smc-1124-10-Simple Interest

Financial Maths, STD1 F1 2024 HSC 21

Jan borrowed $1500 at 6% per annum.

Calculate the simple interest for the first three months. (2 marks)

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$$22.50$

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$\text{Interest}$	$=Prn$
	$=1500\times 0.06\times \dfrac{3}{12}$
	$=$22.50$

Financial Maths, STD1 F2 2023 HSC 24

Bobby invested $5000.

The table shows the progress of his investment over the first 4 months.

What are the values of $A$ and $B$? (2 marks)

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Bobby could have earned simple interest on the investment at 0.62% per month.
How much interest would Bobby have earned over 4 months by choosing this option? (2 marks)

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a. $A=$30.54,\ \ B=$5121.08$

b. $$124$

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a.	$A$	$=5090.54\times\dfrac{0.6}{100}$
		$=$30.54\text{ (2 d.p.)}$

	$B$	$=5090.54+30.54$
		$=$5121.08$

b. $P=$5000,\ \ r=\dfrac{0.62}{100},\ \ n=4$

$I$	$=Prn$
	$=5000\times \dfrac{0.62}{100}\times 4$
	$=$124$

♦♦ Mean mark (b) 31%.

Financial Maths, STD1 F1 2023 HSC 2 MC

An amount of $$2500$ is invested at a simple interest rate of $3%$ per annum.

How much interest is earned in the first two years?

$$75$
$$150$
$$2575$
$$2652$

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$B$

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$I$	$=Prn$
	$=2500 \times\ \dfrac{3}{100} \times\ 2$
	$=$150$

$\Rightarrow B$

Financial Maths, STD1 F1 2019 HSC 16

What is the interest earned when $800 is invested for 7 months at a simple interest rate of 3% per annum? (2 marks)

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`$14`

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♦ Mean mark 38%.

`text(Interest)`	`= Prn`
	`= 800 xx 0.03 xx 7/12`
	`= $14`

Financial Maths, STD2 F1 SM-Bank 2

George makes a single deposit of $9000 into an account that pays simple interest.

After 4 years, George's account has a balance of $10 350.

What simple interest rate did George receive on his investment? (2 marks)

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`3.75text(%)`

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`text(Interest earned)`	`= 10\ 350 – 9000`
	`= $1350`

`text(Using)\ \ I = Prn,`

`1350`	`= 9000 xx r xx 4`
`:. r`	`= 1350/(4 xx 9000)`
	`= 0.0375`
	`= 3.75text(%)`

Financial Maths, STD2 F1 SM-Bank 3 MC

$6000 is invested in an account that earns simple interest at the rate of 3.5% per annum.

The total interest earned in the first four years is

$70
$84
$210
$840

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`D`

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`P = 6000,\ \ r = 3.5text(%),\ \ n = 4`

`I`	`= Prn`
	`= 6000 xx 3.5/100 xx 4`
	`= 840`

`=> D`

Financial Maths, STD2 F1 2005 HSC 25a

Reece is preparing his annual budget for 2006.

His expected income is:

• $90 every week as a swimming coach
• Interest earned from an investment of $5000 at a rate of 4% per annum.

His planned expenses are:

• $30 every week on transport
• $12 every week on lunches
• $48 every month on entertainment.

Reece will save his remaining income. He uses the spreadsheet below for his budget.

Determine the values of `X`, `Y` and `Z`. (Assume there are exactly 52 weeks in a year.) (3 marks)

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At the beginning of 2006, Reece starts saving.

Will Reece have saved enough money during 2006 for a deposit of $2100 on a car if he keeps to his budget? Justify your answer with suitable calculations. (2 marks)

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`X =$200, Y = $1560\ text(and) \ Z = $576`
`text(Reece will have saved have saved enough for a $2100 deposit.)`

Show Worked Solution

i. `text(Interest on Investment) = X`

`X`	`= 5000 xx 4 text(%)`
	`= $200`

`text(Transport =)\ Y`

`Y`	`= 52 × 30`
	`= $1560`

`text(Entertainment =)\ Z`

`Z`	`= 48 × 12`
	`= $576`

ii.	`text(Total Income)`	`= 4680 + 200`
		`= $4880`

`text(Total Expenses)`	`= 1560 + 624 + 576`
	`= $2760`

`text(Savings)`	`= 4880 − 2760`
	`= $2120`

`:.\ text(Reece will have saved enough for)`

`text(a $2100 deposit.)`

Financial Maths, STD2 F1 2007 HSC 23a

Lilly and Rose each have money to invest and choose different investment accounts.

The graph shows the values of their investments over time.

How much was Rose’s original investment? (1 mark)

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At the end of 6 years, which investment will be worth the most and by how much? (2 marks)

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Lilly’s investment will reach a value of $20 000 first.
How much longer will it take Rose’s investment to reach a value of $20 000? (1 mark)

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`$5000`
`text(Rose’s is worth $2000 more.)`
`text(It takes Lilly 14 years to reach $20 000 and it takes)`

`text{Rose 1 year longer (15 years) to reach the same value}`

Show Worked Solution

i. `$5000\ text{(} y text(-intercept) text{)}`

ii. `text(After 6 years,)`

`text(Lilly’s investment)`	`= $9000`
`text(Rose’s investment)`	`= $11\ 000`
`:.\ text(Rose’s is worth $2000 more.)`

iii. `text(It takes Lilly 14 years to reach $20 000 and it)`

`text{takes Rose 1 year longer (15 years) to reach the}`

`text(same value.)`

Financial Maths, STD2 F1 2009 HSC 20 MC

Lou bought a plasma TV which was priced at $3499. He paid $1000 deposit and got a loan for the balance that was paid off by 24 monthly instalments of $135.36.

What simple interest rate per annum, to the nearest percent, was charged on his loan?

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`B`

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♦♦ Mean mark 34%
COMMENT: A multi-step question targeting higher bands that can be a time-trap for many students.

`text(Loan)`	`=\ text(Price – deposit)`
	`= 3499-1000`
	`= $2499`

`text(Total repaid)`	`= 24 xx 135.36`
	`= $3248.64`

`:.\ text(Interest paid)`	`= 3248.64\-2499`
	`= $749.64`

`text(Simple Interest)`	`= Prn`
`749.64`	`= 2499 xx r xx 2`
`:. r`	`= 749.64/(2 xx 2499)`
	`= 0.1499…`
	`~~15 text(%)`

`=> B`

Financial Maths, STD2 F1 2010 HSC 5 MC

Minjy invests $2000 for 1 year and 5 months. The simple interest is calculated at a rate of 6% per annum.

What is the total value of the investment at the end of this period?

$2170
$2180
$3003
$3700

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`A`

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`text(Interest)`	`=Prn`
	`=2000 xx\ text(6%)\ xx 17/12`
	`=$170`

`:.\ text(Value of Investment)`	`=2000+170`
	`=$2170`

`=> A`

Financial Maths, STD2 F1 2013 HSC 13 MC

Polly borrowed $11 000. She repaid the loan in full at the end of two years with a lump sum of $12 000.

What annual simple interest rate was she charged?

4.17%
4.55%
8.33%
9.09%

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`B`

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`text(Total interest paid)=12\ 000-11\ 000=$1000`

`I`	`=Prn`
`1000`	`=11\ 000 xx r xx2`
`r`	`=1000/(22\ 000)`
	`=4.55 text(%)`

`=>\ B`

Financial Maths, STD2 F1 2013 HSC 9 MC

Lynne invests $1000 for a term of 15 months. Simple interest is paid on the investment at a rate of 3.75% per annum.

How much will Lynne's investment be worth at the end of the term?

$1046.88
$1047.09
$1296.88
$1468.75

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`A`

Show Worked Solution

`I=Prn=1000xx3.75/100xx15/12=$46.88`

`:.\ text(Investment worth)\ = 1000+46.88=$1046.88`

`=>\ A`

\(\text{Interest}\)	\(=Prn\)
	\(=1500\times 0.06\times \dfrac{3}{12}\)
	\(=$22.50\)

a.	\(A\)	\(=5090.54\times\dfrac{0.6}{100}\)
		\(=$30.54\text{ (2 d.p.)}\)

	\(B\)	\(=5090.54+30.54\)
		\(=$5121.08\)

\(I\)	\(=Prn\)
	\(=5000\times \dfrac{0.62}{100}\times 4\)
	\(=$124\)

\(I\)	\(=Prn\)
	\(=2500 \times\ \dfrac{3}{100} \times\ 2\)
	\(=$150\)