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]
Question
Pauline owns a piece of land ABCD in the shape of a quadrilateral. The length of BC is \(190{\text{ m}}\) , the length of CD is \(120{\text{ m}}\) , the length of AD is \(70{\text{ m}}\) , the size of angle BCD is \({75^ \circ }\) and the size of angle BAD is \({115^ \circ }\) .
Pauline decides to sell the triangular portion of land ABD . She first builds a straight fence from B to D .
Calculate the length of the fence.[3]
The fence costs \(17\) USD per metre to build.
Calculate the cost of building the fence. Give your answer correct to the nearest USD.[2]
Show that the size of angle ABD is \({18.8^ \circ }\) , correct to three significant figures.[3]
Calculate the area of triangle ABD .[4]
She sells the land for \(120\) USD per square metre.
Calculate the value of the land that Pauline sells. Give your answer correct to the nearest USD.[2]
Pauline invests \(300 000\) USD from the sale of the land in a bank that pays compound interest compounded annually.
Find the interest rate that the bank pays so that the investment will double in value in 15 years.[4]
Answer/Explanation
Markscheme
\({\text{B}}{{\text{D}}^2} = {190^2} + {120^2} – 2(190)(120)\cos {75^ \circ }\) (M1)(A1)
Note: Award (M1) for substituted cosine formula, (A1) for correct substitution.
\(= 197\) m (A1)(G2)
Note: If radians are used award a maximum of (M1)(A1)(A0).[3 marks]
\({\text{cost}} = 196.717 \ldots \times 17\) (M1)
\( = 3344{\text{ USD}}\) (A1)(ft)(G2)
Note: Accept \(3349\) from \(197\).[2 marks]
\(\frac{{\sin ({\text{ABD}})}}{{70}} = \frac{{\sin ({{115}^ \circ })}}{{196.7}}\) (M1)(A1)
Note: Award (M1) for substituted sine formula, (A1) for correct substitution.
\( = {18.81^ \circ } \ldots \) (A1)(ft)
\( = {18.8^ \circ } \) (AG)
Notes: Both the unrounded and rounded answers must be seen for the final (A1) to be awarded. Follow through from their (a). If 197 is used the unrounded answer is \( = {18.78^ \circ } \ldots \)[3 marks]
\({\text{angle BDA}} = {46.2^ \circ }\) (A1)
\({\text{Area}} = \frac{{70 \times (196.717 \ldots ) \times \sin ({{46.2}^ \circ })}}{2}\) (M1)(A1)
Note: Award (M1) for substituted area formula, (A1) for correct substitution.
\({\text{Area ABD}} = 4970{\text{ }}{{\text{m}}^2}\) (A1)(ft)(G2)
Notes: If \(197\) used answer is \(4980\).
Notes: Follow through from (a) only. Award (G2) if there is no working shown and \({46.2^ \circ }\) not seen. If \({46.2^ \circ }\) seen without subsequent working, award (A1)(G2).[4 marks]
\(4969.38 \ldots \times 120\) (M1)
\( = 596 327{\text{ USD}}\) (A1)(ft)(G2)
Notes: Follow through from their (d).[2 marks]
\(300000{\left( {1 + \frac{r}{{100}}} \right)^{15}} = 600000\) or equivalent (A1)(M1)(A1)
Notes: Award (A1) for \(600 000\) seen or implied by alternative formula, (M1) for substituted CI formula, (A1) for correct substitutions.
\(r = 4.73\) (A1)(ft)(G3)
Notes: Award (G3) for \(4.73\) with no working. Award (G2) for \(4.7\) with no working.[4 marks]
Question
Give all your numerical answers correct to two decimal places.
On 1 January 2005, Daniel invested \(30000{\text{ AUD}}\) at an annual simple interest rate in a Regular Saver account. On 1 January 2007, Daniel had \(31650{\text{ AUD}}\) in the account.
On 1 January 2005, Rebecca invested \(30000{\text{ AUD}}\) in a Supersaver account at a nominal annual rate of \(2.5\% \) compounded annually. Calculate the amount in the Supersaver account after two years.[3]
On 1 January 2005, Rebecca invested \(30000{\text{ AUD}}\) in a Supersaver account at a nominal annual rate of \(2.5\% \) compounded annually.
Find the number of complete years since 1 January 2005 it would take for the amount in Rebecca’s account to exceed the amount in Daniel’s account.[3]
On 1 January 2007, Daniel reinvested \(80\% \) of the money from the Regular Saver account in an Extra Saver account at a nominal annual rate of \(3\% \) 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 \(30000{\text{ AUD}}\).[5]
Answer/Explanation
Markscheme
\({\text{Amount}} = 30000{\left( {1 + \frac{{2.5}}{{100}}} \right)^2}\) (M1)(A1)
Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.
\(31518.75{\text{ AUD}}\) (A1)(G2)
OR
\({\text{I}} = 30000{\left( {1 + \frac{{2.5}}{{100}}} \right)^2} – 30000\) (M1)(A1)
Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.
\(31518.75{\text{ AUD}}\) (A1)(G2)[3 marks]
\({\text{Rebecca’s amount}} = 30000{\left( {1 + \frac{{2.5}}{{100}}} \right)^n}\)
\({\text{Daniel’s amount}} = 30000 + \frac{{30000 \times 2.75 \times n}}{{100}}\) (M1)(A1)(ft)
Note: Award (M1) for substitution in the correct formula for the two amounts, (A1) for correct substitution. Follow through from their expressions 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 \(n = 8\) and \(n = 9\) (A1)(ft)
For \(n = 8\) \({\text{CI}} = 36552.09\) \({\text{SI}} = 36600\)
For \(n = 9\) \({\text{CI}} = 37465.89\) \({\text{SI}} = 37425\)
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).
\(n = 9\) (A1)(ft)(G2)
Note: Answer \(8.57\) without working is awarded (G1).
Note: Accept comparison of interests instead of the total amounts in the two accounts.[3 marks]
(i) \(0.80 \times 31650 = 25320\) (M1)(A1)(G2)
Note: Award (M1) for correct use of percentages.
(ii) \(25320{\left( {1 + \frac{3}{{4 \times 100}}} \right)^{4n}} > 30000\) (M1)(M1)(ft)
Notes: Award (M1) for correct left-hand side of the inequality, (M1) for comparison to \(30000\). Accept equation. Follow through from their answer to part (d) (i).
OR
List of values from their \(25320{\left( {1 + \frac{3}{{4 \times 100}}} \right)^{4n}} \) seen (at least 2 terms) (M1)
Their correct values for \(n = 5\) (\(29401.18\)) and \(n = 6\) (\(30293\)) 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).
\(n = 6\) (A1)(ft)(G2)
Note: Award (G1) for answer \(5.67\) with no working.[5 marks]
Question
Cedric wants to buy an €8000 car. The car salesman offers him a loan repayment option of a 25 % deposit followed by 12 equal monthly payments of €600 .
Write down the amount of the deposit.[1]
Calculate the total cost of the loan under this repayment scheme.[2]
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 repay the 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 total amount that Cedric’s mother receives after 24 months is €7100.
Write down a second equation involving x and y.[1]
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 repay the 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 total amount that Cedric’s mother receives after 24 months is €7100.
Write down the value of x and the value of y.[2]
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 repay the 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 total amount that Cedric’s mother receives after 24 months is €7100.
Calculate the number of months it will take Cedric’s mother to receive the €8000.[3]
Cedric decides to buy a cheaper car for €6000 and invests the remaining €2000 at his bank. The bank offers two investment options 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.[5]
Answer/Explanation
Markscheme
2000 (euros) (A1)[1 mark]
\(2000 + 12 \times 600\) (M1)
Note: Award (M1) for addition of two correct terms.
9200 (euros) (A1)(ft)(G2)
Note: Follow through from their part (a).[2 marks]
x + 23y = 7100 (A1)[1 mark]
x = 200, y = 300 (A1)(ft)(A1)(ft)(G2)[2 marks]
\(200 + n \times 300 = 8000\) (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 is incorrect). 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
\(\frac{{8000 – 7100}}{{300}} + 24\) (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]
(i) \(2000{\left( {1 + \frac{8}{{100}}} \right)^3}\) (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) \(2000{\left( {1 + \frac{7.5}{{100 \times 12}}} \right)^{3\times12}}\) (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]
Question
Give all answers in this question correct to two decimal places.
Arthur lives in London. On \({1^{{\text{st}}}}\) August 2008 Arthur paid \({\text{37}}\,{\text{500}}\) euros (\({\text{EUR}}\)) for a new car from Germany. The price of the same car in London was \({\text{34}}\,{\text{075}}\) British pounds (\({\text{GBP}}\)).
The exchange rate on \({1^{{\text{st}}}}\) August 2008 was \({\text{1}}\,{\text{EUR = 0.7234}}\,{\text{GBP}}\).
Calculate, in GBP, the price that Arthur paid for the car.[2]
Write down, in \({\text{GBP}}\), the amount of money Arthur saved by buying the car in Germany.[1]
Between \({1^{{\text{st}}}}\) August 2008 and \({1^{{\text{st}}}}\) August 2012 Arthur’s car depreciated at an annual rate of \(9\%\) of its current value.
Calculate the value, in \({\text{GBP}}\), of Arthur’s car on \({1^{{\text{st}}}}\) August 2009.[3]
Between \({1^{{\text{st}}}}\) August 2008 and \({1^{{\text{st}}}}\) August 2012 Arthur’s car depreciated at an annual rate of \(9\%\) of its current value.
Show that the value of Arthur’s car on \({1^{{\text{st}}}}\) August 2012 was \({\text{18}}\,{\text{600}}\,{\text{GBP}}\), correct to the nearest \({\text{100}}\,{\text{GBP}}\).[3]
Answer/Explanation
Markscheme
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
\(37\,500 \times 0.7234\) (M1)
\( = 27\,127.50\) (A1)(G2)[2 marks]
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
\(6947.50\) (A1)(ft)(G1)
Note: Follow through from part (a) irrespective of whether working is seen.[1 mark]
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
\(27\,127.50 \times 0.91\) (A1)(M1)
Note: Award (A1) for \(0.91\) seen or equivalent, (M1) for their \({\text{27}}\,{\text{127.50}}\) multiplied by \(0.91\)
OR
\(27\,127.50 – 0.09 \times 27\,127.50\) (A1)(M1)
Note: Award (A1) for \(0.09 \times 27\,127.50\) seen, and (M1) for \(27\,127.50 – 0.09 \times 27\,127.50\).
\( = 24\,686.03\) (A1)(ft)(G2)
Note: Follow through from part (a).[3 marks]
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
\(27\,127.50 \times {\left( {1 – \frac{9}{{100}}} \right)^4}\) (M1)(A1)(ft)
Notes: Award (M1) for substituted compound interest formula, (A1)(ft) for correct substitution.
Follow through from part (a).
OR
\(27\,127.50 \times {(0.91)^4}\) (M1)(A1)(ft)
Notes: Award (M1) for substituted geometric sequence formula, (A1)(ft) for correct substitution.
Follow through from part (a).
OR (lists (i))
\({\text{24}}\,{\text{686.03, 22}}\,{\text{464.28…, 20}}\,{\text{442.50…, 18}}\,{\text{602.67…}}\) (M1)(A1)(ft)
Notes: Award (M1) for at least the \({{\text{2}}^{{\text{nd}}}}\) term correct (calculated from their \(({\text{a}}) \times 0.91\)). Award (A1)(ft) for four correct terms (rounded or unrounded).
Follow through from part (a).
Accept list containing the last three terms only (\({\text{24}}\,{\text{686.03}}\) may be implied).
OR (lists(ii))
\(27\,127.50 – (2441.47… + 2221.74… + 2021.79… + 1839.82…)\) (M1)(A1)(ft)
Notes: Award (M1) for subtraction of four terms from \({\text{27}}\,{\text{127.50}}\).
Award (A1) for four correct terms (rounded or unrounded).
Follow through from part (a).
\( = 18\,602.67\) (A1)
\( = 18\,600\) (AG)
Note: The final (A1) is not awarded unless both the unrounded and rounded answers are seen.[3 marks]
Question
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 \(\$60\) each month for two years
Option B \(\$10\) in the first month, \(\$15\) in the second month, \(\$20\) in the third month, increasing by \(\$5\) each month for two years
Option C \(\$15\) in the first month and increasing by \(10\%\) each month for two years
Option D Investing \(\$1500\) at a bank at the beginning of the first year, with an interest rate of \(6\%\) per annum, compounded monthly.
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.[2]
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.[5]
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.[5]
If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.[3]
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.[1]
Another bank guarantees Hugh an amount of \(\$1750\) 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.[3]
Answer/Explanation
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.
\(60 \times 24\) (M1)
Note: Award (M1) for correct product.
\( = 1440\) (A1)(G2)[2 marks]
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.
(i) \(10 + (17 – 1)(5)\) (M1)(A1)
Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.
\( = 90\) (A1)(G2)
(ii) \(\frac{{24}}{2}\left( {2(10) + (24 – 1)(5)} \right)\) (M1)
OR
\(\frac{{24}}{2}\left( {10 + 125} \right)\) (M1)
Note: Award (M1) for correct substitution in arithmetic series formula.
\( = 1620\) (A1)(ft)(G1)
Note: Follow through from part (b)(i).[5 marks]
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.
(i) \(15{(1.1)^{12}}\) (M1)(A1)
Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.
\( = 47\) (A1)(G2)
Note: Award (M1)(A1)(A0) for \(47.08\).
Award (G1) for \(47.08\) if workings are not shown.
(ii) \(\frac{{15({{1.1}^{24}} – 1)}}{{1.1 – 1}}\) (M1)
Note: Award (M1) for correct substitution in geometric series formula.
\( = 1327\) (A1)(ft)(G1)
Note: Follow through from part (c)(i).[5 marks]
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.
\(1500{\left( {1 + \frac{6}{{100(12)}}} \right)^{12(2)}}\) (M1)(A1)
Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.
OR
\(N = 2\)
\(I\% = 6\)
\(PV = 1500\)
\(P/Y = 1\)
\(C/Y = 12\) (A1)(M1)
Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.
OR
\(N = 24\)
\(I\% = 6\)
\(PV = 1500\)
\(P/Y = 12\)
\(C/Y = 12\) (A1)(M1)
Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.
\( = 1691\) (A1)(G2)[3 marks]
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.
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]
\(1750 = 1500{\left( {1 + \frac{r}{{100}}} \right)^2}\) (M1)(A1)
Note: Award (M1) for substituted compound interest formula equated to \(1750\), (A1) for correct substitutions into formula.
OR
\(N = 2\)
\(PV = 1500\)
\(FV = – 1750\)
\(P/Y = 1\)
\(C/Y = 1\) (A1)(M1)
Note: Award (A1) for \(FV = 1750\) seen, (M1) for other correct entries.
\( = 8.01\% {\text{ (8.01234}} \ldots \% ,{\text{ }}0.0801{\text{)}}\) (A1)(G2)[3 marks]
Question
Give your answers to parts (b), (c) and (d) to the nearest whole number.
Harinder has 14 000 US Dollars (USD) to invest for a period of five years. He has two options of how to invest the money.
Option A: Invest the full amount, in USD, in a fixed deposit account in an American bank.
The account pays a nominal annual interest rate of r % , compounded yearly, for the five years. The bank manager says that this will give Harinder a return of 17 500 USD.
Option B: Invest the full amount, in Indian Rupees (INR), in a fixed deposit account in an Indian bank. The money must be converted from USD to INR before it is invested.
The exchange rate is 1 USD = 66.91 INR.
The account in the Indian bank pays a nominal annual interest rate of 5.2 % compounded monthly.
Calculate the value of r.[3]
Calculate 14 000 USD in INR.[2]
Calculate the amount of this investment, in INR, in this account after five years.[3]
Harinder chose option B. At the end of five years, Harinder converted this investment back to USD. The exchange rate, at that time, was 1 USD = 67.16 INR.
Calculate how much more money, in USD, Harinder earned by choosing option B instead of option A.[3]
Answer/Explanation
Markscheme
\(17500 = 14000{\left( {1 + \frac{r}{{100}}} \right)^5}\) (M1)(A1)
Note: Award (M1) for substitution into the compound interest formula, (A1) for correct substitution. Award at most (M1)(A0) if not equated to 17500.
OR
N = 5
PV = ±14000
FV = \( \mp \)17500
P/Y = 1
C/Y = 1 (A1)(M1)
Note: Award (A1) for C/Y = 1 seen, (M1) for all other correct entries. FV and PV must have opposite signs.
= 4.56 (%) (4.56395… (%)) (A1) (G3)[3 marks]
14000 × 66.91 (M1)
Note: Award (M1) for multiplying 14000 by 66.91.
936740 (INR) (A1) (G2)
Note: Answer must be given to the nearest whole number.[2 marks]
\(936740 \times {\left( {1 + \frac{{5.2}}{{12 \times 100}}} \right)^{12 \times 5}}\) (M1)(A1)(ft)
Note: Award (M1) for substitution into the compound interest formula, (A1)(ft) for their correct substitution.
OR
N = 60
I% = 5.2
PV = ±936740
P/Y= 12
C/Y= 12 (A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries.
OR
N = 5
I% = 5.2
PV = ±936740
P/Y= 1
C/Y= 12 (A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries
= 1214204 (INR) (A1)(ft) (G3)
Note: Follow through from part (b). Answer must be given to the nearest whole number.[3 marks]
\(\frac{{1214204}}{{67.16}}\) (M1)
Note: Award (M1) for dividing their (c) by 67.16.
\(\left( {\frac{{1214204}}{{67.16}}} \right) – 17500 = 579\) (USD) (M1)(A1)(ft) (G3)
Note: Award (M1) for finding the difference between their conversion and 17500. Answer must be given to the nearest whole number. Follow through from part (c).[3 marks]