# Math Help - Investment, interest and profit

1. ## Investment, interest and profit

The formula $A=P(1+\frac{R}{100})^{n}$

If I want to reverse this to have minus profit (depreciates) and minus income do I do $A=P(1+\frac{-r}{100})^{n}$???

2. Originally Posted by Mukilab
The formula $A=P(1+\frac{R}{100})^{t}$

If I want to reverse this to have minus profit (depreciates) and minus income do I do $A=P(1+\frac{-r}{100})^{t}$???
You can but you can still use the first formula. It just so happens that $R < 0$

For example if a car worth £14000 depreciates by 4% annually.

P = 14000
R = -4

$A = P \left(1-\frac{R}{100}\right)^t = 14000 \left(1-\frac{4}{100}\right)^t$

In the compound interest formula it's usually better to have time as an exponent (time is denoted by t) although n works fine so use whichever is easier

3. Thanks. Solved

4. Actually I'm confused by the wording

The company bought a new truck
Each year the value of the truck depreciates by 20%
The value of the new truck can be multiplied by a single number to find its value at the end of 4 years.

I did $P=A(1-\frac{20}{100})^{4}=A(0.4096)$

and $P=A(1-\frac{20}{100})^0=A(1)=A$

So I did $\frac{1}{0.496}$

/Don't think it's correct :/

5. Originally Posted by Mukilab
Actually I'm confused by the wording

The company bought a new truck
Each year the value of the truck depreciates by 20%
The value of the new truck can be multiplied by a single number to find its value at the end of 4 years.

I did $P=A(1-\frac{20}{100})^{4}=A(0.4096)$

and $P=A(1-\frac{20}{100})^0=A(1)=A$

So I did $\frac{1}{0.496}$

/Don't think it's correct :/

I believe the number is 0.4096. From what I can gather the question is asking what percentage of the original value the truck has after 4 years. Mathematically:

$P_4 = k\,P_0$ where the constant $k$ is the number in question.

$k = \frac{P_4}{P_0} = \frac{0.4096A}{A} = 0.4096$

6. Originally Posted by e^(i*pi)
I believe the number is 0.4096. From what I can gather the question is asking what percentage of the original value the truck has after 4 years. Mathematically:

$P_4 = k\,P_0$ where the constant $k$ is the number in question.

$k = \frac{P_4}{P_0} = \frac{0.4096A}{A} = 0.4096$

Oh I didn't need to do anything more ^^

THanks again