Sunday

February 1, 2015

February 1, 2015

Posted by **john** on Sunday, September 17, 2006 at 10:40pm.

[(1+ 0.04/12)]^6 x 1000 = ??

is the answer 1020.164?

You got it.

well theres another part of it if you could help me with.

in the same problem find the effective interest rate after six months?

The effective interest rate is what the annual rate would be if the compounding continued at that rate.

Thus for this problem divide the rate by 12 like you did and use 12 for the number periods intead of 6 like this

1000*(1 + .04/12)^12=Amount

Subtract 1000 from that amount and divide it by 1000. That will give the effective rate and it should be slightly greater than .04

it the answer .041

It's very close to that. I got 0.0407415

However I think they want the rate based on the 6 month or 1/2 year amount. I'm not sure what your instructions were asking for.

I may've done that incorrectly. If yoy take the 1020.16 and subtract 1000 = 20.16 and divide that by 1000 you get

.0216 for 6 months, so the effective annual rate based on that is 2*.0216=.0432 since 6 months = 1/2 year.

ok thanks alot roger

If you check this problem again I may still have the wrong formula.

The formula for effective rate I found is (1+ i/n)^n-1

For this problem i=.04 and n=6 so effective rate should be 0.04067

I hope you check this problem again. Sorry about that. I haven't worked the formula for some time, so that's why I deciced to check it again.

Let me work an example so you can see how the calculations are done.

The formula is A=P*(1+r)^n where P is the initial amount, r is the rate expressed as a decimal and n is the number of periods in the compounding period.

For this problem P=1000, r=.04 and n=6. However, the interest is being compounded monthly so we need to divide the .04 by 12 so r=.04/12

Here is an example:

Suppose the amount were P=1200. r=.05 and n=6 and the interest is compounded monthly as it is for this problem.

The monthly interest rate would be r=.05/12=0.00416667

Then A=1200*(1+0.00416667)^6 and A=1230.31

In your problem substitute P=1000 for 1200, r=.0033333 and do the calculations similar to the ones I did.

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