Annuities help!! (1 Viewer)

[littlesmallworld]

1. A person borrows $5000 at 1.5% monthly reducible and pays it off in equal monthly instalments. What should the instalments be in order to pay off the loan at the end of 3 years? Ans: $180.76

2. A firm is established with new equipment and creates a fund to providefoe the replacement of the equipment after 6 years at an estimated cost of $20000. How much should be paid into the fund annually if interest at 11% can be obtained?? Ans:$2528

3. A family borrows $30000 from a building society with interest at 11.5%. If repayments are made at the rate of $4500 per year,
Over what period will the loan be repaid?? Ans: 13.4 years

 

[ChillTime]

I believe there's a formula for it, but the 'derivation method' is more robust and suits the exam style questions.

1. Let A_n denote the amount owing after n months, and M be the installments

A0 = 5000

A1 = 5000*1.015 - M

A2 = A1*1.015 - M
= 5000*1.015^2 - M*1.015 - M

We spot a pattern:

A3 = A2*1.015 - M
= 5000*1.015^3 - M*1.015^2 - M*1.015^1 - M
= 5000*1.015^3 - M(1.015^2 + 1.015 + 1)
= 5000*1.015^3 - M*[a(r^n - 1)/(r - 1)]
{a = 1, r = 1.015, n = 3}
A3 = 5000*1.015^3 - M[(1.015^3 - 1)/(1.015 - 1)]

Generalising to n:

An = 5000*1.015^n - M[(1.015^n - 1)/(1.015 - 1)]

The loan is paid off after 3 years (36 months), amount owing is zero.

A36 = 5000*1.015^36 - M[(1.015^36 - 1)/(1.015 - 1)] = 0
8545.698 - 47.276M = 0 (use 3dp because you'll be rounding off to 2 dp)
8545.698 = 47.276M

M = 8545.698/47.276 = 180.76

Each payment is $180.76

 

[ChillTime]

2. Let n be the no. of years, An be the amount saved after n years, and M be the regular deposit amount

A0 = M
A1 = M*1.11 + M
A5 = M*1.11^5 + ... + M
A6 = M*1.11^6 + ... + M*1.11 (we won't include the last $M because in the 6th year we are withdrawing it all).
= M[1.11^6 + ... +1.11] = M[a(r^n - 1)/(r-1)] = M[1.11(1.11^6 - 1)/(1.11-1)] = M*8.783 = 20000

M = 20000/8.783 = 2277.13

Yearly deposit is $2277.13 (doesn't match book's answer)

Alternative interpretation: There is some ambiguity to this question. What if we did make that 7th deposit (even though it's invested for 0 years before the lot is withdrawn). A well worded questions would specify dates so you know whether or not to include that final deposit.

A6 = M*1.11^6 + ... + M*1.11 + M = M[1.11^6 + ... + 1.11 + 1] = M[a(r^n - 1)/(r-1)] = M[1(1.11^7 - 1)/(1.11-1)] = M*9.783 = 20000

M*9.783 = 20000
M = 2044.36

$2044.36

Alternative interpretation 2: Let's say the first deposit is made after 1 year has elapsed, rather than immediately. Let's also include that 'final deposit' which is invested for immediately before maturity (i.e. for 0 years).

A1 = M
A2 = M*1.11 + M
A6 = M*1.11^5 + ... + M = M[1.11^5 + ... + 1] = M[1(1.11^6-1)/(1.11-1)] = M*7.913 = 2000

M*7.913 = 2000
M = 2527.55

$2527 (Finally! Poorly worded question)

3. Approach like q1 but solve for n instead of M.

 

[littlesmallworld]

Thanks for ur help!!!

 

[ChillTime]

You're welcome.

If you ever get an ambiguously worded question like #2, make sure you clearly state your assumptions at the start, so the marker can give you all the marks your deserve.