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Interest calculation, computation of instalments
Probably the most common mathematical problem in everyday life is the interest calculation of bank transactions. If we deposit a given amount of money in a bank with a certain interest rate for a couple of months or years it is indeed a very important question to determine how much our deposit will increase during that time. Also, when a more expensive item is purchased using a bank loan it is necessary to know that how large the instalments will be given the duration and interest of the loan.
In this lesson we will take a look at problems of this type.
Example 1
At the start of the year we deposit \latex{ 1,000 } euros in a bank account at a yearly interest rate of \latex{ 8 }%. How much will this be worth after \latex{ 4 } years if the bank compounds the interest at the end of each year? (This means the interest is added to the deposit and in the following year this total amount will earn interest.)
Solution
Compounding at the end of the year, adding \latex{ 8 }% interest means the amount is increased by a factor of \latex{ 1.08 }. This happens for four years, which means its value will be
\latex{10^3\times 1.08^4\approx 1360.49}
This is the fifth term of a geometric progression with first term being \latex{ 10^3 } and the quotient being \latex{ 1.08 }.
Example 2
For four years we deposit \latex{ 1,000 } euros in a bank account at the start of each year. How much will this be worth at the end of the fourth year?
Solution
Again, we will suppose that the interest is compounded at the end of each year, thus the deposited amounts increase by a factor of \latex{ 1.08 } every year.
The whole sum is
\latex{10^3\times 1.08+10^3\times 1.08^2+10^3\times 1.08^3+10^3\times 1.08^4=}
\latex{=10^3\times 1.08\times \frac{1.08^4-1}{1.08-1}\approx 4866.6 } euros.
Here we used the formula for the sum of a geometric progression's terms.
Example 3
At the start of \latex{ 2010 }, Andrew deposited \latex{ 1,000 } euros in a bank account at a yearly interest rate of \latex{ 10 }%. He will take out the whole deposit at the end of \latex{ 2015 }. Meanwhile Bob deposits \latex{ b } EUR at the same bank at a yearly interest rate of \latex{ 10 }% at the start of each year from \latex{ 2010 } to \latex{ 2015 }. At the end of \latex{ 2015 } he will also withdraw the whole deposit, which turns out to be equal to Andrew’s. Calculate the value of \latex{ b }.
Solution
First calculate the value of Andrew's capital, which is increased by interest, at the end of \latex{ 2015 }. Since \latex{ 10 }% interest is added each year to the deposit, it will also earn interest in the following year.
This means the amount will increase by a factor of \latex{1+\frac{10}{100}=1.1 } each year.
Thus by the end of \latex{ 2015 }, after \latex{ 6 } years, it will be worth
\latex{a=10^3\times 1.1^6=1771.56} euros.
The first deposit made by Bob in \latex{ 2010 } will earn interest for \latex{ 6 } years.
The next deposit will earn interest for \latex{ 5 } years and so on. The last deposit will earn interest for only 1 year. Therefore the total value of his deposit will be
\latex{b\times 1.1^6+b\times 1.1^5+\dotsc+b\times 1.1=b\times 1.1\times \frac{1.1^6-1}{1.1-1}= }
\latex{=11\times b\times (1.1^6-1)=8.487171\times b} euros.
Since this amount is equal to a,
\latex{b\approx 209} euros.
Example 4
The Smith family wants to buy a new house. They get a loan of \latex{ 100,000 } EUR for \latex{ 20 } years, payable at a yearly interest rate of \latex{ 6 }%. The family will repay the loan and its interest on a yearly basis, and they want to pay the same amount in each year. How large will this amount be?
Solution
Denote the amount needed to pay annually by \latex{ x }. Naturally since they pay at the end of each year, in the first year the whole amount earn interest. Starting from the second year only the decreasing debt's interests and instalment is needed to pay and at the end of the year \latex{ 20 } the remaining amount will be \latex{ 0 }.
At the end of the first year the remaining amount is
\latex{10^5\times1.06-x}.
At the end of the second year the remaining amount is
\latex{10^5\times1.06^2-x\times-x}.
At the end of the third year the remaining amount is
\latex{10^5\times1.06^3-x\times1.06^2-x\times1.06-x},
and so on, finally at the end of the \latex{ 20 }th year
\latex{10^5\times1.06^{20}-x\times(1.06^{19}+1.06^{18}+\dots+1.06+1)=0}.
At the end of the third year the remaining amount is
\latex{10^5\times1.06^3-x\times1.06^2-x\times1.06-x},
and so on, finally at the end of the \latex{ 20 }th year
\latex{10^5\times1.06^{20}-x\times(1.06^{19}+1.06^{18}+\dots+1.06+1)=0}.
The coefficient of \latex{ x } is the sum of the first \latex{ 20 } terms of a geometric progression, therefore
\latex{10^5\times1.06^{20}-x\times\frac{1.06^{20}-1}{1.06-1}=0}.
From this, thus the yearly amount is \latex{ 8,720 } euros.
\latex{x=10^5\times1.06^{20}\times\frac{0.06}{1.06^{20}-1}=6\times10^3\times\frac{1.06^{20}}{1.06^{20}-1}\approx8,720},
Example 5
At the age of \latex{ 40 } a man signs a life insurance agreement with the following conditions. At the start of each year he deposits the same amount to the insurance company and at the age of \latex{ 70 } he gets \latex{ 50 } thousand euros (if he is still alive). The deposited money earns interest of \latex{ 8 }%. How much money does he need to deposit annually?
Solution
Denote the amount annually deposited by \latex{ x }. As he deposits at the start of each year, the amount already deposited will earn interest in the coming year.
The total at the end of the \latex{ 1 }st year is
\latex{x\times1.08} euros,
the total at the end of the \latex{ 2 }nd year is
\latex{x\times1.08^2+x\times1.08} euros,
the total at the end of the \latex{ 3 }rd year is
\latex{x\times1.08^3+x\times1.08^2+x\times1.08=x\times1.08\times\frac{1.08^3-1}{1.08-1}} euros,
and so on, at the end of the \latex{ 30 }th year it is
\latex{x\times1.08\times\frac{1.08^{30}-1}{1.08-1}} euros \latex{=5\times10^4} euros.
Thus the value of \latex{ x } is
\latex{x=\frac{4,000}{1.08\times(1.08^{30}-1)}\approx410} euros.
To reach the goal he needs to deposit \latex{ 410 } euros every year.
Example 6
The price of a car is \latex{ 19,000 } euros and it can be purchased using a loan. The duration of the loan is \latex{ 8 } years, the instalments are \latex{ 400 } euros per month which begins the first month after the purchase. Calculate the interest rate of the loan.
Solution
Denote the car's current price by \latex{ a } and the monthly instalment by \latex{ b }. Let the interest rate be \latex{ p }; we need to determine \latex{ p }. Let's look at the change in the amount to be repaid during the first few months. Since the annual interest rate if \latex{ p }%, the monthly interest rate is \latex{\frac{1}{12}\%}
Let us introduce the following notation: \latex{x=1+\frac{p}{100}\times\frac{1}{12}=1+\frac{p}{1,200}}.
At the end of the first month the amount still to be repaid is
At the end of the first month the amount still to be repaid is
\latex{a+a\times\frac{p}{1,200}=a\times x}.
At the start of the second month \latex{ b } is paid and only the decreased amount earns interest, thus at the end of the second month the amount to be repaid is
\latex{(a\times x-b)\times x=a\times x^2-b\times x}.
Similarly, at the end of the third month it is
\latex{(a\times x^2-b\times x-b)\times x=a\times x^3-b\times(x^2+x)}.
Using the same method, at the of the 8th year, that is, the \latex{ 96 }th month it is
\latex{a\times x^{96}-b\times(x^{95}+x^{94}+\dots+x)=b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\tag 1}
since by this instalment we have repaid the whole debt. The equation (1) can be written as
\latex{a\times x^{96}=b\times(x^{95}+x^{94}+\dots+x+1)}.
In the parenthesis on the right hand side there is the sum of the first 96 terms of a geometric progression with first term being \latex{ 1 } and quotient being \latex{ x }, thus
\latex{a\times x^{96}=b\times\frac{x^{96}-1}{x-1}},
since it is known that \latex{x \gt 1}.
By rearranging the equation we get an equation of degree \latex{ 97 } for \latex{ x }, which we cannot solve using algebraic methods. Therefore we need to use approximations (for example, we can use interval halving [see the Functions chapter from last year's book]). This time we don't perform the exact calculation, the approximation is good enough:
\latex{x=1.0167},
from which \latex{p = 20} follows. Thus the interest rate of the loan is \latex{ 20 }% which is quite high.
Exercises
{{exercise_number}}. Andrew takes out a loan of \latex{ 5,000 } euros at an annual interest rate of \latex{ 12 }%. He has to repay it in two years by making equation monthly instalments. How much is this monthly instalment?
{{exercise_number}}. We would like to begin a construction project, for which we can take out a discount loan of \latex{ 100 } thousand euros at an interest rate of \latex{ 6 }% for a duration of \latex{ 20 } years. We know that we cannot afford more than 800 euros per month to pay the instalments. Should we take the loan or not?