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Compound Interest, SMB-031

Camilla is considering a term deposit that pays 4.8% interest per annum, compounded monthly.

She calculates that if she invests in the term deposit, her money will be worth $9800 in 2 years' time.

Determine the amount that Camilla is planning to invest, giving your answer to the nearest cent.   (2 marks)

Show Answers Only

`$8904.65`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.048}{12} = 0.004\ \ \text{(per month)}`

`\text{Compounding periods}\ (n) = 2 xx 12 = 24`

`FV` `=PV(1+r)^n`
`9800` `= PV(1 + 0.004)^(24)`
`:.PV` `= \frac{9800}{1.004^{24}}`
  `= $8904.65\ \ \text{(nearest cent)}`

Compound Interest, SMB-030

Michelle decided to invest some of her money at a higher interest rate. She deposited $3000 in an account paying 8.2% per annum, compounding half yearly.

  1. Write down an expression involving the compound interest formula that can be used to find the value of Michelle’s $3000 investment at the end of two years. Find this value correct to the nearest cent.  (2 marks)

  2. How much interest will the $3000 investment earn over a four-year period?

     

    Write your answer correct to the nearest cent.  (2 marks)

Show Answers Only
  1. `$3523.09`
  2. `$1137.40`
Show Worked Solution

a.   `text{Interest rate}\ (r) = \frac{8.2%}{2} = 4.1% = 0.041\ \ \text{(per 6 months)}`

`text(Compounding periods)\ (n) = 2 xx 2 = 4`

`FV` `= PV(1+r)^n`
  `= 3000(1.041)^4`
  `= 3523.093…`
  `= $3523.09\ \ text{(nearest cent)}`

 

b.   `text{Compounding periods}\ (n) = 4 xx 2 = 8`

`FV` `= 3000(1.041)^8`
  `= 4137.396…`

 

`:.\ text(Interest earned)` `= 4137.40-3000`
  `= $1137.40\ \ text{(nearest cent)}`

Compound Interest, SMB-029

The golf club’s social committee has $3400 invested in an account which pays interest at the rate of 4.4% per annum compounding quarterly.

  1. Show that the interest rate per quarter is 1.1%.  (1 mark)

  2. Determine the value of the $3400 investment after three years.

     

    Write your answer in dollars correct to the nearest cent.  (1 mark)

  3. Calculate the interest the $3400 investment will earn over six years.

     

    Write your answer in dollars correct to the nearest cent.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `$3876.97`
  3. `$1020.86`
Show Worked Solution

a.   `text{Interest rate}\ = \frac{4.4}{4} = 1.1text{%  (per quarter)}`
 

b.   `text{Compounding periods}\ (n) = 3 xx 4 =12`

`FV` `= PV(1+r)^n`
  `= 3400(1.011)^12`
  `= 3876.973…`
  `= $3876.97`

 

c.   `text{Compounding periods}\ (n) = 6 xx 4 = 24`

`FV` `= 3400(1.011)^24`
  `= 4420.858…`

 
`text(Interest earned over 6 years)`

`= 4420.86-3400`

`= $1020.86\ \ text{(nearest cent)}`

Compound Interest, SMB-028

It is estimated that inflation will average 3.5% per annum over the next eight years.

If a new machine costs $60 000 now, calculate the cost of a similar new machine in eight years time, adjusted for inflation. Assume no other cost change.

Write your answer correct to the nearest dollar.  (2 marks)

Show Answers Only

`$79\ 008.54`

Show Worked Solution
`FV` `= PV(1+r)^n`
  `= 60\ 000(1.035)^8`
  `= $79\ 008.54\ \ \text{(nearest cent)}`

Compound Interest, SMB-027 MC

An amount of $8000 is invested for a period of 4 years.

The interest rate for this investment is 7.2% per annum compounding quarterly.

The interest earned by the investment in the fourth year (in dollars) is given by

  1. `8000 xx 1.018^4 - 8000 xx 1.018^3`
  2. `8000 xx 1.072^4 - 8000 xx 1.072^3`
  3. `8000 xx 1.018^16 - 8000 xx 1.018^12`
  4. `8000 xx 1.072^16 - 8000 xx 1.072^12`
Show Answers Only

 `C`

Show Worked Solution

`text{Interest earned in 4th year}\ = FV\text{(after 4 years)}-FV\text{(after 3 years)}`

`text{Interest rate}\ (r) = \frac{7.2%}{4} = \frac{0.072}{4} = 0.018`

`text{Compounding periods (3 years)}\ = 3 xx 4 = 12`

`text{Compounding periods (4 years)}\ = 4 xx 4 = 16`
 

`FV\text{(after 4 years)}\ = PV(1+r)^n = 8000(1.018)^16`

`FV\text{(after 3 years)}\ = 8000(1.018)^12`

`:.\ text(Interest earned in 4th year) = 8000 xx 1.018^(16)-8000 xx 1.018^(12)`

`=>  C`

Compound Interest, SMB-026 MC

$10 000 is invested for five years. Interest is earned at a rate of 8% per annum, compounding quarterly.

Which one of the following calculations will give the total interest earned, in dollars, by this investment?

  1. `10\ 000 xx 1.02^5-10\ 000`  
  2. `10\ 000 xx 1.02^20-10\ 000`  
  3. `10\ 000 xx 1.08^5 -10\ 000`
  4. `10\ 000 xx 1.08^20-10\ 000`
Show Answers Only

`B`

Show Worked Solution

`text{Interest rate}\ (r) = \frac{8%}{4} =2% = 0.02\ text{(per quarter)}`

`text{Compounding periods}\ (n) =5 xx4=20`

`FV=PV(1+r)^n = 10\ 000(1.02)^20`

`:.\ text(Interest earned)= 10\ 000 xx 1.02^20-10\ 000`

`=>  B`

Compound Interest, SMB-025

$15 000 is invested for 12 months.

For the first six months the interest rate is 6.1% per annum compounding monthly.

After six months the interest rate increases to 6.25% per annum compounding monthly.

Calculate the total interest earned by this investment over 12 months, giving your answer to the nearest dollar.   (4 marks)

Show Answers Only

`$953`

Show Worked Solution

`text{In the first 6 months:}`

`text{Interest rate}\ (r) = \frac{0.061}{12}`

`FV(\text{after 6 months})` `=PV(1+r)^n`  
  `=15\ 000(1 + \frac{0.061}{12})^6`   
  `=$15\ 463.35…`  

 
`text{In the second 6 months:}`

`text{Interest rate}\ (r) = \frac{0.0625}{12}`

`FV(\text{after 2nd 6 months})` `=15\ 463.35(1 + \frac{0.0625}{12})^6`  
  `=$15\ 952.91…`   

 

`:.\ text(Interest)` `= 15\ 952.91-15\ 000`
  `= $953\ \ \text{(nearest dollar)}`

Compound Interest, SMB-024

$9000 is invested at a rate of 8% per annum compounding half yearly.

Determine the value, to the nearest dollar, of this investment after four years.   (2 marks)

Show Answers Only

`$12\ 317`

Show Worked Solution

`text(Interest rate)\ (r) = \frac{8%}{2} = \frac{0.08}{2} = 0.04\ \ \text{(per 6 months)}`

`text{Compounding periods}\ (n) =4 xx 2=8`

`FV` ` = PV(1+r)^n`
  `= 9000 xx (1+0.04)^(8)`
  `= $12\ 317.12`
  `=$12\ 317\ \ \text{(nearest dollar)}`

Compound Interest, SMB-023 MC

$10 000 is invested at a rate of 10% per annum compounding half yearly.

The value, in dollars, of this investment after five years, is given by

  1. `10\ 000 xx 0.10 xx 5`
  2. `10\ 000 xx 0.05 xx 10`
  3. `10\ 000 xx 0.05^10`
  4. `10\ 000 xx 1.05^10`
Show Answers Only

`D`

Show Worked Solution

`text(Interest rate)\ (r) = \frac{10%}{2} = \frac{0.10}{2} = 0.05\ \ \text{(per 6 months)}`

`text{Compounding periods}\ (n) =5 xx 2=10`

`FV` ` = PV(1+r)^n`
  `= 10\ 000 xx (1+0.05)^(10)`
  `=10\ 000(1.05)^10`

 
`=>  D`

Compound Interest, SMB-022 MC

The points on the graph below show the balance of an investment at the start of each quarter for a period of six years.

The same rate of interest applied for these six years.
 

   

In relation to this investment, which one of the following statements is true?

  1. interest is compounding annually and is credited annually
  2. interest is compounding annually and is credited quarterly
  3. simple interest is paid on the opening balance and is credited annually
  4. simple interest is paid on the opening balance and is credited quarterly
Show Answers Only

`A`

Show Worked Solution

`text{By elimination:}`

`text{From the graph, as balance increases after each year, interest is}`

`text{credited annually (eliminate B and D).}`

`text{The difference of the balances between successive years is increasing}`

`text{which indicates that interest is compounding (eliminate D).}`

`=>  A`

Compound Interest, SMB-021

Callum invests $20 000 in a term deposit account that adds 3.8% interest annually, calculated on the account balance at the end of each year.

Calculate the interest paid in the third year of Callum's investment.   (3 marks)

Show Answers Only

`$818.86`

Show Worked Solution

`r=3.8% = 0.038`

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

`text(Value after 2 years)` `= 20\ 000(1.038)^2`
  `= $21\ 548.88`
 `text(Value after 3 years)` `= 20\ 000(1.038)^3`
  `= $22\ 367.74`

 

 `:.\ text(Interest paid in 3rd year)` `= 22\ 367.74-21\ 548.88`
  `= $818.86`

Compound Interest, SMB-020

Tim invests $3000 in a term deposit account that adds 6.5% interest annually, calculated on the account balance at the end of each year.

The interest paid in the fourth year is

  1. `$221.16`
  2. `$235.55`
  3. `$3623.85`
  4. `$3859.40`
Show Answers Only

`B`

Show Worked Solution

`r=6.5% = 0.065`

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

`text(Value after 3rd year)` `= 3000(1.065)^3`
  `= $3623.85`
 `text(Value after 4th year)` `= 3000(1.065)^4`
  `= $3859.40`

 

 `:.\ text(Interest paid in 4th year)` `= 3859.40-3623.85`
  `= $235.55`

`=>  B`

Compound Interest, SMB-019

Kelly invests $8 000 at an interest rate of 7.5% per annum, compounding annually.

After how many years will her investment first be more than double its original value, giving your answer to the nearest year?   (3 marks)

Show Answers Only

`10\ text{years}`

Show Worked Solution

`FV=PV(1+r)^n`

`r=7.5% = 0.075, PV=8000`

`text(Find)\ n\ text(when)\ FV>16\ 000:`

`8000(1+0.075)^n` `> 16\ 000`
`1.075^n` `> 2`

 
`text(Testing possible values:)`

`1.075^10 = 2.06`

`1.075^9 = 1.92`

`:.\ \text{Kelly’s investment will first double after 10 years.}`

Compound Interest, SMB-018 MC

Gerry invests $10 000 at an interest rate of 5.5% per annum, compounding annually.

After how many years will his investment first be more than double its original value?

  1. 12
  2. 13
  3. 14
  4. 15
Show Answers Only

`B`

Show Worked Solution

`FV=PV(1+r)^n`

`r=5.5% = 0.055, PV=10\ 000`

`text(Find)\ n\ text(when)\ FV>20\ 000:`

`10\ 000(1+0.055)^n` `> 20\ 000`
`1.055^n` `> 2`

 
`text(Test answer options:)`

`1.055^12 = 1.90`

`1.055^13 = 2.005`

`=>  B`

Compound Interest, SMB-017

Ekamjot invests $13 000 for two years.

Interest is calculated at the rate of 7.2% per annum, compounding quarterly.

How much interest does Ekamjot earn from this investment, giving your answer to the nearest cent?   (2 marks)

Show Answers Only

`$1994.28`

Show Worked Solution

`text{Annual interest rate}\ = 7.2% = 0.072`

`text(Quarterly interest rate)\ (r) = \frac{0.072}{4} = 0.018`

`text(Compounding periods)\ (n) = 2 xx 4 =8`

`FV(text{after 2 years})` `= PV(1+r)^n`
  `=13\ 000(1 + 0.018)^8`
  `=$14\ 994.278….`
  `=$14\ 994.28`

 
`text{Interest earned}\ =14\ 994.28-13\ 000=$1994.28`

Compound Interest, SMB-016 MC

Mary invests $1200 for two years.

Interest is calculated at the rate of 3.35% per annum, compounding monthly.

The amount of interest she earns in two years is closest to

  1. `$6.71`
  2. `$80.40`
  3. `$81.75`
  4. `$83.03`
Show Answers Only

`D`

Show Worked Solution

`text{Annual interest rate}\ = 3.35% = 0.0335`

`text(Monthly interest rate)\ (r) = \frac{0.0335}{12}`

`text(Compounding periods)\ (n) = 2 xx 12 =24`

`FV(text{after 2 years})` `= PV(1+r)^n`
  `=1200(1 + \frac{0.0335}{12})^24`
  `=$1283.03…`

 
`text{Interest earned}\ =1283.03-1200=$83.03`

`=> D`

Compound Interest, SMB-015

Amy invests  $15 000 for 150 days.

Interest is calculated at the rate of 4.60% per annum, compounding daily.

Assuming that there are 365 days in a year, find the value of her investment after 150 days, giving your answer to the nearest dollar.   (3 marks)

Show Answers Only

`$15\ 286`

Show Worked Solution

`text{Annual interest rate}\ =4.60%=0.046`

`text{Daily interest rate}\ = \frac{0.046}{365}`

`FV` `= 15\ 000 xx (1 + \frac{0.046}{365})^150`
  `= 15\ 000 xx (1.000126…)^150`
  `= 15\ 000 xx (1.01908…)`
  `= $15\ 286\ \ \text{(nearest dollar)}`

Compound Interest, SMB-014

Sam and Charlie each invest $5000 for three years.

Sam’s investment earns simple interest at the rate of 7.5% per annum.

Charlie’s investment earns interest at the rate of 7.5% per annum compounding annually.

At the conclusion of three years, determine which investment has the highest value and by how much.   (4 marks)

Show Answers Only

 `text{Charlie’s investment is worth $86.48 more than Sam’s.}`

Show Worked Solution

`text(Sam’s Investment:)`

`I=Prn = 5000 xx \frac{7.5}{100} xx 3 = $1125`

`text{Investment value}\ = 5000 + 1125 = $6125`

  
`text(Charlie’s Investment:)`

`FV` `= PV(1+r)^n`
  `= 5000 xx 1.075^3`
  `= $6211.48`

 
`text{Difference}\ = 6211.48-6125= $86.48`

`:.\ \text{Charlie’s investment is worth $86.48 more than Sam’s.}`

Compound Interest, SMB-013

What amount must be invested now at 6% per annum, compounded quarterly, so that in eighteen months it will have grown to `$14\ 000`? Give your answer to the nearest cent.  (2 marks)

Show Answers Only

`$12\ 803.59`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.06}{4} = 0.015\ \ \text{(per quarter)}`

`\text{Compounding periods}\ (n) = \frac{18}{3} = 6`
 

`FV` `=PV(1+r)^n`
`14\ 000` `= PV(1 + 0.015)^(6)`
`:.PV` `= (14\ 000)/1.015^(6)`
  `= $12\ 803.59`

Compound Interest, SMB-012

Louise's investment earns 3.6% per annum, compounded quarterly.

She calculates that her investment will be worth $7400 in 4 years.

Determine the amount that Louise initially invests, giving your answer to the nearest cent.   (2 marks)

Show Answers Only

`$6411.70`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.036}{4} = 0.009\ \ \text{(per quarter)}`

`\text{Compounding periods}\ (n) = 4 xx 4 = 16`

`FV` `=PV(1+r)^n`
`7400` `= PV(1 + 0.009)^(16)`
`:.PV` `= \frac{7400}{1.009^{16}}`
  `= $6411.70\ \ \text{(nearest cent)}`

Compound Interest, SMB-011 MC

Marshall's investment earns 5% per annum, compounded annually.

He calculates that his investment will be worth $1100 in 3 years, to the nearest dollar.

The amount Marshall invests now is closest to

  1. $892
  2. $928
  3. $950
  4. $1008
Show Answers Only

`C`

Show Worked Solution

`FV = PV(1 + r)^n`

`r` `=\ text(5%)` `= 0.05\ text(per annum)`
`n` `=3`  

 

`1100` `= PV(1 + 0.05)^(3)`
`:.PV` `= \frac{1100}{1.05^{3}}`
  `= $950`

 
`=> C`

Compound Interest, SMB-010

A bank's Compound Saver Account pays interest at 4% per annum, compounded quarterly.

Sacha deposits $6500 when he opens his Compound Saver Account and makes no further deposits or withdrawals.

What will be the balance in the account at the end of 1.5 years, to the nearest dollar?   (2 marks)

Show Answers Only

`$6900`

Show Worked Solution

`text{Compounding periods}\ (n) =1.5 xx 4=6`

`text{Compounding rate}\ (r) = \frac{0.04}{4} = 0.01`

`FV` `= PV(1 + r)^n`
  `= 6500(1 + 0.01)^6`
  `= 6899.88…`
  `=$6900\ \ \text{(nearest dollar)}`

Compound Interest, SMB-009

Jill opens a Smartsave Account that pays interest at 5% per annum, compounded quarterly.

Jill deposits $700 when she opens the account and makes no further deposits or withdrawals.

What will be the balance in the account at the end of 3 years, to the nearest dollar?   (2 marks)

Show Answers Only

`$813`

Show Worked Solution

`text{Compounding periods}\ (n) =3 xx 4=12`

`text{Compounding rate}\ (r) = \frac{0.05}{4} = 0.0125`

`FV` `= PV(1 + r)^n`
  `= 700(1 + 0.0125)^12`
  `= 812.528…`
  `=$813\ \ \text{(nearest dollar)}`

Compound Interest, SMB-008 MC

A bank's Maxi Saver Account pays interest at 6% per annum, compounded quarterly.

Jen opens a Maxi Saver Account and deposits $3000 into it. 

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

  1. $3120.00
  2. $3360.00
  3. $3379.48
  4. $4781.54
Show Answers Only

`=> C`

Show Worked Solution

`text{Compounding periods}\ (n) =2 xx 4=8`

`text{Compounding rate}\ (r) = \frac{0.06}{4} = 0.015`

`FV` `= PV(1 + r)^n`
  `= 3000(1 + 0.015)^8`
  `= $3379.48`

 
`=> C`

Compound Interest, SMB-007

Shannon invests $25 000 in an account that earns interest at 6% per annum, compounded monthly.

What is the future value of Shannon's investment, to the nearest dollar, after 2 years?   (2 marks)

Show Answers Only

`$28\ 179`

Show Worked Solution

`text{Annual interest rate = 6.0% = 0.06}` 

`text(Monthly interest rate) = \frac(0.06)(12) = 0.005`

`n = 2 xx 12 = 24`
  

`FV` `= PV(1 + r)^n`
  `= 25\ 000 (1 + 0.005)^24`
  `= 28\ 178.99`
  `=$28\ 179\ \ \text{(nearest dollar)}`

Compound Interest, SMB-006

Natalie is saving for a netball hoop and invests $2200 in an account that earns interest at 5.5% per annum, compounded monthly.

What is the future value of Natalie's investment, to the nearest dollar, after 1.5 years?   (2 marks)

Show Answers Only

`$2389`

Show Worked Solution

`text{Annual interest rate = 5.5% = 0.055}` 

`text(Monthly interest rate) = \frac(0.055)(12)`

`n = 1.5 xx 12 = 18`
  

`FV` `= PV(1 + r)^n`
  `= 2200 (1 + frac(0.055)(12))^18`
  `= 2388.74…`
  `=$2389`

Compound Interest, SMB-005 MC

Roger invests $1400. He earns interest at 4% per annum, compounded monthly.

What is the future value of Roger's investment after 2.5 years?

  1. $1540.00
  2. $1546.98
  3. $3080.00
  4. $4540.76
Show Answers Only

`B`

Show Worked Solution

`text{Annual interest rate = 4% = 0.04}` 

`text(Monthly interest rate) = \frac(0.04)(12)`

`n = 2.5 xx 12 = 130`
  

`FV` `= PV(1 + r)^n`
  `= 1400 (1 + frac(0.04)(12))^30`
  `= $1546.98`

 
`=> \ B`

Compound Interest, SMB-004

In Japan, a bowl of ramen cost 1000 Japanese yen in 2005. The cost of ramen has increased by 0.5% per annum since then. 

Determine the cost of the same bowl of ramen in 2020 in Japanese yen, to the nearest yen.   (2 marks)

Show Answers Only

`1078\ text{yen}`

Show Worked Solution

`r =\ text(0.5%)\ = 0.005`

`n = 15\ text(years)`

`FV` `=PV(1+r)^n`  
  `=1000(1 + 0.005)^15`  
  `=1077.68…`  
  `=1078\ text{yen}`  

Compound Interest, SMB-003 MC

At the pump, one litre of diesel fuel cost 98 cents in NSW in 2003. The value of diesel fuel has increased by 4% per annum since then. 

Which expression gives the value, in cents, of one litre of diesel at the pump in 2023?  

  1. `98 xx 20 xx 1.04` 
  2. `98(1 + 0.04)^20`
  3. `98 xx 1.04 xx 20`
  4. `98 xx 20 xx 0.04`
Show Answers Only

`B`

Show Worked Solution

`r =\ text(4%)\ = 0.04`

`n = 20\ text(years)`

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

`text{Value in 2023} =98(1 + 0.04)^20`

`=>  B`

Compound Interest, SMB-001 MC

Resilia opens a bank account and deposits $2000 into it. Interest is paid at 4% 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. $2160.00
  2. $2163.20
  3. $2346.24
  4. $3920.00
Show Answers Only

`=> B`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 2000(1 + 0.04)^2`
  `= $2163.20`

 
`=> B`

Financial Maths, SMB-017

Albert invests $3000 and earns interest at 4% per annum, compounded quarterly.

What is the future value of Albert's investment after 2.5 years?  (3 marks)

Show Answers Only

`$3313.87`

Show Worked Solution

`text{Annual interest rate = 4%}`

`text(Quarterly interest rate) \ = frac(4%)(4)=1text{%}`

`n = 2.5 xx 4 = 10`
  

`FV` `= PV(1 + r)^n`
  `= 3000 (1 + 0.01)^10`
  `=3000(1.01)^10`
  `= $3313.87`

Financial Maths, SMB-016

What amount must be invested now at 3% per annum, compounded annually, so that in two years it will have grown to  $20 000?  (2 marks)

Show Answers Only

`$18\ 851.92`

Show Worked Solution

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

`r = 3text{%}, \ n=2`
 

`20\ 000` `= PV(1 + 0.03)^(2)`
`:.PV` `= (20\ 000)/1.03^(2)`
  `= $18\ 851.92`

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 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 13 MC

The graph show the future values over time of  `$P`, invested at three different rates of compound interest.
 


 

Which of the following correctly identifies each graph?

A. B.
C. D.
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`C`

Show Worked Solution

`text(Values increase quicker)`

`text(- higher compounding interest rate)`

`text(- same rate but more frequent compounding period)`

`:. W = 10text(% quarterly)`

`X = 10text(% annually)`

`Y = 5text(% annually)`

 
`=> C`

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 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 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`