USE in mathematics profile level

The work consists of 19 tasks.
Part 1:
8 tasks with a short answer of the basic level of complexity.
Part 2:
4 tasks with a short answer
7 tasks with a detailed answer of a high level of complexity.

Run time - 3 hours 55 minutes.

Examples of USE assignments

Solving USE tasks in mathematics.

For a standalone solution:

1 kilowatt-hour of electricity costs 1 ruble 80 kopecks.
The electricity meter on November 1 showed 12625 kilowatt-hours, and on December 1 it showed 12802 kilowatt-hours.
How much do you need to pay for electricity in November?
Give your answer in rubles.

Problem with solution:

In a regular triangular pyramid ABCS with a base ABC, the edges are known: AB \u003d 5 roots out of 3, SC \u003d 13.
Find the angle formed by the plane of the base and the straight line passing through the midpoint of the edges AS and BC.

Solution:

1. Since SABC is a regular pyramid, then ABC is an equilateral triangle, and the remaining faces are equal isosceles triangles.
That is, all sides of the base are 5 sqrt(3), and all side edges are 13.

2. Let D be the midpoint of BC, E the midpoint of AS, SH the height from point S to the base of the pyramid, EP the height from point E to the base of the pyramid.

3. Find AD from the right triangle CAD using the Pythagorean theorem. You get 15/2 = 7.5.

4. Since the pyramid is regular, point H is the intersection point of heights / medians / bisectors of triangle ABC, which means it divides AD in a ratio of 2: 1 (AH = 2 AD).

5. Find SH from right triangle ASH. AH = AD 2/3 = 5, AS = 13, by the Pythagorean theorem SH = sqrt(13 2 -5 2) = 12.

6. Triangles AEP and ASH are both right-angled and have a common angle A, hence similar. By assumption, AE = AS/2, hence both AP = AH/2 and EP = SH/2.

7. It remains to consider the right triangle EDP (we are just interested in the angle EDP).
EP = SH/2 = 6;
DP = AD 2/3 = 5;

Angle tangent EDP = EP/DP = 6/5,
Angle EDP = arctg(6/5)

Answer:

USE 2019 in mathematics task 17 with a solution

Demo version of the Unified State Examination 2019 in mathematics

Unified State Examination in Mathematics 2019 in pdf format Basic level | Profile level

Tasks for preparing for the exam in mathematics: basic and profile level with answers and solutions.

17 task of the profile level of the exam in mathematics is a task related to finance, namely, this task can be for interest, part of debts, etc. The difficulty lies in the fact that it is necessary to calculate the interest or part over a long period, so this task is not direct analogy of standard problems with percentages. In order not to talk about the general, let's go directly to the analysis of a typical task.

Analysis of typical options for assignments No. 17 USE in mathematics at the profile level

The first version of the task (demo version 2018)

The conditions for its return are as follows:

• On the 1st of each month, the debt increases by r percent compared to the end of the previous month, where r is an integer;
• from the 2nd to the 14th of each month, part of the debt must be paid;
• On the 15th day of each month, the debt must amount to a certain amount in accordance with the following table.

Find the largest value of r for which the total amount of payments will be less than 1.2 million rubles.

Solution algorithm:

1. We consider what the amount of payments on the loan is monthly.
2. We determine the debt for each month.
3. Find the required percentage.
4. We determine the amount of payments for the entire period.
5. We calculate the percentage r of the amount of debt payments.
6. We write down the answer.

Solution:

1. According to the condition, the debt to the bank should decrease monthly in the following order:

1; 0,6; 0,4; 0,3; 0,2; 0,1; 0.

2. Let k = 1 + r / 100, then the debt each month is:

k; 0.6k; 0.4k; 0.3k; 0.2k; 0.1k.

3. So, payments from the 2nd to the 14th monthly are:

k - 0.6; 0.6k - 0.4; 0.4k - 0.3; 0.3k - 0.2; 0.2k - 0.1; 0.1k

4. The total amount of payments is equal to:

By condition, the entire amount of payments is less than 1.2 million rubles, therefore,

The largest integer solution of the resulting inequality is 7. Then it is the required one - 7.

The second option (from Yaschenko, No. 1)

In July 2020, it is planned to take a loan from a bank in the amount of 300,000 rubles. The conditions for its return are as follows:

• each January, the debt increases by r% compared to the end of the previous year;
• From February to June of each year, part of the debt must be paid in one payment.

Find r if it is known that the loan will be fully repaid in two years, and in the first year 160,000 rubles will be paid, and in the second year - 240,000 rubles.

Algorithm for solving the problem:

1. Determine the amount of debt.
2. We calculate the amount of debt after the first installment.
3. Finding the amount of debt after the second installment
4. Find the required percentage.
5. We write down the answer.

Solution:

1. 300,000 rubles were borrowed. According to the condition, the amount of debt to be repaid increases by r%, which means times. To pay off the debt, you need to give the bank 300,000∙k.

2. After making a payment equal to 160,000 rubles. The balance of the debt is

Today we will digress a little from standard logarithms, integrals, trigonometry, etc., and together we will consider a more vital task from the Unified State Examination in mathematics, which is directly related to our backward Russian resource-based economy. And to be precise, we will consider the problem of deposits, interest and loans. Because it is the tasks with percentages that have recently been added to the second part of the unified state exam in mathematics. I’ll make a reservation right away that for solving this problem, according to the specifications of the Unified State Examination, three primary points are offered at once, i.e. examiners consider this task one of the most difficult.

At the same time, to solve any of these tasks from the Unified State Examination in mathematics, you need to know only two formulas, each of which is quite accessible to any school graduate, however, for reasons I do not understand, these formulas are completely ignored by both school teachers and compilers of various tasks for preparation to the exam. Therefore, today I will not just tell you what these formulas are and how to apply them, but I will deduce each of these formulas literally before your eyes, taking as a basis tasks from the open USE bank in mathematics.

Therefore, the lesson turned out to be quite voluminous, quite meaningful, so make yourself comfortable, and we begin.

Putting money in the bank

First of all, I would like to make a small lyrical digression related to finance, banks, loans and deposits, on the basis of which we will get the formulas that we will use to solve this problem. So, let's digress a little from the exams, from the upcoming school problems, and look into the future.

Let's say you have grown up and are going to buy an apartment. Let's say you are going to buy not some bad apartment on the outskirts, but a good quality apartment for 20 million rubles. At the same time, let's also assume that you got a more or less normal job and earn 300 thousand rubles a month. In this case, for the year you can save about three million rubles. Of course, earning 300 thousand rubles a month, for the year you will get a slightly larger amount - 3,600,000 - but let these 600,000 be spent on food, clothes and other daily household joys. The total input data is as follows: it is necessary to earn twenty million rubles, while we have at our disposal only three million rubles a year. A natural question arises: how many years do we need to put aside three million in order to get these same twenty million. It is considered elementary:

\[\frac(20)(3)=6,....\to 7\]

However, as we have already noted, you earn 300 thousand rubles a month, which means that you are smart people and will not save money "under the pillow", but take it to the bank. And, therefore, annually on those deposits that you bring to the bank, interest will be charged. Let's say you choose a reliable, but at the same time more or less profitable bank, and therefore your deposits will grow by 15% per annum annually. In other words, we can say that the amount on your accounts will increase by 1.15 times every year. Let me remind you the formula:

Let's calculate how much money will be in your accounts after each year:

In the first year, when you just start saving money, no interest will accumulate, that is, at the end of the year you will save three million rubles:

At the end of the second year, interest will already be accrued on those three million rubles that have remained from the first year, i.e. we need to multiply by 1.15. However, during the second year, you also reported another three million rubles. Of course, these three million had not yet accrued interest, because by the end of the second year, these three million had only appeared in the account:

So, the third year. At the end of the third year, interest will be accrued on this amount, that is, it is necessary to multiply this entire amount by 1.15. And again, throughout the year you worked hard and put aside three million rubles:

\[\left(3m\cdot 1.15+3m \right)\cdot 1.15+3m\]

Let's calculate another fourth year. Again, the entire amount that we had at the end of the third year is multiplied by 1.15, i.e. Interest will be charged on the entire amount. This includes interest on interest. And three million more is added to this amount, because during the fourth year you also worked and also saved money:

\[\left(\left(3m\cdot 1.15+3m \right)\cdot 1.15+3m \right)\cdot 1.15+3m\]

And now let's open the brackets and see what amount we will have by the end of the fourth year of saving money:

\[\begin(align)& \left(\left(3m\cdot 1,15+3m \right)\cdot 1,15+3m \right)\cdot 1,15+3m= \\& =\left( 3m\cdot ((1,15)^(2))+3m\cdot 1,15+3m \right)\cdot 1,15+3m= \\& =3m\cdot ((1,15)^(3 ))+3m\cdot ((1,15)^(2))+3m\cdot 1,15+3m= \\& =3m\left(((1,15)^(3))+((1 ,15)^(2))+1,15+1 \right)= \\& =3m\left(1+1,15+((1,15)^(2))+((1,15) ^(3)) \right) \\\end(align)\]

As you can see, in brackets we have elements of a geometric progression, i.e. we have the sum of the elements of a geometric progression.

Let me remind you that if the geometric progression is given by the element $((b)_(1))$, as well as the denominator $q$, then the sum of the elements will be calculated according to the following formula:

This formula must be known and clearly applied.

Please note: the formula n th element sounds like this:

\[((b)_(n))=((b)_(1))\cdot ((q)^(n-1))\]

Because of this degree, many students are confused. In total, we have just n for the sum n- elements, and n-th element has degree $n-1$. In other words, if we now try to calculate the sum of a geometric progression, then we need to consider the following:

\[\begin(align)& ((b)_(1))=1 \\& q=1,15 \\\end(align)\]

\[((S)_(4))=1\cdot \frac(((1,15)^(4))-1)(1,15-1)\]

Let's calculate the numerator separately:

\[((1,15)^(4))=((\left(((1,15)^(2)) \right))^(2))=((\left(1,3225 \right ))^(2))=1.74900625\approx 1.75\]

In total, returning to the sum of the geometric progression, we get:

\[((S)_(4))=1\cdot \frac(1.75-1)(0.15)=\frac(0.75)(0.15)=\frac(75)(15 )=5\]

As a result, we get that in four years of savings, our initial amount will not increase four times, as if we had not deposited money in the bank, but five times, that is, fifteen million. Let's write it separately:

4 years → 5 times

Looking ahead, I’ll say that if we had been saving not for four years, but for five years, then as a result, our amount of savings would have increased by 6.7 times:

5 years → 6.7 times

In other words, by the end of the fifth year, we would have the following amount in the account:

That is, by the end of the fifth year of savings, taking into account interest on the deposit, we would have already received over twenty million rubles. Thus, the total savings account from bank interest would decrease from almost seven years to five years, i.e., by almost two years.

Thus, even despite the fact that the bank charges a fairly low interest on our deposits (15%), after five years these same 15% give an increase that significantly exceeds our annual earnings. At the same time, the main multiplier effect occurs in recent years and even, rather, in the last year of savings.

Why did I write all this? Of course, not to agitate you to carry money to the bank. Because if you really want to increase your savings, then you need to invest them not in a bank, but in a real business, where these same percentages, i.e. profitability in the conditions of the Russian economy, rarely drops below 30%, i.e. twice as much bank deposits.

But what is really useful in all this reasoning is a formula that allows us to find the final amount of the deposit through the amount of annual payments, as well as through the interest that the bank charges. So let's write:

\[\text(Vklad)=\text(platezh)\frac(((\text(%))^(n))-1)(\text(%)-1)\]

By itself, % is calculated using the following formula:

This formula also needs to be known, as well as the basic formula for the amount of the contribution. And, in turn, the main formula can significantly reduce calculations in those problems with percentages where it is required to calculate the contribution.

Why use formulas instead of tables?

Many will probably have a question, why all these difficulties at all, is it possible to simply write each year on a tablet, as they do in many textbooks, calculate each year separately, and then calculate the total amount of the contribution? Of course, you can generally forget about the sum of a geometric progression and count everything using classic tablets - this is done in most collections to prepare for the exam. However, firstly, the volume of calculations increases sharply, and secondly, as a result, the probability of making an error increases.

And in general, using tables instead of this wonderful formula is the same as digging trenches with your hands at a construction site instead of using an excavator standing nearby and fully working.

Well, or the same thing as multiplying five by ten not using the multiplication table, but adding five to itself ten times in a row. However, I have already digressed, so I will repeat the most important idea once again: if there is some way to simplify and shorten the calculations, then this is the way to use.

Interest on loans

We figured out the deposits, so we move on to the next topic, namely, to interest on loans.

So, while you are saving money, carefully planning your budget, thinking about your future apartment, your classmate, and now a simple unemployed person, decided to live for today and just took out a loan. At the same time, he will still tease and laugh at you, they say, he has a credit phone and a used car, taken on credit, and you still ride the subway and use an old push-button phone. Of course, for all these cheap "show-offs" your former classmate will have to pay dearly. How expensive - this is what we will calculate right now.

First, a brief introduction. Let's say your former classmate took two million rubles on credit. At the same time, according to the contract, he must pay x rubles per month. Let's say that he took a loan at a rate of 20% per annum, which in the current conditions looks quite decent. Also, assume that the loan term is only three months. Let's try to connect all these quantities in one formula.

So, at the very beginning, as soon as your former classmate left the bank, he has two million in his pocket, and this is his debt. At the same time, not a year has passed, and not a month, but this is only the very beginning:

Then, after one month, interest will accrue on the amount owed. As we already know, to calculate interest, it is enough to multiply the original debt by a coefficient, which is calculated using the following formula:

In our case, we are talking about a rate of 20% per annum, i.e. we can write:

This is the ratio of the amount that will be charged per year. However, our classmate is not very smart and he did not read the contract, and in fact he was given a loan not at 20% per year, but at 20% per month. And by the end of the first month, interest will be accrued on this amount, and it will increase by 1.2 times. Immediately after that, the person will need to pay the agreed amount, i.e. x rubles per month:

\[\left(2m\cdot 1,2-x\right)\cdot 1,2-x\]

And again, our boy makes a payment in the amount of $x$ rubles.

Then, by the end of the third month, the amount of his debt increases again by 20%:

\[\left(\left(2m\cdot 1,2- x\right)\cdot 1,2- x\right)1,2- x\]

And according to the condition for three months, he must pay in full, that is, after making the last third payment, his amount of debt should be equal to zero. We can write this equation:

\[\left(\left(2m\cdot 1,2- x\right)\cdot 1,2- x\right)1,2 - x=0\]

Let's decide:

\[\begin(align)& \left(2m\cdot ((1,2)^(2))- x\cdot 1,2- x\right)\cdot 1,2- x=0 \\& 2m \cdot ((1,2)^(3))- x\cdot ((1,2)^(2))- x\cdot 1,2- x=0 \\& 2m\cdot ((1,2 )^(3))=\cdot ((1,2)^(2))+\cdot 1,2+ \\& 2m\cdot ((1,2)^(3))=\left((( 1,2)^(2))+1,2+1 \right) \\\end(align)\]

Before us is again a geometric progression, or rather, the sum of the three elements of a geometric progression. Let's rewrite it in ascending order of elements:

Now we need to find the sum of the three elements of a geometric progression. Let's write:

\[\begin(align)& ((b)_(1))=1; \\& q=1,2 \\\end(align)\]

Now let's find the sum of the geometric progression:

\[((S)_(3))=1\cdot \frac(((1,2)^(3))-1)(1,2-1)\]

It should be recalled that the sum of a geometric progression with such parameters $\left(((b)_(1));q \right)$ is calculated by the formula:

\[((S)_(n))=((b)_(1))\cdot \frac(((q)^(n))-1)(q-1)\]

This is the formula we just used. Substitute this formula into our expression:

For further calculations, we need to find out what $((1,2)^(3))$ is equal to. Unfortunately, in this case, we can no longer paint as last time in the form of a double square, but we can calculate like this:

\[\begin(align)& ((1,2)^(3))=((1,2)^(2))\cdot 1,2 \\& ((1,2)^(3)) =1,44\cdot 1,2 \\& ((1,2)^(3))=1,728 \\\end(align)\]

We rewrite our expression:

This is a classic linear expression. Let's go back to the next formula:

In fact, if we generalize it, we will get a formula linking interest, loans, payments and terms. The formula goes like this:

Here it is, the most important formula of today's video lesson, with the help of which at least 80% of all economic tasks from the Unified State Exam in mathematics in the second part are considered.

Most often, in real tasks, you will be asked for a payment, or a little less often for a loan, that is, the total amount of debt that our classmate had at the very beginning of the payments. In more complex tasks, you will be asked to find a percentage, but for very complex ones, which we will analyze in a separate video lesson, you will be asked to find the time frame during which, with the given loan and payment parameters, our unemployed classmate will be able to fully pay off the bank.

Perhaps someone will now think that I am a fierce opponent of loans, finance and the banking system in general. So, nothing like that! On the contrary, I believe that credit instruments are very useful and essential for our economy, but only on the condition that the loan is taken for business development. In extreme cases, you can take out a loan to buy a home, that is, a mortgage or for emergency medical treatment - that's it, there are simply no other reasons to take a loan. And all sorts of unemployed people who take loans to buy "show-offs" and at the same time do not think at all about the consequences in the end and become the cause of crises and problems in our economy.

Returning to the topic of today's lesson, I would like to note that it is also necessary to know this formula connecting loans, payments and interest, as well as the amount of a geometric progression. It is with the help of these formulas that real economic problems from the Unified State Examination in mathematics are solved. Well, now that you know all this very well, when you understand what a loan is and why you should not take it, let's move on to solving real economic problems from the Unified State Examination in mathematics.

We solve real problems from the exam in mathematics

Example #1

So the first task is:

On December 31, 2014, Alexei took a loan of 9,282,000 rubles from the bank at 10% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank accrues interest on the remaining amount of the debt (that is, increases the debt by 10%), then Alexey transfers X rubles to the bank. What should be the amount X for Alexey to pay off the debt in four equal payments (i.e. for four years)?

So, this is a problem about a loan, so we immediately write down our formula:

We know the loan - 9,282,000 rubles.

We will deal with percentages now. We are talking about 10% of the problem. Therefore, we can translate them:

We can make an equation:

We have obtained an ordinary linear equation with respect to $x$, although with quite formidable coefficients. Let's try to solve it. First, let's find the expression $((1,1)^(4))$:

$\begin(align)& ((1,1)^(4))=((\left(((1,1)^(2)) \right))^(2)) \\& 1,1 \cdot 1,1=1,21 \\& ((1,1)^(4))=1,4641 \\\end(align)$

Now let's rewrite the equation:

\[\begin(align)& 9289000\cdot 1,4641=x\cdot \frac(1,4641-1)(0,1) \\& 9282000\cdot 1,4641=x\cdot \frac(0, 4641)(0,1)|:10000 \\& 9282000\cdot \frac(14641)(10000)=x\cdot \frac(4641)(1000) \\& \frac(9282\cdot 14641)(10) =x\cdot \frac(4641)(1000)|:\frac(4641)(1000) \\& x=\frac(9282\cdot 14641)(10)\cdot \frac(1000)(4641) \\ & x=\frac(2\cdot 14641\cdot 1000)(10) \\& x=200\cdot 14641 \\& x=2928200 \\\end(align)\]\[\]

That's it, our problem with percentages is solved.

Of course, this was only the simplest task with percentages from the Unified State Examination in mathematics. In a real exam, there will most likely not be such a task. And if it does, consider yourself very lucky. Well, for those who like to count and do not like to take risks, let's move on to the next more difficult tasks.

Example #2

On December 31, 2014, Stepan borrowed 4,004,000 rubles from a bank at 20% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank accrues interest on the remaining amount of the debt (i.e., increases the debt by 20%), then Stepan makes a payment to the bank. Stepan paid off the entire debt in 3 equal payments. How many rubles less would he give to the bank if he could pay off the debt in 2 equal payments.

Before us is a problem about loans, so we write down our formula:

\[\]\

What do we know? First, we know the total credit. We also know the percentages. Let's find the ratio:

As for $n$, you need to carefully read the condition of the problem. That is, first we need to calculate how much he paid for three years, i.e. $n=3$, and then perform the same steps again but calculate payments for two years. Let's write an equation for the case where the payment is paid for three years:

Let's solve this equation. But first, let's find the expression $((1,2)^(3))$:

\[\begin(align)& ((1,2)^(3))=1,2\cdot ((1,2)^(2)) \\& ((1,2)^(3)) =1,44\cdot 1,2 \\& ((1,2)^(3))=1,728 \\\end(align)\]

We rewrite our expression:

\[\begin(align)& 4004000\cdot 1,728=x\cdot \frac(1,728-1)(0,2) \\& 4004000\cdot \frac(1728)(1000)=x\cdot \frac(728 )(200)|:\frac(728)(200) \\& x=\frac(4004\cdot 1728\cdot 200)(728) \\& x=\frac(4004\cdot 216\cdot 200)( 91) \\& x=44\cdot 216\cdot 200 \\& x=8800\cdot 216 \\& x=1900800 \\\end(align)\]

In total, our payment will be 1900800 rubles. However, pay attention: in the task, we were required to find not a monthly payment, but how much Stepan would pay in total for three equal payments, that is, for the entire period of using the loan. Therefore, the resulting value must be multiplied by three again. Let's count:

In total, Stepan will pay 5,702,400 rubles for three equal payments. That's how much it will cost him to use the loan for three years.

Now consider the second situation, when Stepan pulled himself together, got ready and paid off the entire loan not in three, but in two equal payments. We write down our same formula:

\[\begin(align)& 4004000\cdot ((1,2)^(2))=x\cdot \frac(((1,2)^(2))-1)(1,2-1) \\& 4004000\cdot \frac(144)(100)=x\cdot \frac(11)(5)|\cdot \frac(5)(11) \\& x=\frac(40040\cdot 144\ cdot 5)(11) \\& x=3640\cdot 144\cdot 5=3640\cdot 720 \\& x=2620800 \\\end(align)\]

But that's not all, because now we have calculated only one of the two payments, so in total Stepan will pay exactly twice as much:

Great, now we are close to the final answer. But pay attention: in no case have we yet received a final answer, because for three years of payments Stepan will pay 5,702,400 rubles, and for two years of payments he will pay 5,241,600 rubles, that is, a little less. How much less? To find out, you need to subtract the second payment amount from the first payment amount:

The total final answer is 460,800 rubles. Exactly how much Stepan will save if he pays not three years, but two.

As you can see, the formula linking interest, terms, and payments greatly simplifies calculations compared to classical tables, and, unfortunately, for unknown reasons, tables are still used in most problem collections.

Separately, I would like to draw your attention to the term for which the loan was taken, and the amount of monthly payments. The fact is that this connection is not directly visible from the formulas that we wrote down, but its understanding is necessary for the quick and effective solution of real problems in the exam. In fact, this connection is very simple: the longer the loan is taken, the smaller the amount will be in monthly payments, but the larger the amount will accumulate over the entire period of using the loan. And vice versa: the shorter the term, the higher the monthly payment, but the lower the final overpayment and the lower the total cost of the loan.

Of course, all these statements will be equal only on the condition that the amount of the loan and the interest rate in both cases is the same. In general, for now, just remember this fact - it will be used to solve the most difficult problems on this topic, but for now we will analyze a simpler problem, where you just need to find the total amount of the original loan.

Example #3

So, one more task for a loan and, in combination, the last task in today's video tutorial.

On December 31, 2014, Vasily took out a certain amount from the bank on credit at 13% per annum. The loan repayment scheme is as follows: on December 31 of each next year, the bank accrues interest on the remaining amount of the debt (that is, it increases the debt by 13%), then Vasily transfers 5,107,600 rubles to the bank. What amount did Vasily borrow from the bank if he repaid the debt in two equal installments (for two years)?

So, first of all, this problem is again about loans, so we write down our wonderful formula:

Let's see what we know from the condition of the problem. First, the payment - it is equal to 5,107,600 rubles a year. Secondly, percentages, so we can find the ratio:

In addition, according to the condition of the problem, Vasily took a loan from the bank for two years, i.e. paid in two equal installments, hence $n=2$. Let's substitute everything and also note that the loan is unknown to us, i.e. the amount he took, and let's denote it as $x$. We get:

\[{{1,13}^{2}}=1,2769\]

Let's rewrite our equation with this fact in mind:

\[\begin(align)& x\cdot \frac(12769)(10000)=5107600\cdot \frac(1,2769-1)(0,13) \\& x\cdot \frac(12769)(10000 )=\frac(5107600\cdot 2769)(1300)|:\frac(12769)(10000) \\& x=\frac(51076\cdot 2769)(13)\cdot \frac(10000)(12769) \ \& x=4\cdot 213\cdot 10000 \\& x=8520000 \\\end(align)\]

That's it, this is the final answer. It was this amount that Vasily took on credit at the very beginning.

Now it’s clear why in this problem we are asked to take out a loan for only two years, because double-digit interest rates appear here, namely 13%, which, squared, already gives a rather “brutal” number. But this is not the limit - in the next separate lesson, we will consider more complex tasks where it will be required to find the loan term, and the rate will be one, two or three percent.

In general, learn to solve problems for deposits and loans, prepare for exams and pass them "excellently". And if something is not clear in the materials of today's video lesson, then do not hesitate - write, call, and I will try to help you.

financial mathematics

For the correct completed task without errors you will receive 3 points.

Approximately 35 minutes.

To solve task 17 in mathematics of a profile level, you need to know:

1. The task is divided into several types:
   • tasks related to banks, deposits and loans;
   • tasks for the optimal choice.
2. The formula for calculating the monthly payment: S credit = S/12t
3. Formula for calculating simple interest: S=α (1 + tp/m)
4. Formula for calculating compound interest: C \u003d x (1 + a%)n

Percent - is one hundredth of a value.

• x*(1 + p/100) - value x increased by p%
• x*(1 - k/100) - value x decreased by k%
• x*(1 + p/100) k - value x increased by p% k once
• x*(1 + p/100)*(1 - k/100) – value X first increased by p%, and then decreased by k%

Tasks for repaying the loan in equal installments:

The loan amount is taken as X. Bank interest - a. Loan repayment - S.

One year after the accrual of interest and payment of the amount S debt - x * (1 + a/100), p = 1 + a/100

• Debt after 2 years: (xp-S)p-S
• Debt after 3 years: ((xp - S)p - S)p - S
• The amount of debt through n years: xp n - S(p n-1 + ... + p 3 + p 2 + p + 1)