QCAA General Mathematics Loans, investments and annuities 2

Fortnightly payment = $\frac{0.0494}{26} \times 50 000 =＄95$

The effective annual rate of interest is 5.06% p.a.

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.

Value of regular contributions
$M = 2500$
$i = \frac{3.6}{400} = 0.009$
$n = 6 \times 4 = 24$

$A = M \left( \frac{(1+i)^n - 1}{i} \right)$
$= 2500 \left( \frac{(1.009)^{24} - 1}{0.009} \right)$
$= 66\,639.94$

Value of extra payment
$P = 10\,000$
$i = \frac{3.6}{400} = 0.009$
$n = 2 \times 4 = 8$

$A = P(1+i)^n$
$= 10\,000(1.009)^8$
$= 10\,743.09$

Total value $= 66\,639.94 + 10\,743.09$
$= 77\,383.03$
$= ＄77\,383$

Perpetuity — find the size of the savings
$M = 3600$
$i = \frac{0.0576}{12} = 0.0048$
$A = ?$

$A = \frac{M}{i}$
$= \frac{3600}{0.0048}$
$= 750\,000$

Use the total savings to find the size of the monthly payment
$A = 750\,000$
$M = ?$
$i = \frac{0.042}{12} = 0.0035$
$n = 20 \times 12 = 240$

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

$750\,000 = M \times 375.13...$

$M = 1999.281$

The monthly savings were ＄1999.29.