| Reducible loans. |
1. (i) Barbara's monthly interest rate:  .
(ii) After Barbara's first repayment:
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2. (i) Monthly interest rate = .
(ii) 
(iii) 
(iv) Interest paid after 2 repayments:
Total repayments = 2× 539 = $1,078.
Principal has reduced by $30000 - $29070.84 = $929.16
∴ $1078 - $929.16 = $148.84 interest. |
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3.
| Amount borrowed: |
$300,000 |
This table assumes the same number of days each month.
Simple interest for each calculation:
I = Pr or I = P×(r/12) |
| Annual interest rate (r): |
7% |
| Monthly repayment (R): |
$2000 |
| Month (n) |
Principal (P) |
Interest (I) |
P + I |
P + I - R |
| 1 |
$300,000.00 |
$1,750.00 |
$301,750.00 |
$299,750.00 |
| 2 |
$299,750.00 |
$1,748.54 |
$301,498.54 |
$299,498.54 |
| 3 |
$299,498.54 |
A |
B |
C |
(i) 
(ii) At end of 1st month, balance = 300,000 + $1,750 - $2,000 = $299,750.
(iii) The interest for month 2 is less than that for month 1 because the principal (i.e. amount yet to pay off) is less.
(iv) The payment in month 2 was $2,000 and the interest which was due was $1,748. Hence the difference - $252 - is the amount by which the loan decreases.
(v) For Month 3:
(vi) By the end of month 3, the loan has been reduced by
$300,000 - $299,245.61 = $754.39
(and $6,000 has been paid to do that!!!).. |
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4.
| Amount borrowed: |
$45,000 |
This table assumes each month is one-twelfth. of a year.
Simple interest for each calculation:
I = Pr or I = P×(r/12) |
| Annual interest rate (r): |
8% |
| Monthly repayment (R): |
$500 |
| Month (n) |
Principal (P) |
Interest (I) |
P + I |
P + I - R |
| 1 |
$45,000.00 |
$300.00 |
$45,300.00 |
$44,800.00 |
| 2 |
$44,800.00 |
$298.67 |
$45,098.67 |
$44,598.67 |
| 3 |
$44,598.67 |
$297.32 |
$44,895.99 |
$44,395.99 |
(i) A simply copies the last value in the previous row.
(ii) By the end of Month 3, Danielle has reduced her loan by
$45,000 - $44,395.99 =
$604.01
(iii) Danielle has paid 3 × $500 in three months = $1,500.
Her loan has reduced by $604.01 so her interest ill has been the difference between these two amounts - $895.99.
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| Loan repayment tables. |
5. Will is considering a loan to enable him to do various things in the future. Some of his plans are short term while other activities are longer term.
The lending institution provides the following table showing the fortnightly repayments required by Will to repay a loan at 11.5% p.a. for various periods.
| Amount borrowed |
2 years |
3 years |
4 years |
5 years |
| $12,000 |
$269 |
$190 |
$141 |
$127 |
| $16,000 |
$358 |
$252 |
$201 |
$170 |
| $20,000 |
$427 |
$316 |
$251 |
$212 |
| $24,000 |
$536 |
$379 |
$301 |
$254 |
| $28,000 |
$581 |
$411 |
$326 |
$275 |
| $30,000 |
$679 |
$474 |
$376 |
$317 |
Assume there are 26 fortnights in each year.
(i) In 5 years there are 5 × 26 = 130 fortnights.
In the $28,000 row and the 5 years column, the repayment is $275.
∴ 130 × $275 = $35,750 is paid by Will to borrow $28,000 over 5 years. Hence he will pay $7,750 in interest
(ii) Total repayment for a $12,000 loan for 5 years instead than a $12,000 loan for 2 years?
(iii) If Will takes out a loan for $24,000 for 2 years, he will repay
= $536 × 52 =
$27,872. Hence the interest = $3,872.
If the loan was for $12,000, repayments total $269 × 52 = $13,988 making the interest bill $1,988.
So the interest for the $24,000 loan is just less (by about $100) compared to double the interest for the $12,000 loan.
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