smc-4334-10-Find FV

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.

Calculate the future value of Mila's investment after 1.5 years if she chooses Option A. (2 marks)
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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)
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`$45\ 264.08`
`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 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)}`

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.

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)

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

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

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`$3523.09`
`$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.

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

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

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

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

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

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`text(See Worked Solutions)`
`$3876.97`
`$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)

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`$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-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)

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`$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

`10\ 000 xx 0.10 xx 5`
`10\ 000 xx 0.05 xx 10`
`10\ 000 xx 0.05^10`
`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-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)

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`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?

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-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)

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`$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-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)

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`$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)

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`$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?

$3120.00
$3360.00
$3379.48
$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)

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`$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)

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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?

$1540.00
$1546.98
$3080.00
$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)

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`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?

`98 xx 20 xx 1.04`
`98(1 + 0.04)^20`
`98 xx 1.04 xx 20`
`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-002

Samanda opens a bank account and deposits $5000 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 three years? (2 marks)

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`$5624.32`

Show Worked Solution

`FV`	`= PV(1 + r)^n`
	`= 5000(1 + 0.05)^3`
	`= $5624.32`

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?

$2160.00
$2163.20
$2346.24
$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)

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`$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, 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.

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

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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)

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Show Answers Only

`$37\ 158.72`
`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 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?

$209.07
$209.19
$279.51
$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?

$1070.00
$1071.23
$1822.50
$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 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?

`35\ 000(1 + 0.03)^25`
`35\ 000(1 + 3)^25`
`35\ 000 xx 25 xx 0.03`
`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`