Money has become so firmly entrenched in our lives that all of us, regardless of age, gender and method of earning income, from time to time find ourselves in situations where we are forced to make decisions that require financial calculations. And then our ability to operate with specific financial categories depends on how profitable the option we choose will be. In this article, we will look at the main categories of financial mathematics and show how to use them to take right decisions in a variety of situations.

Interest. Compound interest. Interest Capitalization (Compaunding)

Interest refers to income received as payment for lending money in any form. Percentages can be expressed in absolute and relative form. The absolute form is a specific amount for a certain period. Relative - in the form of an interest rate tied to a specified period (year, month or day). To calculate the accumulated amount (S), by which we mean the principal amount plus accrued interest, you must use the following formula:

(1) S = P * (1 + i * n),
where P is the amount on which interest is accrued, i is the interest rate, N is the number of accrual periods.

Example
You gave a friend a loan of $10,000 for 3 months, according to which he promises to pay you 2% per month. It is necessary to calculate the amount that you will receive at the end of the loan term. We get 10,000 * (1 + 2% * 3) = $10,600.

Often you can find a situation where interest is not paid, but is added to the invested amount, and from the new period, accrual is already made on the amount, taking into account the previously attached interest. Such interest is called compound interest, and the process of calculating interest on interest is called interest capitalization. In the case of compound interest, the accrued amount is calculated differently:

(2) S = P * (1 + i) ^ n,
where the meaning of the letters is the same as in the formula above, and the “^” sign means exponentiation.

What is the difference between compound and simple interest? If the growth of simple interest occurs linearly (by the same amount each period), then compound interest grows exponentially (each subsequent period the amount of interest is greater than in the previous one). Due to this effect, the amount placed at compound interest for a long time exceeds the growth of the amount placed at simple interest many times over. Below are the results of deposit growth (6% per annum) with simple and compound interest. If at first the difference remains small, then later one reaches a critical value. So, for the year 80, a deposit with a simple interest will reach $58,000, while a deposit with a complex one will reach $1,057,960.

In practice, there is often a practice in which the interest calculation period differs from an integer. In such a situation, the formula for calculating the accrued amount with simple interest takes the form:

(3) S = P * (1 + i * d / 365),
where d is the interest calculation period expressed in days.

There are also situations when the interest rate is expressed in annual terms, but interest is calculated monthly. In such cases, the formula for calculating the accrued amount (as a rule, compound interest is used in this case) will look like:

(4) S = P * (1 + i/m) ^ (n*m),
where m is the number of interest periods within a period (usually 12 is used for the number of months in a year).

And finally, let's pay attention that, regardless of the type of interest, all formulas for calculating the accumulated amount can be reduced to general view:

(5) S = P * k,
where k is the accrual factor, which is calculated in various ways depending on the type of interest used. This conclusion will greatly facilitate our understanding of subsequent mathematical operations.

Discounting and its essence

The concept of interest that we discussed above reflects the time value of money. In other words, due to the fact that the money that we own today, tomorrow can bring us income as a result of their placement at a certain percentage, future cash receipts have a lower present value. This principle is based mathematical operation, which is called discounting. Discounting means bringing future payments to the current value and, in terms of meaning, is an operation that is the opposite of increasing interest. That is, discounting considers future payments as an accrued amount (S) and the task of the investor is to calculate their current value (P) based on the interest rate available to him (i). Depending on the type of interest, the discount formula will look like this: or

(6) P = S / (1 + i * n)

(7) P = S / (1 + i)^n

The task of discounting is to show us how much the money we will receive in the future is worth today in order not to overpay for future payments in terms of the investment alternative available to us. Let's take a look at a few common transactions that use discounting.

Acquisition of a stream of future payments (accounting transactions)
A bond with a nominal value of $ 1,000 with an interest rate of 6% per annum is offered for purchase, interest on which is paid quarterly, and redemption is at the end of the year. The task is to calculate the present value of the liability based on the discount rate 15% per annum.

Solution
Calculate the quarterly interest income and plotin a programme excel cash flow table. Let's find the value of the present value using the built-in NPV formula. Thus, at a discount rate of 15% per annum, the present value of this financial liability is $916.22

Note

2) In the NPV formula, instead of the interest rate, we put the annual percentage divided by 12

financial equivalence
The parties agree on the terms of payment for office space. The price of the premises is $24,000. The seller agrees to installment payment on the following terms: 8,000$ immediately, the rest in equal parts within 4 months. However, he is ready to consider a longer installment period if the seller offers him a large amount for the premises being sold.

Solution
Let's reflect the initial installment terms in the form of a table in the Excel program. Let's simulate in the same table an offer with increasing monthly payments, as a result of which the price of the premises will increase to $24,400. Let's calculate the current cost of each option to compare their equivalence based on the interest rate equal to 10% per annum. The calculation shows that the second option, even with a higher purchase price, is more profitable for the buyer than the first one.

Consolidation of payments
Consolidation of payments is the operation of combining several payment obligations into one payment (S0) at a certain time (T0). The peculiarity of this operation is that all payments, the receipt of which is expected before the given date, are calculated by accretion, and those that are expected after it, by discounting. Depending on the type of interest used, the consolidation formula has the following form:

(8) S = ∑ Pn * (1 + i * (T0 - Tn))

(9) S = ∑ Pn* (1 + i) ^ (T0 - Ta))

Example
You opened a bank deposit of $10,000 for 12 months at 10% per annum. How much money do you need to put into the account for 14 months so that after 3 years you have $15,000 in your account.

Solution
Let's imagine the problem as a consolidation of payments, where the existing contribution will be expressed as a positive number, and the amount expected in the future will be negative. Considering that the interest is charged at the compound interest rate, we get the following calculation = -1,264$.

Determination of the internal rate of return

In business and investing, there are often situations when an investor knows future payments and the amount of investments, and he needs to calculate the accumulation factor at which the amount of future payments reduced to the current value will be numerically equal to the amount of investments. The accrual coefficient for which this condition is met is called the internal rate of return (IRR, in English - IRR, internal return of return). To calculate the internal rate of return, the built-in Excel function, IRR, is used.

Example
The investor is considering an investment proposal, which is an equity participation in the opening of a pizzeria (see here). We know: a) the amount of the requested investment; b) financial plan (cash flow forecast); c) a scheme for the distribution of cash flows. The summary of the investment proposal (see table) contains 6 yield options. It is necessary to determine the overall profitability of the investment proposal forcomparisons with other investment options.

Solution
Let's build in the Excel program a table of cash flows that the investor will receive according to the financial plan (see table). We calculate the internal rate of return using the built-in IRR formula, where we indicate all payment values, including the initial investment, as the range of values. The resulting value of the internal rate of return (IRR) = 38.47%. Thus, the total expected profitability of the considered investment proposal is 38.47% per annum.

Note
1) In periods when there are no payments, we put "0".
2) To obtain the annual rate of IRR, we multiply the obtained value by 12.

Annuity (financial rent)
The flow of payments, all components of which are positive values, and the time intervals between payments are the same, is called an annuity or financial rent. For example, an annuity is a sequence of receiving interest on a bond, payments on a consumer loan, regular contributions under endowment insurance contracts, and payment of pensions. Annuities are characterized by the following parameters: 1) the value of each individual payment; 2) the interval between payments; 3) duration of payments (there are perpetual annuities); 4) interest rate. Due to the complexity of the calculation formula, it is best to use the built-in formulas of the Excel program to calculate the various components of the annuity. Let's consider the main ones.

When calculating the loan, the following formulas are used: PMT (calculates the amount of a monthly payment), OSPLT (calculates the amount of repayment of the principal debt as part of a specific monthly payment), IPMT (calculates the amount of interest as part of a specific monthly payment).

Example
It is necessary to calculate the monthly payment and draw up a payment schedule for the loan, the amount is $ 10,000, the interest rate is 20%, the term is 20 months.

Solution
We use the PMT formula to calculate the payment. We substitute the monthly value (annual value divided by 12) in place of the interest rate, indicate the loan amount as the present value, indicate 0 as the future value. We use the same values ​​for the OSPL and HPMT formulas, in which only the serial number of the period changes. We present the obtained values ​​in the form of a table:

The same PMT formula can be used to calculate monthly installments to accumulate an amount by a given point in time. To do this, we put the amount of the down payment in place of the present value, and the required amount in place of the future value.

Example
You are 25 years old. You opened a retirement savings account with an interest rate of 6% per annum and deposited $10,000 of your savings into it. Let's calculate the amount of the monthly payment that you need to set aside in order to receive the amount of $ 100,000 by the age of 45.

Solution
We use the PMT function. We indicate 6% / 12 as the interest rate, the number of periods is 20 * 12, the present value is $10,000, the future value is $100,000. In this case, the completed formula will look like this = PMT (6% / 12; 20 * 12; 10000; 100,000). We get the amount of the monthly fee in the amount of $ 288.

As you noticed, in the above examples, we calculated the amount of the monthly payment, we knew other parameters of the annuity. Excel allows us to calculate other annuity parameters - present value, future value, number of periodic payments. Let's look at examples of how these formulas work.

Present Value Calculation Example
On your son's 10th birthday, you decide to open a savings account so that on his 18th birthday you can save $10,000. What initial deposit do you need to make into this account if the planned monthly installments are $50?

Solution
We use the PS function. We indicate 6% / 12 as the interest rate, the number of payments is 8 * 12, the periodic payment is $50, the future value is minus $10,000. In this case, the completed formula will look like this = PS (6% / 12; 8 * 12; 50; -10000). The resulting value of the down payment is $ 2390.

Note
A negative value in the PS and BS formulas means “I will receive”, a positive value means “I pay”.

An example of calculating the future value and number of payments
Two friends decided to secure an additional pension for themselves. To do this, each of them opened a savings account with a yield of 6% per annum, one made an initial contribution of $ 3,000 to it, and the second - $ 5,000. The first is 25 years old, the second is 30, both want to retire by the age of 45. Both are ready to deduct $50 per month. It is necessary to calculate the amount of their pension savings and the number of months of accrual of pensions from the accumulated funds, if pension payments are planned in the amount of $150.

Solution
First, we calculate the amount of pension savings. To do this, we use the BS formula. In the first case, the number of payments will be 20 * 12, in the second - 15 * 12, the present value in the first case is $3,000, in the second - $5,000, the interest rate in both cases will be 6% / 12, and the periodic payment will be $50 . The assembled formula in the first case will look like = BS (6% / 12; 20 * 12; 50; 3000), in the second = BS (6% / 12; 15 * 12; 50; 5000). In the first case, pension savings will amount to $33,032, in the second - $26,811. Now let's calculate the period during which the accumulated amount can provide the above pension payments. To do this, we use the NPER function, where we indicate 6% / 12 as the interest rate, set $ 150 as the payment amount, and substitute the obtained values ​​​​as the present value. We get the amount in months - 149 for the first and 128 for the second.

Note
A negative value in the formula indicates that we are receiving payments, in case the formula is used to calculate payments that need to be paid, the resulting value will be positive.

Perpetual annuity (perpetuity) and the Gordon model

A special case of an annuity is a sequence of payments, the duration of which is not conditionally determined, in connection with which this annuity is considered eternal. An example of a perpetual annuity can be consoles - a type of securities (bonds) on which interest is accrued indefinitely, but the return of the face value is not made. In practice, such securities are quite rare. A more common example of a perpetual annuity is the long-term dividend payments that some companies make to their shareholders. The Gordon model is used to calculate the cost of a perpetual annuity:

(10) S = P * (1+g) / (r - g) , where S is the cost of the annuity, P is the current payment, g is the growth rate of the current payment, r is the rate of return.

The above formulas are the main list of tools for calculations of various kinds and allow you to make calculations in relation to any situation. In the comments to this article, you can describe situations that require financial calculations, and I will try to show how the above mathematical apparatus will help you in solving them.

In preparing the article, materials from study guide"Financial Mathematics" Shirshova E.V., N.I. Petrik, Tutygina A.G., Menshikova T.V., Moscow, ed. "Knorus", 2010

Percentages in mathematics. Interest tasks.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

Percentages in mathematics.

What's happened percentages in mathematics? How to decide interest tasks? These questions pop up, alas, suddenly ... When a graduate reads the USE task. And they confuse him. But in vain. These are very simple concepts.

The only thing you need to remember ironically - what is one percent . This concept is master key to solving problems with percentages, and to working with percentages in general.

One percent is one hundredth of a number . And that's it. There is no more wisdom.

Reasonable question - a hundredth part what date ? And here is the number that is being discussed in the assignment. If it talks about price, one percent is one hundredth of the price. When it comes to speed, one percent is one hundredth of the speed. And so on. It is clear that the number itself in question is always 100%. And if there is no number itself, then the percentages have no meaning ...

Another thing is that in complex problems the number itself will be hidden so that you won’t find it. But we are not aiming at the complex yet. Dealing with percentages in mathematics.

I do not in vain emphasize the words one percent, one hundredth. Remembering what is one percent, you can easily find two percent, and thirty-four, and seventeen, and one hundred and twenty-six! As much as you need, so much you will find.

And this, by the way, is the main skill for solving problems with percentages.

Shall we try?

Let's find 3% of 400. First we find one percent. This will be one hundredth, i.e. 400/100 = 4. One percent is 4. How many percent do we need? Three. So let's multiply 4 by 3. We get 12. Everything. Three percent of 400 is 12.

5% of 20 is 20 divided by 100 (one hundredth is 1%), and multiplied by five (5%):

5% of 20 will be 1. That's it.

Easier nowhere. Let's quickly, before we forget, let's practice!

Find how much will be:
5% from 200 rubles.
8% of 350 kilometers.
120% of 10 liters.
15% off 60 degrees.
4% of excellent students from 25 students.
10% losers out of 20 people.

Answers (in complete disarray): 9, 10, 2, 1, 28, 12.

These numbers are the number of rubles, degrees, students, etc. I did not write how much of what, to make it more interesting to decide ...

What if we need to write X% from some number, for example, from 50? Yes, everything is the same. One percent of 50 is how much? That's right, 50/100 = 0.5. And we have these percentages - X. So let's multiply 0.5 by X! We get that X% from 50 it is 0.5x.

I hope that is percentages in mathematics you figured it out. And you can easily find any percentage of any number. It's simple. You are now capable of about 60% of all tasks for interest! Already more than half. So, let's get the rest, shall we? Okay, whatever you say!

In problems with percentages, the reverse situation is often encountered. We are given quantities (whatever), but you need to find interest . Let's master this simple process.

3 people out of 120 - how many percent? Do not know? Well then, let it be X percent.

Compute X% from 120 people. In people. This is what we can do. 120 divided by 100 (calculate 1%) and multiply by X(calculate X%). We get 1.2 X.

Let's consider the result.

X percent from 120 people, this is 1.2 X Human . And we have three such people. It remains to compare:

We recall that for x we ​​took the number of percent. So 3 people out of 120 people is 2.5%.

That's all.

It is possible in another way. Get by with simple ingenuity, without any equations. Thinking how many times 3 people less than 120? We divide 120 by 3 and get 40. So 3 is less than 120 by 40 times.

The desired number of people in percent will be the same amount less than 100%. After all, 120 people - this is 100%. Divide 100 by 40, 100/40 = 2.5

That's all. Got 2.5%.

There is another method of proportions, but this is, in essence, the same thing in an abbreviated version. All of these methods are correct. As you feel more comfortable, more familiar, more understandable - so consider.

We are training again.

Calculate what percentage is:
3 people out of 12.
10 rubles from 800.
4 textbooks of 160 books.
24 correct answers to 32 questions.
2 correct answers to 32 questions.
9 hits out of 10 shots.

Answers (in disarray): 75%, 25%, 90%, 1.25%, 2.5%, 6.25%.

In the process of calculations, you may well encounter fractions. Including uncomfortable ones, such as 1.333333... And who told you to use a calculator? yourself? No need. Consider without calculator , as written in the "Fractions" topic. There are all sorts of percentages...

So we have mastered the transition from values ​​​​to percentages and vice versa. You can take on tasks.

Interest tasks.

IN USE tasks on interest are very popular. From the simplest to the most complex. In this section, we work with simple tasks. IN simple tasks, as a rule, you need to move from percentages to those quantities that are discussed in the problem. To rubles, kilograms, seconds, meters, and so on. Or vice versa. This we already know. After that, the problem becomes clear and easily solved. Don't believe? See for yourself.
Let us have such a problem.

“The bus ride costs 14 rubles. During the school holidays, a 25% discount was introduced for students. How much is the bus fare during the school holidays?

How to decide? If we find out how much 25% in rubles- then there is nothing to decide. Subtract the discount from the original price - and that's it!

But we already know how to do it! How much will one percent from 14 rubles? One hundredth part. That is, 14/100 = 0.14 rubles. And we have 25 such percentages. So we multiply 0.14 rubles by 25. We get 3.5 rubles. That's all. We have set the discount amount in rubles, it remains to find out the new fare:

14 – 3,5 = 10,5.

Ten and a half rubles. This is the answer.

As soon as they switched from interest to rubles, everything became simple and clear. This is a general approach to solving percentage problems.

Of course, not all tasks are equally elementary. There are more difficult ones. Think! We will solve them now. The problem is that it's the other way around. We are given some values, but we need to find percentages. For example, such a task:

“Before, Vasya solved two problems correctly for percentages out of twenty. After studying the topic on one useful site, Vasya began to solve 16 problems out of 20 correctly. By what percentage did Vasya become smarter? We consider 20 solved tasks as one hundred percent intelligence.

Since the question is about percentages (and not rubles, kilograms, seconds, etc.), then we turn to percentages. Find out how many percent Vasya solved before wondering how many percent after - and it's in the hat!

We consider. Two problems out of 20 - how many percent? 2 is less than 20 by 10 times, right? So the number of tasks in percentages will be 10 times less than 100%. So 100/10 = 10.

10%. Yes, Vasya decided a little ... There is nothing to do at the Unified State Examination. But now he has grown wiser, and solves 16 tasks out of 20. We consider how many percent it will be? How many times is 16 less than 20? You can’t say offhand ... You’ll have to share.

5/4 times. Well, now we divide 100 by 5/4:

Here. 80% is already solid. And most importantly - there is no limit!

But that's not the answer! We read the problem again, so as not to make a mistake out of the blue. Yes, we are asked how much percent wiser Vasya? Well, it's simple. 80% - 10% = 70%. At 70%.

70% is the correct answer.

As you can see, in simple tasks it is enough to convert the given values ​​into percentages, or the given percentages into values, as everything becomes clear. It is clear that there may well be additional bells and whistles in the problem. Which, often, have nothing to do with interest at all. Here, most importantly, carefully read the condition and step by step, slowly, unfold the problem. We will talk about this in the next topic.

But there is one serious ambush in problems for percentages! Many fall into it, yes ... This ambush looks quite innocent. For example, here is a problem.

“A beautiful notebook cost 40 rubles in the summer. Before the start of the school year, the seller raised the price by 25%. However, notebooks began to buy so badly that he reduced the price by 10%. Still don't take it! He had to reduce the price by another 15%. Here comes the trade! What was the final price of the notebook?”

Well, how? Elementary?

If you promptly and joyfully gave the answer “40 rubles!”, Then you were ambushed ...

The trick is that percentages are always calculated from something .

Here we consider. How much rubles did the seller raise the price? 25% from 40 rubles - it's 10 rubles. That is, the increased price of a notebook began to cost 50 rubles. It's understandable, right?

And now we need to drop the price by 10% from 50 rubles. From 50, not 40! 10% of 50 rubles is 5 rubles. Consequently, after the first reduction in price, the notebook began to cost 45 rubles.

We consider the second reduction in price. 15% from 45 rubles ( from 45, not 40, or 50! ) is 6.75 rubles. So, the final price of the notebook:

45 - 6.75 = 38.25 rubles.

As you can see, the ambush lies in the fact that interest is calculated each time from the new price. From the last. This happens almost always. If the task for a sequential increase-decrease in a value is not explicitly stated, from what to calculate percentages, it is necessary to calculate them from the last value. And that's true. How does the seller know how many times this notebook went up and down in price before him and how much it cost at the very beginning ...

By the way, now you may think why the last phrase is written in the puzzle about clever Vasya? This one: " We consider 20 solved problems to be one hundred percent smart”? It seems that everything is clear anyway ... Uh-uh ... How to say. If this phrase does not exist, Vasya may well count his initial successes as 100%. That is two solved problems. And 16 tasks - eight times more. Those. 800%! Vasya will be able to justifiably talk about his own wisdom as much as 700%!

And you can also take 16 tasks for 100%. And get a new answer. Also correct...

Hence the conclusion: the most important thing in tasks for percentages is to clearly define what this or that percentage should be calculated from.

This, by the way, is necessary in life. where interest is used. In stores, banks, at all sorts of promotions. And then you wait for a 70% discount, but you get 7%. And not discounts, but rises in price ... And everything is honest, he himself miscalculated.

Well, you got an idea about percentages in mathematics. Let's note the most important.

Practical Tips:

1. In problems with percentages, we move from percentages to specific values. Or, if necessary, from specific values ​​to percentages. Read the task carefully!

2. We study very carefully, from what interest must be calculated. If this is not stated directly, then it is necessarily implied. When changing a value successively, percentages are assumed from the last value. We carefully read the task!

3. Having finished solving the problem, we read it again. It is possible that you have found an intermediate answer, and not a final one. We carefully read the task!

Solve several percentage problems. For reinforcement, so to speak. In these puzzles, I tried to collect all the main difficulties that await the decisive ones. Those rakes that are most often stepped on. Here they are:

1. Elementary logic in the analysis of simple problems.

2. Right choice the value from which the percentage is to be calculated. How many people stumbled on this! But there is a very simple rule...

3. Interest on interest. A trifle, but confuses great ...

4. And one more pitchfork. Relationship of percentages with fractions and parts. Translation of them into each other.

“50 people took part in the Mathematics Olympiad. 68% of students solved few problems. 75% of the rest solved the average, and the rest - a lot of problems. How many people solved many problems?

Clue. If you get fractional students, this is wrong. Read the problem carefully, there is one important word there ... Another problem:

“Vasya (yes, the same one!) loves donuts with jam very much. Which are baked at the bakery, one stop from the house. Donuts cost 15 rubles apiece. Having 43 rubles available, Vasya went to the bakery by bus for 13 rubles. And in the bakery there was an action "Discount for everything - 30%!!!". Question: how many additional donuts Vasya could not buy because of his laziness (he could have walked on foot, right?)”

Short puzzles.

What percentage is 4 less than 5?

How many percent is 5 greater than 4?

Long task...

Kolya got a simple job related to the calculation of interest. During the interview, the boss with a sly smile offered Kolya two options for remuneration. According to the first option, Kolya was immediately assigned a rate of 15,000 rubles per month. According to the second Kolya, if he agrees, the first 2 months will be paid a salary reduced by 50%. Kind of like a newbie. But then they will increase his reduced salary by as much as 80%!

Kolya visited one useful site on the Internet ... Therefore, after thinking for six seconds, he chose the first option with a slight smile. The boss smiled back and set Kolya a fixed salary of 17,000 rubles.

Question: How much money per year (in thousands of rubles) did Kolya win at this interview? When compared with the most unfortunate option? And one more thing: why are they smiling all the time!?)

Another short puzzle.

Find 20% of 50%.

And again long.)

Fast train No. 205 "Krasnoyarsk - Anapa" made a stop at the station "Syzran-gorod". Vasily and Kirill went to the railway station shop for ice cream for Lena and a hamburger for themselves. When they bought everything they needed, the store cleaner said that their train had already left ... Vasily and Kirill quickly ran and managed to jump into the car. Question: under these conditions, would the world champion in running have time to jump into the car?
We believe that under normal conditions the world champion runs 30% faster than Vasily and Kirill. However, the desire to catch up with the car (it was the last one), treat Lena with ice cream and eat a hamburger increased their speed by 20%. And ice cream with a hamburger in the hands of a champion and flip-flops on his feet would reduce his speed by 10%...

But the problem without interest ... I wonder why she is here?)

Determine how much 3/4 of an apple weighs if the whole apple weighs 200 grams?

And the last one.

In the fast train No. 205 "Krasnoyarsk - Anapa", fellow travelers were solving a scanword. Lena guessed 2/5 of all the words, and Vasily guessed one third of the rest. Then Cyril connected and solved 30% of the entire scanword! Seryozha guessed the last 5 words. How many words were in the scanword? Is it true that Lena guessed the most words?

Answers in the traditional disorder and without unit names. Where are the donuts, where are the students, where are the rubles with interest - it's you yourself ...

10; 50; Yes; 4; 20; No; 54; 2; 25; 150.

So how is it? If everything fits together, congratulations! Interest is not your problem. You can safely go to work in a bank.)

Is there something wrong? Does not work? Don't know how to quickly calculate percentages of a number? Do not know very simple and clear rules? From what to count percentages, for example? Or how to convert fractions to percentages?

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Consider an example:

The price of a refrigerator in a store has increased by. What was the price if the original refrigerator cost a ruble?

Solution:

To begin with, let's determine how many rubles have changed (in this case, increased) the cost of the refrigerator.

By condition - on.

But from what?

Of course, from the very initial cost of the refrigerator - rubles.

It turns out that we need to find from rub:

Now we know that the price has increased by Rs.

It remains only, according to the rule, to add to the initial cost the amount of change:

The new price of rubles.

Another example(try to figure it out on your own):

The book "Mathematics for Dummies" in the store costs rubles. During the promotion, all books are sold at a discount.

How much will you pay for this book now?

Solution:

What is a discount, you probably know? Discount in means that the cost of the goods has been reduced by

How much has the price of the book decreased (in rubles)?

It is necessary to find from its initial cost in rubles:

The price has decreased, so you need to subtract from the initial cost how much it decreased:

The new price of rubles.

Is it really simple?

But there is a way to make this solution even simpler and shorter!

Consider an example:

Increase the number by.

What are from?

As we have already found out before, it will be.

Now let's increase the number x itself by this amount:

It turns out that as a result, we added to the decimal notation and multiplied by a number.

Let's generalize this rule:

Suppose we need to increase the number by.

from the number is.

Then the new number will be: .

For example, let's increase the number by:

Now try it yourself:

1. Increase number by
2. Increase number by
3. By what percent is the number greater than the number?

Solutions:

3) Let the required quantity percent equals.

This means that if the number is increased by, you get:

Answer to.

If the number x needs to be reduced by, everything is similar:

So the rule is:

Examples:

1) Decrease the number by.

2) On how many percent number less than number?

3) The price of the discounted item in is equal to p. What is the price without discount?

Solutions:

2) Number decreased by x percent and got:

Answer to.

3) Let the price without discount be equal. It turns out that x was reduced by and got:

Finally, let's consider one more type of tasks that often cause bewilderment.

Solving complex problems with interest

The number is greater than the number by. On how many percent number less than number?

What a strange question: of course!

Right?

And here it is not.

If, for example, the mass of one cabinet is 25 kg more than the mass of the other, then, without a doubt, the mass of the second cabinet is 25 kg less than the mass of the first.

Nose percent it won't work like that!

Indeed, in the first case, when we say that the number is greater than the number, we count from the number; and in the second case, when we say that the number is less than the number, we count from the number. And since the numbers and are different, then these numbers will be different!

To solve this problem correctly, let's write the condition in the form of an equation:

The number is greater than the number by. This means that if the number is increased by, we get the number:

Now, in the same form, we write the question: if the number a is reduced by percent, we get the number:

Let's express the number from equality (1):

And substitute into (2):

From this it follows that:

So, we get that the number is less than the number!

Problems like this often come up in the exam.

For example:

On Monday, the company's shares rose by a certain number. percent, and on Tuesday they fell by the same number percent. As a result, they began to cost less than at the opening of trading on Monday. On how many percent did the company's shares rise in price on Monday?

Solution:

Let the share price on Monday be equal, and the desired quantity percent, written as a decimal fraction (that is, already divided by), is equal to.

Let's write down the formula, what is the value of the stock after the rise in price:

It is known that this final price is less than the initial price. That is, if we reduce by, we get:

Let's substitute what was expressed earlier:

According to common sense, only a positive solution is suitable:

Recall now that this is so far only a decimal notation of the desired amount percent, that is, this quantity percent divided by. To translate to interest, you need to multiply by 100%:

Where do we use interest in life?

Well, for example, in banking products: deposits, loans, mortgages, etc.

If you understand well what interest is and know how to solve equations, then you can easily calculate, for example, the amount of a monthly loan payment.

Or how much you will have to overpay by taking a mortgage. There is such a task in the exam at number 17.

Interest. Briefly about the main

One percent of any number is one hundredth of that number.

1. Percentages and decimals

2. Change the number by a certain percentage

Let's say you want to increase the number by.

from the number is.

Then, the new number will be: .

To increase a number by, you need to multiply it by.

If the number needs to be reduced by, then:

To decrease a number by some value means to subtract this value from it:

To decrease a number by, you need to multiply it by.

The concept of interest has a wide practical application, so it is obligatory part school curriculum mathematics. Schoolchildren should learn how to solve basic problems with percentages, represent them in the form of decimal and ordinary fractions.

Traditionally, the topic "Percentage" is studied in the framework of the lower grades of the middle link. There are several approaches to the study of this topic.

First approach. Consideration of percentages is conducted as a separate topic, without relying on fractions. Finding a few percent of a number is done in two steps. The study of fractions is a separate topic, much later than tasks for percentages. Thus, learning goes from the particular to the general, which is less effective and provides fewer opportunities for the development of the student.

Second approach. Problems with percentages are mastered as a special case of problems with fractions and all methods of solving are transferred to them, that is, the study goes from the general case - problems with fractions, to a particular one. Most modern textbooks implement the second approach.

Let us consider in more detail the study of this topic in some modern textbooks recommended by the Ministry of Education of Russia for 2003/2004 academic year for teaching mathematics in elementary school.

According to textbooks, the topic "Percentage" is studied in grade V. Before introducing the concept of "percentage", the author suggests considering examples:

“A hundredth of a centner is called a kilogram, a hundredth of a meter is called a centimeter, a hundredth of a hectare is called an acre. It is customary to call a hundredth of any value a percentage.

There are three main problems for interest:

View task K1.

Example 1: A team of workers repaired 40% of a road with a length of 120 m in a day. How many meters of road was repaired by a team in a day?

120m is 100%

1) 120:100 = 1.2 m is 1%.

2) m repaired by the brigade in a day.

Answer: During the day the brigade repaired 48 meters of the road.

View task K2.

Example 2: The student has read 72 pages, which is 30% of the total number of pages in the book. How many pages are in the book?

Unknown number - 100%.

1) 72:30=2.4 pages is 1%.

2) pages is 100%.

Answer: There are 240 pages in the book.

View task P1.

Example 3: In a class of 40 students, 32 solved the problem correctly. What percentage of the students solved the problem correctly?

40 students make up 100%.

1) 40:100=0.4 is 1%.

2) 32:0.4=80; 32 students make up 80%.

Answer: 80% of the students solved the problem correctly.

However, these types of problems are not distinguished, since the method of reduction to unity is adopted as the main method for solving problems for percentages. It has certain advantages:

a) easier to perform calculations;

b) teaches students to highlight the number taken as 100%;

c) requires appropriate reasoning in the process of solving a specific problem, which does not include memorizing the rules for solving one or another type of problems for interest.

The textbook suggests solving some percentage problems using equations. This recommendation applies essentially to two kinds of problems: finding a number given the number of its percentages, and finding the percentage of two numbers. The experience of teaching mathematics in the 5th grade shows that students face certain difficulties in the process of solving problems with percentages, which is mainly due to the lack of awareness by students of the method of reduction to unity. Therefore, working out the essence of this method in two steps is of decisive importance in teaching how to solve problems with percentages, especially at the initial stage of mastering knowledge. The problems considered in examples 2 and 3 can be solved using equations. In the 5th grade, solving problems with the help of equations causes significant difficulties for students.

This topic is one of the last in the V class course. Further, the authors specifically do not return to the topic. This is not very successful, since the topic is objectively difficult.

A slightly different approach to this topic in textbooks. The study of interest begins at the end of the fifth grade. The authors define percentage as another name for one hundredth. “We know that one second is otherwise called half, one fourth - a quarter, three fourths - three quarters. One hundredth has a special name: one hundredth is called a percentage. Students consider only two types of problems:

View task K1.

Example 4. There are 800 students in a school, 15% of them received fives in mathematics in a quarter. How many students got A's in math?

Let us first find one percent, or one hundredth, of the number those who are

800: 100=8.

To find 15%, you need to multiply:

Answer: 120 students received fives.

Much attention is paid to the relationship between fractions (decimal and ordinary) and percent.

View task P1.

Example 5. What percentage of 1 m is 1 cm, 9 cm, 0.15 m?

In the VI grade, the authors return to this topic again. Students repeat the material studied in the fifth grade, and new tasks are considered. At the same time, for each type of problem, an analogy is made with actions on decimal and ordinary fractions, the rule is formulated:

For a task of the form K1.

2) multiply the given number by this fraction

And also for a problem of the form K2.

“1) express interest as an ordinary or decimal fraction;

2) divide the given number by this fraction

Example 6. For test in mathematics, 9 students received a mark of "4". This is 36% of all students in the class. How many students are in the class?

Let's express the percentage as an ordinary or decimal fraction: 36% \u003d \u003d 0.36.

Let's use the rule for finding a number by its fraction:

9:==25 or 9:0.36=25

Answer: There were 25 students in the class.

First, students consider the expression of the quotient of two numbers as a percentage: "to express the quotient as a percentage, you need to multiply the quotient by 100 and attribute the percent sign to the resulting product."

Only then do they move on to problem solving. P1.

"For this you need

1) divide the first number by the second;

2) express the resulting quotient as a percentage "

Example 7. There are 25 students in a class, 20 of them are pioneers. What percentage are pioneers?

To solve, you need to express the quotient as a percentage. =0.8=80%.

Answer: pioneers make up 80%.

At the end of the topic, a problem of the form P2 And P3.

“... in order to find out by what percentage a given value has increased or decreased, it is necessary to find:

1) by how many units this value has increased or decreased;

2) how many percent is the resulting difference from the initial value of the value "

Example 8. Before the price reduction, the refrigerator cost 250 rubles, after the reduction - 230 rubles. By what percent did the cost of the refrigerator decrease?

We will find out by how many rubles the price of the refrigerator has changed: 250-230 \u003d 20 rubles.

Let's find how many percent the received difference from the initial cost of the refrigerator is: \u003d 0.08 \u003d 8%

Answer: the cost of the refrigerator decreased by 8%.

Rules limit students, do not allow them to reason over the solution. Therefore, each percentage problem becomes an algorithm and causes difficulties if the rule is forgotten. The solution of problems in this course is arithmetic. The use of equations in solving begins only at the end of the year only in complex problems. Therefore, not every student will be able to master this skill. Therefore, you need to include tasks for percentages when studying equations.

In textbooks, the concept of percentage is also studied at the end of grade V. Before introducing the definition, examples of the use of the concept of "percentage" are considered:

“Seed germination is 98 percent; 65 percent of voters took part in the presidential elections in Russia ... ”. A percentage is defined as a hundredth. In class V, the authors consider only two types of problems: problems of the form K1 And K2. The solution of these problems is carried out in an arithmetic way. Much attention is paid to the question of what value to take as 100%.

Further, the topic "Interest" is studied in the VI grade. The same types of problems are considered here, but the solution is already carried out in an algebraic way (drawing up linear equations). The authors formulate the rules for finding a part from a whole and a whole from its part:

“1) to find a part of a whole, you need to multiply the whole (the number corresponding to it) by the fraction (corresponding to this part);

2) to find a whole by its part, it is necessary to divide the part (the number corresponding to this part) by the fraction corresponding to it.

After that, the topic is not considered.

A slightly different approach in textbooks,. Interest begins to be studied at the beginning of the sixth grade. The concept of percentage is introduced as one hundredth of a number (value). Three types of problems are considered:

a) finding percentages of a given number K1.

First, finding 1% of a given number is considered. Then - finding an arbitrary number of percent.

b) finding a number for a given number of its percentages K2.

It also primarily discusses how to find a number whose 1% is known. This problem is then considered for any arbitrary number of percent.

c) finding the percentage of two numbers P1. The authors formulate the rule “To express the ratio of two numbers as a percentage, you can multiply this ratio by 100”

All three types of problems are solved first in an arithmetic way, and then they are solved based on the properties of proportionality.

Example 9. Find 8% of 35.

Solution: Let x be the desired number, then:

Answer: 2

Problems are also considered in which it is necessary to increase (decrease) the number by a certain number of percent K3 And K4. Percentages are also used when studying charts.

Example 10

The price of a product was increased by 10%, then another 10%. By what percentage did the price of the product increase twice?

Problems involving mixtures and alloys are also considered here (this section is marked as a section of increased difficulty). It seems to me that tasks of this type are difficult for sixth graders. Therefore, not every teacher will want to consider such complex tasks with the whole class, and a very important layer of tasks will remain unconsidered. But this is very important tasks which should be given due attention, perhaps at an older age.

This set also pays attention to working with a calculator when solving percentage problems. A separate paragraph is devoted to this issue and a system of exercises has been developed.

In high school, the topic of interest is considered within the framework of tasks for repetition and tasks of increased difficulty. In high school, operations with interest become the prerogative of chemistry, which introduces its own view of interest. Therefore, questions of the universality of interest and the diversity of their areas of application are gradually forgotten by students.

We will show how it is proposed to study this material in the teaching kits in mathematics for grades V-VI, ed. G.V. Dorofeeva and I.F. Sharygin and for grades VII - IX, ed. G.V. Dorofeeva.

First of all, it should be noted that when presenting the topic “Interest”, many general methodological features that are characteristic of the course as a whole are implemented. The topic unfolds in a spiral and is studied in several stages from grades VI to IX inclusive. With each pass, students return to percentages at a new level, their knowledge is replenished, new types of problems and solutions are added. Such repeated appeal to the concept leads to the fact that it is gradually assimilated firmly and consciously. It becomes possible to include tasks that cannot be considered in current textbooks simply because of the age characteristics of schoolchildren.

Questions related to percentages allow you to make the course practice-oriented, show students that the mathematical knowledge they acquire is applied in Everyday life. Interest is also largely supported by the content of the tasks, the plots of which are close to modern topics and to the life experience of children, and then adolescents. This serves as a sufficiently strong motive for solving the proposed tasks.

The introduction of percentages is based on the subject-practical activities of schoolchildren, on geometric clarity and geometric modeling. Drawings and drawings are widely used to help understand the problem and see the solution.

As in all the main sections of the course, when presenting this topic, ample opportunities for differentiated learning of students are implemented. Tasks are offered in a wide range of complexity - from basic to quite difficult. The teacher can choose the material that suits the abilities of each student.

When learning to solve problems with percentages, students are introduced to different ways of solving problems, and many of the techniques are wider than is usually the case. The student masters various ways of reasoning, enriching his arsenal of techniques and methods. But at the same time, it is also important that he has the opportunity to choose and can use the method that seems more convenient to him.