Problems from Compound interest

A client deposits $$25.000$$ € in an account at $$6,5\%$$ annual interest of face-value for $$10$$ years. If the interest accumulates every month in the account: what will the total interest obtained at the end of the period be?

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Development:

The first thing that has to be observed is that the liquidation period is monthly, therefore it will be necessary to calculate the monthly yield and the quantity of periods of liquidation that take place in $$10$$ years.

We begin by the yield:

$$$(1+r_m)^12=1+r \Rightarrow 1+r_m=(1+r)^{\frac{1}{12}} \Rightarrow$$$ $$$\Rightarrow r_m=(1+r)^{\frac{1}{12}} -1$$$

So, we note that the yield of $$12$$ months has to be equal to the annual yield, and from that point we isolate the monthly yield, called $$r_m$$.

Now we just have to find out what it is:

$$$r_m=(1+0,065)^{\frac{1}{12}} -1=1,065^{\frac{1}{12}} -1=1,0052-1=0,0052$$$

Concerning the liquidation periods, if in one year there are $$12$$ periods, i.e. monthly, in $$10$$ years there will be:

$$$10\cdot 12= 120$$$

With this data already calculated, now we can use the formula of the compound interest:

$$$C_f=C(1+r_m)^t$$$

$$$C_f=25.000\cdot(1+0,0052)^{120}=25.000\cdot(1,0052)^{120}=25.000\cdot 1,863=46.575€$$$

Namely, after $$10$$ years the balance of the account will have increased to $$46.575$$ €.

To know what fraction of this money corresponds to the interests it is only necessary to subtract the final quantity and the initial:

$$$46.575-25.000=21.575€$$$

So, the money has almost doubled.

Solution:

$$21.575$$ €

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