Now we will learn to solve problems related to simple interest, which happen among clients, banks and savings banks.

For example, if a bank says to offer a saving account with an interest of $$6\%$$ per year, theoretically, and leaving aside the small letter, what they mean is that for every $$100$$ € deposited in the above mentioned account the bank offers a benefit of $$6$$ €.

In this sense, if $$3.000$$ € settled, after one year the interests will be:

$$$\dfrac{100}{6}=\dfrac{3000}{x} \Rightarrow 100x=3000\cdot 6 \Rightarrow $$$

$$$ \Rightarrow 100x=18.000 \Rightarrow x=\dfrac{18.000}{100}=180€$$$

At the end of the year in the account there will be the $$3.000$$ initial € plus $$180$$ € generated by the interests, which results in a total of $$3.180$$ €.

What would happen if this $$3.000$$ € are left deposited in the account during $$5$$ years?

If every year $$180$$€ interest is generated, in $$5$$ years there will be:

$$$180 \cdot 5 = 900€$$$

So, at the end of the period the $$3.000$$ € will have turned into $$3.900$$ €, that is to say, the initial capital plus the interests.

This is a typical problem of simple interest, where the generated interests are not added to the entire quantity deposited in the account to generate new interests.

A real example would be as if having a saving account where we withdraw the interests whenever they are generated, so that the entire quantity never increases. From now on the retreat of the interests will be considered as a fact , therefore, it will not be mentioned.

A quick way to calculate the simple interest that will generate a quantity of money in a certain period of time consists of applying the following relation:

$$$I=C \cdot \dfrac{i}{100}\cdot t$$$

where $$I$$ is the total interest generated, $$C$$ is the deposited initial quantity, $$\dfrac{i}{100}$$ is the amount as a fraction of unity of interest for the time unit, usually $$1$$ year, and $$t$$ is the time that has already gone by.

Usually, the amount as a fraction of unity of interest is called a yield and expresses itself as $$r$$, so that:

$$$\dfrac{i}{100}=r \Rightarrow I=C\cdot r \cdot t$$$

Once we have applied the above mentioned relation to the previous example, the interest generated after any period of time is obtained straight away, for example $$5$$ years:

$$$I=3000\cdot\dfrac{6}{100}\cdot 5 = 3000\cdot 0,06 \cdot 5 = 900€$$$

If the same $$3.000$$ € were left in the account for $$15$$ years, at the end the generated interest would be:

$$$I=3000\cdot 0,06 \cdot 15 = 2.700€$$$

So, at the end of the period in the account there would be $$$3.000 + 2.700 = 5.700 €$$$

Another typical case of problem on simple interest is to ask about the necessary time to obtain a certain quantity. For example:

How long will it take for the $$20.000$$ € deposited in an account at $$5\%$$ per year to double?

To solve it, it is necessary to isolate the t of the relation of the simple interest:

$$$I=C\cdot r \cdot t \Rightarrow t=\dfrac{I}{C\cdot r} $$$

The generated interest will be $$20.000$$ €, since what we want is that the initial quantity doubles, that is also $$20.000$$ €, and the yield will be $$0,05$$. Replacing these values in the expression, we get:

$$$t=\dfrac{20.000}{20.000\cdot 0,05}=\dfrac{1}{0,05}=20 \ \mbox{years}$$$

We can also find out the ideal interest to get some concrete aims. For instance:

Which annual interest must offer an account so that $$6.000$$ € turn into $$10.000$$ € in a period of $$10$$ years?

Again, it is necessary to use the relation, but in this case it is necessary to isolate the yield:

$$$I=C\cdot r \cdot t \Rightarrow r=\dfrac{I}{C\cdot t} $$$

Before we have to calculate what the interest is that it will be necessary to generate in those $$10$$ years:

$$$10.000-6.000=4.000€$$$

So that:

$$$r=\dfrac{4.000}{6.000\cdot 10}=\dfrac{4.000}{60.000}=0,0666$$$

Converting it into a percentage, this number corresponds to an interest of:

$$$0,0666\cdot 100= 6,66\%$$$

Finally, the relation of the simple interest is also applicable to the ambience of the loans, although in this case instead of a benefit the interest supposes a cost. For example:

A client signs with its office a loan of $$30.000$$ € at an interest per year of $$3,5\%$$ to be returned in $$10$$ years. What quantity will the client have returned to the bank when the period finishes?

It is necessary to look for the interests generated by the above mentioned quantity in this period of time, so that:

$$$I=C\cdot r \cdot t = 30.000\cdot 0,035 \cdot 10=10.500 €$$$

$$10.500$$ € of interest has been generated during these $$10$$ years, so that the client will have to return the $$30.000$$ € requested plus the interests:

$$$30.000+10.500=40.500€$$$