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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 2021 HSC 26

Mila plans to invest $42 000 for 1.5 years. She is offered two different investment options.

Option A:  Interest is paid at 5% per annum compounded monthly.

Option B:  Interest is paid at `r` % per annum simple interest.

  1. Calculate the future value of Mila's investment after 1.5 years if she chooses Option A. (2 marks)
  2. Find the value of `r` in Option B that would give Mila the same future value after 1.5 years as for Option A. Give your answer correct to two decimal places. (2 marks)
Show Answers Only
  1. `$45\ 264.08`
  2. `5.18text(%)`
Show Worked Solution
a.   `r` `= text(5%)/12 = text(0.4167%) = 0.004167\ \text(per month)`
  `n` `= 12 × 1.5 = 18`
`FV` `= PV(1 + r)^n`
  `= 42\ 000(1 + 0.004167)^{18}`
  `= $45\ 264.08`

 

b.   `I` `= Prn`
  `3\ 264.08` `= 42\ 000 × r × 1.5`
  `r` `= 3\ 264.08 / (42\ 000 × 1.5)`
    `= 0.0518…`
    `= 5.18\ \text{% (to 2 d.p.)}`

v1 Financial Maths, STD2 F4 2024 HSC 25

Priya and Leo each invest $2500 for 6 years.

    • Priya's investment earns simple interest at a rate of 5.8% per annum.
    • Leo's investment earns interest at a rate of 4.5% per annum, compounding half-yearly.

By calculating the interest earned over the 6 years, determine who will have the greater amount.   (3 marks)

Show Answers Only

\(\text{Priya’s investment:}\)

\(\text{Interest} = Prn = 2500 \times 0.058 \times 6 = \$870\)

 

\(\text{Leo’s investment:}\)

\(r = \dfrac{4.5\%}{2} = 2.25\% \text{ per half-year}\)

\(\text{Compounding periods} = 6 \times 2 = 12\)

\(FV = PV(1+r)^n = 2500(1+0.0225)^{12} = \$3211.83\)

\(\text{Total interest} = FV-PV = 3211.83-2500 = \$711.83\)

 

\(\text{Priya’s interest } > \text{ Leo’s interest.}\)

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

Show Worked Solution

\(\text{Priya’s investment:}\)

\(\text{Interest} = Prn = 2500 \times 0.058 \times 6 = \$870\)

 

\(\text{Leo’s investment:}\)

\(r = \dfrac{4.5\%}{2} = 2.25\% \text{ per half-year}\)

\(\text{Compounding periods} = 6 \times 2 = 12\)

\(FV = PV(1+r)^n = 2500(1+0.0225)^{12} = \$3211.83\)

\(\text{Total interest} = FV-PV = 3211.83-2500 = \$711.83\)

 

\(\text{Priya’s interest } > \text{ Leo’s interest.}\)

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

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, STD2 F1 2024 NHT1 24*

Jarryd invested $14 000 into an account earning compound interest at a fixed rate per time period.

The graph below shows the balance of the account for four of the first five time periods after the initial investment. The information for time period 3 is not shown.
 

 

Immediately after the interest was calculated for time period 3, Jarryd added an extra one-off amount into the account.

Determine the value of Jarrod's extra one off amount, giving your answer correct to the nearest cent.   (3 marks)

Show Answers Only

\(\$224.03 \)

Show Worked Solution

\(\text{Increase factor between periods}\ = \dfrac{15\,120}{14\,000}=1.08\)

\(\text{At time period 3:}\)

\(\text{Balance (before extra payment)}\ = 14\,000 \times 1.08^{3} = 17\,635.97 \)

\(\text{Let}\ V = 17\,635.97 +\ \text{extra payment}\)

\(V \times 1.08 = 19\,288.80\ \ \Rightarrow\ \ V=17\,860.00\)

\(\therefore \ \text{Extra payment}\ = 17\,860.00-17\,635.97=\$224.03 \)

v1 Financial Maths, STD2 F4 2023 HSC 10 MC

An amount of $18 000 is invested for five years. Interest is earned at a rate of 6% per annum, compounding monthly.

Which expression gives the value of the investment after 5 years, in dollars?

  1. `18\ 000 xx 1.005^{60}`
  2. `18\ 000 xx 1.06^{5}`
  3. `18\ 000 xx 1.005^{5}`
  4. `18\ 000 xx 1.06^{60}`
Show Answers Only

`A`

Show Worked Solution

`text{Interest rate}\ = 6/12=0.5%\ text{per month}`

`text{Compounding periods}\ = 5xx12=60`

`:.FV=18\ 000 xx 1.005^{60}`

`=>A`

v1 Financial Maths, STD2 F4 2022 HSC 11 MC

In eight years, the future value of an investment will be $120 000. The interest rate is 6% per annum, compounded half-yearly.

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

  1. `PV=(120\ 000)/((1+0.06)^(8))`
  2. `PV=(120\ 000)/((1+0.03)^(8))`
  3. `PV=(120\ 000)/((1+0.03)^(16))`
  4. `PV=(120\ 000)/((1+0.06)^(16))`
Show Answers Only

`C`

Show Worked Solution

`text{Compounding periods} = 8 xx 2 = 16`

`text{Compounding rate} = (6text{%}) / 2 = 3text{%} = 0.03`

`PV = (120\ 000) / ((1 + 0.03) ^{16})`

`=> C`

Financial Maths, STD2 F4 2024 HSC 25

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

\(\text {Alex’s investment:}\)

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

\(\text {Jun’s investment:}\)

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

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

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

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

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

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

Financial Maths, STD2 F4 2023 HSC 10 MC

An amount of $25 000 is invested for six years. Interest is earned at a rate of 8% per annum, compounding quarterly.

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

  1. `25\ 000 xx 1.02^{24}`
  2. `25\ 000 xx 1.02^{6}`
  3. `25\ 000 xx 1.08^{24}`
  4. `25\ 000 xx 1.08^{6}`
Show Answers Only

`A`

Show Worked Solution

`text{Interest rate}\ = 8/2=2%\ text{per quarter}`

`text{Compounding periods}\ = 6xx4=24`

`:.FV=25\ 000 xx 1.02^{24}`

`=>A`

Financial Maths, STD2 F4 2022 HSC 11 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 `(PV)` 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`

Financial Maths, STD2 F4 2021 HSC 26

Nina plans to invest $35 000 for 1 year. She is offered two different investment options.

Option A:  Interest is paid at 6% per annum compounded monthly.

Option B:  Interest is paid at `r` % per annum simple interest.

  1. Calculate the future value of Nina's investment after 1 year if she chooses Option A.  (2 marks)

  2. Find the value of `r` in Option B that would give Nina the same future value after 1 year as for Option A. Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `$37\ 158.72`
  2. `6.17text(%)`
Show Worked Solution
a.   `r` `= text(6%)/12= text(0.5%) = 0.005\ text(per month)`
  `n` `=12`

 

`FV` `= PV(1 + r)^n`
  `= 35\ 000(1 + 0.005)^(12)`
  `= $37\ 158.72`

♦♦ Mean mark part (b) 36%.
b.   `I` `=Prn`
  `2158.72` `=35\ 000 xx r xx 1`
  `r` `=2158.72/(35\ 000)`
    `=0.06167…`
    `=6.17 text{% (to 2 d.p.)}`

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, 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, STD2 F4 2020 HSC 4 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)`

`n \ = \ 1.5 xx 12 = 18`
  

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

 
`=> \ B`

Financial Maths, STD2 F4 2019 HSC 3 MC

Chris opens a bank account and deposits $1000 into it. Interest is paid at 3.5% per annum, compounding annually.

Assuming no further deposits or withdrawals are made, what will be the balance in the account at the end of two years?

  1. $1070.00
  2. $1071.23
  3. $1822.50
  4. $2070.00
Show Answers Only

`=> B`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 1000(1 + 0.035)^2`
  `= $1071.23`

 
`=> B`

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`