quick math prob

[Shyguyelite]

Suppose you have 1000. You subtract 10% from it. You do it 3 times.

You could brute force it and do it manually, however...

The question is how do you subtract y% from x, z number of times? Or rather, what would the final formula be?

For example: Suppose you have $20,000 in an account. 10% is taken every month. How much remains after 33 months?

In case you are wondering... I am just curious. I graduated college with a chem degree last Dec., starting new job in two weeks. Was driving today and thought of this, and realized I didn't know how to solve it.

 

[acropolis]

Subtracting 10% is the same as multiplying by 90%, which is equal to 1-(.1). In general, it is (1-r), where r is the negative interest rate. It would be (1+r) for a positive interest rate.

Compounding your example once would be 20000(1-.1) = 20000(.9). The next month would take the amount from the first, and multiply it again by .9, which is equal to 20,000(.9)(.9), which is also equal to 20,000(.9)^2. The third month would be 20,000(.9)^3 etc., so the general form for n number of months would be 20,000(.9)^n.

The solution to the example is then 20,000(.9)^33 = 618.06

The general equation for simple interest is, using your variables, would be x(1+y)^z = final amount.

 

[Shyguyelite]

Yay. Thanks.