smc-6927-10-Interest on Purchases

Financial Maths, STD2 EQ-Bank 32

Yasmin is planning to purchase a Newcastle Knights corporate box for $5000. She intends to repay the entire amount 90 days after making the purchase.

She is comparing two credit options.

Credit card: compound interest charged daily at 18% per annum, with a 30-day interest-free period from the date of purchase.
Personal loan: simple interest charged at 9.5% per annum.

Determine which option Yasmin should choose, showing calculations to support your answer. (4 marks)

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$\text{Credit card (daily compound interest):}$

$\text{Days accruing interest}\ =90-30=60$

$\text{Amount owing}$	$= 5000\left(1+\dfrac{0.18}{365}\right)^{60}$
	$= 5150.118\ldots$
	$= \$5150.12\ \text{(nearest cent)}$

$\text{Interest} = 5150.12-5000 = \$150.12$

$\text{Personal loan (simple interest):}$

$I = Prn$	$=5000 \times 0.095 \times \dfrac{90}{365}$
	$= 117.123\ldots$
	$= \$117.12\ \text{(nearest cent)}$

$\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}$

Show Worked Solution

$\text{Days accruing interest}\ =90-30=60$

$\text{Credit card (daily compound interest):}$

$\text{Amount owing}$	$= 5000\left(1+\dfrac{0.18}{365}\right)^{60}$
	$= 5150.118\ldots$
	$= \$5150.12\ \text{(nearest cent)}$

$\text{Interest} = 5150.12-5000 = \$150.12$

$\text{Personal loan (simple interest):}$

$I = Prn$	$=5000 \times 0.095 \times \dfrac{90}{365}$
	$= 117.123\ldots$
	$= \$117.12\ \text{(nearest cent)}$

$\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}$

Financial Maths, STD2 EQ-Bank 28

Sam needs to borrow $3000 and plans to repay it in full after 30 days. He is comparing two options.

Credit card: compound interest charged daily at 19.9% per annum, with no interest-free period.
Personal loan: simple interest charged at 11.5% per annum.

Calculate the interest Sam would be charged on the credit card over the 30 days. (2 marks)

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How much does Sam save by choosing the personal loan. (2 marks)

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a. $\$49.46$

b. $\text{Saving =}\ \$21.10$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.199}{365}$

$\text{Amount owing}$	$= 3000\left(1+\dfrac{0.199}{365}\right)^{30}$
	$= 3049.458\ldots$
	$= \$3049.46\ \text{(nearest cent)}$

$\therefore \text{Interest} = 3049.46-3000 = \$49.46$

b. $\text{Personal loan (simple interest):}$

$I = Prn$	$=3000 \times 0.115 \times \dfrac{30}{365}$
	$= 28.356\ldots$
	$= \$28.36\ \text{(nearest cent)}$

$\therefore \text{Saving} = 49.46-28.36 = \$21.10$

Financial Maths, STD2 EQ-Bank 27

A credit card account has no interest-free period. Compound interest is calculated daily at a rate of 18.5% per annum and charged to the account up to and including the statement date.

A single purchase of $1500 was made and interest accrued on the purchase for 28 days, including the statement date. There were no other purchases on the account.

Calculate the closing balance on the statement date. (2 marks)

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The minimum payment is calculated as $30 or 3% of the closing balance, whichever is greater. Calculate the minimum payment due. (2 marks)

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a. $\$1521.43$

b. $\$45.64$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.185}{365}$

$\text{Closing balance}$	$= 1500\left(1+\dfrac{0.185}{365}\right)^{28}$
	$= 1521.433\ldots$
	$= \$1521.43$

b. $3\% \text{ of closing balance} = 0.03 \times 1521.43= \$45.64$

$\text{Since } \$45.64 > \$30, \text{ the minimum payment due is } \$45.64.$

Financial Maths, STD2 F4 2023 HSC 32

Ali has a credit card which has no interest-free period. Interest is charged at 13.5% per annum, compounding daily, on the amount owing.

During the month, Ali made only one purchase of $450 using the credit card. The full amount owing was repaid 21 days later.

Calculate the amount of interest charged on the purchase, assuming that interest is charged for the 21 days. (2 marks)

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What percentage of the full amount repaid is the interest? Give the answer to two decimal places. (2 marks)

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a. `$3.51`

b. `0.77%`

Show Worked Solution

a. `text{Daily interest rate}\ (r) = 13.5/365 % = 0.135/365`

`n=\ text{21 days}`

`FV`	`=PV(1+r)^n`
	`=450(1+0.135/365)^21=$453.51`

`:.\ text{Interest charged}\ = 453.51-450=$3.51`

♦ Mean mark (a) 45%.

b.	`text{Interest as % total repaid}`	`=3.51/453.51 xx 100`
		`=0.007739…`
		`=0.77%\ text{(to 2 d.p.)}`

♦ Mean mark (b) 49%.

Financial Maths, STD2 F4 2016 HSC 17 MC

Ariana is charged compound interest at the rate of 0.036% per day on outstanding credit card balances. She has $780 outstanding for 24 days.

How much compound interest is she charged?

$6.74
$6.77
$786.74
$786.77

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`B`

Show Worked Solution

`text(Total owing)`

♦ Mean mark 38%.
COMMENT: Credit card problems consistently produce sub-50% mean marks. Important review area.

`= P(1+r)^n`

`= 780(1 + 0.036/100)^24= 786.77`

`:.\ text(Interest charged)= 786.77-780= $6.77`

`=> B`

Financial Maths, STD2 F4 2015* HSC 29a

On 20 August, tickets were purchased for $425 using a credit card. No other purchases were made using this card in August. Compound interest was charged daily at a rate of 18.25% per annum. There was no interest‑free period. The period for which interest was charged included the date of purchase and the date of payment.

What amount was paid when the account was paid in full on 31 August? (2 marks)

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`$427.56`

Show Worked Solution

`text(Days of interest)\ = 12`

♦ Mean mark 35%.

`text(Daily interest rate)\ = 0.1825/365=0.0005`

`text(Total Paid)\ (FV)`	`= PV(1+r)^n`
	`= 425(1.0005)^12`
	`= 427.557…=$427.56\ \ text{(nearest cent)}`

Financial Maths, STD2 F4 2004 HSC 19 MC

Kerry has a credit card. She is charged 0.05% compound interest per day on outstanding balances.

How much interest is Kerry charged on an amount of $250, which is outstanding on her credit card for 30 days?

$3.75
$3.78
$253.75
$253.78

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`B`

Show Worked Solution

`text(Using)\ A = P(1 + r)^n`

`A`	`= 250 (1 + 0.0005)^30`
	`= 250 (1.0151…)= 253.777…`

`:.\ text(Amount of interest charged)`

`= 253.777…-250= $3.777…`

`=> B`

Financial Maths, STD2 F4 2011* HSC 10 MC

A television was purchased for $2100 on 12 April 2011 using a credit card. Compound interest was charged daily at a rate basis 19.71% per annum for purchases on this credit card. There were no other purchases on this credit card account.

There was no interest-free period. The period for which interest was charged included the date of purchase and the date of payment.

What amount was paid when the account was paid in full on 20 May 2011?

$2143.09
$2143.53
$2144.23
$2144.68

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`D`

Show Worked Solution

♦♦ Mean mark 32%
COMMENT: Make sure you know how to adjust an annual rate to a daily rate, as shown in the Worked Solutions.

`text(Days for interest)\ (n) =19+20=39`

`text(Daily interest rate)\ (r) = 0.1971/365 = 0.00054`

`text{Total paid}\ (FV)`	`= PV (1+r)^n`
	`=2100(1.00054)^39=$2144.68`

`=>D`

Financial Maths, STD2 F4 2012* HSC 26c

Heather used her credit card to purchase a plane ticket valued at $1990 on 28 January 2011. She made no other purchases on her credit card account in January. She paid the January account in full on 19 February 2011.

The credit card account has no interest free period. Compound interest is charged daily at the rate of 21.9% per annum, including the date of purchase and the date the account is paid.

How much interest did she pay, to the nearest cent? (2 marks)

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`$27.64`

Show Worked Solution

`text(Days of interest)\ (n) = 4+19=23`

`text(Daily interest rate)\ (r) = 0.219/365=0.0006`

♦ Mean mark 38%
MARKER’S COMMENT: Too many students calculated the # days as 22 instead of 23!

`text(Total Paid)\ (FV)`	`= PV(1+r)^n`
	`= 1990(1.0006)^23`
	`=2017.644…=$2017.64`

`:.\ text(Interest Paid)=2017.64-1990=$27.64`

\(\text{Amount owing}\)	\(= 5000\left(1+\dfrac{0.18}{365}\right)^{60}\)
	\(= 5150.118\ldots\)
	\(= \$5150.12\ \text{(nearest cent)}\)

\(I = Prn\)	\(=5000 \times 0.095 \times \dfrac{90}{365}\)
	\(= 117.123\ldots\)
	\(= \$117.12\ \text{(nearest cent)}\)

\(\text{Amount owing}\)	\(= 5000\left(1+\dfrac{0.18}{365}\right)^{60}\)
	\(= 5150.118\ldots\)
	\(= \$5150.12\ \text{(nearest cent)}\)

\(I = Prn\)	\(=5000 \times 0.095 \times \dfrac{90}{365}\)
	\(= 117.123\ldots\)
	\(= \$117.12\ \text{(nearest cent)}\)

\(\text{Amount owing}\)	\(= 3000\left(1+\dfrac{0.199}{365}\right)^{30}\)
	\(= 3049.458\ldots\)
	\(= \$3049.46\ \text{(nearest cent)}\)

\(\text{Closing balance}\)	\(= 1500\left(1+\dfrac{0.185}{365}\right)^{28}\)
	\(= 1521.433\ldots\)
	\(= \$1521.43\)