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Financial Maths, STD2 EO-Bank 29

Dao takes out a reducing balance loan of $5000. The loan has an interest rate of 12% per annum, compounded monthly, and Dao makes monthly repayments of $900.

The spreadsheet shown models the first 4 months of the loan.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. By continuing the spreadsheet, determine the number of months it takes Dao to repay the loan in full, and calculate the value of the final repayment.   (3 marks)
Show Answers Only

a.    \(\text{C9} = \$41.50\)

b.    \(\text{The loan is repaid in}\ 6\ \text{months.}\)

\(\text{Final repayment} = \$670.78\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{12\%}{12} = 1\%\)

\(\text{C9 (Month 2 interest)} = 4150 \times 0.01 = \$41.50\)
  

b.    \(\text{Continue the schedule using}\ \ P+I-R\ \ \text{each month:}\)

\(\text{Month 5:}\ 1548.65+1548.65 \times 0.01-900=\$664.14\)

\(\text{Month 6:}\ 664.14+664.14 \times 0.01=\$670.78\)
 

\(\text{The Month 6 balance owing}\ (\$670.78)\ \text{is less than the}\)

\(\text{usual}\ \$900\ \text{repayment, so this final repayment clears the loan.}\)

\(\therefore\ \text{The loan is repaid in}\ 6\ \text{months, with a final}\)

\(\text{repayment of}\ \$670.78.\)

Financial Maths, STD2 EO-Bank 3 MC

Leon borrows $2500 from a short-term loan company. The terms of the loan are:

  • Establishment fee: $120 charged when the loan starts
  • Monthly account-keeping fee: $45
  • Weekly repayments of $90 over 4 months (17 weeks)

What is the total amount Leon will repay over the term of the loan?

  1. \(\$1530\)
  2. \(\$1650\)
  3. \(\$1710\)
  4. \(\$1830\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Account-keeping fees} = 45 \times 4 = \$180\)

\(\text{Weekly repayments} = 90 \times 17 = \$1530\)

\(\text{Total repaid} = 120+180+1530 = \$1830\)
  

\(\Rightarrow D\)

Financial Maths, STD2 EO-Bank 28

Talia takes out a reducing balance loan of $12 000 to buy a boat. The loan has an interest rate of 7.2% per annum and Talia makes monthly repayments of $300.

The spreadsheet shown models the first 3 months of the loan.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. From the end of month 3, Talia increases her monthly repayment from $300 to $500. Calculate the balance owing at the end of month 6, and determine how much less Talia owes compared to keeping repayments at $300.   (3 marks)
Show Answers Only

a.    \(\text{C9} = \$70.63\)

b.    \(\text{Balance at end of month 6} = \$10\,007.71\)

\(\text{Talia owes}\ \$603.61\ \text{less than at the standard repayment.}\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{7.2\%}{12} = 0.6\%\)

\(\text{C9 (Month 2 interest)} = 11\,772 \times 0.006 = \$70.63\)
 

b.    \(\text{Using}\ \ P+I-R\ \ \text{from end of month 3 balance}\ \$11\,311.89:\)

\(\text{Increased repayments of}\ \$500\ \text{from month 4:}\)

\(\text{Month 4:}\ 11\,311.89+11\,311.89 \times 0.006-500=\$10\,879.76\)

\(\text{Month 5:}\ 10\,879.76+10\,879.76 \times 0.006-500=\$10\,445.04\)

\(\text{Month 6:}\ 10\,445.04+10\,445.04 \times 0.006-500=\$10\,007.71\)
 

\(\text{Standard repayments of}\ \$300\ \text{from month 4:}\)

\(\text{Month 4:}\ 11\,311.89+11\,311.89\times 0.006-300=\$11\,079.76\)

\(\text{Month 5:}\ 11\,079.76+11\,079.76\times 0.006-300=\$10\,846.24\)

\(\text{Month 6:}\ 10\,846.24+10\,846.24\times 0.006-300=\$10\,611.32\)
 

\(\text{Difference}=10\,611.32-10\,007.71=\$603.61\)

\(\therefore\ \text{Talia owes}\ \$603.61\ \text{less by increasing her repayments.}\)

Financial Maths, STD2 EO-Bank 20

Noah takes out a reducing balance loan of $10 000 to renovate his bathroom. The loan has an interest rate of 6% per annum and Noah makes monthly repayments of $400.

The spreadsheet shown models the first 4 months of the loan.
  

  1. Complete the missing values for cells C10, B11 and E11.   (3 marks)
  2. Calculate the total interest Noah pays over the first 4 months of the loan.   (1 mark)
Show Answers Only

a.    \(\text{C10} = \$46.49\)

\(\text{B11} = \$8944.74\)

\(\text{E11} = \$8589.46\)

b.    \(\$189.46\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{6\%}{12} = 0.5\%\)

\(\text{C10 (Month 3 interest)} =9298.25 \times 0.005 = \$46.49\)

\(\text{B11 (Month 4 start)}=9298.25+46.49-400=\$8944.74\)

\(\text{E11 (Month 4 end)}=8944.74+44.72-400=\$8589.46\)
  

b.    \(\text{Total interest} = 50+48.25+46.49+44.72=\$189.46\)

Financial Maths, STD2 EO-Bank 24

Ravi takes out a short-term loan to buy a laptop with a cash price of $1800.

The loan has the following terms:

  • Establishment fee: $110 charged when the loan starts
  • Monthly account-keeping fee: $40
  • Weekly repayments of $75 over 6 months (26 weeks)
  1. Calculate the total amount Ravi will pay back over the term of the loan.   (2 marks)

  2. How much more than the cash price does Ravi pay for the laptop?   (1 mark)

Show Answers Only

a.    \(\$2300\)

b.    \(\$500\)

Show Worked Solution

a.    \(\text{Account-keeping fees} = 40 \times 6 = \$240\)

\(\text{Weekly repayments} = 75 \times 26 = \$1950\)

\(\text{Total paid} = 110+240+1950= \$2300\)
  

b.    \(\text{Extra paid} = 2300-1800 = \$500\)

Financial Maths, STD2 EQ-Bank 20_3

Mei uses a buy now, pay later payment option to make a purchase of $1600. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Mei misses her final payment and is charged a late fee of $68. Mei's payment schedule is shown, with her balance totalling $468.
 

  1. Find the total amount Mei pays for her purchase if repaying in full on 28 September 2026.   (1 mark)
  2. Mei's bank offers short-term loans where simple interest is charged at 18% per annum.
  3. Suppose Mei had borrowed $1600 from the bank to make this purchase on 3 August 2026 and repaid it in full 9 weeks later.
  4. How much would Mei have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($1668\)

b.    \($18.29\)

Show Worked Solution

a.    \(\text{If total owing paid on 28 September:}\)

\(\text{Total paid} = 400+400+400+468=$1668\)
 

b.    \(r=18\%=0.18,\ \ n=\dfrac{9 \times 7}{365} = \dfrac{63}{365}\)

\(I=Prn=1600 \times 0.18 \times \dfrac{63}{365} = 49.709… = $49.71 \)

\(\text{Amount saved} = 68-49.71=$18.29\)

Financial Maths, STD2 EQ-Bank 20_2

Tane uses a buy now, pay later payment option to make a purchase of $200. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Tane misses his final payment and is charged a late fee of $22. Tane's payment schedule is shown, with his balance totalling $72.

  1. Find the total amount Tane pays for his purchase if repaying in full on 1 June 2026.   (1 mark)
  2. Tane's bank offers short-term loans where simple interest is charged at 14% per annum.
  3. Suppose Tane had borrowed $200 from the bank to make this purchase on 6 April 2026 and repaid it in full 7 weeks later.
  4. How much would Tane have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($222\)

b.    \($18.24\)

Show Worked Solution

a.    \(\text{If total owing paid on 1 June:}\)

\(\text{Total paid} = 50+50+50+72=$222\)
 

b.    \(r=14\%=0.14,\ \ n=\dfrac{7 \times 7}{365} = \dfrac{49}{365}\)

\(I=Prn=200 \times 0.14 \times \dfrac{49}{365} = 3.758… = $3.76 \)

\(\text{Amount saved} = 22-3.76=$18.24\)

Financial Maths, STD2 EQ-Bank 20_4

Idris uses a buy now, pay later payment option to make a purchase of $440. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Idris misses his final payment and is charged a late fee of $26. Idris's payment schedule is shown, with his balance totalling $136.
 

  1. Find the total amount Idris pays for his purchase if repaying in full on 30 November 2026.  (1 mark)
  2. Idris's bank offers short-term loans where simple interest is charged at 12% per annum.
  3. Suppose Idris had borrowed $440 from the bank to make this purchase on 5 October 2026 and repaid it in full 8 weeks later.
  4. How much would Idris have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($466\)

b.    \($17.90\)

Show Worked Solution

a.    \(\text{If total owing paid on 30 November:}\)

\(\text{Total paid} = 110+110+110+136=$466\)
 

b.    \(r=12\%=0.12,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}\)

\(I=Prn=440 \times 0.12 \times \dfrac{56}{365} = 8.100… = $8.10 \)

\(\text{Amount saved} = 26-8.10=$17.90\)