Main types of interest challenge

I. Finding a part of the whole

To find part (%) from the whole, the number is necessary to multiply into part (interest translated into the decimal fraction).

EXAMPLE: In class 32 student. During test work There was no 12.5% \u200b\u200bof students. Find how many students were absent?
Solution 1: Inteid in this problem - the total number of students (32).
12,5% = 0,125
32 · 0.125 \u003d 4
Solution 2: Let x students were absent, which is 12.5%. If 32 student -
total number of students (100%), then
32 student - 100%
x Pupils - 12.5%

ANSWER: There were no 4 student in the class.

II. Finding a whole by its part

To find an integer in its part (%), it is necessary to divide the number to part (interest translated into a decimal fraction).

EXAMPLE: Kohl I spent 120 kroons in the amusement park, which amounted to75% of all his pocket money. How much did your pocket money from Kohl before coming to the amusement park?
Solution 1:In this task, you need to find an integer if this part and value is known.
of this part.
75% = 0,75
120: 0,75 = 160

Solution 2: Let x of the crowns were in the case, which is a whole, that is, 100%. If he spent 120 kroons, which amounted to 75%, then
120 kroons - 75%
x crowns - 100%

ANSWER:Kolya had 160 kroons.

III. Expression as a percentage of the relationship of two numbers

Typical question:
How much% is one value from another?

EXAMPLE: The width of the rectangle is 20m, and the length is 32m. How much% is the width of the length? (Length is the basis for comparison)
Solution 1:

Solution 2: In this problem, the length of the rectangle 32m is 100%, then the width of 20m is x%. We will make and solve the proportion:
20 meters - x%
32 meters - 100%

ANSWER: The width is from the length of 62.5%.

NB! Pay attention to how the solution changes depending on the change in the question.

EXAMPLE: The width of the rectangle is 20m, and the length is 32m. How much% is the length of the width? (Width is the basis for comparison)
Solution 1:

Solution 2: In this problem, the width of the rectangle 20m is 100%, then the length of 32m is x%. We will make and solve the proportion:
20 meters - 100%
32 meters - x%

ANSWER: The length is from the width of 160%.

IV. Expression as a percentage of changes in magnitude

Typical question:
How much has changed (increased, decreased) the initial value?

To find a change in% in%:
1) Find as much as the value (without%)
2) divide the resulting value from claim 1) by the value that is the basis for comparison
3) Translate the result in% (by performing multiplication by 100%)

EXAMPLE: The price of the dress fell from 1250 kroons to 1000 kroons. Find how much percent declined the price of the dress?
Solution 1:


2) the basis for comparison here is 1250 kroons (that is, what was originally)
3)

Answer: The dress price has decreased by 20%.

NB! Pay attention to how the solution changes depending on the change in the question.

EXAMPLE: The price of the dress rose from 1000 kroons to 1250 kroons. Find how much percent increased the price of the dress?
Solution 1:

1) 1250 -1000 \u003d 250 (cr) the price has changed so much
2) the basis for comparison here is 1000 kroons (i.e. what was originally)
3)
Solving the problem with one action:

Solution 2:
1250 -1000 \u003d 250 (CR) the price has changed so much
In this problem, the initial price of 1000 kroons is 100%, then changing the price of 250 kroons is x%. We will make and solve the proportion:
1000 CZK - 100%
250 kroons - x%

x \u003d
ANSWER: Dress price increased by 25%.

V. Sequential change in size (number)

EXAMPLE: The number was reduced by 15%, and then increased by 20%. Find how much percent changed the number?

The most common error: the number increased by 5%.

Solution 1:
1) although the initial number is not given, it is possible to take it for 100 for simplicity (i.e. one or 1)
2) If the number decreased by 15%, the resulting number will be 85%, or from 100 it would be 85.
3) now the result obtained should be increased by 20%, i.e.
85 – 100%
a new number x - 120% (because increased by 20%)

x \u003d
4) Thus, as a result of the changes, the number 100 (initial) has changed and has become 102, and this means that the initial number increased by 2%

Solution 2:
1) Let the initial number x
2) if the number decreased by 15%, the resulting number will be 85% of x, i.e. 0.85x.
3) now the resulting number should be increased by 20%, i.e.
0.85х - 100%
and the new number? - 120% (because increased by 20%)

? =
4) Thus, as a result of the changes, the number X (initial) is the basis for comparison, and the number 1.02x (obtained), (see IV type of solving tasks), then

ANSWER: The number increased by 2%.

§ 1 Rules for finding a part of the whole and the whole of its part

In this lesson, we formulate the rules for finding a part of the whole and the whole to its part, as well as consider solving problems using these rules.

Consider two tasks:

How many kilometers took place tourists on the first day, if the entire tourist route is 20 km.?

Find the length of the entire path of tourists.

Compare these tasks - in both of the whole, the whole path is accepted. In the first task, the whole is known - 20 km, and in the second - unknown. In the first task, it is necessary to find a part of the whole, and in the second - the integer in its part. The value known in the first task is 20 km, unknown in the second task, and on the contrary, known in the second task - 8 km, in the first one must be found. Such tasks are called mutually reverse, as they are known and the desired values \u200b\u200bare changed in places.

Consider the first task:

The denominator 5 shows how many parts were divided, i.e. If the whole 20 is divided into 5, we learn how many kilometers is one part, 20: 5 \u003d 4 km. Numerator 2 shows that tourists have passed 2 parts of the path, which means 4 must be multiplied by 2, it will work out 8 km. On the first day, tourists took place 8 km.

It turned out expression 20: 5 ∙ 2 \u003d 8.

We turn to the second task.

Consequently, one part will be equal to private 8 and 2, it will be 4, denominator 5, which means the entire parts 5.

4 Multiply to 5, it will turn out 20. The answer is 20 km the length of the entire path.

We write the expression: 8: 2 ∙ 5 \u003d 20

Using the meaning of multiplying and dividing the number to the fraction, the rules for finding a part from the whole and the whole part of it can be formulated as follows:

To find a part of the whole, the number corresponding to the whole, multiply by the fraction corresponding to this part;

to find an integer in its part, it is necessary a number corresponding to this part, divide into the corresponding part of the fraction.

Accordingly, the task solution can be recorded now differently:

for the first task 20 ∙ 2/5 \u003d 8 (km),

for the second task 8: 2/5 \u003d 20 (km).

In order not to be difficult, the solution to such tasks is written as follows:

In order: all the way, it is known - 20 km.

Answer: 8 km.

A whole: all the way is unknown.

Answer: 20 km.

§ 2 Algorithm for solving problems on finding a whole according to its part and part of the whole

We will make an algorithm for solving such tasks.

First, we analyze the condition and question of the task: find out what is the whole, it is known or not, then find out how the part of the whole is presented and what to find.

If it is necessary to find a part of the whole, then the whole multiply by the fraction corresponding to this part, if you need to find an integer in its part, then the number corresponding to the part split into the fraction corresponding to this part. As a result, we get an expression. Next, we find the value of the expression and write the answer by reading before it once again the issue of the task.

So, before you solve such tasks, you need to answer the following questions:

What is the magnitude of pleasant for the whole?

Is this magnitude known?

What is required to find: part of the whole or integer by part of it?

Let us summarize: In this lesson, you got acquainted with the rules for finding a part of the whole and the whole of its part, and also learned how to solve problems on these rules.

List of references:

1. Mathematics. Grade 6: Pounding plans for the textbook I.I. Zubareva, A.G. Mordkovich // Author-compiler L.A. Topil. Mnemozina, 2009.
2. Mathematics. Grade 6: Textbook for students of general educational institutions. I.I. Zubareva, A.G. Mordkovich. - M.: Mnemozina, 2013.
3. Mathematics. Grade 6: Textbook for general education institutions / GV. Dorofeev, I.F. Sharygin, S.B. Suvorov et al. / Edited by G.V. Dorofeeva, I.F. Sharygin; Ros.Akad. Nauk, Ros.Akad.Formation, M.: Education, 2010.
4. Mathematics. Grade 6: studies. For general education. Institutions / N.I. Vilenkin, V.I. Zhokhov, A.S. Chesnokov, S.I. Schwartzbord. - M.: Mnemozina, 2013.
5. Mathematics. 6 cl.: Tutorial / G.K. Muravin, O.V. Moravin. - M.: Drop, 2014.

Topic: Finding a part of the whole and a whole to its part

Purpose: Systematize, expand, summarize and consolidate the knowledge gained on the topic "Finding a part of the whole and a whole part. Informatics among us "
Tasks:
Intensify the knowledge of students about the concepts of fraction, solving tasks on the fraction.
Teach students to solve problems on the topic, be able to distinguish ways to solve problems.
The use of the resulting theoretical knowledge in solving practical problems.
Expand the horizons of students in the field of computer science.
Stages of the lesson.

Goaling - 2 min.
Actualization of reference knowledge - 8 min.
Fastening and generalization of the material. - 23 min.
Summing up the lesson and setting homework. - 5 minutes.

Expected results: Students must learn to apply the necessary methods Solutions for a particular task must be able to solve problems, be able to perform fractions.

During the classes:

Organizing time. - 2 minutes.
Greetings students.
Goaling - 2 min.
Guess the rebus.

What word is encrypted here? True, the Internet.
What topic are we studying with you now? (right, "Finding a part of the whole and the whole of its part")
How will the Internet be associated with this topic? (We will solve the tasks on this topic on the knowledge of the Internet0
Who can formulate the topic of today's lesson? (Internet among us)
Do you know what is the Internet? (Discard their versions)
Internet - (from Lat. Inter - between and Net - network), a global computer network connecting both users of computer networks and users of individual (including domestic) computers.
Actualization of reference knowledge - 8 min.
Perform orally:
A) Find a part from the number:
3/4 of 16;
2/5 from 80;
7/10 from 120;
3/5 from 150;
6/11 from 121;
5/6 from 108.

B) Find the number if:
3/8 it is equal to 15;
2/5 of it is equal to 30;
5/8 of its 45;
4/9 it is equal to 36;
7/10 it is equal to 42;
2/11 It is equal to 99.

Fixing and generalization of the material. - 23 min.
What do you think, where and when did the Internet appear? (express opinions)
In 1957, after launch Soviet Union The first artificial satellite of the Earth, the US Department of Defense considered that in case of the US War I need reliable system Information transfer. Agency for promising defense research and development of the United States proposed to develop a computer network for this.

Now we will solve several tasks.

At Alena N. personal page On the site "Odnoklassniki" downloaded 140 photos. 2/7 of the number of all photos is loaded into the album "Personal photos", 1/4 - in the album "Hobbies", 3/35 - in the album "Rest", 5/28 - in the album "Family", and the rest - "on Photo of friends. " How many photos of Alena in each album?
140: 7 * 2 \u003d 40 (f) "Personal photos"
140: 4 * 1 \u003d 35 (f) "Hobby"
140: 35 * 3 \u003d 12 (f) "Rest"
140: 28 * 5 \u003d 25 (f) "Family"
140 - 40 - 35 - 12 - 25 \u003d 28 (f) "In the photo of friends"

Misha B. e-mail 276 letters, which is 3/5 from the number of letters in email. How many letters have more than Misha?
276: 3 * 5 = 460
460 – 276 = 184.

On the flash card designed for 4g byte (1g byte \u003d 1024 m bytes) there are various files. Photo occupy 3/16 all memory, movies - on 1/8 part (from all over memory) more than a photo, text documents - on 5/64 part (from all over memory) more than a photo. How many m bytes fall on each of the files?
4 * 1024 = 4096
4096: 16 * 3 \u003d 768 (m byte) in the photo
4096: 8 * 1=512
768 + 512 \u003d 1280 (m byte) on movies
4096: 64 *5 = 320
320 +768 \u003d 1088 (m byte) on text documents.

Guys, why do you need the Internet?
Communication;
Information;
Games.
What are you known social networks? (express your opinion)
Let's call "pros" and "minuses" of social networks:
"Pros":
Communication;
Information.
"Minuses":
Negative impact on health;
Internet - addiction;
Immersion in the virtual world;
Danger from strangers.

Let's decide the following task.

Among students of 5 classes, one of the schools passed the survey on the topic "Social Schools and Children". The question "How much time a day you spend on the Internet", 3/10 of the number of all the surveyed schoolchildren answered "5 - 6 hours." How many schoolchildren spend this time on the Internet daily, if 150 children participated in the survey?
150: 10 * 3 \u003d 45 (children).
45 children! This is very big number! After all, every day they spend so much time wasted, sitting at the computer.
Guys, how do you think, how harm to health can be long-term pastime on the Internet?
Possible replies of students:
Impairment;
Reduction of motor activity;
Psychological overvoltage;
A person loses the ability to communicate;
Rachiocampsis;
Headaches;
Sleep disturbance.

Here you see how much the negative you can earn, sitting for a few hours on the Internet!

5. Summing up the lesson and housekeeping. - 5 minutes.
What's new you learned today in class?
What do you think is optimal for spending on the Internet daily?
What will you basically use the Internet?
Do you think that 5 - 6 hours on the Internet every day - is it a norm?
Homework: Prepare a message on the topic "The history of the occurrence of the Internet"
Announcement of estimates.
Thank you for the lesson!

§ 20. Speak of parts from the whole and a whole, but its part is a textbook on mathematician Grade 5 (Zubareva, Mordkovich)

Short description:

It happens that we need to find some part of the number, for example, from a certain number of potatoes only to be cleaned about it. Or, on the contrary, when we are told that only a quarter of a class has come to an excursion, we need to find out what the total number of class disciples. Knowing the whole, you can find some kind of specified part from it, in the same way, knowing the part, you can determine which there was an integer. About this today you will learn from this paragraph of the textbook.
The definition of a part of the whole, and on the contrary, directly related to simple fractionsthat you have already studied. Actions in this case occur not with two numbers that are designated, but with one fraction and one integer. For example, to find 1/2 of 16 will mean multiplying 16 per 1/2, in this case the denominator of the number 16 \u003d 1 and the expression can be written as: 1/2 16/1 \u003d 16/2 \u003d 8.
To find an integer in its part, we use the inverse method, and multiply the known number on the inverted fraction (that is, we divide it). Otherwise, this can be explained as follows: In order to find an integer from its part, it is necessary that the known number that corresponds to its part, divide into the numerator and multiplied by the denomoter, which denotes this part (which is the action of fractions, or multiplication On an inverted fraction - you can remember the most convenient way for you in solving such tasks). Thus, to find an integer, 3/4 of which are 12, you need 12: 3/4 \u003d 12 4/3 \u003d 48/3 \u003d 16. Or Method No. 2, which removes extra mathematical actions - the number X, 2/5 From which 20: x \u003d 20: 2 5 \u003d 50.
Check yourself when performing tasks from the textbook and do not forget to see the material to better master it and remember!

§ 20. Speak of parts from the whole and a whole, but its part is a textbook on mathematician Grade 5 (Zubareva, Mordkovich)

Short description:

It happens that we need to find some part of the number, for example, from a certain number of potatoes only to be cleaned about it. Or, on the contrary, when we are told that only a quarter of a class has come to an excursion, we need to find out what the total number of class disciples. Knowing the whole, you can find some kind of specified part from it, in the same way, knowing the part, you can determine which there was an integer. About this today you will learn from this paragraph of the textbook.
The definition of a part of the whole, and vice versa, directly connected with the simple fractions that you have already studied. Actions in this case occur not with two numbers that are designated, but with one fraction and one integer. For example, to find 1/2 of 16 will mean multiplying 16 per 1/2, in this case the denominator of the number 16 \u003d 1 and the expression can be written as: 1/2 16/1 \u003d 16/2 \u003d 8.
To find an integer in its part, we use the inverse method, and multiply the known number on the inverted fraction (that is, we divide it). Otherwise, this can be explained as follows: In order to find an integer from its part, it is necessary that the known number that corresponds to its part, divide into the numerator and multiplied by the denomoter, which denotes this part (which is the action of fractions, or multiplication On an inverted fraction - you can remember the most convenient way for you in solving such tasks). Thus, to find an integer, 3/4 of which are 12, you need 12: 3/4 \u003d 12 4/3 \u003d 48/3 \u003d 16. Or Method No. 2, which removes extra mathematical actions - the number X, 2/5 From which 20: x \u003d 20: 2 5 \u003d 50.
Check yourself when performing tasks from the textbook and do not forget to see the material to better master it and remember!