2, 3, …
Calculate the value of his investment at the end of the third year for each investment option, correct to two decimal places.[8]
Determine Daniel’s best investment option.[1]
Calculate u1, the first term in the sequence.[2]
Show that the common difference is 7.[2]
Sn is the sum of the first n terms of the sequence.
Find an expression for Sn. Give your answer in the form Sn = An2 + Bn, where A and B are constants.[3]
The first term, v1, of a geometric sequence is 20 and its fourth term v4 is 67.5.
Show that the common ratio, r, of the geometric sequence is 1.5.[2]
Tn is the sum of the first n terms of the geometric sequence.
Calculate T7, the sum of the first seven terms of the geometric sequence.[2]
Tn is the sum of the first n terms of the geometric sequence.
Use your graphic display calculator to find the smallest value of n for which Tn > Sn.[2]
Answer/Explanation
Markscheme
Option 1: Amount \( = 25\,000{\left( {1 + \frac{5}{{100}}} \right)^3}\) (M1)(A1)
= \(28\,940.63\) (A1)(G2)
Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution. Give full credit for use of lists.
Option 2: Amount \( = 25\,000{\left( {1 + \frac{{4.8}}{{12(100)}}} \right)^{3 \times 12}}\) (M1)
= \(28\,863.81\) (A1)(G2)
Note: Award (M1) for correct substitution in the compound interest formula. Give full credit for use of lists.[8 marks]
Option 1 is the best investment option. (A1)(ft)[1 mark]
u1 = 135 + 7(1) (M1)
= 142 (A1)(G2)[2 marks]
u2 = 135 + 7(2) = 149 (M1)
d = 149 – 142 OR alternatives (M1)(ft)
d = 7 (AG)[2 marks]
\({S_n} = \frac{{n[2(142) + 7(n – 1)]}}{2}\) (M1)(ft)
Note: Award (M1) for correct substitution in correct formula.
\( = \frac{{n[277 + 7n]}}{2}\) OR equivalent (A1)
\( = \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\) (= 3.5n2 + 138.5n) (A1)(G3)[3 marks]
20r3 = 67.5 (M1)
r3 = 3.375 OR \(r = \sqrt[3]{{3.375}}\) (A1)
r = 1.5 (AG)[2 marks]
\({T_7} = \frac{{20({{1.5}^7} – 1)}}{{(1.5 – 1)}}\) (M1)
Note: Award (M1) for correct substitution in correct formula.
= 643 (accept 643.4375) (A1)(G2)[2 marks]
\(\frac{{20({{1.5}^n} – 1)}}{{(1.5 – 1)}} > \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\) (M1)
Note: Award (M1) for an attempt using lists or for relevant graph.
n = 10 (A1)(ft)(G2)
Note: Follow through from their (c).[2 marks]
Question
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 invests her 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.[2]
Show that Ying will have 7545 USD in the bank at the end of 10 years.[3]
Find the number of complete years it will take for Ying’s investment to first exceed 6500 USD.[3]
Find the number of complete years it will take for Ying’s investment to exceed Ruby’s investment.[3]
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.[4]
Ruby returns to the USA for a short holiday. She converts 800 Euros at a bank in Chicago and receives 1006.20 USD. The bank advertises an exchange rate of 1 Euro = 1.29 USD.
Calculate the percentage commission Ruby is charged by the bank.[5]
Answer/Explanation
Markscheme
5000 + 3 × 230 = 5690 (M1)(A1)(G2)
Note: Accept alternative method.[2 marks]
\({\text{A}} = 5000{\left( {1 + \frac{{4.2}}{{100}}} \right)^{10}}\) or equivalent (M1)(A1)
\( = 7544.79 \ldots \) (A1)
\( = 7545{\text{ USD}}\) (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]
5000(1.042)n > 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]
5000(1.042)n > 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]
6610 × 0.735 (M1)
= 4858.35 (A1)
4858.35 × 0.982(= 4770.8997…) (M1)
= 4771 Euros (A1)(ft)(G3)
Note: Accept alternative method.[4 marks]
800 × 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)
\(25.8 \times \frac{100}{1032} \%\) (A1)(M1)
Note: Award (A1) for 25.8 seen, (M1) for multiplying by \(\frac{100}{1032}\).
OR
\(\frac{1006.20}{1032} = 0.975\) (M1)(A1)
OR
\(\frac{1006.20}{1032} \times 100 = 97.5\) (M1)(A1)
\( = 2.5{\text{ }}\% \) (A1)(G3)
Notes: If working not shown award (G3) for 2.5.
Accept alternative method.[5 marks]
Question
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.
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.
Find, in BRL, the amount of money Estela has after commission.[2]
Find, in GBP, the amount of money Estela receives.[2]
After her trip to the United Kingdom Estela has 400 GBP left. At the airport she changes this money back into BRL. The exchange rate is now 1.00 BRL = 0.3125 GBP.
Find, in BRL, the amount of money that Estela should receive.[2]
Estela actually receives 1216.80 BRL after commission.
Find, in BRL, the commission charged to Estela.[1]
The commission rate is t % . Find the value of t.[2]
Show that after three years Daniel will have $1109.70 in his account, correct to two decimal places.[3]
Write down the interest Daniel receives after three years.[1]
Answer/Explanation
Markscheme
4000 × 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]
3880 × 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]
\(\frac{{400}}{{0.3125}}\) (M1)
= 1280.00 (1280) (A1)(G2)
Note: Award (M1) for division of correct numbers.[2 marks]
63.20 (A1)(ft)
Note: Follow through (their (c) –1216.80).[1 mark]
\(t = \frac{{63.20 \times 100}}{{1280}}\) (M1)
t = 4.94 (A1)(ft)(G2)
Note: Follow through from their answers to parts (c) and (d).[2 marks]
\({\text{A}} = 1000{\left( {1 + \frac{{3.5}}{{2 \times 100}}} \right)^6} = 1109.7023…\) (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 not seen award at most (M1)(A1)(A0).
OR
\({\text{I}} = 1000{\left( {1 + \frac{{3.5}}{{2 \times 100}}} \right)^6} – 1000 = 109.7023\) (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]
109.70 (A1)
Note: No follow through here.[1 mark]