Fundamentals of financial mathematics. Problems on percentages Learn percentages

The concept of interest is broad practical use, so it is a required part school curriculum mathematics. Students should learn to solve basic problems involving percentages and represent them as decimals and ordinary fractions.

Traditionally, the topic “Percentage” is studied in lower secondary classes. There are several approaches to studying this topic.

First approach. Percentages are discussed 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 percentage problems. Thus, training goes from the specific to the general, which is less effective and provides fewer opportunities for the development of the student.

Second approach. Problems involving percentages are mastered as special case problems on fractions and all solving techniques are transferred to them, that is, the study goes from the general case - problems on fractions - to the 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 primary schools.

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 quintal is called a kilogram, a hundredth of a meter is a centimeter, a hundredth of a hectare is an acre. It is customary to call a hundredth of any quantity a percentage.”

Three main problems involving percentages are considered:

Problem of the form 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 were repaired by the team in a day?

120m is 100%

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

2) m repaired by the team in a day.

Answer: During the day, the team repaired 48 m of the road.

Problem of the form K2.

Example 2: A student 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: The book has 240 pages.

Problem of the form P1.

Example 3: In a class of 40 students, 32 solved the problem correctly. What percentage of 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 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 of solving problems involving percentages. It has certain advantages:

a) easier to perform calculations;

b) accustoms students to highlighting 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 problem involving percentages.

The textbook involves solving some percentage problems using equations. This recommendation essentially applies to two types of problems: finding a number from a given number of its percentages and finding the percentage of two numbers. The experience of teaching mathematics in the fifth grade shows that students encounter certain difficulties in the process of solving problems with percentages, which is mainly due to students’ lack of awareness of the method of reduction to one. Therefore, practicing the essence of this method in two steps is crucial in learning to solve problems involving percentages, especially at the initial stage of mastering knowledge. The problems discussed in examples 2 and 3 can be solved using equations. In grade V, solving problems using equations causes significant difficulties for students.

This topic is one of the last in the V class course. Further, the authors do not specifically 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 percentages begins at the end of grade V. The authors define percentage as another name for one hundredth. “We know that one second is otherwise called a half, one fourth is a quarter, three quarters is three quarters. One hundredth also has a special name: one hundredth is called a percentage.” Students consider only two types of problems:

Problem of the form K1.

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

Let's first find one percent, or one hundredth, of the number current ones.

800: 100=8.

To find 15%, you need to multiply:

Answer: 120 students received A's.

Much attention is paid to the connection between fractions (decimal and ordinary) and percentages.

Problem of the form P1.

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

In grade VI, the authors return to this topic again. Students repeat material learned in grade V, and new problems are considered. At the same time, for each type of problem, an analogy is drawn with operations on decimal and ordinary fractions, and the rule is formulated:

For a problem of the form K1.

2) multiply this number by this fraction"

And also for a problem of the form K2.

“1) express percentages as ordinary or decimal fractions;

2) divide this number by this fraction"

Example 6. For test In mathematics, 9 students received a grade of “4”. This represents 36% of all students in the class. How many students are in the class?

Let's express the percentages as an ordinary or decimal fraction: 36%= =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 add a percent sign to the resulting product.”

Only after this do they move on to solving the problem. 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 is considered P2 And P3.

“... to find out by what percentage a given value increased or decreased, you need to find:

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

2) what percentage is the resulting difference from the initial value of the quantity?

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

Let's find out how many rubles the price of the refrigerator has changed: 250-230 = 20 rubles.

Let's find the percentage of the resulting difference from the initial cost of the refrigerator: =0.08=8%

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

Rules limit students and prevent them from reasoning about a solution. Therefore, every percentage problem becomes an algorithm and causes difficulties if the rule is forgotten. Solving 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 percentage problems when learning 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 “percentage” are considered:

“Seed germination rate is 98 percent; 65 percent of voters took part in the Russian presidential elections..." Percentage is defined as the designation of hundredths. In grade V, the authors consider only two types of problems: problems of the form K1 And K2. These problems are solved using an arithmetic method. Much attention is paid to the question of what value to take as 100%.

Next, the topic “Percentage” is studied in grade VI. The same types of problems are considered here, but the solution is carried out in an algebraic way (composing 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 from its part, you need to divide the part (the number corresponding to this part) by the fraction corresponding to it.”

After this, the topic is not considered.

A slightly different approach in textbooks. Percentages begin to be studied at the beginning of class VI. The concept of percentage is introduced as one hundredth of a number (quantity). Three types of problems are considered:

a) finding the percentage of a given number K1.

First, we consider finding 1% of a given number. Then - finding an arbitrary number of percentages.

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

It also first discusses how to find a number of which 1% is known. This problem is then considered for any arbitrary number of percentages.

c) finding the percentage ratio 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 required number, then:

Answer: 2

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

Example 10.

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

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

This kit also focuses on using a calculator to solve percentage problems. This issue A separate paragraph is devoted and a system of exercises has been developed.

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

We will show how it is proposed to study this material in educational 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 common methodological features, characteristic of the course as a whole. 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 reference to a concept leads to the fact that it is gradually assimilated firmly and consciously. It becomes possible to include tasks that currently cannot be considered in existing textbooks simply due to age characteristics schoolchildren.

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

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 path to solution.

As in all main sections of the course, when presenting this topic, ample opportunities for differentiated learning for students are implemented. The tasks are offered in a wide range of difficulty - from basic to quite difficult. The teacher can select material that matches the capabilities of each student.

When learning to solve problems involving percentages, students become familiar with different ways solving problems, and many techniques are broader than is usually the case. The student masters a variety of 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 technique that seems more convenient to him.

When sending their child to school, many parents worry that they will not be able to help them solve a simple problem, thereby falling in the eyes of their children. There is no need to be afraid of this, and in order to avoid such situations, you will have to remember the knowledge you once acquired, and maybe learn in a new way. If the tasks proposed in primary school, you can still decide, then not everyone can cope with the fifth grade program, and it is at this stage that the child will have to learn what percentages are, and you will have to think about how to explain percentages in mathematics to your child. Having rummaged through their memory, many will find a solution to the problem, but if you have forgotten how to calculate percentages, you will have to sit down to textbooks.

Teaching your child to calculate percentages

A mathematics teacher knows exactly how to explain percentages in mathematics to a child, and he will also teach other arithmetic operations, but not all children are endowed with the ability to perceive information by ear or from books on their own. In this case, they will turn to their parents, who must explain how to calculate the percentage of something. If you don’t know how to explain percentages to a student, try to turn the lesson into a fun game. You might have to draw 100 shapes to do this, but it's worth it because you can explain everything clearly. You must tell that all one hundred figures are 100%, and if you paint 50 figures in any color, then exactly half of the unpainted figures will remain, and half is 50%.

Most likely, the child will like this game, and you have room for maneuver - you can color any number of shapes, inviting the child to count them. After all, everything is simple here - 30 painted figures - 30% and so on. After your child understands what percentages are through visual examples, you can decide how to calculate the percentage of a quantity. If you don’t know how to explain to your child the topic of percentages in grades 5 and 6, ask him to solve simple task, calculating 50 percent of any number of people. To do this, he just needs to divide 50 by 100 and multiply by the total number of people. There are other possibilities, but do not forget the somewhat forgotten proportions that are best suited for calculating the percentage.

Applying percentages in life

In order for your child to better master percentages, and if you have not yet understood how to explain percentage problems to your child in grades 5 and 6, first try to explain why he needs it, in principle. To do this you will have to be creative. Take, for example, a child to the bank and try to explain to him what interest is using the example of the interest rate on a loan. The child should be interested in this, and he will understand that knowing percentages is important, and now you can calmly begin to study percentages. You can use remembering percentages in other life situations, the main thing is that the child finds it interesting, and he understands that if he does not understand percentages, he will lose a lot.

The first thing a child must learn is that a percentage is a hundredth of a number. You can convert percentages to a decimal by dividing the required number by 100, but to convert a decimal to a percentage, you need to do the opposite - multiply a fractional number by 100. If your child is interested in learning percentages, invite him to memorize a table that shows the relationships between fractions and percentages, making it easier to learn the information with the help of interesting pictures.

As students enter fifth grade, they are faced with a new type of math problem—percentage problems. For many of them this topic can be quite difficult. How to explain finding percentages?

Instructions

The child usually quickly understands prime number problems. For example, if there are 100 kopecks in one ruble, 50 kopecks is 50 percent. It is much more difficult to explain that percentages can be found on any value. Having dealt with simple quantities: grams and kilograms, centimeters and meters, move on to more complex questions.

1200 suits – 100%

X suits – 30%

X (1200 * 30)/100.
You just need to multiply the numbers crosswise and solve the resulting equation. Don't worry if you think your child is solving things mechanically. While he doesn’t need to think deeply about the essence, the most important thing is that he remembers the algorithm of actions, this will be enough to solve school problems. Be patient and don't yell at your child or get angry with him. After all, it seems to him that this information is very complex, incomprehensible and completely unnecessary. Try to offer him practical tasks, for example, for the family budget.

Moving into fifth grade, schoolchildren are faced with a new type of mathematical problems- problems involving percentages. For many of them this topic can be quite difficult. How to explain finding percentages?

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Instructions

Tell your child the story of how the word percentage came to be. It comes from the Latin “pro centum”, which translates as “hundredth part”. Later, in Mathieu de la Porte's textbook on commercial arithmetic, a typo was made, which is why the % sign appeared. Thus, the most important thing is to understand that a percentage is one hundredth of any number.

The child usually quickly understands the tasks prime numbers. For example, if there are 100 kopecks in one ruble, 50 kopecks is 50 percent. It is much more difficult to explain that percentages can be found on any value. Having dealt with simple quantities: grams and kilograms, centimeters and meters, move on to more complex questions.

If a child cannot understand the very essence of percentages, teach him to solve problems using an algorithm, making sure that he does not skip a single step of the solution. For example, the task: a clothing factory produced 1200 suits in a year. Of these, 30% are costumes of blue color. How many blue suits did the factory produce? First find how many suits make up 1%. To do this, divide the total by 100. 1200/100 = 12. That is, every 12 suits is 1 percent. Then multiply 12 by 30% to get your answer.

You can use the old “grandfather’s” method of proportion. For some reason, they rarely show it in schools now, but it works flawlessly. From the same task:

1200 suits – 100%
X suits – 30%
X (1200 * 30)/100.

You just need to multiply the numbers crosswise and solve the resulting equation. Don't worry if you think your child is solving things mechanically. While he doesn’t need to think deeply about the essence, the most important thing is that he remembers the algorithm of actions, this will be enough to solve school problems. Be patient and don't yell at your child or get angry with him. After all, it seems to him that this information is very complex, incomprehensible and completely unnecessary. Try to offer him practical tasks, for example, for the family budget.

How simple

I hope that you, dear reader, were pleased to find yourself in an article devoted to the study of percentages and learn something new and useful for yourself.

Specific problems involving percentages will be discussed in a separate article.

Please leave your comment on this issue below.

9B grade student

Head: Olga Sergeevna Drobkova, mathematics teacher

INTRODUCTION

Percentage is one of the most difficult topics in mathematics, and many students find it difficult or even unable to solve percentage problems. An understanding of percentages and the ability to make percentage calculations are necessary for every person. I believe that this topic is relevant in our time. After all, percentages are found in almost all areas of human activity. You cannot do without the concept of “percentage” either in accounting, or in finance, or in statistics. To calculate an employee's salary, you need to know the percentage of tax deductions; to open an account in a savings bank or take out a loan, our parents are interested in the amount of interest charged on the deposit amount and the interest on the loan; To know the approximate rise in prices next year, we are interested in the percentage of inflation. In trading, the concept of “interest” is used most often. We can very often hear about discounts, markups, markdowns, profits, credits, etc. - it's all interest. A modern person needs to navigate well in a large flow of information and make the right decisions in different life situations. To do this, you need to do percentage calculations well.

Thus, by studying this topic, we will find out what significance percentages have in our lives.

Purpose of the study: show the breadth of application of percentage calculations in real life .

Tasks:study literature on this topic; consider the need to use interest; explore areas of human activity in which percentages are used.

THE CONCEPT OF PERCENTAGE

A percentage is one hundredth of a number. Percentage is written using the % sign.

To convert a percentage to a fraction, remove the % sign and divide the number by 100.

To convert a decimal fraction to a percentage, you need to multiply the fraction by 100 and add the % sign.

To convert a fraction to a percentage, you must first convert it to a decimal, then multiply by 100 and add the % sign.

As you understand, percentages are closely related to ordinary and decimal fractions. Therefore, it is worth remembering a few simple equalities. In everyday life, you need to know about the numerical relationship between fractions and percentages. So, half - 50%, a quarter - 25%, three quarters - 75%, one fifth - 20%, and three fifths - 60%.

Knowing the relationships from the table below by heart will make it easier for you to solve many problems.



Interest

2. BASIC TYPES OF PERCENTAGE PROBLEMS

The main tasks for interest are the following:

Example 1. The school has 940 students. Of these, 15% study at a music school. How many students attend music school?

Solution : since 15% = 0.15, then to solve the problem you need to multiply 940 by 0.15. We get,

This means that 141 students attend the music school.

Answer: 141 students.

Finding a number by percentage
Example 2. IN school library 2100 textbooks, which is 40% of all books. How many books are in the school's library collection?

Solution: Let's denote the total number of books by x - this is 100%. According to the condition, 40% are textbooks, there are 2100 of them. Let's make a proportion: So,

Answer: 5250 books are in the school library.

Example 3. The school has 800 students, 16 of them are excellent students. What percentage of school students get "5" grades?

Solution: There are 800 students in the school - that's 100%. The percentage of students studying with “5” grades will be designated as x. Let's make a proportion. Means,

Answer: 2% of students are excellent students.

3 . RESEARCH ON THE TOPIC "INTEREST"

In order to find out what place percentages occupy in our lives, we decided to find out where we can find percentages:

1. In stores during the holidays, discounts appear, which are expressed as a percentage, for example, in a clothing store, when you buy 2 items, there is a 10% discount, etc.

Task . During a seasonal sale, an outerwear store reduced prices on fur coats, first by 20%, and then by another 10%. How many rubles can you save when buying a fur coat, if before the price reduction they cost 18,000 rubles?

Solution:

1 way to solve:

The cost of a fur coat is 18,000 rubles - that’s 100%. Let's find how many rubles a 20% discount will be: , So, rub. Thus, the price of a fur coat will be 18000-3600 = 14400 rubles.After the second markdown, the new price of fur coats decreased by another 10%, which will amount to 1,440 rubles. As a result, fur coats fell in price by 5,040 rubles;

2 way to solve:

18000-18000●0.2=14400 (rub) - the price of a fur coat after a 20% discount

14400-14400●0.1=12960 (rub) - the price of a fur coat after the second 10% discount

18000-12960=5040 (rub) - the buyer will save.

2. The composition of the fabric is indicated as a percentage, for example, when buying a suit, which contains 60% cotton and 40% synthetics, etc.;

3. Various statistical data on the population, on the output of certain products, etc. are expressed as percentages;

4. When purchasing any product on credit, you must be able to calculate interest;

5. At school, the progress and quality of knowledge of students is calculated as a percentage;

6.Accountants when calculating wages. For example, here in the village of Shira, there is an additional payment of 30% for northern and 30% for rural ones.

Task . When hiring, the director of the enterprise offers you a salary of 14,000 rubles. What amount will you receive after additional payments: 30% northern and 30% rural, and withholding personal income tax?

Solution:

1 way to solve:

IN This additional payment is 60%, i.e.. Means, rubles make up the allowances. Thus, the accrual with additional payments will be equal to 14000 + 8400 = 22400 (14000 * 1.6 = 22400). Now let’s calculate how much you will receive after withholding personal income tax (this tax is 13%) :

rub. - compiles the tax

22400-2912=19488 rubles.

2 way to solve:

in accounting,

in everyday life, etc.

It is difficult to name an area where percentages are not used. It is very difficult to fully consider the use of interest calculations in life, since interest is used in all spheres of human life.

In my work, I showed the use of the concept of percentage in solving various problems, and examined the main types of problems involving percentages.

This topic leaves a wide field for further research. Problems involving percentages are of great practical importance and the acquired knowledge, I hope, will help me in later life. I plan to develop the topic I started and look at interest rates in the banking sector in more detail. To be a modern person, you need to be able to calculate possible loan payments yourself or at least know approximately whether it is worth taking out a loan or a loan.

BIBLIOGRAPHY

Borovskikh A. What is interest? / A. Borovskikh, N. Rozov // Mathematics. - 2012. - No. 1. - pp. 23-25;
Valieva Yu. Interest in the past and present / Yu. Valieva // Mathematics. - 2012. - No. 9. - pp. 13-15;
Dyatlov V. Technologies for solving problems. Lecture 15. Text problems involving interest and fractional content / V. Dyatlov // Mathematics. - 2013. - No. 11. - pp. 44-49;
Zubareva I.I. Mathematics. 5th grade: educational. for general education students. institutions / I.I. Zubareva, A.G. Mordkovich. - 12th edition, rev. and additional - M.: Mnemosyne, 2012. - 270 p.;
Petrova I.N. Interest for all occasions / I.N. Petrova. - M., Education, 2006;
Tumasheva O.V. Mathematics lesson in grades 5-6: teaching aid / O.V. Tumasheva; Krasnoyarsk State Ped. University named after V.P. Astafieva. - Krasnoyarsk, 2007 - 104 p.

We continue to study elementary problems in mathematics. This lesson is about percentage problems. We will look at several problems, and also touch on those points that we did not mention earlier when studying percentages, considering that at first they create difficulties for learning.

Most percentage problems boil down to finding a percentage of a number, finding a number using a percentage, expressing some part as a percentage, or expressing it in percentage relationship between several objects, numbers, quantities.

Preliminary Skills Lesson content

Methods for finding percentage

Percentage can be found in various ways. The most popular way is to divide the number by 100 and multiply the result by the desired percentage.

For example, to find 60% of 200 rubles, you must first divide these 200 rubles by one hundred equal parts:

200 rubles: 100 = 2 rubles.

When we divide a number by 100, we find one percent of that number. So, dividing 200 rubles into 100 parts, we automatically found 1% of two hundred rubles, that is, we found out how many rubles are per part. As can be seen from the example, one part (one percent) accounts for 2 rubles.

1% of 200 rubles - 2 rubles

Knowing how many rubles are in one part (1%), you can find out how many rubles are in two parts, three, four, five, etc. That is, you can find any number of percentages. To do this, just multiply these 2 rubles by the required number of parts (percents). Let's find sixty pieces (60%)

2 rubles × 60 = 120 rubles.

2 rubles × 5 = 10 rubles.

Let's find 90%

2 rubles × 90 = 180 rubles.

We will find 100%

2 rubles × 100 = 200 rubles.

100% is all one hundred parts and they make up all 200 rubles.

The second way is to represent the percentage as common fraction and find this fraction from the number from which you want to find the percentage.

For example, let’s find the same 60% of 200 rubles. First, let's represent 60% as a fraction. 60% is sixty parts out of one hundred, that is, sixty hundredths:

Now the task can be understood as « find from 200rubles " . This is what we studied earlier. Let us remind you that to find a fraction of a number, you need to divide this number by the denominator of the fraction and multiply the resulting result by the numerator of the fraction

200: 100 = 2

2 × 60 = 120

Or multiply the number by a fraction ():

The third way is to represent the percentage as a decimal and multiply the number by the decimal.

For example, let’s find the same 60% of 200 rubles. To begin, represent 60% as a fraction. 60% percent is sixty parts out of one hundred

Let's do the division in this fraction. Let's move the decimal point in the number 60 two digits to the left:

Now we find 0.60 from 200 rubles. To find the decimal fraction of a number, you need to multiply this number by the decimal fraction:

200 × 0.60 = 120 rub.

The above method of finding a percentage is the most convenient, especially if a person is used to using a calculator. This method allows you to find the percentage in one step.

As a rule, expressing a percentage in decimal fractions is not difficult. It is sufficient to add “zero integer” before the percentage if the percentage represents two-digit number, or add “zero integer” and another zero if the percentage is a single digit number. Examples:

60% = 0.60 - added zero integers before the number 60, since the number 60 is two-digit

6% = 0.06 - added zero integers and one more zero before the number 6, since the number 6 is single-digit.

When dividing by 100, we used the method of moving the decimal point two digits to the left. In the answer 0.60, the zero after the number 6 was retained. But if you do this division with a corner, the zero disappears - you get the answer 0.6

We must remember that decimals 0.60 and 0.6 are equal to the same value:

0,60 = 0,6

In the same “corner” you can continue the division indefinitely, each time adding a zero to the remainder, but this will be a meaningless action:

You can express percentages as a decimal fraction not only by dividing by 100, but also by multiplying. The percentage symbol (%) itself replaces the 0.01 multiplier. And if we take into account that the number of percents and the percent sign are written together, then between them there is an “invisible” multiplication sign (×).

So, the 45% entry actually looks like this:

Replace the percent sign with a factor of 0.01

This multiplication by 0.01 is performed by moving the decimal point two digits to the left:

Problem 1. The family budget is 75 thousand rubles per month. Of these, 70% is money earned by dad. How much did mom earn?

Solution

The total is 100 percent. If dad earned 70% of the money, then mom earned the remaining 30% of the money.

Problem 2. The family budget is 75 thousand rubles per month. Of these, 70% is money earned by dad, and 30% is money earned by mom. How much money did each person make?

Solution

We will find 70 and 30 percent of 75 thousand rubles. This way we will determine how much money each person earned. For convenience, we write 70% and 30% as decimal fractions:

75 × 0.70 = 52.5 (thousand rubles dad earned)

75 × 0.30 = 22.5 (thousand rubles earned by mother)

Examination

52,5 + 22,5 = 75

75 = 75

Answer: 52.5 thousand rubles. Dad earned 22.5 rubles. Mom made money.

Problem 3. When cooling, bread loses up to 4% of its mass as a result of water evaporation. How many kilograms will evaporate when 12 tons of bread cools?

Solution

Let's convert 12 tons to kilograms. One ton contains a thousand kilograms, and 12 tons contains 12 times more:

1000 × 12 = 12,000 kg

Now let’s find 4% of 12000. The result obtained will be the answer to the problem:

12,000 × 0.04 = 480 kg

Answer: When 12 tons of bread cool, 480 kilograms will evaporate.

Problem 4. When dried, apples lose 84% of their mass. How many dried apples will you get from 300 kg of fresh ones?

Let's find 84% of 300 kg

300: 100 × 84 = 252 kg

300 kg of fresh apples will lose 252 kg of their mass as a result of drying. To answer the question how many dried apples you will get, you need to subtract 252 from 300

300 − 252 = 48 kg

Answer: from 300 kg of fresh apples you will get 48 kg of dried ones.

Problem 5. Soybean seeds contain 20% oil. How much oil is contained in 700 kg of soybeans?

Solution

Let's find 20% of 700 kg

700 × 0.20 = 140 kg

Answer: 700 kg of soybeans contain 140 kg of oil

Problem 6. Buckwheat contains 10% proteins, 2.5% fats and 60% carbohydrates. How many of these products are contained in 14.4 kg of buckwheat?

Solution

Let's convert 14.4 centners to kilograms. There are 100 kilograms in one centner, 14.4 times more in 14.4 centners

100 × 14.4 = 1440 kg

Let's find 10%, 2.5% and 60% of 1440 kg

1440 × 0.10 = 144 (kg proteins)

1440 × 0.025 = 36 (kg fat)

1440 × 0.60 = 864 (kg carbohydrates)

Answer: 14.4 kg of buckwheat contains 144 kg of protein, 36 kg of fat, 864 kg of carbohydrates.

Problem 7. Schoolchildren collected 60 kg of oak, acacia, linden and maple seeds for the tree nursery. Acorns made up 60%, maple seeds 15%, linden seeds 20% of all seeds, and the rest were acacia seeds. How many kilograms of acacia seeds were collected by schoolchildren?

Solution

Let's take oak, acacia, linden and maple seeds as 100%. Let us subtract from these 100% the percentages expressing the seeds of oak, linden and maple. This is how we find out what percentage of acacia seeds are:

100% − (60% + 15% + 20%) = 100% − 95% = 5%

Now we find acacia seeds:

60 × 0.05 = 3 kg

Answer: Schoolchildren collected 3 kg of acacia seeds.

Examination:

60 × 0.60 = 36

60 × 0.15 = 9

60 × 0.20 = 12

60 × 0.05 = 3

36 + 9 + 12 + 3 = 60

60 = 60

Problem 8. A man bought groceries. Milk costs 60 rubles, which is 48% of the cost of all purchases. Determine the total amount of money spent on groceries.

Solution

This is a task of finding a number by its percentage, that is, by its known part. This problem can be solved in two ways. The first is to express a known number of percents as a decimal fraction and find the unknown number from this fraction

Express 48% as a decimal

48% : 100 = 0,48

Knowing that 0.48 is 60 rubles, we can determine the amount of all purchases. To do this, you need to find the unknown number by decimal fraction:

60: 0.48 = 125 rubles

This means that the total amount of money spent on groceries is 125 rubles.

The second way is to first find out how much money is per one percent, then multiply the result by 100

48% is 60 rubles. If we divide 60 rubles by 48, we will find out how many rubles account for 1%

60: 48% = 1.25 rubles

1% accounts for 1.25 rubles. The total is 100 percent. If we multiply 1.25 rubles by 100, we get the total amount of money spent on products

1.25 × 100 = 125 rubles

Problem 9. Fresh plums yield 35% dried plums. How many fresh plums do you need to get 140 kg of dried plums? How many dried plums will you get from 600 kg of fresh ones?

Solution

Let's express 35% as a decimal fraction and find the unknown number using this fraction:

35% = 0,35

140: 0.35 = 400 kg

To get 140 kg of dried plums, you need to take 400 kg of fresh ones.

Let's answer the second question of the problem - how many dried plums will you get from 600 kg of fresh ones? If 35% of dried plums come out of fresh plums, then it is enough to find these 35% of 600 kg of fresh plums

600 × 0.35 = 210 kg

Answer: to get 140 kg of dried plums, you need to take 400 kg of fresh ones. From 600 kg of fresh plums you will get 210 kg of dried ones.

Problem 10. The absorption of fats by the human body is 95%. Over the course of a month, the student consumed 1.2 kg of fat. How much fat can his body absorb?

Solution

Convert 1.2 kg to grams

1.2 × 1000 = 1200 g

Let's find 95% of 1200 g

1200 × 0.95 = 1140 g

Answer: 1140 g of fat can be absorbed by the student's body.

Expressing numbers as percentages

Percentage, as mentioned earlier, can be expressed as a decimal fraction. To do this, just divide the number of these percentages by 100. For example, imagine 12% as a decimal fraction:

Comment. We now do not find the percentage of something, but simply write it as a decimal fraction.

But it is also possible reverse process. A decimal fraction can be represented as a percentage. To do this, you need to multiply this fraction by 100 and put a percent sign (%)

Let's represent the decimal fraction 0.12 as a percent

0.12 × 100 = 12%

This action is called expressing a number as a percentage or expressing numbers in hundredths.

Multiplication and division are inverse operations. For example, if 2 × 5 = 10, then 10: 5 = 2

In the same way, division can be written as reverse order. If 10: 5 = 2, then 2 × 5 = 10:

The same thing happens when we express a decimal as a percentage. So, 12% was expressed as a decimal as follows: 12: 100 = 0.12 but then the same 12% was “returned” using multiplication, writing the expression 0.12 × 100 = 12%.

Similarly, you can express any other numbers, including integers, as percentages. For example, let’s express the number 3 as a percentage. Multiply this number by 100 and add a percentage sign to the result:

3 × 100 = 300%

Large percentages like 300% can be confusing at first because people are used to thinking of 100% as the maximum percentage. From additional information about fractions, we know that one whole object can be denoted by one. For example, if there is a whole uncut cake, then it can be designated by 1

The same cake can be designated as 100% cake. In this case, both one and 100% will mean the same whole cake:

Let's cut the cake in half. In this case, the unit will turn into decimal number 0.5 (since it is half of one), and 100% becomes 50% (since 50 is half of a hundred)

We will return the whole cake, one unit and 100%

Let's depict two more such cakes with the same notations:

If one cake is a unit, then three cakes are three units. Each cake is 100% whole. If you add up these three hundred you get 300%.

Therefore, when converting whole numbers to percentages, we multiply these numbers by 100.

Problem 2. Express the number 5 as a percentage

5 × 100 = 500%

Problem 3. Express the number 7 as a percentage

7 × 100 = 700%

Problem 4. Express the number 7.5 as a percentage

7.5 × 100 = 750%

Problem 5. Express the number 0.5 as a percentage

0.5 × 100 = 50%

Problem 6. Express the number 0.9 as a percentage

0.9 × 100 = 90%

Example 7. Express the number 1.5 as a percentage

1.5 × 100 = 150%

Example 8. Express the number 2.8 as a percentage

2.8 × 100 = 280%

Problem 9. George is walking home from school. In the first fifteen minutes he walked 0.75 of the way. The rest of the time he walked the remaining 0.25 of the way. Express the percentage of the distance traveled by George.

Solution

0.75 × 100 = 75%

0.25 × 100 = 25%

Problem 10. John was treated to half an apple. Express this half as a percentage.

Solution

Half an apple is written as a fraction 0.5. To express this fraction as a percentage, multiply it by 100 and add a percent sign to the result.

0.5 × 100 = 50%

Analogues in the form of fractions

A value expressed as a percentage has its analogue in the form common fraction. So, the analogue for 50% is the fraction. Fifty percent can also be called "half".

The equivalent for 25% is a fraction. Twenty-five percent can also be called a quarter.

The equivalent for 20% is a fraction. Twenty percent can also be referred to as a fifth.

The analogue for 40% is a fraction.

The analogue for 60% is a fraction

Example 1. Five centimeters is 50% of a decimeter, or just half. In all cases we are talking about the same value - five centimeters out of ten

Example 2. Two and a half centimeters is 25% of a decimeter or just a quarter

Example 3. Two centimeters is 20% of a decimeter or

Example 4. Four centimeters is 40% of a decimeter or

Example 5. Six centimeters is 60% of a decimeter or

Decrease and increase in interest

When increasing or decreasing a value expressed as a percentage, the preposition “to” is used.

Examples:

Increase by 50% means increase the value by 1.5 times;
Increase by 100% means increase the value by 2 times;
Increase by 200% means increase by 3 times;
Reduce by 50% means reduce the value by 2 times;
Reduce by 80% means reduce by 5 times.

Example 1. Ten centimeters increased by 50%. How many centimeters did you get?

To solve such problems, you need to take the initial value as 100%. The original value is 10 cm. 50% of them are 5 cm

The original 10 cm was increased by 50% (by 5 cm), which means it turned out to be 10+5 cm, that is, 15 cm

The equivalent of increasing ten centimeters by 50% is a multiplier of 1.5. If you multiply 10 cm by it you get 15 cm

10 × 1.5 = 15 cm

Therefore, the expressions “increase by 50%” and “increase by 1.5 times” mean the same thing.

Example 2. Five centimeters increased by 100%. How many centimeters did you get?

Let's take the original five centimeters as 100%. One hundred percent of these five centimeters will be 5 cm themselves. If you increase 5 cm by the same 5 cm, you will get 10 cm

The analogue of an increase of five centimeters by 100% is a factor of 2. If you multiply 5 cm by it, you get 10 cm

5 × 2 = 10 cm

Therefore, the expressions “increase by 100%” and “increase by 2 times” mean the same thing.

Example 3. Five centimeters increased by 200%. How many centimeters did you get?

Let's take the original five centimeters as 100%. Two hundred percent is two times one hundred percent. That is, 200% of 5 cm will be 10 cm (5 cm for every 100%). If you increase 5 cm by these 10 cm, you get 15 cm

The equivalent of increasing five centimeters by 200% is a factor of 3. If you multiply 5 cm by it, you get 15 cm

5 × 3 = 15 cm

Therefore, the expressions “increase by 200%” and “increase by 3 times” mean the same thing.

Example 4. Ten centimeters reduced by 50%. How many centimeters are left?

Let's take the original 10 cm as 100%. Fifty percent of 10 cm is 5 cm. If you reduce 10 cm by these 5 cm, you will be left with 5 cm

An analogue of reducing ten centimeters by 50% is divisor 2. If you divide 10 cm by it, you get 5 cm

10: 2 = 5 cm

Therefore, the expressions “reduce by 50%” and “reduce by 2 times” mean the same thing.

Example 5. Ten centimeters were reduced by 80%. How many centimeters are left?

Let's take the original 10 cm as 100%. Eighty percent of 10 cm is 8 cm. If you reduce 10 cm by this 8 cm, you will be left with 2 cm

An analogue of reducing ten centimeters by 80% is divisor 5. If you divide 10 cm by it, you get 2 cm

10:5 = 2 cm

Therefore, the expressions “reduce by 80%” and “reduce by 5 times” mean the same thing.

When solving problems involving decreasing and increasing percentages, you can multiply/divide the value by the factor specified in the problem.

Problem 1. By what percentage did the value change if it increased by 1.5 times?

The value referred to in the problem can be designated as 100%. Next, multiply this 100% by a factor of 1.5

100% × 1.5 = 150%

Now from the received 150% we subtract the original 100% and get the answer to the problem:

150% − 100% = 50%

Problem 2. By what percentage did the value change if it decreased by 4 times?

This time the value will decrease, so we will perform division. Let us denote the value mentioned in the problem as 100%. Next, divide this 100% by a divisor of 4

From the initial 100%, subtract the resulting 25% and get the answer to the problem:

100% − 25% = 75%

This means that when the value decreases by 4 times, it decreases by 75%.

Problem 3. By what percentage did the value change if it decreased by 5 times?

Let us denote the value mentioned in the problem as 100%. Next, divide this 100% by a divisor of 5

From the initial 100%, subtract the resulting 20% and get the answer to the problem:

100% − 20% = 80%

This means that when the value decreases by 5 times, it decreases by 80%.

Problem 4. By what percentage did the value change if it decreased by 10 times?

Let us denote the value mentioned in the problem as 100%. Next, divide this 100% by a divisor of 10

From the initial 100%, subtract the resulting 10% and get the answer to the problem:

100% − 10% = 90%

This means that when the value decreases by 10 times, it decreases by 90%.

Problem finding a percentage

To express something as a percentage, you first need to write down a fraction showing what part the first number is of the second, then divide in this fraction and express the resulting result as a percentage.

For example, let there be five apples. In this case, two apples are red, three are green. Let's express red and green apples as a percentage.

First you need to find out what part are red apples. There are five apples in total, and two red ones. This means that two out of five or two fifths are red apples:

There are three green apples. This means that three out of five or three fifths are green apples:

We have two fractions and . Let's do the division in these fractions

We received decimal fractions 0.4 and 0.6. Now let's express these decimal fractions as a percentage:

0.4 × 100 = 40%

0.6 × 100 = 60%

This means that 40% are red apples, 60% are green.

And all five apples make up 40%+60%, that is, 100%

Problem 2. My mother gave my two sons 200 rubles. My mother gave my younger brother 80 rubles, and my older brother 120 rubles. Express as a percentage the money given to each brother.

Solution

The younger brother received 80 rubles out of 200 rubles. We write the fraction eighty two hundredths:

The elder brother received 120 rubles out of 200 rubles. We write the fraction one hundred twenty two hundredths:

We have fractions and . Let's do the division in these fractions

Let us express the results obtained as percentages:

0.4 × 100 = 40%

0.6 × 100 = 60%

This means that the younger brother received 40% of the money, and the older brother received 60%.

Some fractions that show what part the first number is of the second can be reduced.

This is how the fractions could be reduced. This would not change the answer to the problem:

Problem 3. The family budget is 75 thousand rubles per month. Of these, 52.5 thousand rubles. - money earned by dad. 22.5 thousand rubles. - money earned by mom. Express as a percentage the money earned by mom and dad.

Solution

This task, like the previous one, is a task of finding a percentage.

Let's express the money dad earned as a percentage. He earned 52.5 thousand rubles out of 75 thousand rubles

Let's do the division in this fraction:

0.7 × 100 = 70%

This means that dad earned 70% of the money. Further, it is not difficult to guess that the remaining 30% of the money was earned by my mother. After all, 75 thousand rubles is 100% money. Let's check to be sure. Mom earned 22.5 thousand rubles. out of 75 thousand rubles. We write down the fraction, perform the division and express the result as a percentage:

Problem 4. A schoolboy is training to do pull-ups on a bar. Last month he could do 8 pull-ups per set. This month he can do 10 pull-ups per set. By what percentage did he increase the number of pull-ups?

Solution

Let's find out how many more pull-ups a student does in the current month than in the past

Let's find out what part two pull-ups make up from eight pull-ups. To do this, find the ratio 2 to 8

Let's do the division in this fraction

Let's express the result as a percentage:

0.25 × 100 = 25%

This means that the student increased the number of pull-ups by 25%.

This problem can be solved in a second, more quick method— find out how many times 10 pull-ups are greater than 8 pull-ups and express the result as a percentage.

To find out how many times ten pull-ups are greater than eight pull-ups, you need to find the ratio of 10 to 8

Let's divide the resulting fraction

Let's express the result as a percentage:

1.25 × 100 = 125%

The pull-up rate for the current month is 125%. This statement must be understood exactly as "is 125%", not how “the indicator increased by 125%”. These are two different statements expressing different quantities.

The statement “is 125%” should be understood as “eight pull-ups that make up 100% plus two pull-ups that make up 25% of the eight pull-ups.” Graphically it looks like this:

And the statement “increased by 125%” should be understood as “to the current eight pull-ups, which were 100%, another 100% (8 more pull-ups) plus another 25% (2 pull-ups) were added.” That's a total of 18 pull-ups.

100% + 100% + 25% = 8 + 8 + 2 = 18 pull-ups

Graphically this statement looks like this:

In total it turns out to be 225%. If we find 225% of eight pull-ups, we get 18 pull-ups

8 × 2.25 = 18

Problem 5. Last month the salary was 19.2 thousand rubles. This month it amounted to 20.16 thousand rubles. By what percentage did the salary increase?

This problem, like the previous one, can be solved in two ways. The first is to first find out how many rubles the salary has increased. Next, find out what part of this increase is from last month’s salary

Let's find out how many rubles the salary increased:

20.16 − 19.2 = 0.96 thousand rubles.

Let's find out what part of 0.96 thousand rubles. ranges from 19.2. To do this, we find the ratio 0.96 to 19.2

Let's do the division in the resulting fraction. Along the way, let's remember:

Let's express the result as a percentage:

0.05 × 100 = 5%

This means that the salary increased by 5%.

Let's solve the problem in the second way. Let's find out how many times 20.16 thousand rubles. more than 19.2 thousand rubles. To do this, we find the ratio 20.16 to 19.2

Let's divide the resulting fraction:

Let's express the result as a percentage:

1.05 × 100 = 105%

The salary is 105%. That is, this includes 100%, which amounted to 19.2 thousand rubles, plus 5%, which amounted to 0.96 thousand rubles.

100% + 5% = 19,2 + 0,96

Problem 6. The price of a laptop has increased by 5% this month. What is its price if last month it cost 18.3 thousand rubles?

Solution

Let's find 5% of 18.3:

18.3 × 0.05 = 0.915

Let's add this 5% to 18.3:

18.3 + 0.915 = 19.215 thousand rubles.

Answer: the price of the laptop is 19,215 thousand rubles.

Problem 7. The price of a laptop has decreased by 10% this month. What is its price if last month it cost 16.3 thousand rubles?

Solution

Let's find 10% of 16.3:

16.3 × 0.10 = 1.63

Subtract this 10% from 16.3:

16.3 − 1.63 = 14.67 (thousand rubles)

Such tasks can be written briefly:

16.3 − (16.3 × 0.10) = 14.67 (thousand rubles)

Answer: The price of the laptop is 14.67 thousand rubles.

Problem 8. Last month the price of a laptop was 21 thousand rubles. This month the price increased to 22.05 thousand rubles. By what percentage did the price increase?

Solution

Let's determine how many rubles the price has increased

22.05 − 21 = 1.05 (thousand rubles)

Let's find out what part of 1.05 thousand rubles. is from 21 thousand rubles.

Let's express the result as a percentage

0.05 × 100 = 5%

Answer: laptop price increased by 5%

Problem 8. The worker was supposed to produce 600 parts according to the plan, but he produced 900 parts. To what percentage did he fulfill the plan?

Solution

Let's find out how many times more 900 parts are than 600 parts. To do this, find the ratio 900 to 600

The value of this fraction is 1.5. Let's express this value as a percentage:

1.5 × 100 = 150%

This means that the worker fulfilled the plan by 150%. That is, he completed it 100%, producing 600 parts. Then he made another 300 parts, which is 50% of the original plan.

Answer: the worker completed the plan by 150%.

Comparison of percentage values

We have already compared quantities many times in various ways. Our first tool was difference. So, for example, to compare 5 rubles and 3 rubles, we wrote down the difference 5−3. Having received the answer 2, one could say that “five rubles is two rubles more than three rubles.”

The answer obtained as a result of subtraction in everyday life is called not “difference”, but “difference”.

So, the difference between five and three rubles is two rubles.

The next tool we used to compare values was the ratio. The ratio allowed us to find out how many times the first number is greater than the second (or how many times the first number contains the second).

So, for example, ten apples are five times more than two apples. Or to put it another way, ten apples contain two apples five times. This comparison can be written using the relation

But the values can also be compared as percentages. For example, compare the price of two goods not in rubles, but evaluate how much the price of one product is more or less than the price of the other as a percentage.

To compare percentage values, one of them must be designated as 100%, and the second based on the conditions of the problem.

For example, let’s find out what percentage ten apples are more than eight apples.

100% is the value with which we compare something. We are comparing 10 apples to 8 apples. So, for 100% we denote 8 apples:

Now our task is to compare what percentage are 10 apples greater than these 8 apples. 10 apples is 8+2 apples. This means that by adding two more apples to eight apples, we will increase 100% by another number of percentages. To find out which one, let’s determine what percentage of eight apples is two apples

Adding this 25% to eight apples gives us 10 apples. And 10 apples are 8+2, that is, 100% and another 25%. Total we get 125%

This means that ten apples are 25% greater than eight apples.

Now let's solve the inverse problem. Let's find out how many percent eight apples are less than ten apples. The answer immediately suggests itself: eight apples are 25% smaller. However, it is not.

We are comparing eight apples to ten apples. We agreed that we will take for 100% what we compare with. Therefore, this time we take 10 apples for 100%:

Eight apples are 10−2, that is, by reducing 10 apples by 2 apples, we will reduce them by a certain percentage. To find out which one, let’s determine what percentage of ten apples is two apples

Subtracting this 20% from ten apples, we get 8 apples. And 8 apples are 10−2, that is, 100% and minus 20%. Total we get 80%

This means that eight apples are 20% less than ten apples.

Problem 2. By what percentage is 5,000 rubles more than 4,000 rubles?

Solution

Let's take 4000 rubles for 100%. 5 thousand is more than 4 thousand by 1 thousand. This means that by increasing four thousand by one thousand, we will increase four thousand by a certain amount of percent. Let's find out which one. To do this, we determine what part one thousand is from four thousand:

Let's express the result as a percentage:

0.25 × 100 = 25%

1000 rubles from 4000 rubles is 25%. If you add this 25% to 4000, you get 5000 rubles. This means that 5,000 rubles is 25% more than 4,000 rubles

Problem 3. What percentage is 4000 rubles less than 5000 rubles?

This time we compare 4000 with 5000. Let's take 5000 as 100%. Five thousand is more than four thousand by one thousand rubles. Find out what part one thousand is from five thousand

One thousand out of five thousand is 20%. If we subtract this 20% from 5,000 rubles, we get 4,000 rubles.

This means that 4000 rubles is less than 5000 rubles by 20%

Problems on concentration, alloys and mixtures

Let's say you want to make some juice. We have water and raspberry syrup at our disposal.

Pour 200 ml of water into a glass:

Add 50 ml of raspberry syrup and stir the resulting liquid. As a result, we will get 250 ml of raspberry juice (200 ml water + 50 ml syrup = 250 ml juice)

What part of the resulting juice is raspberry syrup?

Raspberry syrup makes up the juice. Let's calculate this ratio and get the number 0.20. This number shows the amount of dissolved syrup in the resulting juice. Let's call this number syrup concentration.

The concentration of a solute is the ratio of the amount of a solute or its mass to the volume of a solution.

Concentration is usually expressed as a percentage. Let's express the syrup concentration as a percentage:

0.20 × 100 = 20%

Thus, the concentration of syrup in raspberry juice is 20%.

Substances in solution may be heterogeneous. For example, mix 3 liters of water and 200 g of salt.

The mass of 1 liter of water is 1 kg. Then the mass of 3 liters of water will be 3 kg. Let's convert 3 kg to grams, we get 3 kg = 3000 g.

Now add 200 g of salt to 3000 g of water and mix the resulting liquid. The result will be a saline solution, the total mass of which will be 3000 + 200, that is, 3200 g. Let us find the concentration of salt in the resulting solution. To do this, find the ratio of the mass of dissolved salt to the mass of the solution

This means that when mixing 3 liters of water and 200 g of salt, you will get a 6.25% salt solution.

Similarly, the amount of a substance in an alloy or mixture can be determined. For example, an alloy contains tin weighing 210 g and silver weighing 90 g. Then the mass of the alloy will be 210 + 90, that is, 300 g. The alloy will contain tin and silver. The percentage of tin will be 70% and silver 30%

When two solutions are mixed, a new solution is obtained, consisting of the first and second solutions. The new solution may have a different concentration of the substance. A useful skill is the ability to solve problems involving concentration, alloys and mixtures. In general, the point of such tasks is to monitor the changes that occur when solutions of different concentrations are mixed.

Mix two raspberry juices. The first 250 ml juice contains 12.8% raspberry syrup. And the second juice, 300 ml, contains 15% raspberry syrup. Pour these two juices into a large glass and mix. As a result, we get a new juice with a volume of 550 ml.

Now let's determine the concentration of syrup in the resulting juice. The first 250 ml drained juice contained 12.8% syrup. And 12.8% of 250 ml is 32 ml. This means that the first juice contained 32 ml of syrup.

The second drained juice of 300 ml contained 15% syrup. And 15% of 300 ml is 45 ml. This means that the second juice contained 45 ml of syrup.

Let's add up the quantities of syrups:

32 ml + 45 ml = 77 ml

This 77 ml of syrup is contained in the new juice, which has a volume of 550 ml. Let's determine the concentration of syrup in this juice. To do this, find the ratio of 77 ml of dissolved syrup to the volume of juice 550 ml:

This means that when mixing 12.8% raspberry juice with a volume of 250 ml and 15% raspberry juice with a volume of 300 ml, the result is 14% raspberry juice with a volume of 550 ml.

Problem 1. There are 3 solutions of sea salt in water: the first solution contains 10% salt, the second contains 15% salt and the third contains 20% salt. Mix 130 ml of the first solution, 200 ml of the second solution and 170 ml of the third solution. Determine what percentage of sea salt is in the resulting solution.

Solution

Let's determine the volume of the resulting solution:

130 ml + 200 ml + 170 ml = 500 ml

Since the first solution contained 130 × 0.10 = 13 ml of sea salt, the second solution contained 200 × 0.15 = 30 ml of sea salt, and the third contained 170 × 0.20 = 34 ml of sea salt, then the resulting solution will contain contains 13 + 30 + 34 = 77 ml of sea salt.

Let us determine the concentration of sea salt in the resulting solution. To do this, find the ratio of 77 ml of sea salt to a solution volume of 500 ml

This means that the resulting solution contains 15.4% sea salt.

Problem 2. How many grams of water must be added to 50 g of a solution containing 8% salt to obtain a 5% solution?

Solution

Note that if water is added to the existing solution, the amount of salt in it will not change. Only its percentage will change, since adding water to the solution will lead to a change in its mass.

We need to add such an amount of water that eight percent salt becomes five percent.

Let's determine how many grams of salt are contained in 50 g of solution. To do this, find 8% of 50

50 g × 0.08 = 4 g

8% of 50 grams is 4 grams. In other words, eight parts out of a hundred equals 4 grams of salt. Let's make sure that these 4 grams come not from eight parts, but from five parts, that is, 5%

4 grams - 5%

Now knowing that there are 4 grams per 5% solution, we can find the mass of the entire solution. To do this you need:

4 g: 5 = 0.8 g
0.8 g × 100 = 80 g

80 grams of solution is the mass at which 4 grams of salt will be per 5% solution. And to get these 80 grams, you need to add 30 grams of water to the original 50 grams.

This means that to obtain a 5% salt solution, you need to add 30 g of water to the existing solution.

Problem 2. Grapes contain 91% moisture, and raisins - 7%. How many kilograms of grapes are required to produce 21 kilograms of raisins?

Solution

Grapes are composed of moisture and pure matter. If fresh grapes contain 91% moisture, then the remaining 9% will be the pure substance of this grape:

Raisins contain 93% pure substance and 7% moisture:

Note that in the process of turning grapes into raisins, only the moisture of these grapes disappears. The pure substance remains unchanged. After the grapes turn into raisins, the resulting raisins will have 7% moisture and 93% pure matter.

Let's determine how much pure substance is contained in 21 kg of raisins. To do this, we will find 93% of 21 kg

21 kg × 0.93 = 19.53 kg

Now let's go back to the first drawing. Our task was to determine how many grapes needed to get 21 kg of raisins. A pure substance weighing 19.53 kg will account for 9% of the grapes:

Now knowing that 9% pure substance is 19.53 kg, we can determine how many grapes are required to produce 21 kg of raisins. To do this, you need to find the number by its percentage:

19.53 kg: 9 = 2.17 kg
2.17 kg × 100 = 217 kg

This means that to obtain 21 kg of raisins you need to take 217 kg of grapes.

Problem 3. In an alloy of tin and copper, copper makes up 85%. How much alloy must be taken so that it contains 4.5 kg of tin?

Solution

If copper makes up 85% of the alloy, then the remaining 15% will be tin:

The question is how much alloy must be taken so that it contains 4.5 tin. Since the alloy contains 15% tin, 4.5 kg of tin will account for this 15%.

And knowing that 4.5 kg of alloy makes up 15%, we can determine the mass of the entire alloy. To do this, you need to find the number by its percentage:

4.5 kg: 15 = 0.3 kg
0.3 kg × 100 = 30 kg

This means that you need to take 30 kg of alloy so that it contains 4.5 kg of tin.

Problem 4. We mixed a certain amount of a 12% solution of hydrochloric acid with the same amount of a 20% solution of the same acid. Find the concentration of the resulting hydrochloric acid.

Solution

Let's depict the first solution in the figure as a straight line and highlight 12% on it.

Since the number of solutions is the same, the same figure can be drawn next to each other, illustrating the second solution with a hydrochloric acid content of 20%

We have two hundred parts of solution (100% + 100%), thirty-two parts of which are hydrochloric acid (12% + 20%)

Let's determine what part 32 parts make up of 200 parts

This means that when mixing a 12% solution of hydrochloric acid with the same amount of a 20% solution of the same acid, the result is a 16% solution of hydrochloric acid.

To check, imagine that the mass of the first solution was 2 kg. The mass of the second solution will also be 2 kg. Then, when mixing these solutions, you will get 4 kg of solution. In the first solution of hydrochloric acid there was 2 × 0.12 = 0.24 kg, and in the second - 2 × 0.20 = 0.40 kg. Then in the new solution of hydrochloric acid there will be 0.24 + 0.40 = 0.64 kg. The concentration of hydrochloric acid will be 16%

Problems to solve independently

on , we will find 60% of the number

Now let’s increase the number by the found 60%, i.e. per number

Answer: the new value is

Task 12. Answer the following questions:

1) Spent 80% of the amount. What percent of this amount is left?
2) Men make up 75% of all factory workers. What percentage of the plant's employees are women?
3) Girls make up 40% of the class. What percentage of the class are boys?

A Solution

Let's use a variable. Let P this is the original number referred to in the problem. Let's take this initial number P for 100%

Let's reduce this original number P by 50%

Now the new number is 50% of the original number. Find out how many times the original number is P more than the new number. To do this, find the ratio of 100% to 50%

The original number is twice the new one. This can be seen even from the drawing. And to make the new number equal to the original one, it needs to be doubled. And doubling a number means increasing it by 100%.

This means that the new number, which is half the original number, needs to be increased by 100%.

When considering the new number, it is also taken as 100%. So, in the above figure, the new number is half of the original number and is labeled as 50%. In relation to the original number, the new number is half. But if we consider it separately from the original one, it must be taken as 100%.

Therefore, in the figure, the new number, which is represented by a line, was first designated as 50%. But then we designated this number as 100%.

Answer: To get the original number, the new number must be increased by 100%.

Problem 16. Last month there were 15 accidents in the city.
This month this figure dropped to 6. By what percentage did the number of accidents decrease?

Solution

Last month there were 15 accidents. This month there are 6. This means that the number of accidents has decreased by 9.
Let's take 15 accidents as 100%. By reducing 15 road accidents by 9, we will reduce them by a certain percentage. To find out which one, we will find out what part of 9 accidents is from 15 accidents

Answer: the concentration of the resulting solution is 12%.

Problem 18. We mixed a certain amount of an 11% solution of a certain substance with the same amount of a 19% solution of the same substance. Find the concentration of the resulting solution.

Solution

The mass of both solutions is the same. Each solution can be taken as 100%. After adding the solutions, you get a 200% solution. The first solution contained 11% of the substance, and the second solution contained 19% of the substance. Then the resulting 200% solution will contain 11% + 19% = 30% of the substance.

Let us determine the concentration of the resulting solution. To do this, we find out what part thirty parts of a substance make up of two hundred parts of a substance:

1,10. This means that the price for the first month will be 1.10.

In the second month the price also increased by 10%. Add ten percent of this price to the current price of 1.10, we get 1.10 + 0.10 × 1.10. This amount is equal to the expression 1.21 . This means that the price for the second month will become 1.21.

In the third month the price also increased by 10%. Add ten percent of this price to the current price of 1.21, we get 1.21 + 0.10 × 1.21. This amount is equal to the expression 1.331 . Then the price for the third month will become 1.331.

Let's calculate the difference between the new and old prices. If the initial price was equal to 1, then it increased by 1.331 − 1 = 0.331. Let's express this result as a percentage, we get 0.331 × 100 = 33.1%

Answer: over 3 months, food prices increased by 33.1%.

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, a series of articles about personal finance.

Today we'll talk about interest.

It is impossible to invest without understanding what interest is and how profitability is calculated.

As a rule, there are no problems with simple interest; anyone who has ever kept money on deposit in a bank understands that, for example, the interest rate is 10% per annum on a deposit of 50,000 rubles. will give 5000 income per year.

It is more difficult to understand the effect of compound interest, but it is very important in long-term investing, i.e. when investments are made with the goal of achieving financial freedom.

In essence, with compound interest, interest income is reinvested, increasing the size of the deposit. Here's an example, let's say you have 100,000 rubles. and on them you receive 10% of income, i.e. 10,000 rub. in year.

In the first year you received 10,000 rubles. and your contribution increased by these 10,000, amounting to 110,000 rubles.

In the second year, your income will already be 10% of 110,000 rubles, i.e. 11,000 rubles, which you also add to the deposit, which becomes 110,000 + 11,000 = 121,000 rubles.

Third year: Your 121 thousand rubles again brings 10%, which is 12,100 rubles in rubles, and your contribution at the end of the third year will be 121,000 + 12,100 = 133,100 rubles.

Etc.

In formalized form, compound interest is written as follows:

FV = PV (1 + r)^n

Where F.V.– future value of the deposit;PV– initial cost of the deposit;r– rate of return (profitability);n– number of periods.

Well, check the formula using our example FV = 10,000 (1 + 0.1)^3 = 133,100 rubles. As you can see, everything came together :)

When you invest for the long term, then the importance of compound interest increases greatly.

Imagine this example: if milk goes up in price by 10% per year, how much will it cost in 20 years? If today milk costs 30 rubles per liter, then allowing the cost of milk to increase by 10% per year, in 20 years milk will cost FV = 30 (1+0.1)^20 = 201 rubles 82 kopecks!

This example, by the way, very well shows the need to invest and preserve one’s capital, since it also depreciates according to the compound interest formula.

This formula is also called the “Rothschild formula”, “the devil’s formula”, and in English and in financial circles it is called “compounding”.

Everything on earth changes according to the formula of compound interest: inflation, increased consumption of oil or wheat, the population of the earth changes, etc.

When you invest the percentage works for you, here is an exampleI mentioned earlier about pensions:

How much money will the average Russian be able to save if he invests 3,000 rubles? per month for 30 years? Let's assume that the growth of his investments will be 5% per year, and the return on investment will be equal to 17% per annum.

After 30 years, 32,022,812 rubles will have accumulated. This is how compound interest works for you, acting as a lever that increases your contribution.

But it also works against it when you take out loans, for example.

In principle, there are programs that allow you to calculate compound interest and the annuity formulas associated with them (an annuity is considered a series of payments that are the same (or change according to a pattern) and are spaced from each other for the same period of time; the example with the accumulation of 3,000 rubles in a month is also considered an annuity. month higher and equal monthly loan payments over time).

You can try it yourself, I use itlike this program for iPad , it's free, and they have options for Android as well.

The figure shows an example of calculating the amount of loan payments using this program.

There you can also try other financial calculations, for example, calculating compound interest and annuities.

Try it, the main thing is to understand the principle itself.