smc-604-20-Compound interest

Financial Maths, GEN1 2024 NHT 24 MC

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.

This amount was closest to

$ 224.03
$ 225.97
$ 228.62
$ 229.38
$ 231.46

Show Answers Only

$A$

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 $

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

$V \times 1.08 = 19\,288.80\ \ \Rightarrow\ \ V=17\,860$

$\therefore \ \text{Extra payment}\ = 17\,860-17\,635.97=$224.03 $

$\Rightarrow A$

Financial Maths, GEN1 2024 NHT 19-20 MC

Edo invests $10 000 in an account earning 3% interest per annum compounding monthly.

Question 19

The value, $V_n$, of Edo's investment after $n$ months is given by

$V_n=10\,000 \times 1.0025^n$
$V_n=10\,000 \times 1.003^n$
$V_n=10\,000 \times 1.03^n$
$V_n=10\,000 \times 1.003^{12 n}$
$V_n=10\,000 \times 1.03^{12 n}$

Question 20

The effective interest rate for Edo's investment is closest to

2.96%
2.98%
3.00%
3.02%
3.04%

Show Answers Only

$\text{Question 19:}\ A$

$\text{Question 20:}\ E$

Show Worked Solution

$\text{Question 19}$

$\text{Interest rate}\ = \dfrac{3.0\%}{12}=0.25\%\ \text{per month}$

$\text{After \(n$ months:}\)

$V_n=10\,000 \times 1.0025^{n}$

$\Rightarrow A$

$\text{Question 20}$

$\text{Let \(r$ = annual bank rate = 3.0%} \)

$r_{effective} = \left[\left(1+\dfrac{3.0}{100 \times 12}\right)^{12}-1\right] \times 100\%=3.04\% $

$\Rightarrow E$

Financial Maths, GEN1 2024 VCAA 20 MC

Dainika invested $2000 for three years at 4.4% per annum, compounding quarterly.

To earn the same amount of interest in three years in a simple interest account, the annual simple interest rate would need to be closest to

4.60%
4.68%
4.84%
4.98%

Show Answers Only

$B$

Show Worked Solution

$\text{Interest rate (per compounding period)} = \dfrac{4.4}{4} = 1.1 \%$

$\text{Compounding periods}\ =3\times4=12$

$FV=PV(1+r)^{n} = 2000(1+0.011)^{12}=$2280.57$

$\text{Annual interest}\ = \dfrac{280.57}{3}=$93.52$

$\text{S.I. rate}\ =\dfrac{93.52}{2000}\times 100\%=4.676\dots\%$

$\Rightarrow B$

♦ Mean mark 48%.

CORE, FUR1 2020 VCAA 27 MC

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

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

Show Answers Only

`B`

Show Worked Solution

`text(Find)\ n\ text(such that:)`

`10\ 000 xx 1.055^n`	`> 20\ 000`
`1.055^n`	`> 2`

`text(Testing some answer options:)`

`1.055^12 = 1.90`

`1.055^13 = 2.005`

`=> B`

CORE, FUR1 2020 VCAA 26 MC

Ray deposited $5000 in an investment account earning interest at the rate of 3% per annum, compounding quarterly.

A rule for the balance, `R_n` , in dollars, after `n` years is given by

`R_n = 5000 xx 0.03^n`
`R_n = 5000 xx 1.03^n`
`R_n = 5000 xx 0.03^(4n)`
`R_n = 5000 xx 1.0075^n`
`R_n = 5000 xx 1.0075^(4n)`

Show Answers Only

`E`

Show Worked Solution

`text(Quarterly rate) = 0.03/4 = 0.0075`

♦♦♦ Mean mark 20%.

`text(Compounding periods) = 4n`

`:.\ text(Balance after)\ n\ text(years)`

`= 5000 xx 1.0075^(4n)`

`=> E`

CORE, FUR1 2018 VCAA 19 MC

Daniel borrows $5000, which he intends to repay fully in a lump sum after one year.

The annual interest rate and compounding period for five different compound interest loans are given below:

• Loan I – 12.6% per annum, compounding weekly
• Loan II – 12.8% per annum, compounding weekly
• Loan III – 12.9% per annum, compounding weekly
• Loan IV – 12.7% per annum, compounding quarterly
• Loan V – 13.2% per annum, compounding quarterly

When fully repaid, the loan that will cost Daniel the least amount of money is

Loan I.
Loan II.
Loan III.
Loan IV.
Loan V.

Show Answers Only

`D`

Show Worked Solution

`text(Weekly compounding loans ⇒ Loan I is cheaper)`

`text(Repayment of Loan I)`	`= 5000 xx (1 + 12.6/(52 xx 100))^52`
	`= $5670.55`

`text(Quarterly compounding loans ⇒ Loan IV is cheaper)`

`text(Repayment of Loan IV)`	`= 5000 xx (1 + 2.7/(12 xx 100))^12`
	`= $5136.68`

`:.\ text(Loan IV is cheapest.)`

`=> D`

CORE, FUR2 2017 VCAA 6

Alex sends a bill to his customers after repairs are completed.

If a customer does not pay the bill by the due date, interest is charged.

Alex charges interest after the due date at the rate of 1.5% per month on the amount of an unpaid bill.

The interest on this amount will compound monthly.

Alex sent Marcus a bill of $200 for repairs to his car.

Marcus paid the full amount one month after the due date.

How much did Marcus pay? (1 mark)

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Alex sent Lily a bill of $428 for repairs to her car.

Lily did not pay the bill by the due date.

Let `A_n` be the amount of this bill `n` months after the due date.

Write down a recurrence relation, in terms of `A_0`, `A_(n + 1)` and `A_n`, that models the amount of the bill. (2 marks)

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Lily paid the full amount of her bill four months after the due date.

How much interest was Lily charged?

Round your answer to the nearest cent. (1 mark)

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Show Answers Only

`$203`
`A_o = 428,qquadA_(n + 1) = 1.015A_n`
`$26.26\ \ (text(nearest cent))`

Show Worked Solution

a.	`text(Amount paid)`	`= 200 + 200 xx 1.5text(%)`
		`= 1.015 xx 200`
		`= $203`

♦ Mean mark part (b) 47%.
MARKER’S COMMENT: A recurrence relation has the initial value written first. Know why `A_n=428 xx 1.015^n` is incorrect.

b. `A_o = 428,qquadA_(n + 1) = 1.015A_n`

c.	`text(Total paid)\ (A_4)`	`= 1.015^4 xx 428`
		`= $454.26`

♦♦ Mean mark part (c) 29%.

`:.\ text(Total Interest)`	`= 454.26-428`
	`= $26.26\ \ (text(nearest cent))`

CORE*, FUR2 2006 VCAA 3

The company prepares for this expenditure by establishing three different investments.

$7000 is invested at a simple interest rate of 6.25% per annum for eight years.
Determine the total value of this investment at the end of eight years. (2 marks)
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$10 000 is invested at an interest rate of 6% per annum compounding quarterly for eight years.
Determine the total value of this investment at the end of eight years.
Write your answer correct to the nearest dollar. (1 mark)
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$500 is deposited into an account with an interest rate of 6.5% per annum compounding monthly.
Deposits of $200 are made to this account on the last day of each month after interest has been paid.
Determine the total value of this investment at the end of eight years.
Write your answer correct to the nearest dollar. (1 mark)
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Show Answers Only

`$10\ 500`
`$16\ 103`
`$25\ 935`

Show Worked Solution

a.	`I`	`= (PrT)/100`
		`= (7000 xx 6.25 xx 8)/100`
		`= $3500`

`:.\ text(Total value of investment)`

`= 7000 + 3500`

`= $10\ 500`

b. `text(Compounding periods) = 8 xx 4 = 32`

`text(Interest rate)`	`= (text(6%))/4`
	`= 1.5text(% per quarter)`

`:.\ text(Total value of investment)`

`= PR^n`

`= 10\ 000(1.015)^32`

`= 16\ 103.24…`

`= $16\ 103\ \ text{(nearest $)}`

c. `text(By TVM Solver,)`

`N`	`= 8 xx 12 = 96`
`I(text(%))`	`= 6.5`
`PV`	`= 500`
`PMT`	`= 200`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −25\ 935.30…`

`:.\ text(Total value of investment is $25 935.)`

CORE*, FUR2 2006 VCAA 2

It is estimated that inflation will average 2% 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. (1 mark)

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Show Answers Only

`$70\ 300\ \ text{(nearest $)}`

Show Worked Solution

`text(Find value)\ (A)\ text(in 8 years.)`

`A`	`= PR^n`
	`= 60\ 000(1.02)^8`
	`= 70\ 299.562…`
	`= $70\ 300\ \ text{(nearest $)}`

CORE*, FUR2 2008 VCAA 2

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. (1 mark)

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Show Answers Only

`$3523.09`
`$1137.40`

Show Worked Solution

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

`text(Interest per half year) = 8.2/2 = 4.1text(%)`

`:. A`	`= PR^n`
	`= 3000(1.041)^4`
	`= 3523.093…`
	`= $3523.09\ \ text{(nearest cent)}`

b. `text(After 4 years)\ (n = 8),`

MARKER’S COMMENT: A TVM calculator could also be used to solve this question.

`A`	`= 3000(1.041)^8`
	`= 4137.396…`

`:.\ text(Interest)`	`= 4137.396-3000`
	`= 1137.396…`
	`= $1137.40\ \ text{(nearest cent)}`

CORE*, FUR2 2009 VCAA 3

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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Show Answers Only

`text(See Worked Solutions)`
`$3876.97`
`$1020.86`

Show Worked Solution

a. `text(Interest rate per quarter)`

`= 4.4/4`

`= 1.1text(% …as required.)`

b. `text(Compounding periods = 12)`

`A`	`= PR^n`
	`= 3400(1.011)^12`
	`= 3876.973…`
	`= $3876.97`

c. `text(Compounding periods) = 6 xx 4 = 24`

`A`	`= 3400(1.011)^24`
	`= 4420.858…`

`text(Interest earned over 6 years)`

`= 4420.86- 3400`

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

CORE*, FUR2 2010 VCAA 3

Simple Saver is a simple interest investment in which interest is paid annually.

Growth Plus is a compound interest investment in which interest is paid annually.

Initially, $8000 is invested with both Simple Saver and Growth Plus.

The graph below shows the total value (principal and all interest earned) of each of these investments over a 15 year period.

The increase in the value of each investment over time is due to interest

Which investment pays the highest annual interest rate, Growth Plus or Simple Saver?

Give a reason to justify your answer. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
After 15 years, the total value (principal and all interest earned) of the Simple Saver investment is $21 800.

Find the amount of interest paid annually. (1 mark)

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After 15 years, the total value (principal and all interest earned) of the Growth Plus investment is $24 000.
1. Write down an equation that can be used to ﬁnd the annual compound interest rate, `r`. (1 mark)
  
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2. Determine the annual compound interest rate.
  
  Write your answer as a percentage correct to one decimal place. (1 mark)
  
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Show Answers Only

`text(Simple Saver has the highest annual)`
`text(interest rate because after 1 year,)`
`text(the value of investment is higher.)`
`$920`
1. `24\ 000 = 8000 (1 + r/100)^15`
2. `7.6text{% (1 d.p.)}`

Show Worked Solution

a. `text(Simple Saver has the highest annual)`

♦♦♦ Part (a) was “very” poorly answered although exact data unavailable.
MARKER’S COMMENT: Most students ignored the word “rate” and instead referred to the eventual return of each investment.

`text(interest rate because after 1 year,)`

`text(the value of investment is higher.)`

b. `text(Total interest earned)`

`= 21\ 800-8000`

`= $13\ 800`

`:.\ text(Interest paid annually)`

`= (13\ 800)/15`

`= $920`

c.i. `text(Using)\ A = PR^n,`

`24\ 000 = 8000 (1 + r/100)^15`

c.ii.	`(1 + r/100)^15`	`= 3`
	`1 + r/100`	`= 1.0759…`
	`:. r`	`= 0.0759…`
		`= 7.6text{% (1 d.p.)}`

CORE*, FUR2 2011 VCAA 3

Tania purchased a house for $300 000.

She will have to pay stamp duty based on this purchase price.

Stamp duty rates are listed in the table below.

Calculate the amount of stamp duty that Tania will have to pay. (1 mark)

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Assuming that her house will increase in value at a rate of 3.17% per annum, what will the value of Tania's house be after 5 years?

Write your answer to the nearest thousand dollars. (1 mark)

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Tania bought her house at the start of 2011.

If the rate of increase in value remains at 3.17% per annum, at the start of which year will the value of Tania's house first exceed $450 000? (2 marks)

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Show Answers Only

`$13\ 070`
`$351\ 000`
`2024`

Show Worked Solution

a.	`text(Stamp duty)`	`= 2870 + 6text(%) xx (300\ 000-130\ 000)`
		`= $13\ 070`

♦♦ Part (a) was “poorly answered” although exact data is unavailable.
MARKER’S COMMENT: A majority of students did not understand how to use the table.

b. `text(Using)\ \ \ A = PR^n`

`text(Value)`	`= 300\ 000(1.0317)^5`
	`= 350\ 661.7…`
	`= $351\ 000\ \ text{(nearest $1000)}`

c. `text(Find)\ n\ text(when)\ \ \ A > $450\ 000`

`300\ 000 xx 1.0317^n`	`= 450\ 000`
`n`	`~~12.99…`

MARKER’S COMMENT: Many students who correctly found `n` lost a mark by failing to identify the exact year.

`:.\ text(After 13 years, in 2024, the house value)`

`text(will be over)\ $450\ 000.`

CORE*, FUR2 2014 VCAA 2

A sponsor of the cricket club has invested $20 000 in a perpetuity.

The annual interest from this perpetuity is $750.

The interest from the perpetuity is given to the best player in the club every year, for a period of 10 years.

What is the annual rate of interest for this perpetuity investment? (1 mark)

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After 10 years, how much money is still invested in the perpetuity? (1 mark)

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The average rate of inflation over the next 10 years is expected to be 3% per annum.
1. Michael was the best player in 2014 and he considered purchasing cricket equipment that was valued at $750.
2. What is the expected price of this cricket equipment in 2015? (1 mark)
  
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3. What is the 2014 value of cricket equipment that could be bought for $750 in 2024? Write your answer, correct to the nearest dollar. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`3.75`
`$20\ 000`
1. `$772.50`
2. `$558\ \ text{(nearest dollar)}`

Show Worked Solution

a.	`20\ 000 xx r`	`= 750`
	`:. r`	`= 750/(20\ 000)`
		`= 0.0375`

`:.\ text(Annual interest rate = 3.75%)`

b. `$20\ 000`

`text{(A perpetuity’s balance remains constant.)}`

c.i. `text(Expected price in 2015)`

`= 750 xx (1 + 3/100)`

`= 750 xx 1.03`

`= $772.50`

c.ii. `text(Value in 2014) xx (1.03)^10 = 750`

`:.\ text(Value in 2014)\ `	`= 750/((1.03)^10)`
	`= 558.07…`
	`= $558\ \ text{(nearest dollar)}`

CORE*, FUR1 2015 VCAA 4 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

A. `$6.71`

B. `$40.82`

C. `$80.40`

D. `$81.75`

E. `$83.03`

Show Answers Only

`E`

Show Worked Solution

`text(Monthly interest rate)\ = 3.35/12 = 0.27916…`

`text(Total after 2 years)`	`= 1200(1 + (0.27916…)/100)^24`
	`=$1283.03…`

`:.\ text(Interest earned)`	`= 1283.03 – 1200`
	`= $83.03`

`=> E`

CORE*, FUR1 2008 VCAA 9 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

A. `4 xx (7.2/100 xx 8000)`

B. `8000 xx 1.018^4 - 8000 xx 1.018^3`

C. `8000 xx 1.018^16 - 8000 xx 1.018^12`

D. `8000 xx 1.072^4 - 8000 xx 1.072^3`

E. `8000 xx 1.072^16 - 8000 xx 1.072^12`

Show Answers Only

`C`

Show Worked Solution

`text(4th year interest)`	`=\ text(Value after 4 years) -`
	`text(Value after 3 years.)`

`text(Using)\ \ A = PR^n,`

♦♦ Mean mark 33%.
MARKERS’ COMMENT: The most common errors were not converting and applying the interest rate to a quarterly one.

`text(where)\ R = 1 + 7.2/(100 xx 4) = 1.018`

`text(Value after 4 years:)`

`A_4`	`= 8000 xx 1.018^((4 xx 4))`
	`= 8000 xx 1.018^(16)`

`text(Value after 3 years:)`

`A_3`	`= 8000 xx 1.018^((4 xx 3))`
	`= 8000 xx 1.018^(12)`

`:.\ text(4th year interest) = 8000 xx 1.018^(16) − 8000 xx 1.018^(12)`

`=> C`

CORE*, FUR1 2008 VCAA 6 MC

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, correct to the nearest cent, Sam will have

A. $86.48 less than Charlie.

B. $86.48 more than Charlie.

C. $132.23 less than Charlie.

D. $132.23 more than Charlie.

E. the same as Charlie.

Show Answers Only

`A`

Show Worked Solution

`text(Sam’s Investment,)`

`I`	`= (PrT)/100`
	`= ((5000)(7.5)(3))/100`
	`= 1125`

`:.\ text(Total amount)`	`= 5000 + 1125`
	`= $6125`

`text(Charlie’s Investment,)`

`A`	`= PR^n`
	`= (5000)(1.075)^3`
	`= $6211.48`

`text(Difference)`	`= 6211.48-6125`
	`= $86.48`

`=> A`

CORE*, FUR1 2007 VCAA 6 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

A. `10\ 000 xx 0.10 xx 5`

B. `10\ 000 xx 0.05 xx 10`

C. `10\ 000 xx 0.05^10`

D. `10\ 000 xx 1.05^10`

E. `10\ 000 xx 1.10^5`

Show Answers Only

`D`

Show Worked Solution

`text(Interest rate)\ = \frac{10%}{2} = 5%\ \ \text{(per 6 months)}`

♦ Mean mark 42%.

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

`:.A`	` = PR^n`
	`= 10\ 000 xx 1.05^(10)`

`=> D`

CORE*, FUR1 2006 VCAA 8 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?

A. interest is compounding annually and is credited annually

B. interest is compounding annually and is credited quarterly

C. interest is compounding quarterly and is credited quarterly

D. simple interest is paid on the opening balance and is credited annually

E. simple interest is paid on the opening balance and is credited quarterly

Show Answers Only

`A`

Show Worked Solution

`text(From the graph, as balance increases after)`

♦ Mean mark 37%.
MARKERS’ COMMENT: Graphical analysis of financial situations is a requirement of the study design and deserves attention.

`text(each year, interest is credited annually.)`

`:.\ text(Eliminate)\ B, C\ text(and)\ E.`

`text(The difference of the balances between successive)`

`text(years is increasing which indicates that interest is)`

`text(compounding.)`

`:.\ text(Eliminate)\ D.`

`=> A`

CORE*, FUR1 2005 VCAA 6 MC

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

A. `$195.00`

B. `$221.16`

C. `$235.55`

D. `$3623.85`

E. `$3859.40`

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`text(Total after 3rd year)`	`= 3000(1.065)^3 \ \ \ \ \ (n =3)`
	`= $3623.848875…`
	`= $3623.85 \ \ (text(2 d.p.))`
`text(Total after 4th year)`	`= 3000(1.065)^4 \ \ \ \ \ (n = 4)`
	`= $3859.399…`
	`= $3859.40 \ \ (text(2 d.p.))`

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

`=> C`

CORE*, FUR1 2011 VCAA 7 MC

Anthony invested $15 000 in an account. It earned `rtext(%)` interest per annum, compounding monthly.

The amount of interest that is earned in the third year of the investment is given by

A. `15\ 000 (1 + r / 1200)^3 - 15\ 000 (1 + r / 1200)^2`

B. `15\ 000 (1 + r / 1200)^36 - 15\ 000 (1 + r / 1200)^24`

C. `15\ 000 (1 + r / 100)^3 - 15\ 000 (1 + r / 100)^2`

D. `15\ 000 (1 + r / 100)^36 - 15\ 000 (1 + r / 100)^24`

E. `15\ 000 (1 + r / 1200)^4 - 15\ 000 (1 + r / 1200)^3`

Show Answers Only

`B`

Show Worked Solution

`text(Interest in 3rd year)`

♦♦ Mean mark 33%.

`=\ text(Value after 3 yrs) – text(Value after 2 years)`

`text(Using)\ \ A = PR^n,\ \ text(where)\ \ A = 15\ 000, and`

`R`	`= (1 + r / {12 xx 100})`
	`= (1 + r / 1200)`

`text(In first 3 years:)`

`A_3`	`= 15\ 000 (1 + r / 1200)^(12 xx 3)`
	`= 15\ 000 (1 + r / 1200)^36`

`text(In first 2 years:)`

`A_2`	`= 15\ 000 (1 + r / 1200)^(12 xx 2)`
	`= 15\ 000 (1 + r / 1200)^24`

`:.\ text(Interest in 3rd year)`

`= 15\ 000 (1 + r/1200)^36 – 15\ 000 (1 + r/1200)^24`

`=> B`

CORE*, FUR1 2011 VCAA 2 MC

An amount of $22 000 is invested for three years at an interest rate of 3.5% per annum, compounding annually.

The value of the investment at the end of three years is closest to

A. `$2310`

B. `$9433`

C. `$24\ 040`

D. `$24\ 392`

E. `$31\ 433`

Show Answers Only

`D`

Show Worked Solution

`P = 22\ 000,\ \ R = 1.035,\ \ n = 3`

`A`	`=PR^n`
	`= 22\ 000 xx 1.035^3`
	`= 24\ 391.79…`

`=> D`

CORE*, FUR1 2012 VCAA 8 MC

$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.

The total interest earned by this investment over 12 months is closest to

A. $926

B. $935

C. $941

D. $953

E. $965

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ A = PR^n,`

♦ Mean mark 42%.

`text(Value after the 1st 6 months)`

`= 15\ 000 xx (1 + 6.1 / {12 xx 100})^6`

`=$15\ 463.35…`

`text(Value after the 2nd 6 months)`

`=15\ 463.35… xx (1 + 6.25 / {12 xx 100})^6`

`=$15\ 952.91…`

`:.\ text(Interest)`	`= 15\ 952.91 – 15\ 000`
	`= $952.91`

`=> D`

CORE*, FUR1 2012 VCAA 3 MC

The transaction details for a savings account for the month of July 2012 are shown below.

Interest is calculated and paid monthly on the minimum monthly balance.

The annual rate of interest paid on this account is closest to

A. `3.5text(%)`

B. `4.3text(%)`

C. `4.7text(%)`

D. `4.9text(%)`

E. `5.2text(%)`

Show Answers Only

`E`

Show Worked Solution

`text(Interest) = 21.99`

♦ Mean mark 39%.

`text(Minimum balance) = 5101.82`

`text{Rate of interest (monthly)}`	`= 21.99 / 5101.82`
	`= 0.0043…`

`:.\ text(Yearly rate of interest)`

`= 0.0043… xx 12`

`= 0.0517…`

`= 5.17text(%)`

`=> E`

CORE*, FUR1 2013 VCAA 3 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?

A. `10\ 000 xx 1.02^5-10\ 000`

B. `10\ 000 xx 1.02^20-10\ 000`

C. `10\ 000 xx 1.08^5 -10\ 000`

D. `10\ 000 xx 1.08^20-10\ 000`

E. `10\ 000 xx 1.02^20`

Show Answers Only

`B`

Show Worked Solution

`text{Quarterly interest rate (R)} = 8/4=2%=0.02`

♦ Mean mark 40%.

`text(5 years)\ =5 xx4=20\ text(quarters),\ \ n=20`

`text(Using)\ A = PR^n,`

`A`	`=10\ 000 xx (1 + 0.02)^20`
	`=10\ 000xx (1.02)^20`

`:.\ text(Interest)`	`=\ text(Final amount) – text(original investment)`
	`= 10\ 000 xx 1.02^20 – 10\ 000`

`=> B`

CORE*, FUR1 2014 VCAA 3 MC

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, the value of her investment after 150 days is closest to

A. `$15\ 279`

B. `$15\ 284`

C. `$15\ 286`

D. `$15\ 690`

E. `$16\ 776`

Show Answers Only

`C`

Show Worked Solution

`text(Annual rate)`	`= 4.60text(%)`
`text(Daily rate)`	`= (4.60text{%})/365=0.046/365`

`text(Using)\ \ A = PR^n`

`text(where)\ R`	`= 1 + 0.0460/365`
	`= 1.0001260…`

`:.\ text(Investment)`	`= 15\ 000 xx (1.0001260…)^150`
	`= 15\ 000 xx 1.01908…`
	`= $15\ 286.24`

`=> C`

`N`	`=120`
`Itext{(%)}`	`= 2.8`
`PMT`	`=0`
`PV`	`= ?`
`FV`	`= 686\ 904.09`
`text(PY)`	`= text(CY) =12`

`N`	`=120`
`Itext{(%)}`	`= 3.2`
`PMT`	`=0`
`PV`	`= ?`
`FV`	`= 509\ 320.3`
`text(PY)`	`= text(CY) =12`

`V_0`	`= 500`
`V_1`	`= V_0 xx (1 + r)^n`
`580`	`= 500 (1 + r)^1`
`1 + r`	`= 1.16`
`r`	`= 16text(%)`

Financial Maths, GEN1 2024 NHT 24 MC

Financial Maths, GEN1 2024 NHT 19-20 MC

Financial Maths, GEN1 2024 VCAA 20 MC

CORE, FUR1 2020 VCAA 30 MC

CORE, FUR1 2020 VCAA 27 MC

CORE, FUR1 2020 VCAA 26 MC

CORE, FUR1 2019 VCAA 21 MC

CORE, FUR1 2018 VCAA 19 MC

CORE, FUR2 2017 VCAA 6

CORE*, FUR2 2006 VCAA 3

CORE*, FUR2 2006 VCAA 2

CORE*, FUR2 2008 VCAA 2

CORE*, FUR2 2009 VCAA 3

CORE*, FUR2 2010 VCAA 3

CORE*, FUR2 2011 VCAA 3

CORE*, FUR2 2014 VCAA 2

CORE*, FUR1 2015 VCAA 4 MC

CORE*, FUR1 2008 VCAA 9 MC

CORE*, FUR1 2008 VCAA 6 MC

CORE*, FUR1 2007 VCAA 6 MC

CORE*, FUR1 2006 VCAA 8 MC

CORE*, FUR1 2005 VCAA 6 MC

CORE*, FUR1 2011 VCAA 7 MC

CORE*, FUR1 2011 VCAA 2 MC

CORE*, FUR1 2012 VCAA 8 MC

CORE*, FUR1 2012 VCAA 3 MC

CORE*, FUR1 2013 VCAA 3 MC

CORE*, FUR1 2014 VCAA 3 MC

`text(Balance)`	`= [I(1 + 3.2/(12 xx 100))^120 + 10\ 000](1 + 2.8/(12 xx 100))^120`
	`= $686\ 904.09`

`:. b`	`= 500 (1.16)^2`
	`= 672.80`