Loans, investments and annuities: reducing balance loans and present value (QCE General Mathematics Unit 4)

Step 1: rate and number of periods (1 mark)

$i = \frac{0.048}{12} = 0.004$ per month and $n = 25 \times 12 = 300$ months.

Step 2: substitute into the present value annuity formula (1 mark)

The loan amount is the present value of the repayments, so $720\,000 = d\left(\frac{1 - 1.004^{-300}}{0.004}\right)$.

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

The bracket evaluates to about $174.521$, so $d = \frac{720\,000}{174.521} \approx 4125.58$.

Step 4: round and state (1 mark)

The monthly repayment is about $4126.

Keep full calculator precision in the bracket and only round at the end; rounding the bracket to 174.5 moves the repayment by about 50 cents.

a) Amount borrowed (3 marks). Parameters (1 mark): $i = \frac{0.024}{12} = 0.002$, $n = 15 \times 12 = 180$, $d = 993.14$. Substitute into the present value annuity formula (1 mark):

Evaluate (1 mark): the bracket is about $151.036$, so $A_{PV} \approx 150\,000.29$. They borrowed about $150 000.

b) Recurrence relation (2 marks). A reducing balance loan multiplies by $r = 1 + i$ each month and then subtracts the repayment (1 mark for the form $A_{n+1} = rA_n - d$). Substituting (1 mark): $A_{n+1} = 1.002A_n - 993.14$, with $A_0 = 150\,000$.

Part b) rewards both the general structure and substituting the actual monthly multiplier and repayment.

a) Advertised rate (2 marks). The multiplier is $r = 1 + i$, so $1.00375 = 1 + i$ gives $i = 0.00375$ per month (1 mark). The advertised annual rate is $12 \times 0.00375 = 0.045$, which is 4.5% p.a. compounding monthly (1 mark).

b) Months to exceed $51 000 (2 marks). Iterate the recurrence (1 mark): $A_1 = 50\,187.50$, $A_2 = 50\,375.70$, $A_3 = 50\,564.61$, $A_4 = 50\,754.23$, $A_5 = 50\,944.56$, $A_6 = 51\,135.60$. The balance first passes $51 000 at $n = 6$, so it takes 6 months (1 mark).

This is a compound interest investment with no payments ($d = 0$), which is why the recurrence has no subtracted term.

(a) $i = \frac{0.084}{4} = 0.021$, so $r = 1.021$: $A_{n+1} = 1.021A_n - d$, $A_0 = 60\,000$, where $n$ counts quarters. (1 mark)

(b) $n = 5 \times 4 = 20$. The loan is the present value of the repayments:

so $d = \frac{60\,000}{16.19473} \approx 3704.91$. The quarterly repayment is $3704.91. (2 marks: substitution, answer)

(c) Total repayments $= 20 \times 3704.91 = 74\,098.20$, so total interest $= 74\,098.20 - 60\,000 = 14\,098.20$. (1 mark)

(a) $i = \frac{0.042}{12} = 0.0035$, $n = 300$. $380\,000 = d\left(\frac{1 - 1.0035^{-300}}{0.0035}\right) = d \times 185.5486$, so $d \approx 2047.98$. The pension is $2047.98 per month. (2 marks)

(b) $A_{PV} = 2500\left(\frac{1 - 1.0035^{-300}}{0.0035}\right) \approx 463\,871.53$. Sam would need about $463 871.53, which is $83 871.53 more. (2 marks)

(c) Total payments $= 300 \times 2500 = 750\,000$. Interest $= 750\,000 - 463\,871.53 = 286\,128.47$. About $286 128 of the pension comes from interest earned while the balance is drawn down. (1 mark)

(a) $i = 0.005$, $n = 300$: $d = \frac{320\,000 \times 0.005}{1 - 1.005^{-300}} \approx 2061.76$. (1 mark)

(b) The amount owing is the present value of the 240 repayments still to come:

About $287 782 is still owed. (2 marks) Running the recurrence $A_{n+1} = 1.005A_n - 2061.76$ for 60 months gives $287 782.99; the 94-cent difference comes from rounding the repayment to the nearest cent.

(c) New $i = \frac{0.054}{12} = 0.0045$, $n = 240$: $d = \frac{287\,782.05 \times 0.0045}{1 - 1.0045^{-240}} \approx 1963.40$. The repayment falls by about $98.36 per month. (2 marks)

(a) $r = 1.005$ so $i = 0.005$: 0.5% per month, $0.5 \times 12 = 6\%$ p.a. compounding monthly. The repayment is $800 per month. (1 mark)

(b) $A_1 = 1.005 \times 40\,000 - 800 = 40\,200 - 800 = 39\,400$. $A_2 = 1.005 \times 39\,400 - 800 = 39\,597 - 800 = 38\,797$. (1 mark)

$i = \frac{0.066}{12} = 0.0055$, so $r = 1.0055$. (1 mark)

$A_{n+1} = 1.0055A_n - 490$, $A_0 = 25\,000$. (1 mark)

Total repayments $= 16 \times 2476.78 = 39\,628.48$. (1 mark)

Total interest $= 39\,628.48 - 35\,000 = 4628.48$. (1 mark)

$i = \frac{0.078}{12} = 0.0065$, $n = 60$. (1 mark)

$A_{PV} = 600\left(\frac{1 - 1.0065^{-60}}{0.0065}\right) = 600 \times 49.55202 \approx 29\,731.21$. The present value is $29 731.21. (1 mark)

$i = \frac{0.06}{4} = 0.015$, $n = 16$. (1 mark)

$35\,000 = d\left(\frac{1 - 1.015^{-16}}{0.015}\right) = d \times 14.13126$. (1 mark)

$d = \frac{35\,000}{14.13126} \approx 2476.78$. The repayment is $2476.78 per quarter. (1 mark)

(a) $i = 0.003$, $n = 180$: $d = \frac{240\,000 \times 0.003}{1 - 1.003^{-180}} \approx 1727.53$. (1 mark)

(b) The balance is the present value of the 120 payments still to come: $A_{PV} = 1727.53\left(\frac{1 - 1.003^{-120}}{0.003}\right) \approx 173\,874.50$, about $173 874. (2 marks)

(c) Total payments $= 180 \times 1727.53 = 310\,955.40$, so interest earned $= 310\,955.40 - 240\,000 = 70\,955.40$. (1 mark)

Lender A

$i = 0.006$, $n = 60$: $d = \frac{30\,000 \times 0.006}{1 - 1.006^{-60}} \approx 596.87$. Total repaid $= 60 \times 596.87 = 35\,812.20$. (1 mark)

Lender B

Amount borrowed $= 30\,400$, $i = 0.00575$: $d = \frac{30\,400 \times 0.00575}{1 - 1.00575^{-60}} \approx 600.52$. Total repaid $= 60 \times 600.52 = 36\,031.20$. (2 marks)

Decision

Lender A costs less by $36\,031.20 - 35\,812.20 = 219.00$, so by $219. The lower rate does not make up for the fee on a loan this size. (1 mark)