0

This I assume is a very simple question but can't really wrap my head around it.

So the question is:

If 100 dollars is deposited at time t = 0 into an account earning 10% interest and $20 is withdrawn at t = 1 and 2, then how much can be withdrawn at t = 3?

I tried getting the answer by doing this: $$((100(1.1) - 20)(1.1)-20) = 79$$

But the solutions state:

$$100 (1.10)^ 3 − 20 (1.10)^ 2 − 20 (1.10) = 86.90$$

I have no idea of how this is correct.

p.s. This is my first post here. If there was something I did wrong, please let me know.

Timothy
  • 3

1 Answers1

1

Just go period by period.

$t=0$: you put $\fbox {100}$ in the bank.

$t=1$ the $100$ has grown to $110$. You take out $20$, leaving you with $\fbox {90}$ in the bank.

$t=2$: the $90$ has grown to $99$. You take out $20$ leaving you with $\fbox {79}$ in the bank

$t=3$: The $79$ has grown to $79\times (1+.1)=\fbox {86.9}$

lulu
  • 70,402
  • Thanks you. can't believe I made that silly mistake – Timothy Mar 02 '16 at 21:26
  • No problem. I always find that breaking it down period by period (by hand if it's easy, on a spreadsheet otherwise) clarifies things. Closed formulas are nice, but they can be hard to apply consistently. – lulu Mar 02 '16 at 21:28