Slow and steady interest

Given figures:

1.) Future value (FV) = $100.00

2.) Time (t) = 10 years

3.) Interest rate (r) = 1% p. a. = 0.01

Present value (PV) = ?

Solution:

• PV = FV / (1 + r)^t

• PV = 100 / (1 + 0.01)^10

• PV = 100 / (1.01)^10

Therefore, PV ≈ $90.53 (close to $90.50)

If the initial investment is $x$ , then in ten years the balance will be $x \times 1.01^{10}$ .

So let's solve:

$x \times 1.01^{10} = 100$

$x = \frac{100}{1.01^{10}}$

$x = \frac{100}{1.104622}$

$x = 90.5287$

The amount accumulated after n periods from a principal amount $A_0$ and with an interest rate $r$ is given by:

$A_n = A_0 (1+r)^n\quad \Rightarrow A_0 = \dfrac {A_n}{(1+r)^n}$

For $A_{10} = \$ 100$ and $r= 1\%$ we have $A_0 = \dfrac {\$100}{1.01^{10}} = \$\boxed{90.53}$

Formula: $F=P(1+i)^n$

We have,

$100=P(1+0.01)^{10}$

$P\approx 90.52869547$

So the desired answer among the choices is $90.50$ .

x = Amount place in a bank
After 10 years it will be = 100
x = (100^11)/(101^10) = 90.52869