Loan Repayment/Annuties (1 Viewer)

[reflectia]

Sorry for imposing on your thread xD But I actually have a question in regards to this too:
How do you differentiate between a "loan repayment" type question and "superannuation" question? I mean, I get the simple ones but sometimes I get confused as to what to do such as in this question:

Justin invested $10 000 into an investment account on 1 January 2000. The investment earned interest at a fixed rate of 8% per annum, compounding each year.
i) How much would be in the investment account after the interest payment made on 1 January 2010? (Got this one, so that's okay)
ii) Suppose that Justin had in fact added $1000 to his account on 1 January 2001. How much would be in the account on 1 January 2010 after the payment of interest, including his deposit? The same rate of interest applies as in part i.

 

[BenBrownTown]

Despite it taking Keelan 50 years to pay off his $100 loan, he goes to Harvey Norman to buy a high end TV despite the warning of the issues with Interest free periods from his favourite television show, Today Tonight, of which he somehow watches without a tv. He chooses the $5000 TV, and will pay the tv off monthly over 5 years, at an interest rate of 12%p.a. compounding monthly. His first 3 months are free from interest, however he must still make monthly repayments.
i) Determine how much Price remains after the 3 months interest free in terms of M

ii) Determine the formula for Price remaining (P) for n months.

iii) Find the value of M

i) A3 = 5000-3M
ii)An= 5000x1.01^(n-3) - M[1+1.01+...+3x1.01^(n-3)]
iii)M=$112.90

Is that right?

 

[Keelan134]

Sorry for imposing on your thread xD But I actually have a question in regards to this too:
How do you differentiate between a "loan repayment" type question and "superannuation" question? I mean, I get the simple ones but sometimes I get confused as to what to do such as in this question:

Justin invested $10 000 into an investment account on 1 January 2000. The investment earned interest at a fixed rate of 8% per annum, compounding each year.
i) How much would be in the investment account after the interest payment made on 1 January 2010? (Got this one, so that's okay)
ii) Suppose that Justin had in fact added $1000 to his account on 1 January 2001. How much would be in the account on 1 January 2010 after the payment of interest, including his deposit? The same rate of interest applies as in part i.

Superannuation is an annuity as it is an ongoing investment. Typically the question wants to know how much money they will have after a period of time after a series of investments. One off investments only require the compound interest formula.


Loan Repayments are any instance where money is deposited once off, and deductions are being made(for example, Keelan places 1000$ in an investment fund to pay for perpetual prizes annually of value $72, compounding annually at 6%, or it says they borrow. The question typically will have an M value for their repayments or asks for how long until no money/debt remains.

The question you have posted is just a compound interest question.
This is how we'd go about it, Im doing part i) for others who read this

i) = Principal * 1+Interest ^ Compounds
= 10000 x 1.08^10 = 21589.25

ii)
From 2000-2001 Assuming Interest is paid before he invests another 1000
Principal * 1+Interest ^ Compounds
10000 x 1.08^1 = 10800

Assuming interest is paid after the investment of 1000
11000 x 1.08^1 =11880

then add a compound interest formula for the remaining years from your new balance

So for if interest was paid before investment,
10800 +1000 = 11800
11800 x 1.08 ^9 = 23588.25

If interest was paid after the investment,
11880 x 1.08^9 = 23748.17

Which answer is correct? I hate how these questions don't specify enough detail such as if the investment is made before or after the compound of interest...

 

[Keelan134]

i) A3 = 5000-3M
ii)An= 5000x1.01^(n-3) - M[1+1.01+...+3x1.01^(n-3)]
iii)M=$112.90

Is that right?

i) Yes
ii) I put it as (5000-3m)*1.01^(n-3) - M (1.01^(n-3) -1 (ALL OVER) .01) (Yours is correct too, except the principal, gotta have that 5000-3m. Also I would put your last set of brackets into the GP formula as it could be the final step that gives you the mark or not )

iii) (5000-3m)*1.01^(n-3) - M (1.01^(n-3) -1 (ALL OVER) .01) = 0
(5000-3m)*1.01^(n-3) = M (1.01^(n-3) -1 (ALL OVER) .01)
(5000-3m)*1.01^(n-3) = 100M (1.01^(n-3) -1)
(5000*1.01^(n-3) - 3m*1.01^(n-3) = 100M (1.01^(n-3) -1)
(5000*1.01^(n-3) = 100M (1.01^(n-3) -1) + 3m*1.01^(n-3)
(5000*1.01^(n-3) = M [100(1.01^(n-3) -1) + 3(1.01^(n-3))]
(5000*1.01^(n-3) = M [103*1.01^(n-3) -100)]

n = 5*12 =60
n-3= 57

M = $108.02

 

[reflectia]

Superannuation is an annuity as it is an ongoing investment. Typically the question wants to know how much money they will have after a period of time after a series of investments. One off investments only require the compound interest formula.


Loan Repayments are any instance where money is deposited once off, and deductions are being made(for example, Keelan places 1000$ in an investment fund to pay for perpetual prizes annually of value $72, compounding annually at 6%, or it says they borrow. The question typically will have an M value for their repayments or asks for how long until no money/debt remains.

The question you have posted is just a compound interest question.
This is how we'd go about it, Im doing part i) for others who read this

i) = Principal * 1+Interest ^ Compounds
= 10000 x 1.08^10 = 21589.25

ii)
From 2000-2001 Assuming Interest is paid before he invests another 1000
Principal * 1+Interest ^ Compounds
10000 x 1.08^1 = 10800

Assuming interest is paid after the investment of 1000
11000 x 1.08^1 =11880

then add a compound interest formula for the remaining years from your new balance

So for if interest was paid before investment,
10800 +1000 = 11800
11800 x 1.08 ^9 = 23588.25

If interest was paid after the investment,
11880 x 1.08^9 = 23748.17

Which answer is correct? I hate how these questions don't specify enough detail such as if the investment is made before or after the compound of interest...

Whoops sorry! I missed out a section of the question x.x
It was meant to be: ii) Suppose that Justin had in fact added $1000 to his account on 1 January, each year beginning on 1 January 2001. How much would be in the account on 1 January 2010 after the payment of interest, including his deposit? The same rate of interest applies as in part i.