SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Financial Maths, STD1 F2 2025 HSC 25

Bobbie plans to invest $25 000 for 10 years and is offered two investment options.

Option \(A\):  Earns interest at a rate of 5% per annum, compounded monthly.

Option \(B\):  Earns simple interest at a rate of 8% per annum.

Which investment option will provide Bobbie with the best return value at the end of 10 years? Justify your answer with calculations.   (4 marks)

Show Answers Only

\(\text{Option A:}\)

\(PV=$25\ 000,\ \ \text{monthly interest rate: }r=\dfrac{0.05}{12},\ \ n=12\times 10=120\)

\(FV=25\ 000\Bigg(1+\dfrac{0.05}{12}\Bigg)^{120}\approx $41\ 175\)

  
\(\text{Option B:}\)

\(\text{Interest}=25\ 000\times 0.08\times 10=$20\ 000\)

\(\text{Total}=25\ 000+20\ 000=$45\ 000\)

\(\therefore\ \text{Option B gives the best return.}\)

Show Worked Solution

\(\text{Option A:}\)

\(PV=$25\ 000,\ \ \text{monthly interest rate:}\ r=\dfrac{0.05}{12},\ \ n=12\times 10=120\)

\(FV=25\ 000\Bigg(1+\dfrac{0.05}{12}\Bigg)^{120}\approx $41\ 175\)

  
\(\text{Option B:}\)

\(\text{Interest}=25\ 000\times 0.08\times 10=$20\ 000\)

\(\text{Total}=25\ 000+20\ 000=$45\ 000\)

\(\therefore\ \text{Option B gives the best return.}\)


♦♦ Mean mark 46%.

Financial Maths, STD1 F3 2025 HSC 21

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest is charged on purchases made, compounding daily at a rate of 13.74% per annum, from and including the day following the interest-free period.

Concert tickets were purchased for a total of $392 using this credit card.

Full payment was made on the 68th day from the date of purchase. There were no other purchases on this credit card.

What was the total interest charged when the account was paid in full?   (3 marks)

Show Answers Only

\(\text{Interest charged}\ =\$ 3.41\)

Show Worked Solution

\(\text{Day 1-45: no interest is charged}\)

\(\text{Day 46-68: interest charged (23 days)}\)

\(\text{Daily interest rate}=\dfrac{13.74}{365} \%=\dfrac{13.74}{365 \times 100}\)

\(\text{Amount owing}=392\left(1+\dfrac{13.74}{365 \times 100}\right)^{23}=\$ 395.41\)

\(\text{Interest charged}=395.41-392=\$ 3.41\)


♦♦ Mean mark 33%.

Financial Maths, STD1 F2 2025 HSC 9 MC

An amount of $90 000 is invested at 4% per annum, compounded quarterly.

Which expression gives the value of this investment, in dollars, after 6 years?

  1. \(90\ 000(1+0.04)^6\)
  2. \(90\ 000(1+0.04)^{24}\)
  3. \(90\ 000(1+0.01)^6\)
  4. \(90\ 000(1+0.01)^{24}\)
Show Answers Only

\(D\)

Show Worked Solution

\(PV=90\ 000,\  r=\dfrac{4\%}{4}=1\%=0.01,\  n=4\times 6=24\)

\(FV\) \(=PV(1+r)^n\)
  \(=90\ 000(1+0.01)^{24}\)

  
\(\Rightarrow D\)


♦♦♦ Mean mark 22%.

v1 Financial Maths, STD2 F4 2014 HSC 30a

Jordan wants to accumulate $15 000 in a savings account over 10 years to buy a new car.

The account pays interest at 4% per annum compounded monthly.

Calculate how much Jordan must deposit now to achieve this goal. (3 marks)

Show Answers Only

`$10\ 110\ \ \text{(nearest $)}`

Show Worked Solution
♦ Mean mark 52%

`FV = 15\ 000,\ \ n = 10 \times 12 = 120,`

`r = 0.04 / 12 = 0.003333…`

`FV` `= PV (1 + r)^n`
`15\ 000` `= PV (1 + 0.003333…)^{120}`
`PV` `= \frac{15\ 000}{(1.003333…)^{120}}`
  `= 10\ 109.88…`

`∴ \ \text{Jordan must deposit} \ $10\ 110\ \text{(nearest $)}`

v1 Financial Maths, STD2 F4 2015 HSC 26d

A laptop currently costs $850.

Assuming a constant annual inflation rate of 3.2%, calculate the cost of the same laptop in 4 years’ time.  (2 marks)

Show Answers Only

`$962.38\ \text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 850(1.032)^4`
  `= 850(1.132216)`
  `= 962.3836…`
  `= $962.38\ \text{(nearest cent)}`

v1 Financial Maths, STD2 F4 2008 HSC 24c

Daniel’s funds in a retirement account are projected to have a future value of $600 000 in 15 years’ time. The interest rate is 5% per annum, with earnings calculated six-monthly.

What single amount could be invested now to produce the same result over the same period of time at the same interest rate? (3 marks)

Show Answers Only

`$288\ 629.97`

Show Worked Solution
`FV` `= PV(1 + r)^n`
`600\ 000` `= PV(1 + 2.5/100)^30`
`:. PV` `= (600\ 000)/((1.025)^30)`
  `= 288\ 629.966…`
  `= $288\ 629.97`

Financial Maths, STD1 F2 2024 HSC 28

Alex and Jun each invest $1800 for 5 years.

  • Alex's investment earns simple interest at a rate of 7.5% per annum.
  • Jun's investment earns interest at a rate of 6.0% per annum, compounding quarterly.

By calculating the interest earned over the 5 years, determine who will have the greater amount.   (3 marks)

Show Answers Only

\(\text {Alex’s investment:}\)

\(\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675\)
 

\(\text {Jun’s investment:}\)

\(r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}\)

\(\text {Compounding periods }=5 \times 4=20\)

\(F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34\)

\(\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34\)
 

\(\text {Alex’s interest }>\text { Jun’s interest.}\)

\(\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}\)

Show Worked Solution

\(\text {Alex’s investment:}\)

\(\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675\)

♦♦ Mean mark 36%.

\(\text {Jun’s investment:}\)

\(r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}\)

\(\text {Compounding periods }=5 \times 4=20\)

\(F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34\)

\(\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34\)
 

\(\text {Alex’s interest }>\text { Jun’s interest.}\)

\(\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}\)

Financial Maths, STD1 F2 2024 HSC 8 MC

Three years ago, the price of a uniform was $180.

Due to inflation, the price increased annually by 2.5%.

What is the price of this uniform now?

  1. $180.14
  2. $ 181.35
  3. $ 193.50
  4. $ 193.84
Show Answers Only

\(D\)

Show Worked Solution

\(r=2.5 \%=\dfrac{2.5}{100}=0.025\)

  \(FV\) \(=PV(1+r)^n\)
    \(=180(1.025)^3\)
    \(=193.84\)

 
\(\Rightarrow D\)

♦♦ Mean mark 38%.

Financial Maths, STD1 F2 2023 HSC 25

An artwork is currently valued at $ 15000. It appreciates at a rate of 5.3% per annum.

What will the value of the artwork be in 8 years time?  (2 marks)

Show Answers Only

\($22\ 673.48\text{ (2 d.p.)}\)

Show Worked Solution

\(PV=$15\ 000,\ \ r=\dfrac{5.3}{100},\ \ n=8\)

\(FV\) \(=PV(1+r)^n\)
  \(=15\ 000(1+\dfrac{5.3}{100})^8\)
  \(=$22\ 673.48242\)
  \(=$22\ 673.48\text{ (2 d.p.)}\)

Financial Maths, STD1 F2 2023 HSC 21

An amount of $12 000 is invested in an account that pays 1 % interest per quarter, compounding quarterly for five years.

What is the future value of this investment?  (3 marks)

Show Answers Only

\($14\ 642.28\)

Show Worked Solution

\(PV=$12\ 000,\  n=5\times 4=20, \ r=\dfrac{1}{100}\)
 

\(FV\) \(=PV(1+r)^n\)
  \(=12\ 000 \Big(1+\dfrac{1}{100}\Big)^{20}\)
  \(=$14\ 642.28048\)
  \(\approx $14\ 642.28\text{ (2 d.p.)}\)

♦ Mean mark 42%.

Financial Maths, STD1 F2 2022 HSC 31

A watch is currently worth $6100. It has appreciated by 5.8% per annum since purchase.

What was its value 10 years ago?  (2 marks)

Show Answers Only

`$3471.15`

Show Worked Solution

`FV=$6100`

`text{Find}\ PV\ text{given}\ n=10, \ r=5.8text{%}=0.058:`
  

`FV` `=PV(1+r)^n`  
`6100` `=(PV)(1+0.058)^10`  
`PV` `=6100/(1.058)^10`  
  `=$3471.148…`  
  `=$3471.15\ \ text{(nearest cent)}`  

♦♦ Mean mark 37%.

Financial Maths, STD1 F2 2022 HSC 9 MC

In ten years, the future value of an investment will be $150 000. The interest rate is 4% per annum, compounded half-yearly.

Which equation will give the present value `(P V)` of the investment?
 

  1. `PV=(150\ 000)/((1+0.04)^(10))`
  2. `PV=(150\ 000)/((1+0.04)^(20))`
  3. `PV=(150\ 000)/((1+0.02)^(10))`
  4. `PV=(150\ 000)/((1+0.02)^(20))`
Show Answers Only

`D`

Show Worked Solution

`text{Compounding periods} = 10 xx 2 = 20`

`text{Compounding rate} = (4text{%})/2 = 2text{%} = 0.02`

`PV=(150\ 000)/((1+0.02)^(20))`

`=>D`


♦♦♦ Mean mark 16%.

Financial Maths, STD2 F2 2021 HSC 5 MC

Peter currently earns $21.50 per hour. His hourly wage will increase by 2.1% compounded each year for the next four years.

What will his hourly wage be after four years?

  1. `21.50(1.21)^4`
  2. `21.50(1.021)^4`
  3. `21.50 + 21.50 xx 0.21 xx 4`
  4. `21.50 + 21.50 xx 0.021 xx 4`
Show Answers Only

`B`

Show Worked Solution

`text(Wage after 1 year) = 21.50 xx 1.021`

`text(Wage after 2 years) = 21.50 xx 1.021 xx 1.021 = 21.50(1.021)^2`

`vdots`

`text(Wage after 4 years) = 21.50(1.021)^4`

`=>  B`

Financial Maths, STD1 F2 2020 HSC 13

Taro needs $1000 in 5 years time. He is going to invest some money today in an account earning 3% per annum compounded annually. He will make no further deposits or withdrawals.

How much money does he need to invest today?   (3 marks)

Show Answers Only

`$862.61`

Show Worked Solution

♦♦ Mean mark 29%.
`FV` `= PV (1 + r)^n`
`1000` `= PV (1 + frac{3}{100})^5`
`1000` `= PV (1.0.3)^5`
`:. PV` `= frac{1000}{(1.03)^5}`
  `= $862.61`

Financial Maths, STD1 F2 2020 HSC 8 MC

Joan invests $200. She earns interest at 3% per annum, compounded monthly.

What is the future value of Joan's investment after 1.5 years?

  1. $209.07
  2. $209.19
  3. $279.51
  4. $311.93
Show Answers Only

`B`

Show Worked Solution

`text(Monthly interest rate) \ = frac(0.03)(12)`

♦♦ Mean mark 33%.

`n \ = \ 1.5 xx 12 = 18`
  

`FV` `= PV(1 + r)^n`
  `= 200 (1 + frac(0.03)(12))^18`
  `= $209.19`

 
`=> \ B`

Financial Maths, STD2 F4 2020 HSC 21

The inflation rate over the year from January 2019 to January 2020 was 2%.

The cost of a school jumper in January 2020 was $122.

Calculate the cost of the jumper in January 2019 assuming that the only change in the cost of the jumper was due to inflation.   (2 marks)

Show Answers Only

`$119.61`

Show Worked Solution
`FV` `=PV(1+r)^n`
`122` `=C_(2019)(1+0.02)^1`
`C_2019 xx 1.02` `= 122`
`C_2019` `= frac(122)(1.02)`
  `= $119.61`

Financial Maths, STD1 F2 2019 HSC 35

A bank offers two different savings accounts.

Account `X` offers simple interest of 7% per annum.
Account `Y` offers compound interest of 6% per annum compounded yearly.

The table displays the future values of $20 000 invested in each account for the first 2 years.
 


  

  1. How much more money is there in Account `X` than in Account `Y` at the end of 2 years?  (1 mark)

  2. Show that there would be more money in Account `Y` than in Account `X` at the end of 8 years.  (3 marks)

Show Answers Only
  1. `$328`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.    `text(Extra money in)\ \ X` `= 22\ 800 – 22\ 472`
    `= $328`

 

b.   `text(Account)\ X:`

♦ Mean mark part (b) 30%.

`I` `= Prn`
  `= 20\ 000 xx 7/100 xx 8`
  `= 11\ 200`

 
`=> text(Balance)\ X = 20\ 000 + 11\ 200 = $31\ 200`
 

`text(Account)\ Y:`

`FV` `= PV(1 + r)^n`
  `= 20\ 000(1 + 6/100)^8`
  `= $31\ 876.96`

 
`:. text(After 8 years, there’s more money in Account)\ Y.`

Financial Maths, STD2 F4 2008 HSC 24c

Heidi’s funds in a superannuation scheme have a future value of  $740 000  in 20 years time. The interest rate is 4% per annum and earnings are calculated six-monthly.

What single amount could be invested now to produce the same result over the same period of time at the same interest rate?  (3 marks)

Show Answers Only

`$335\ 138.91`

Show Worked Solution
`FV` `= PV(1 + r)^n`
`740\ 000` `= PV(1 + 2/100)^40`
`:. PV` `= (740\ 000)/((1.02)^40)`
  `= 335\ 138.907…`
  `= $335\ 138.91`

Financial Maths, STD2 F4 2017 HSC 10 MC

A single amount of $10 000 is invested for 4 years, earning interest at the rate of 3% per annum, compounded monthly.

Which expression will give the future value of the investment?

  1. `10\ 000 xx (1 + 0.03)^4`
  2. `10\ 000 xx (1 + 0.03)^48`
  3. `10\ 000 xx (1 + 0.03/12)^4`
  4. `10\ 000 xx (1 + 0.03/12)^48`
Show Answers Only

`D`

Show Worked Solution

`text(Compounding rate)\ = 3/100 ÷ 12= 0.03/12`

`text(Compounding periods)` `= 4 xx 12=48`

 
`:.\ text(FV) = 10\ 000 xx (1 + 0.03/12)^48`

\(\Rightarrow D\)

Financial Maths, STD2 F4 2015 HSC 26d

A family currently pays $320 for some groceries.

Assuming a constant annual inflation rate of 2.9%, calculate how much would be paid for the same groceries in 5 years’ time.  (2 marks)

Show Answers Only

`$369.17\ \ text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 320(1.029)^5`
  `= $369.1703…`
  `= $369.17\ \ text{(nearest cent)}`

Financial Maths, STD2 F4 2015 HSC 17 MC

What amount must be invested now at 4% per annum, compounded quarterly, so that in five years it will have grown to  $60 000?

  1. $8919
  2. $11 156
  3. $49 173
  4. $49 316
Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ FV = PV(1 + r)^n`

`r` `= text(4%)/4` `= text(1%) = 0.01\ text(per quarter)`
`n` `= 5 xx 4` `= 20\ text(quarters)`

 

`60\ 000` `= PV(1 + 0.01)^(20)`
`:.PV` `= (60\ 000)/1.01^(20)`
  `= $49\ 172.66…`

`⇒ C`

Financial Maths, STD2 F4 2014 HSC 30a

Chandra and Sascha plan to have $20 000 in an investment account in 15 years time for their grandchild’s university fees.

The interest rate for the investment account will be fixed at 3% per annum compounded monthly.

Calculate the amount that they will need to deposit into the account now in order to achieve their plan.   (3 marks)

Show Answers Only

`$12\ 760\ \ text{(nearest $)}`

Show Worked Solution
♦ Mean mark 49%

`FV = $20\ 000,\ \ n = 15xx 12=180,`

`r = 0.03 /12=0.0025`
 

`FV` `= PV (1 + r)^n`
`20\ 000` `=PV (1 + 0.0025)^180`
`PV` `=(20\ 000)/(1.0025)^180`
  `=12\ 759.73…`

 

`:.\ text(They need to deposit) \ \ $12\ 760\ \ text{(nearest $)}`

Financial Maths, STD2 F4 2009 HSC 6 MC

A house was purchased in 1984 for $35 000. Assume that the value of the house has increased by 3% per annum since then. 

Which expression gives the value of the house in 2009?  

  1. `35\ 000(1 + 0.03)^25`
  2. `35\ 000(1 + 3)^25` 
  3. `35\ 000 xx 25 xx 0.03`
  4. `35\ 000 xx 25 xx 3`
Show Answers Only

`A`

Show Worked Solution

`r =\ text(3%)\ = 0.03`

`n = 25\ text(years)`

`text(Using)\ \ FV = PV(1 + r)^n`

` :.\ text(Value in 2009) = 35\ 000(1+0.03)^25` 

`=>  A`