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