QCAA General Mathematics Loans, investments and annuities 1

Using the financial app with
N = 40, I = 3.26, PV = -112 000, PMT = 970, P/Y = C/Y = 12
FV = 83 910.19

He would owe ＄83 910.19 after the 40th repayment

Step 1. Using the financial app
N = 16, I = 3.26, PV = –112 000, PMT = 970, P/Y = C/Y = 12
FV = ＄101 128.46
New PV = ＄101 128.46 − 1600 = ＄99 528.46

Step 2. Using financial app
N = 24(40 − 16), I = 3.26, PV = −99 528.46, PMT = 970, P/Y = C/Y = 12
FV = ＄82 202.54

New amount owing after the 40th repayment is ＄82 202.54

$r = 1 + \frac{i}{n}$
$1.00375 = 1 + \frac{i}{12}$
$0.00375 = \frac{i}{12}$
$i = 0.045$

Therefore, the annual interest rate is 4.5% p.a. compounding monthly.

$A_0 = 50\ 000$
$A_1 = 50\ 187.50$
$A_2 = 50\ 375.70$
$A_3 = 50\ 564.61$
$A_4 = 50\ 754.23$
$A_5 = 50\ 944.56$
$A_6 = 51\ 135.60$
Therefore, the investment would exceed ＄51 000 at 6 months.

$A = 720\,000$

$M = ?$
$i = \frac{0.048}{12} = 0.004$
$n = 25 \times 12 = 300$

$A = M\left(\frac{1-(1+i)^{-n}}{i}\right)$

$A = M\left(\frac{1-(1+0.004)^{-300}}{0.004}\right)$

$720\,000 = M \times 174.520 ...$

$M = \frac{720\,000}{174.520 ...}$

$M = 4125.578 ...$

The monthly repayment will be ＄4126 each month for 25 years.

Compound interest investment
$A = P(1+i)^n$
$= 100\,000 \left( 1 + \frac{3.8}{12 \times 100} \right)^{5 \times 12}$
$= 120\,888.66$
The balance of the investment account is ＄120 888.66.

Perpetuity
$M = A \times i$
$6000 = A \times 0.04$
$A = \frac{6000}{0.04}$
$= 150\,000$
The present value of the perpetuity needs to be ＄150 000.
$120\,888.66 < 150\,000$

The compound interest investment will not provide enough money to finance the perpetuity.

$i = \frac{2.4}{1200} = 0.002$
$n = 15 \times 12 = 180$
$M = 993.14$

$A = M \left( \frac{1 - (1+i)^{-n}}{i} \right)$
$= 993.14 \left( \frac{1 - (1+i)^{-180}}{i} \right)$
$= 150\ 000.29$

They borrowed ＄150 000.

$A_{n+1} = rA_n - R$
$A_{n+1} = \left( 1 + \frac{2.4}{1200} \right) A_n - 993.14$

$A_{n+1} = 1.002A_n - 993.14$

4.95%

It does not take into account the fortnightly compounding.

Option A: $i = 0.056, n = 12$
$i_{\text{effective}} = \left( 1 + \frac{i}{n} \right)^n - 1$
$= \left( 1 + \frac{0.056}{12} \right)^{12} - 1$
$\approx 0.05745...$

Option B: $i = 0.0562, n = 4$
$i_{\text{effective}} = \left( 1 + \frac{i}{n} \right)^n - 1$
$= \left( 1 + \frac{0.0562}{4} \right)^4 - 1$
$\approx 0.05739...$

$0.05745 > 0.05739$

Ngarra's decision is reasonable because option A has a higher effective interest rate.

Interest for the first month = $(8992.80 + 290) - 9200 = 82.80$

Annual interest rate = $\frac{82.80}{9200} \times 100 \times 12 = 10.8\%$

$T_{n+1} = 1.009T_n - 290, T_0 = 9200$

＄8783.74 (Term 2 in the sequence)

$82.80 + (0.009 \times 8992.80) + (0.009 \times 8783.74) = ＄242.79$

＄7489.24 (Term 8 in the sequence)

38 months

$T_{37} = 150.64$
Therefore, the final repayment is $＄150.64 \times 1.009 = ＄152$

Total amount repaid = $37 \times 290 + 152 = ＄10\,882$

$10\,882 - 9200 = ＄1682$

Interest each fortnight $\frac{10.8}{26} = 0.41538\%$ (5 d.p.)

$A_{n+1} = \left(1 + \frac{0.108}{26}\right)A_n - 145, A_0 = 9200$

74 repayments and total repaid = $74 \times 145 - 28.92 = ＄10\,701.08$

Sonia should use fortnightly repayments to save $10\,882 - 10\,701.08 = ＄180.92$

She would also pay off the loan quicker.

Option 1
$i_{e1} = (1 + \frac{i}{n})^n - 1$
$= (1 + \frac{0.07}{4})^4 - 1$
$\approx 0.07186$

Option 2
$i_{e2} = (1 + \frac{i}{n})^n - 1$
$= (1 + \frac{0.068}{12})^{12} - 1$
$\approx 0.07016$

Option 1 is better because it has a slightly higher effective interest rate.

＄430 000

＄3200

0.0055 × 12 = 6.6%

245 months

＄3005.90

Balance after 7 years (84 months) = ＄341 150.33
New recursive rule:
$T_{n+1} = 1.0055T_n - 3300, \quad T_0 = 291150.33 \quad (341150.33 - 50000)$

New loan = 121 × 3300 + 240.33(final payment: 3300 – 3059.67) + 84 × 3200 + 50 000 = 718 340.33

Original loan = 244 × 3200 + 3005.90 = 783 805.90

Therefore, a saving of 783 805.90 – 718 340.33 = ＄65 465.57