How much money should I save in an account paying 5% interest compounded monthly if I want to have $6,000 in 6 months?

Question

How much money should I save in an account paying 5% interest compounded monthly if I want to have $ 6,000 in 6 months ?

Are you saving each month, too, or are you putting a lump sum into the account at the beginning of the six months?

Amzoti · Answer 1 · 2013-01-18 21:50:40Z

up vote 2 down vote

Using the formula:

$$A = p(1 + \frac{r}{n})^{nt}$$

where:

$\bullet$ $A$ = Amount = 6000

$\bullet$ $r$ = Rate = 5 % = 0.05

$\bullet$ $t$ = Time = 6 months

$\bullet$ $n$ = Monthly = 12

$\bullet$ $p$ = Principal = Unknown

We have:

$$6000 = p(1 + \frac{.05}{12})^{72}$$

Solving yields $p = \$4447.68$

Regards.

edited Jan 18 at 21:50

answered Jan 18 at 21:38

Amzoti
14.1k51243

2

Actually, in the US at least, the "5% compounded monthly" usually means $n=12$. But that might be a regionalism – Thomas Andrews Jan 18 at 21:40

@ThomasAndrews: I started with it that way and changed it! Thanks and regards! – Amzoti Jan 18 at 21:48

Seems students get tripped up with these kinds of questions when first learning interest/compounding, etc.! Good work! – amWhy May 8 at 1:10

@amWhy: Agreed and the financial industry doesn't help by creating every imaginable permutation! ;-) – Amzoti May 8 at 1:12

Shaktal · Answer 2 · 2013-01-18 21:29:41Z

Recall the formula for compound interest:

$$\text{Initial}\times(1+\text{Increase})^{n},$$

Where $n$ is the number of months. So in this case we have $\text{Increase}=0.05$, $n=6$ and we are trying to find $\text{Initial}$. Algebraically:

$$\text{Initial}\times1.05^{6}=\$6,000 \implies \text{Initial}=\frac{\$6,000}{1.05^{6}}=\$4,477.29$$

I hope this helps!

Usually, "5% compounded each month" really means "5%/12 each month, compunded." That is, the "5%" is a yearly number. At least here in the USA.

Ron Gordon · Answer 3 · 2013-01-18 21:36:09Z

up vote 1 down vote

$$ 6000 = P (1.05)^6 \implies P = \frac{6000}{(1.05)^6} $$

Run that through your calculator.

answered Jan 18 at 21:36

Ron Gordon
26.6k12663

3 Answers