# Financial Maths

#### edwardkiely

Hi, could somebody help me with part 1 of this sheet. I attempted it and would like to know if my answers are correct.
My answers for part 1(i) 5.66%, (ii) 5.81 (iii) 5.87 (iv) and i couldn't get an answer for this part

Thanks
Phelim Neenan.

#### DenisB

On part 4: 12,849.00 is the required present value.
In bank statement format, looks like:
Code:
YEAR  PAYMENT  INTEREST  BALANCE
0                      12849.00
1   -1000.00  1284.90  13133.90
2   -1057.69  1313.39  13389.60
3   -1118.71  1338.96  13609.85
...
18   -2594.82   680.81   4894.07
19   -2744.52   489.41   2638.96
20   -2902.85   263.89       .00
Need anything else on that one?

#### edwardkiely

Can you do part iv of question 1?

#### DenisB

Sorry, thought you meant question #4...

Your problem states: "for a rate of interest of 6% per annum
effective, calculate the following equivalent rates".

6% effective means 100 is worth 106 after 1 year,
no matter how often the interest is paid.
An equivalent rate is calculated and used to achieve
this; as example, 1.467% per quarter.

Please explain what you mean by "equivalent".

Also, what is the difference between "rate of interest"
and "rate of discount"?

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#### edwardkiely

Thanks a million, could you show me the workings of how you got question 4?

thanks again
Phelim

#### DenisB

Look at it as 2 bank accounts:
one in which the payments are deposited
one where the initial amount is deposited
Their values in 20 years must equal each other: OK?

i = .10 (annual rate)
j = .057692 (payment annual increase)
n = 20 (number of years)
f = 1000 (first payment)
p = present (or purchase) value (?)
u = future value of p (?)
v = future value of payments (?)

u = p(1+i)^n

v = f[(1+i)^n - (1+j)^n] / [(1+i) - (1+j)]

Since u = v:
p = v / (1+i)^n

Substitute the values in there: result will be p = 12849.005132....

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