Question
Juliana plans to invest money for 10 years in an account paying 3.5% interest, compounded
annually. She expects the annual inflation rate to be 2% per year throughout the 10-year period.
Juliana would like her investment to be worth a real value of $4000, compared to current values, at the end of the 10-year period. She is considering two options.
Option 1: Make a one-time investment at the start of the 10-year period.
Option 2: Invest $1000 at the start of the 10-year period and then invest $ x into the account at the end of each year (including the first and last years).
(a) For option 1, determine the minimum amount Juliana would need to invest. Give your
answer to the nearest dollar.
(b) For option 2, find the minimum value of x that Juliana would need to invest each year.
Give your answer to the nearest dollar.
▶️Answer/Explanation
Ans:
(a) METHOD 1 – (With FV=4000)
EITHER
N= 10
I=1.5
FV= 4000
P/Y= 1
C/Y= 1
OR
4000 = A \((1+0.015)^{-10}\)
THEN
(PV=) $ 3447
METHOD 2 – (With FV including inflation)
calculate FV with inflation
\(4000 \times 1.02^{10} = PV \times 1.035^{10}\)
OR
N= 10
I= 3.5
FV= 4875.977…
P/Y= 1
C/Y= 1
THEN
(PV=) $3457
METHOD 3 – (Using formula to calculate real rate of return)
(real rate of return =) 1.47058…(%)
EITHER
\(4000 = PV \times 1.0147058…^{10}\)
OR
N= 10
I= 1.47058…
FV= 4000
P/Y= 1
C/Y= 1
THEN
(PV=) $3457
(b) METHOD 1 – (Finding the future value of the investment using PV from part (a))
N= 10
I=3.5
PV= 3446.66…(from Method 1) OR 3456.67…(from Methods 2, 3)
P/Y= 1
C/Y= 1
(FV=) $4861.87 OR $4875.97
so payment required (from TVM) will be $294 OR $295
METHOD 2 – (Using FV)
N= 10
I=3.5
PV= −1000
FV= 4875.977…
P/Y= 1
C/Y= 1
(PMT =) $295 (295.393)
Question
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.
▶️Answer/Explanation
Markscheme
\({\text{Ben Amount}} = 6500{\left( {1 + \frac{{4.5}}{{100}}} \right)^6}\) (M1)(A1)
\( = $8464.69\) (A1)
(M1)(A1)(A0) if interest only found (=$1964.69) (C3)[3 marks]
Question
Charles invests \(3000{\text{ USD}}\) in a bank that offers compound interest at a rate of \(3.5\% \) per annum, compounded half-yearly.
Calculate the number of years that it takes for Charles’s money to double.
▶️Answer/Explanation
Markscheme
\(6000 = 3000{\left( {1 + \frac{{3.5}}{{200}}} \right)^{2n}}\) (M1)(A1)
Note: (M1) for substituting values into a compound interest formula, (A1) for correct values with a variable for the power.
\(n = 20{\text{ years}}\) (A1) (C3)
Note: If \(n\) used in formula instead of \(2n\), can allow as long as final answer is halved to get \(20\).[3 marks]
Question
Eva invests \({\text{USD}}2000\) at a nominal annual interest rate of \(8\% \) compounded half-yearly.
a.Calculate the value of her investment after \(5\) years, correct to the nearest dollar.[3]
b.Toni invests \({\text{USD}}1500\) at an annual interest rate of \(7.8\% \) compounded yearly.
Find the number of complete years it will take for his investment to double in value.[3]
▶️Answer/Explanation
Markscheme
\(2000{(1.04)^{10}}\) (M1)(A1)
Note: (M1) for substitution into CI formula. (A1) for correct substitution.
\(2960\) (A1)
Note: Award the final A1 for rounding their answer correctly to the nearest Yuan.
OR
\(2000{\left( {1 + \frac{8}{{200}}} \right)^{10}} – 2000\) (M1)(A1)
Note: (M1) for substitution into CI formula. (A1) for correct substitution.
\(2960\) (A1) (C3)
Note: Award the final A1 for rounding their answer correctly to the nearest Yuan.
\(1500{(1.078)^n} = 3000\) (M1)(M1)
Note: (M1) for correct substitution in CI formula, (M1) for \(3000\) seen.
\(n = 10{\text{ years}}\) (\(n = 9.23{\text{ years}}\) not accepted) (A1)
OR
\(1500{(1.078)^n} – 1500 = 1500\) (M1)(M1)
Note: (M1) for correct substitution in CI formula, (M1) for \(1500\) seen.
\(n = 10{\text{ years}}\) (\(n = 9.23{\text{ years}}\) not accepted) (A1)
OR
(M2) for list or graph. (M2)
\(n = 10{\text{ years}}\) (\(n = 9.23{\text{ years}}\) not accepted) (A1)
Note: If simple interest formula is used in both parts (a) and (b) then award (M0)(M0)(A0) in (a) and
(b) \(1500 = \frac{{1500(7.8)n}}{{100}}\) (M1)(A1)
(M1) for substitution in SI formula or lists, (A1) for correct substitution.
\(n = 13\) (A1) (C3)
Note: Correct answer only. If \(9.23\) seen without working award (A2).
Note: If calculator notation is used in either part with correct unrounded answer award (A1)(d) only if (FP) is applied in (a) or (AP) in (b). Otherwise (A2)(d) if penalty has already been applied in a previous question.