Annuities and superannuation: the future value of an annuity (QCE General Mathematics Unit 4)

Treat the two streams separately and add them.

Step 1: parameters (1 mark)

$i = \frac{0.036}{4} = 0.009$ per quarter. The regular contributions of $2500 run for the full 6 years, so $n = 6 \times 4 = 24$.

Step 2: future value of the regular contributions (1 mark)

$A_{FV} = 2500\left(\frac{1.009^{24} - 1}{0.009}\right) \approx 66\,639.94$.

Step 3: grow the one-off $10 000 (1 mark)

It is added at the end of year 4, so it compounds for the remaining 2 years, $n = 8$ quarters: $A = 10\,000(1.009)^8 \approx 10\,743.09$.

Step 4: add (1 mark)

$66\,639.94 + 10\,743.09 = 77\,383.03$.

Step 5: state (1 mark)

The annuity is worth about $77 383 at the end of 6 years.

The key is timing: the regular deposits accumulate over all 24 quarters, but the lump sum only grows for the 8 quarters between when it is added and the end.

This is a bivariate data question set in a superannuation context.

Step 1: assign each person to an industry (1 mark)

A higher $R^2$ means age explains more of the variation, so Leigh (40) is in industry B ($R^2 = 0.96$) and Tony (50) is in industry A.

Step 2: line for industry B (1 mark)

The least-squares line of balance on age is $y = 9570x - 243\,440$.

Step 3: line for industry A (1 mark)

$y = 7910x - 205\,520$.

Step 4: Leigh (1 mark)

$9570 \times 40 - 243\,440 = 139\,360$.

Step 5: Tony (1 mark)

$7910 \times 50 - 205\,520 = 189\,980$.

Step 6: difference (1 mark)

$189\,980 - 139\,360 = 50\,620$.

Step 7: communicate (1 mark)

Tony's predicted balance is about $50 620 more than Leigh's. Both predictions are interpolations (ages 40 and 50 lie within 22 to 72), so they are reasonably reliable.

Step 1: grow the investment (1 mark)

$i = \frac{0.038}{12}$, $n = 60$: $A = 100\,000\left(1 + \frac{0.038}{12}\right)^{60}$.

Step 2: evaluate (1 mark)

$A \approx 120\,888.66$.

Step 3: amount a perpetuity needs (1 mark)

$A = \frac{d}{i} = \frac{6000}{0.04}$.

Step 4: evaluate (1 mark)

$150 000.

Step 5: advise (1 mark)

The balance of $120 888.66 is less than the $150 000 required, so the investment is not large enough. The retiree could accept a smaller payment ($120\,888.66 \times 0.04 \approx 4835.55$ a year) or save more first.

(a) $i = \frac{0.042}{12} = 0.0035$, $r = 1.0035$: $A_{n+1} = 1.0035A_n + 350$, $A_0 = 0$. (1 mark)

(b) $n = 72$: $A_{FV} = 350\left(\frac{1.0035^{72} - 1}{0.0035}\right) = 350 \times 81.72288 \approx 28\,603.01$. The balance is $28 603.01. (2 marks)

(c) Deposits $= 72 \times 350 = 25\,200$, so interest $= 28\,603.01 - 25\,200 = 3403.01$. (1 mark)

(a) $i = 0.005$, $n = 300$: $A_{FV} = 1200\left(\frac{1.005^{300} - 1}{0.005}\right) = 1200 \times 692.99396 \approx 831\,592.75$. (2 marks)

(b) The pension is the present value of an annuity: $i = 0.0035$, $n = 240$. $d = \frac{831\,592.75 \times 0.0035}{1 - 1.0035^{-240}} \approx 5127.36$ per month. (2 marks)

(c) Contributions $= 300 \times 1200 = 360\,000$. Pension received $= 240 \times 5127.36 = 1\,230\,566.40$, more than three times the contributions, because the money earns interest throughout both stages. (1 mark)

Step 1: grow the opening balance (1 mark)

$i = 0.004$, $n = 60$: $5000(1.004)^{60} \approx 6353.20$.

Step 2: target for the deposits (1 mark)

The monthly deposits must supply $40\,000 - 6353.20 = 33\,646.80$.

Step 3: solve for $d$ (1 mark)

$33\,646.80 = d\left(\frac{1.004^{60} - 1}{0.004}\right) = d \times 67.66018$.

Step 4: answer (1 mark)

$d \approx 497.29$. Tom must deposit about $497.29 a month.

A common error is to ignore the $5000 (giving $591.19) or to subtract $5000 without growing it first.

(a) $i = 0.0025$ per month, so $0.25 \times 12 = 3\%$ p.a. compounding monthly; the deposit is $150 a month. (1 mark)

(b) $A_1 = 1.0025 \times 0 + 150 = 150.00$; $A_2 = 1.0025 \times 150 + 150 = 300.38$; $A_3 = 1.0025 \times 300.375 + 150 = 451.13$. (1 mark)

$i = \frac{0.052}{4} = 0.013$, $n = 32$. (1 mark)

$A_{FV} = 1500\left(\frac{1.013^{32} - 1}{0.013}\right) = 1500 \times 39.37135 \approx 59\,057.02$. (1 mark)

Deposits $= 32 \times 1500 = 48\,000$. (1 mark)

Interest $= 59\,057.02 - 48\,000 = 11\,057.02$. (1 mark)

$i = \frac{0.039}{26} = 0.0015$, $n = 3 \times 26 = 78$. (1 mark)

$A_{FV} = 200\left(\frac{1.0015^{78} - 1}{0.0015}\right) = 200 \times 82.68059 \approx 16\,536.12$. (1 mark)

$i = 0.004$, $n = 36$. $15\,000 = d\left(\frac{1.004^{36} - 1}{0.004}\right) = d \times 38.63811$. (1 mark)

$d = \frac{15\,000}{38.63811} \approx 388.22$. (1 mark)

Deposits $= 36 \times 388.22 = 13\,975.92$, so interest is about $15\,000 - 13\,975.92 = 1024.08$. (1 mark)

$i = 0.00375$, $n = 60$.

Opening balance: $12\,000(1.00375)^{60} \approx 15\,021.55$. Deposits: $250\left(\frac{1.00375^{60} - 1}{0.00375}\right) \approx 16\,786.39$. (1 mark each)

Balance $\approx 15\,021.55 + 16\,786.39 = 31\,807.94$. Interest $= 31\,807.94 - 12\,000 - 60 \times 250 = 4807.94$. (1 mark)

Ana: $200\left(\frac{1.005^{480} - 1}{0.005}\right) \approx 398\,298.15$. Ben: $400\left(\frac{1.005^{240} - 1}{0.005}\right) \approx 184\,816.36$. (2 marks)

Ana ends with about $213 482 more, more than double Ben's balance, from the same total deposits. Her early deposits compound for up to 40 years, and compound growth is geometric, so time in the account matters more than the size of each deposit. (1 mark)

$i = 0.01$. Trial: $1000\left(\frac{1.01^{40} - 1}{0.01}\right) \approx 48\,886.37$ and $1000\left(\frac{1.01^{41} - 1}{0.01}\right) \approx 50\,375.24$. (1 mark)

The balance first exceeds $50 000 after 41 quarters (10 years and 1 quarter). (1 mark)