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Amiel · Nov 27, 2010, 05:41 PM

I REALLY need help with these three questions... let me know if any of this makes sense

1)
Mr Smith deposity Â£900 at the end of each month into a savings account
IF Mr Jones wishes to have the same amount in hos account as Mr Smith after 6 years, what single sum should he invest now?

Assume both accounts earn interest at 6% p.a. compound monthly.

2)
Find the present value of an annuity which pays Â£900
at the end of each month for 5 years, where the nominal discount rate is 6% p.a. compounded monthly

3)
Find the present value of an annuity which pays Â£1000 at the end of each year for 15 years, where the interest rate is 7% per annum compounded yearly?

Thanks for this in advance

Unknown008 · Nov 28, 2010, 12:20 PM

The total sum is given by the formula:

$Money = P(1+ \frac{r}{n})^{nt}$

The sum of money of Mr Jones is then:

$Money = P(1+ \frac{0.06}{72})^{72(6)}$

That of Mr. Smith now.

After a month, he has {900(1 + 0.06/12) + 900}
After two months, he has [{900(1 + 0.06/12) + 900}(1 + 0.06/12) + 900]
After three months [{900(1 + 0.06/12) + 900}(1 + 0.06/12) + 900](1+0.06/12) + 900

The pattern is:
0. P
1. $P(1+\frac{r}{12}) + P$
2. $(P(1+\frac{r}{12}) + P)(1+\frac{r}{12}) + P = P(1+\frac{r}{12})^2 + P(1+\frac{r}{12}) + P$
3. $(P(1+\frac{r}{12})^2 + P(1+\frac{r}{12}) + P)(1+\frac{r}{12}) + P = P(1+\frac{r}{12})^3 + P(1+\frac{r}{12})^2 + P(1+\frac{r}{12}) + P$

Which gives:

$Money\ after\ n\ months = P(1 + (1+\frac{r}{12})^1 + (1+\frac{r}{12})^2 + (1+\frac{r}{12})^3 + ... + (1+\frac{r}{12})^n) = P$\frac{1(1-0.06^{n})}{1-0.06}$$

Which is a geometric progression!

So, after 72 months (6 years), the money of Mr. Smith will be:

$Money = 900$\frac{1(1-0.06^{72})}{1-0.06}$$

Now, this equals to the money of Mr Jones, so we get:

$P(1+ \frac{0.06}{72})^{72(6)} = 900$\frac{1(1-0.06^{72})}{1-0.06}$$

Solve for P.

Post what you tried for the other questions as well as the answer you got for this one.