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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

\(\text{Option A:}\)

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

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

  
\(\text{Option B:}\)

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

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

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

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

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

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

  
\(\Rightarrow D\)

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`