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Theory on the topic: "Problem solving for interest."

Type 1: Convert percent to decimal. percent  fraction A%  A divided by 100 Tasks: 20%; 75%; 125%; 50%; 40%; 1%; 70%; 35%; 80%.... Fill in the table 1% 5% 10% 20% 25% 50% 75% 100%

Type 2: Converting a fraction to a percentage. number  percentages A  A times 100% Convert fractions to percentages: 3/4; 0.07; 2.4. (GIA, thematic tasks) Match the fractions that express the shares of a certain value, and the percentages corresponding to them. A.1/4; B) 3/5; C) 0.5; D) 0.05 1) 5%; 2) 25%; 3) 50%; 4) 60% Answer: A B C D

Type 3: Finding a percentage of a number. X% of A 1) X% is represented as a decimal fraction 2) The number A is multiplied by the decimal fraction. The task is an example. In a month, the company produced 500 devices. 20% of manufactured devices failed to pass quality control. How many devices failed quality control? Solution. You need to find 20% of the total number of manufactured devices (500). 20% = 0.2. 500 * 0.2 = 100. 100 of the total number of manufactured devices did not pass the quality control.

Type 4: Find a number by its percentage. And this is X%: 1) X% is represented as a decimal fraction 2) A is divided by a decimal fraction. The task is an example. Preparing for the exam, the student solved 38 tasks from the manual for self-study. Which is 25% of the number of all tasks in the manual. How many tasks are collected in this self-study manual? Solution. We do not know how many tasks are in the manual. But on the other hand, we know that 38 tasks is 25% of their total number. 25%=0.25 38/0.25 = 152. There are 152 problems in this collection.

Type 5: Find the percentage of two numbers. A and B numbers. What % is B of A? 1) B / A 2) Multiply the resulting quotient by 100% The task is a sample. There are 30 students in the class. 15 of them are girls. What percentage of girls are in the class? Solution. To find out what percentage is one number from another, you need the number that you want to find, divide by the total number and multiply by 100%. So, 1) 15 / 30 = 0.5 2) 0.5 * 100% = 50% The task is a sample. For 1 hour, the automatic machine produced 240 parts. After the reconstruction of this machine, he began to produce 288 of the same parts per hour. By what percent did the productivity of the machine increase? Solution. The productivity of the machine has increased by 288-240=48 parts per hour. You need to find out what percentage of 240 parts are 48 parts. In order to find out how many percent of the number 48 is from the number 240, you need to divide the number 48 by 240 and multiply the result by 100%. 48/240 *100% =20% Answer: machine productivity increased by 20%

Type 6: Increase the number by a percentage. Decrease the number by a percentage. A is a number; increase by X%, then it has increased by (1 + x / 100) times. : 1) the number A is multiplied by 2) (1 + x / 100). The task is an example. . In last year's math exam, 140 high school students got A's. This year the number of excellent students has increased by 15%. How many people got A's on their math exam this year? Solution. 140 * (1 + 15/100) = 161. A - number; we decrease by X%, then it decreased by (1 - x / 100) times. : 1) the number A is multiplied by 2) (1 - x / 100). The task is an example. A year ago, 100 children graduated from school. And this year there are 25% fewer graduates. How many graduates this year? Solution. 100 * (1 - 25/100) = 75.

Type7: Solution Concentration. The task is an example. A kilogram of salt was dissolved in 9 liters of water. What is the concentration of the resulting solution? (The mass of 1 liter of water is 1 kg) (Peterson 6 cells) Solution 1) The mass of the solute is 1 kg 2) The mass of the entire solution 1 + 9 \u003d 10 (kg) 9 kg is the mass of water in the solution (not to be confused with the total mass of the solution ) 3) 1/10 * 100% \u003d 10% 10% - solution concentration

Type 8: The percentage of metal in the alloy. Task - sample 1. There is a piece of an alloy of copper and tin with a total mass of 12 kg, containing 45% copper. How much pure tin must be added to this piece of alloy so that the resulting alloy contains 40% copper? Solution.1)12 . 0.45= 5.4 (kg) - pure copper in the first alloy; 2) 5.4: 0.4= 13.5 (kg) - weight of the new alloy; 3) 13.5- 12= 1.5 (kg) tin. Answer: you need 1.5 kg of tin.

Task - sample 2. There are two alloys, consisting of copper, zinc and tin. It is known that the first alloy contains 40% tin, and the second - 26% copper. The percentage of zinc in the first and second alloys is the same. Having melted 150 kg of the first alloy and 250 kg of the second, a new alloy was obtained, in which 30% zinc turned out to be. Determine how many kilograms of tin are contained in the resulting new alloy. Since the percentage of zinc in the first and second alloys is the same and in the third alloy it turned out to be 30%, then in the first and second alloys the percentage of zinc is 30%. 250 * 0.3 \u003d 75 (kg) - zinc in the second alloy; 250 * 0.26 \u003d 65 (kg) - copper in the second alloy; 250-(75+65)= 110 (kg) tin in the second alloy; 150 . 0.4= 60 (kg) - tin in the first alloy; 110 + 60 = 170 (kg) - tin in the third alloy. Answer: 170 kg. 1 alloy 2 alloy New alloy (3) Copper 26% Zinc 30% 30% 30% Tin 40% ?kg weight 150kg 250kg 150+250=400

Type 9: On "dry matter". Almost any product - apples, watermelons, mushrooms, potatoes, cereals, bread, etc. consists of water and dry matter. Moreover, both fresh and dried foods contain water. During the drying process, only water evaporates, and the mass of dry matter does not change. A.G. Mordkovich “Mathematics 6” Problem No. 362 The problem is a sample. Fresh mushroom contains 90% water, and dried - 15%. How many dried mushrooms will be obtained from 17 kg of fresh ones? How many fresh mushrooms do you need to take to get 3.4 kg of dried ones? Solution. Let's make a table: 1 part of the problem: substance Mass of substance (kg) Percentage of water Percentage of dry matter Mass of dry matter (kg) Fresh mushroom 17kg 90% 10% 17*0.1=1.7 Dried mushroom X kg 15% 85% X * o.85 \u003d 0.85x Since the mass of dry matter in dry and fresh mushrooms remains unchanged, we get the equation: 0.85x \u003d 1.7, x \u003d 1.7: 0.85, x \u003d 2.

Part 2 of the problem: Substance Mass of the substance (kg) Percentage of water Percentage of water Mass of dry matter (kg) Fresh mushroom х 90% 10% 0.1х Dried mushroom 3.4 15% 85% 3.4*0.85=2 .89 0.1x = 2.89, x = 2.89: 0.1, x = 28.9. Answer: from 17 kg of fresh mushrooms you get 2 kg of dried ones; to get 3.4 kg of dried mushrooms, you need to take 28.9 kg of fresh ones.

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%), in 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 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 telephone. 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 that connects 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 rather 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 classic tables and, unfortunately, for unknown reasons, most of the problem collections, however, still use tables.

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.

Solving problems in mathematics on the application of basic concepts of interest.

Problems with percentages are taught to solve from the 5th grade.

Solving problems of this type is closely related to three algorithms:

1. finding a percentage of a number
2. finding a number by its percentage,
3. finding a percentage.

In the lessons with students, they understand that a hundredth of a meter is a centimeter, a hundredth of a ruble is a penny, a hundredth of a centner is a kilogram. People have long noticed that hundredths of values ​​are convenient in practice. Therefore, a special name was coined for them - percentage.

So one penny is one percent of one ruble, and one centimeter is one percent of one meter.

One percent is one hundredth of a number. Mathematically, one percent is written as follows: 1%.

The definition of one percent can be written as: 1% \u003d 0.01. a

5%=0.05, 23%=0.23, 130%=1.3 etc.

How to find 1% of a number?

Since 1% is one hundredth, you need to divide the number by 100. Dividing by 100 can be replaced by multiplying by 0.01. Therefore, to find 1% of a given number, you need to multiply it by 0.01. And if you need to find 5% of the number, then multiply this number by 0.05, etc.

Example. Find: 25% of 120.

1. 25% = 0,25;
2. 120 . 0,25 = 30.

Rule 1. To find a given number of percents of a number, you need to write the percentages as a decimal fraction, and then multiply the number by this decimal fraction.

Example. The turner turned 40 parts in an hour. Using a cutter made of stronger steel, he began to turn 10 more parts per hour. By what percent did labor productivity increase?

To solve this problem, we need to find out how many percent are 10 parts from 40. To do this, we first find what part is the number 10 from the number 40. We know that we need to divide 10 by 40. It will turn out 0.25. Now let's write it down as a percentage - 25%.

Answer: Turner productivity increased by 25%.

Rule 2. To find how many percent one number is from another, you need to divide the first number by the second and write the resulting fraction as a percentage.

Example. With a planned target of 60 vehicles per day, the plant produced 66 vehicles. By what percentage did the plant fulfill the plan?

66: 60 \u003d 1.1 - this part is made up of manufactured cars from the number of cars according to the plan. Let's write in percentage = 110%.

Answer: 110%.

Example. Bronze is an alloy of tin and copper. What percentage of the alloy is copper in a piece of bronze, consisting of 6 kg of tin and 34 kg of copper?

1. 6+ 34 \u003d 40 (kg) - the mass of the entire alloy.
2. 34: 40 = 0.85 = 85 (%) - the alloy is copper.

Answer: 85%.

Example. The baby elephant lost 20% in the spring, then gained 30% in the summer, again lost 20% in the fall, and gained 10% in the winter. Has his weight remained the same this year? If changed, by what percentage and in what direction?

1. 100 - 20 = 80 (%) - after spring.
2. 80 + 80 . 0.3 = 104 (%) - after the summer.
3. 104-104. 0.2 = 83.2 (%) - after autumn.
4. 83.2 + 83.2. 0.1 = 91.52 (%) - after winter.

Answer: lost weight by 8.48%.

Example. We left for storage 20 kg of gooseberries, the berries of which contain 99% water. The water content in the berries has decreased to 98%. How many gooseberries will be the result?

1. 100 - 99 \u003d 1 (%) \u003d 0.01 - the proportion of dry matter in gooseberries first.
2. twenty . 0.01 \u003d 0.2 (kg) - dry matter.
3. 100 - 98 \u003d 2 (%) \u003d 0.02 - the proportion of dry matter in gooseberries after storage.
4. 0.2: 0.02 \u003d 10 (kg) - gooseberries became.

Answer: 10 kg.

Example. What happens to the price of a product if it is first increased by 25% and then lowered by 25%?

Let the price of the product be x rubles, then after the increase the product costs 125% of the previous price, i.e. 1.25x, and after a decrease of 25%, its value is 75% or 0.75 of the increased price, i.e.

0.75 .1.25x = 0.9375x,

then the price of the goods went down by 6.25%.

x - 0.9375x = 0.0625x;
0,0625 . 100% = 6,25%

Answer: The original price of the product has decreased by 6.25%.

Rule 3. To find the percentage of two numbers A and B, you need to multiply the ratio of these numbers by 100%, that is, calculate (A: B). 100%.

Example. Find a number if 15% of it is 30.

1. 15% = 0,15;
2. 30: 0,15 = 200.

x is a given number;
0.15 . x = 300;
x = 200.

Answer: 200.

Example. Raw cotton produces 24% fiber. How much raw cotton should be taken to get 480 kg of fiber?

Let's write 24% as a decimal fraction of 0.24 and get the problem of finding a number from its known part (fraction).
480: 0.24= 2000 kg = 2 t

Answer: 2 t.

Example. How many kg of porcini mushrooms must be harvested to obtain 1 kg of dried mushrooms, if 50% of their mass remains during processing of fresh mushrooms, and 10% of the mass of processed mushrooms remains during drying?

1 kg of dried mushrooms is 10% or 0.01 part of processed, i.e.
1 kg: 0.1=10 kg of processed mushrooms, which is 50% or 0.5 of harvested mushrooms, i.e.
10 kg: 0.05=20 kg.

Answer: 20 kg.

Example. Fresh mushrooms contained 90% water by weight, and dry 12%. How many dry mushrooms will be obtained from 22 kg of fresh ones?

1. 22. 0.1 = 2.2 (kg) - mushrooms by weight in fresh mushrooms; (0.1 is 10% dry matter);
2. 2.2: 0.88 \u003d 2.5 (kg) - dry mushrooms obtained from fresh (the amount of dry matter has not changed, but its percentage in mushrooms has changed and now 2.2 kg is 88% or 0.88 dry mushrooms ).

Answer: 2.5 kg.

Rule 4. To find a number given its percentages, you need to express the percentages as a fraction, and then divide the percentage value by this fraction.

In problems for bank calculations, simple and compound interest are usually found. What is the difference between simple and compound interest growth? With simple growth, the percentage is calculated each time based on the initial value, and with complex growth, it is calculated from the previous value. With simple growth, 100% is the initial amount, and with complex growth, 100% is new each time and equal to the previous value.

Example. The bank pays an income of 4% per month from the amount of the deposit. 300 thousand rubles were put into the account, income is accrued every month. Calculate the value of the contribution after 3 months.

1. 100 + 4 = 104 (%) = 1.04 - the share of the increase in the deposit compared to the previous month.
2. 300 . 1.04 \u003d 312 (thousand rubles) - the amount of the contribution after 1 month.
3. 312 . 1.04 \u003d 324.48 (thousand rubles) - the amount of the contribution after 2 months.
4. 324.48. 1.04 = 337.4592 (thousand r) = 337 459.2 (r) - the value of the contribution after 3 months.

Or you can replace paragraphs 2-4 with one, repeating the concept of degree with the children: 300.1.043 \u003d 337.4592 (thousand rubles) \u003d 337,459.2 (r) - the amount of the contribution after 3 months.

Answer: 337,459.2 rubles

Example. Vasya read in the newspaper that over the past 3 months, food prices have increased by an average of 10% per month. By what percent did prices increase in 3 months?

Example. Money invested in shares of a well-known company brings in 20% of income annually. In how many years will the investment double?

Let's consider a similar task plan using specific examples.

Example. (Option 1 No. 16. OGE-2016. Mathematics. Typical test tasks_ed. Yashchenko_2016 -80s)

Sports store is running a promotion. Any jumper costs 400 rubles. When buying two jumpers - 75% discount on the second jumper. How many rubles will I have to pay for the purchase of two jumpers during the promotion period?

According to the condition of the problem, it turns out that the first jumper is bought for 100% of its original cost, and the second for 100 - 75 = 25 (%), i.e. in total, the buyer must pay 100 + 25 = 125 (%) of the original cost. The solution can then be considered in three ways.

1 way.

We accept 400 rubles as 100%. Then 1% contains 400: 100 = 4 (rubles), and 125%
four . 125 = 500 (rubles)

2 way.

A percentage of a number is found by multiplying the number by the fraction corresponding to the percentage, or by multiplying the number by the given percentage and dividing by 100.
400 . 1.25 = 500 or 400. 125/100 = 500.

3 way.

Applying the proportion property:
400 rub. - 100 %
x rub. - 125%, we get x \u003d 125. 400 / 100 = 500 (rubles)

Answer: 500 rubles.

Example. (Option 4 No. 16. OGE-2016. Mathematics. Typical test tasks_ed. Yashchenko_2016 -80s)

The average weight of boys of the same age as Gosha is 57 kg. Gosha's weight is 150% of the average weight. How many kilograms does Gosha weigh?

Similarly to the example discussed above, you can make a proportion:

57 kg - 100%
x kg - 150%, we get x \u003d 57. 150 / 100 = 85.5 (kg)

Answer: 85.5 kg.

Example. (Option 7 No. 16. OGE-2016. Mathematics. Typical test tasks_ed. Yashchenko_2016 - 80s)

After the markdown of the TV, its new price was 0.52 of the old one. By what percent did the price decrease as a result of the markdown?

1 way.

Let us first find the share of price reduction. If the original price is taken as 1, then 1 - 0.52 = 0.48 is the share of the price reduction. Then we get, 0.48. 100% = 48%. Those. the price decreased by 48% as a result of the markdown.

2 way.

If the initial cost is taken as A, then after the markdown, the new price of the TV will be 0.52A, i.e. it will decrease by A - 0.52A = 0.48A.

Let's make a proportion:
A - 100%
0.48A - x%, we get x = 0.48A. 100 / A = 48 (%).

Answer: the price decreased by 48% as a result of the markdown.

Example. (Option 9 No. 16. OGE-2016. Mathematics. Typical test tasks_ed. Yashchenko_2016 - 80s)

The product on sale was reduced by 15%, while it began to cost 680 rubles. How much did the item cost before the sale?

Before the price decrease, the product was worth 100%. The price of the product after the sale decreased by 15%, i.е. became 100 - 15 = 85 (%), in rubles this value is equal to 680 rubles.

1 way.

680: 85 = 8 (rubles) - in 1%
eight . 100 \u003d 800 (rubles) - the cost of the goods before the sale.

2 way.

This is the problem of finding a number by its percentage, it is solved by dividing the number by the corresponding percentage and by converting the resulting fraction into a percentage, multiplying by 100, or by dividing by the fraction obtained by converting from percentages.
680:85. 100 \u003d 800 (rubles) or 680: 0.85 \u003d 800 (rubles)

3 way.

With proportion:
680 rub. - 85%
x rub. - 100%, we get x = 680. 100 / 85 = 800 (rubles)

Answer: 800 rubles cost the goods before the sale.

Solving problems for mixtures and alloys, using the concepts of "percentage", "concentration", "% solution".

The simplest tasks of this type are listed below.

Example. How many kg of salt in 10 kg of salt water if the percentage of salt is 15%.

ten . 0.15 = 1.5 (kg) salt.

Answer: 1.5 kg.

The percentage of a substance in a solution (eg 15%), sometimes referred to as a % solution (eg 15% saline solution).

Example. The alloy contains 10 kg of tin and 15 kg of zinc. What is the percentage of tin and zinc in the alloy?

The percentage of a substance in an alloy is the part that the weight of a given substance makes up from the weight of the entire alloy.

1. 10 + 15 = 25 (kg) - alloy;
2. 10:25 a.m. 100% = 40% - percentage of tin in the alloy;
3. 15:25. 100% = 60% - percentage of zinc in the alloy.

Answer: 40%, 60%.

In tasks of this type, the concept of "concentration" is the main one. What is it?

Consider, for example, a solution of an acid in water.

Let the vessel contain 10 liters of a solution, which consists of 3 liters of acid and 7 liters of water. Then the relative (in relation to the entire volume) acid content in the solution is equal. This number determines the concentration of the acid in the solution. Sometimes they talk about the percentage of acid in the solution. In the given example, the percentage will be as follows: . As you can see, the transition from concentration to percentage and vice versa is very simple.

So, let a mixture of mass M contain some substance of mass m.

• the concentration of a given substance in a mixture (alloy) is a quantity;
• the percentage of a given substance is called c × 100%;

It follows from the last formula that at known concentrations of a substance and the total mass of a mixture (alloy), the mass of a given substance is determined by the formula m=c×M.

Problems on mixtures (alloys) can be divided into two types:

1. For example, two mixtures (alloys) with masses m1 and m2 and concentrations of some substance in them equal to c1 and c2, respectively, are given. Mixtures (alloys) are drained (fused). It is required to determine the mass of this substance in a new mixture (alloy) and its new concentration. It is clear that in the new mixture (alloy) the mass of the given substance is equal to c1m1+c2m2, and the concentration.
2. A certain volume of the mixture (alloy) is set, and from this volume they begin to cast (remove) a certain amount of the mixture (alloy), and then add (add) the same or a different amount of the mixture (alloy) with the same concentration of this substance or with a different concentration. This operation is carried out several times.

When solving such problems, it is necessary to establish control over the amount of a given substance and its concentration at each ebb, as well as at each addition of the mixture. As a result of such control, we obtain a resolving equation. Let's consider specific tasks.

If the concentration of a substance in a compound by mass is P%, then this means that the mass of this substance is P% of the mass of the entire compound.

Example. The concentration of silver in an alloy of 300 g is 87%. This means that pure silver in the alloy is 261 g.

300 . 0.87 = 261 (g).

In this example, the concentration of a substance is expressed as a percentage.

The ratio of the volume of a pure component in solution to the total volume of the mixture is called the volumetric concentration of this component.

The sum of the concentrations of all components that make up the mixture is 1.

If the percentage of a substance is known, then its concentration is found by the formula:
K \u003d P / 100%,
where K is the concentration of the substance;
P is the percentage of the substance (in percent).

Example. (Option 8 No. 22. OGE-2016. Mathematics. Typical test tasks_ed. Yashchenko_2016 - 80s)

Fresh fruits contain 75% water, while dried fruits contain 25%. How much fresh fruit is required to prepare 45 kg of dried fruit?

If fresh fruits contain 75% water, then the dry matter will be 100 - 75 = 25 (%), and dried - 25%, then the dry matter in them will be 100 - 25 = 75 (%).

When solving a problem, you can use the table:

Fresh fruit x 25% = 0.25 0.25. X

Dried fruits 45 75% = 0.75 0.75. 45 = 33.75

Because the mass of dry matter for fresh and dried fruits does not change, we get the equation:

0.25 . x = 33.75;
x = 33.75: 0.25;
x = 135 (kg) - fresh fruit is required.

Answer: 135 kg.

Example. (Option 8 No. 11. Unified State Examination-2016. Mathematics. Typical. Test. Tasks. Ed. Yashchenko 2016 -56s)

By mixing 70% and 60% acid solutions and adding 2 kg of pure water, a 50% acid solution was obtained. If, instead of 2 kg of water, 2 kg of a 90% solution of the same acid were added, then a 70% solution of the acid would be obtained. How many kilograms of a 70% solution were used to make the mixture?

Total weight, kg | Dry matter concentration | Dry matter mass
I x 70% \u003d 0.7 0.7. X
II in 60% = 0.6 0.6. at
water 2 - -
I + II + water x + y + 2 50% \u003d 0.5 0.5. (x + y + 2)
III 2 90% = 0.9 0.9. 2 = 1.8
I + II + III x + y + 2 70% \u003d 0.7 0.7. (x + y + 2)

Using the last column from the table, we will compose 2 equations:

0.7. x + 0.6. y = 0.5. (x + y + 2) and 0.7. x + 0.6. y + 1.8 = 0.7. (x + y + 2).

Combining them into a system, and solving it, we get that x = 3 kg.

Answer: 3 kilograms of a 70% solution was used to obtain a mixture.

Example. (Option 2 No. 11. Unified State Examination-2016. Mathematics. Typical. Test. Assignments. Ed. Yashchenko 2016 -56s)

Three kilograms of cherries cost the same as five kilograms of cherries, and three kilograms of cherries cost the same as two kilograms of strawberries. By what percent is a kilogram of strawberries cheaper than a kilogram of cherries?

From the first sentence of the problem, we obtain the following equalities:

3h = 5v,
3v = 2k.
From which we can express: h \u003d 5v / 3, k \u003d 3v / 2.

Thus, you can make a proportion:
5v/3 - 100%
3v / 2 - x%, we get x \u003d (3. 100. c.3) / (2. 5. c), x \u003d 90% is the cost of a kilogram of strawberries from the cost of a kilogram of cherries.

So, by 100 - 90 = 10 (%) - a kilogram of strawberries is cheaper than a kilogram of cherries.

Answer: a kilogram of strawberries is 10 percent cheaper than a kilogram of cherries.

Solving problems for "compound" interest, using the concept of an increase (decrease) coefficient.

To increase the positive number A by p percent, multiply the number A by the increase factor K = (1 + 0.01p).

To reduce the positive number A by p percent, multiply the number A by the reduction factor K = (1 - 0.01p).

Example. (Option 29 No. 22. OGE-2015. Mathematics. Typical examination options: 36 options / edited by Yashchenko, 2015 - 224c)

The price of a commodity was reduced twice by the same percentage. By what percent did the price of the goods decrease each time if its initial cost was 5,000 rubles and the final cost was 4,050 rubles?

1 way.

Because the price of a commodity decreased by the same number of %, let's denote the number of % as x. Let the price of the product be lowered by x% for the first and second time, then after the first decrease the price of the product has become (100 - x)%.

Let's make a proportion
5000 rub. - 100%
at rub. - (100 - x)%, we get y \u003d 5000. (100 - x) / 100 = 50 . (100 - x) rubles - the cost of the goods after the first reduction.

Let's make a new proportion for the new price:
fifty . (100 - x) rub. - 100%
z rub. - (100 - x)%, we get z \u003d 50. (100 - x) (100 - x) / 100 = 0.5. (100 - x) 2 rubles - the cost of the goods after the second reduction.

We get the equation 0.5. (100 - x) 2 \u003d 4050. Having solved it, we get that x \u003d 10%.

2 way.

Because the price of a commodity decreased by the same number of%, let us denote the number of% as x, x% = 0.01 x.

Using the concept of the reduction factor, we immediately obtain the equation:
5000 . (1 - 0.01x) 2 = 4050.

Answer: the price of the goods decreased by 10% each time.

Example. (Option 30 No. 22. OGE-2015. Mathematics. Typical examination options: 36 options / edited by Yashchenko, 2015 - 224c)

The price of a commodity was increased twice by the same percentage. By what percent did the price of the goods increase each time if its initial cost was 3,000 rubles and the final cost was 3,630 rubles?

Because the price of a good increased by the same number of %, let us denote the number of % by x, x % = 0.01 x.

Using the concept of the magnification factor, we immediately obtain the equation:
3000 . (1 + 0.01x) 2 = 3630.

Solving it, we get that x = 10%.

Answer: 10% increase in the price of goods each time.

Example. (Option 4 No. 11. Unified State Examination-2016. Mathematics. Typical. Test. Ass. ed. Yashchenko 2016 -56s)

On Thursday, the company's shares rose in price by a certain number of percent, and on Friday they fell in price by the same number of percent. As a result, they began to cost 9% cheaper than at the opening of trading on Thursday. By what percent did the company's shares rise in price on Thursday?

Let the company's shares rise and fall in price by x%, x% = 0.01 x, and the initial value of the shares was A. Using all the conditions of the problem, we obtain the equation:

(1 + 0.01 x) (1 - 0.01 x) A \u003d (1 - 0.09) A,
1 - (0.01 x) 2 \u003d 0.91,
(0.01 x)2 = (0.3)2,
0.01 x \u003d 0.3,
x = 30%.

Answer: The company's shares rose 30 percent on Thursday.

Solving "banking" problems in the new version of the USE-2016 in mathematics.

Example. (Option 2 No. 17. Unified State Exam-2016. Mathematics. 50 types. rev. ed. Yashchenko 2016)

On January 15, it is planned to take a loan from the bank for 15 months. The conditions for its return are as follows:

It is known that the eighth payment amounted to 108 thousand rubles. How much must be repaid to the bank during the entire loan term?

From the 2nd to the 14th, A/15 +0.01A is paid.

After that, the amount of debt will be 1.01A - A / 15 - 0.01A \u003d 14A / 15.

After 2 months we get: 1.01. 14A/15.

Second payment A/15 + 0.01. 14A/15.

Then the debt after the second payment is 13A/15.

Similarly, we get that the eighth payment will look like:

A/15 + 0.01. 8A/15 = A/15. (1 + 0.08) = 1.08A / 15.

And according to the condition, it is equal to 108 thousand rubles. So, we can write and solve the equation:

1.08A / 15 \u003d 108,

A=1500 (thousand rubles) - the initial amount of the debt.

2) To find the amount that needs to be repaid to the bank during the entire loan period, we must find the amount of all payments on the loan.

The sum of all payments on the loan will look like:

(A / 15 + 0.01A) + (A / 15 + 0.01. 14A / 15) + (A / 15 + 0.01. 13A / 15) + ... + (A / 15 + 0.01. A /15) \u003d A + 0.01A / 15 (15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) \u003d A + (0.01. 120A)/15 = 1.08A.

So 1.08. 1500 \u003d 1620 (thousand rubles) \u003d 1620000 rubles must be returned to the bank during the entire loan period.

Answer: 1620000 rubles.

Example. (Option 6 No. 17. Unified State Exam-2016. Mathematics. 50 types. rev. ed. Yashchenko 2016)

On January 15, it is planned to take a loan from the bank for 24 months. The conditions for its return are as follows:

• On the 1st of each month, the debt increases by 1% compared to the end of the previous month;
• 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 be the same amount less than the debt on the 15th day of the previous month.

It is known that for the first 12 months it is necessary to pay 177.75 thousand rubles to the bank. How much are you planning to borrow?

1) Let A be the loan amount, 1% = 0.01.

Then 1.01A debt after the first month.

From the 2nd to the 14th, A/24 +0.01A is paid.

After that, the amount of debt will be 1.01A - A / 24 - 0.01A \u003d A - A / 24 \u003d 23A / 24.

Under this scheme, the debt becomes the same amount less than the debt on the 15th day of the previous month.

After 2 months we get: 1.01. 23A/24.

Second payment A/24 + 0.01. 23A/24.

Then the debt after the second payment is 1.01. 23A/24 - A/24 - 0.01. 23A / 24 \u003d 23A / 24 (1.01 - 0.01) - A / 24 \u003d 23A / 24 - A / 24 \u003d 22A / 24.

Thus, we get that for the first 12 months you need to pay the bank the following amount:
A/24 +0.01A. 24/24 + A/24 + 0.01. 23A/24 + A/24 + 0.01. 22A/24 + ... + A/24 + 0.01. 13A/24 = 12A/24 + 0.01A/24 (24+23+22+21+20+19+18+17+16+15+14+13) = A/2 + 222A/2400 = 711A/1200 .

And according to the condition, it is equal to 177.375 thousand rubles. So, we can write and solve the equation:
711A / 1200 \u003d 177.75,
A = 300 (thousand rubles) = 300,000 rubles - it is planned to take a loan.

Answer: 300,000 rubles.

To be able to correctly and quickly solve text problems with percentages is necessary not only for students who are about to pass the exam in mathematics at a basic or specialized level, but also for all adults, since such tasks are constantly encountered in everyday life. Raising prices, planning a family budget, profitable investment of funds and many other issues cannot be resolved without these skills. In preparation for passing the certification test, it is imperative to repeat how to solve problems for percentages: in the USE in mathematics, they are found both in the basic and in the profile level.

Need to remember

A percentage is \(\frac(1)(100)\) part of some number. Denotes the proportion of something in relation to the whole. The written character is \(\%\) . When preparing for the Unified State Examination on the topic "Interest", schoolchildren both in Moscow and in other parts of the Russian Federation must remember the following formula:

\

How to apply it?

In order to solve a simple task with percentages in the exam in mathematics, you need:

1. Divide the given number by \(100\) .
2. Multiply the resulting value by the number \(\%\) to be found.

For example, in order to calculate \(10\%\) from the number \(300\) , you need to find \(1\) percentage by dividing \(300:100=3\) . And the number \(3\cdot10=30\) obtained from the previous action. Answer: \(30\).

These are the simplest tasks. 11th grade students in the exam are faced with the need to solve complex problems with percentages. As a rule, they are talking about bank deposits or payments. You can get acquainted with the formulas and the rules for their application by going to the "Theoretical Reference" section. Here you can not only repeat the basic definitions, but also get acquainted with the options for solving complex problems for interest on a bank loan, as well as with exercises from other sections of algebra, for example,

Job type: 11
Topic: Tasks for percentages

Condition

Elena made a deposit in the bank in the amount of 5500 rubles. Interest on the deposit is calculated once a year and added to the current deposit amount. A year later, Natalia deposited the same amount in the same bank and on the same terms. A year later, Elena and Natalya simultaneously closed their deposits and took the money. As a result, Elena received 739.2 rubles more than Natalya received. Find what percentage per annum the bank charged on deposits?

Show Solution

Solution

Let the percentage per annum be x, then after a year Elena's contribution was:

5500 + 0.01x \cdot 5500 = 5500(1 + 0.01x) rubles, and a year later - 5500(1 + 0.01x)^2 rubles. Natalia's deposit was in the bank for only a year, so it is equal to 5500(1 + 0.01x) rubles. And the difference between the resulting contributions of Elena and Natalia amounted to 739.2 rubles.

Let's make and solve the equation:

5500(1+0.01x)^2-5500(1+0.01x)= 739,2,

(1+0.01x)^2-(1+0.01x)=0.1344,

x^2+100x-1344=0,

x_1=-112,\enspace x_2=12.

The bank charged 12% per annum.

Answer

Job type: 11
Topic: Tasks for percentages

Condition

Entrepreneur Petrov made a profit of 12,000 rubles in 2005. Each subsequent year, his profit increased by 110% compared to the previous year. How many rubles did Petrov earn in 2008?

Show Solution

Solution

In 2005, the profit was 12\,000 rubles, each next year it increased by 110\%, that is, it became 210\% \u003d 2.1 from the previous year. In three years it will be 12\,000 \cdot 2,1^3 = 111\,132 ruble.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Tasks for percentages

Condition

There are two alloys. The first alloy contains 12% iron, the second - 28% iron. The mass of the second alloy is greater than the mass of the first by 2 kg. From these two alloys, a third alloy was made with an iron content of 21% . Find the mass of the third alloy. Give your answer in kilograms.

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Solution

Let us denote the mass of the first alloy as x kg. Then the mass of the second alloy is (x + 2) kg. The iron content in the first alloy is 0.12x kg, in the second alloy - 0.28(x + 2) kg. The third alloy has a mass x + x + 2 = 2x + 2 (kg), and its iron content is 2(x + 1) \cdot 0.21 = 0.42(x + 1) kg.

Let's make and solve the equation:

0.12x+ 0.28(x + 2) = 0.42(x+1),

6x + 14(x + 2) = 21(x + 1),

X = 7.

The third alloy has a mass of 2 \cdot 7 + 2 = 16 (kg).

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Tasks for percentages

Condition

The price of a TV set in the store is reduced quarterly (in a quarter - three months) by the same number of percent from the previous price. It is known that a TV worth 50,000 rubles was sold two quarters later for 41,405 rubles. Find the percentage by which the cost of the TV decreased quarterly.

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Solution

The price of the TV was originally 50,000 rubles. A quarter later she became 50\,000-50\,000\cdot0,01x = 50\,000(1-0.01x) rubles, where x is the percentage by which the price of the TV is reduced quarterly. After two quarters, its price became

50\,000(1-0.01x)(1-0.01x)=50\,000(1-0.01x)^2.

Let's make and solve the equation:

50\,000(1-0.01x)^2=41\,405,

(1-0.01x)^2=0.8281,

1-0.01x=0.91,

x=9.

So, the price of the TV decreased by 9 percent quarterly.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Tasks for percentages

Condition

In 2005, 55,000 people lived in the village. In 2006, as a result of the construction of new houses, the number of residents increased by 6%, and in 2007 - by 10% compared to 2006. Find the number of inhabitants of the village in 2007.

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Solution

In 2006, the number of residents of the village increased by 6%, i.е. became 106%, which is equal to 55\,000 \cdot 1.06 = 58\,300 (inhabitants). In 2007, the number of residents of the village increased by 10% (became 110%) compared to 2006, i.е. the number of inhabitants of the village became 58\,300 \cdot 1,1 = 64\,130 people.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Tasks for percentages

Condition

Show Solution

Solution

3 liters of a 14% aqueous solution contains 3 \ cdot0.14 \u003d 0.42 liters. some substance. Added 4 liters of water, it became 7 liters of solution. In these 7 liters of a new solution - 0.42 liters of some substance. Let's find the concentration of the new solution: 0.42:7\cdot100=6%.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Tasks for percentages

Condition

Construction companies established a company with an authorized capital of 150 million rubles. The first firm contributed 20% of the authorized capital, the second firm - 22.5 million rubles, the third - 0.3 of the authorized capital, the fourth firm contributed the rest.