So, let us be given some integer a. We need to find half of this number. You can do this with ordinary fractions:
- Let's denote the integer as one, then half of the unit is 1/2. So we need to find 1/2 of the number a.
- To find 1/2 of the number a, we must multiply the number a by the part we need to find, that is, perform the action: a * 1/2 = a/2. That is, half of the number a is a / 2.
- Moreover, if we are looking for a part of an integer, then the result will be less than the original number.
There may be different tasks on finding a part of the whole: if you need to find, for example, a quarter of the number a, then you need a * 1/4 = a/4. If you want to find 1/8 of the number a, then you need a * 1/8 = a/8. Finding any part of a whole is done by multiplying the given integer by the part you want to find.
Consider an example.
How to find the third part of the number 75
We are given an integer - the number 75. We need to find the third part of it, otherwise we need to find 1/3. Let's perform the action of multiplying the whole by the part: 75 * 1/3 = 25. So the third part of the number 75 is the number 25. You can also say this: the number 25 is three times less than the number 75. Or: the number 75 is three times greater than the number 25.
§ 20. Finding a part of the whole and the whole but its parts - Mathematics Textbook Grade 5 (Zubareva, Mordkovich)
Short description:
It happens that we need to find some part of a number, for example, peel only a third of a potato from a certain number. Or vice versa, when we are told that only a quarter of the class came on an excursion, we need to find out what is the total number of students in the class. Knowing the whole, you can find some given part of it, in the same way, knowing the part, you can determine what the whole was. You will learn about this today from this paragraph of the textbook. The definition of a part of a whole, and vice versa, is directly related to simple fractions, which you have already studied. Actions in this case do not occur with two numbers, which are denoted by a fraction, but with one fraction and one integer. For example, finding 1/2 of 16 would mean multiplying 16 by 1/2, in which case the denominator of 16 = 1 and the expression can be written as: 1/2 16/1 = 16/2 = 8.
To find an integer by its part, we use the reverse method, and multiply the known number by the inverted fraction (that is, divide by it). In another way, this can be explained as follows: in order to find a whole from its part, you need the known number that corresponds to its part, divide by the numerator and multiply by the denominator of the fraction that denotes this part (which is the action of dividing the fraction, or multiplying to an inverted fraction - you can remember the most convenient way for you to solve such problems). Thus, to find an integer whose 3/4 are equal to 12, you need 12: 3/4 = 12 4/3 = 48/3 = 16. Or method number 2, which removes unnecessary mathematical operations - the number x, 2/5 from which are 20: x = 20: 2 5 = 50.
Test yourself with the tasks from the textbook and do not forget to review the material to better master and remember it!
Lesson topic:"Finding a part of a whole and a whole by its part."
The purpose of the lesson:
- Learn how to find a fraction from a number and a number from its fraction.
- Generalize the concept of an ordinary fraction and actions with ordinary fractions.
Equipment: Multimedia projector, Power Point presentation ( Application ).
DURING THE CLASSES
I. Organizational moment
Students are seated in groups (5-6 people). You can suggest diagnosing your mood at the stages of the lesson. Each student is given a card on which he highlights the "character" of his mood.
II. Knowledge update
We are already familiar with the concept of an ordinary fraction.
What does the numerator of a fraction show? (Into how many parts the whole is divided).
What does the denominator of a fraction show? (How many parts did you take).
- Look at the picture and answer the questions:
Students are encouraged to reproduce it.
III. Verbal counting. (Best Counter)
Each team on the screen is offered a task. The teams take turns doing the task.
1st team
2nd team
3rd team
4th team
The result is summed up - which team is the best counter.
IV. Dictation
Dictation is carried out with subsequent self-examination. It is possible to perform a carbon copy, students hand over one copy to the teacher for verification.
1. Instead of x, insert the missing number:
2. Reduce the fraction:
3. Arrange fractions in descending order:
4. Follow the steps:
5. Giant turtles live on the islands of the Pacific Ocean. They are of such size that children can ride while sitting on their shell. The following task will help us find out the name of the largest turtle in the world.
After submitting the solution, students check the answers.
V. New material
The teacher offers to solve problems (5-7 minutes are given for their thinking)
1. There were 12 birds sitting on a branch. Then they flew away. How many birds have flown?
2. In your class in mathematics for the third quarter, 6 people received a mark of "5". This is the number of all students in the class. How many students are in the class?
Then the solution, which is shown on the slide, is checked.
1 way: 12: 3 2 = 8 (birds)
2 way: 12 = 8 (birds)
2 task. 6: = 6 = 34 (people)
The teacher draws attention to the fact that two types of tasks can be distinguished:
1. To find part of a number, expressed as a fraction, you need this number multiply for this fraction.
2. To find number by its frequency and, expressed as a fraction, you need divide to this fraction the number corresponding to it.
Students are encouraged to memorize this rule right in class and in pairs retell each other.
The teacher focuses on the following: for those who find it difficult to determine the type of task, I advise you to pay attention to prepositions what , this is . These prepositions are found in the problems of finding numbers by its fraction.
VI. Fixing new material
There are six tasks on the slide and students are asked to sort them into two columns by type.
1. The store accepted 156 kg of fish for sale. 1/3 of all fish was carp. How many kg of carp did the shop receive?
2. Conducted 18 experiments, this amounted to 2/9 of the entire series of experiments. How many experiments should be done?
3. The teacher checked 20 notebooks. This amounted to 4/5 of all notebooks. How many notebooks does a teacher need to check?
4. Out of 72 fifth-graders, 3/8 go in for athletics. How many students are involved in this sport?
5. 30 paintings were selected for the exhibition. This amounted to 2/3 of the paintings in the museum. How many paintings are in the exhibition?
6. From a rope, 18 m long, cut off 3/4 of its length. How many meters of rope are left?
VII. Lesson summary
The teacher returns students to the goal of the lesson, suggests highlighting two types of tasks for fractions and algorithms for solving them. Collected leaflets with mood diagnostics.
VIII. Homework: P. 9.6, No. 1050, 1058, 1060.
§ 20. Finding a part of the whole and the whole but its parts - Mathematics Textbook Grade 5 (Zubareva, Mordkovich)
Short description:
It happens that we need to find some part of a number, for example, peel only a third of a potato from a certain number. Or vice versa, when we are told that only a quarter of the class came on an excursion, we need to find out what is the total number of students in the class. Knowing the whole, you can find some given part of it, in the same way, knowing the part, you can determine what the whole was. You will learn about this today from this paragraph of the textbook.
The definition of a part of a whole, and vice versa, is directly related to simple fractions, which you have already studied. Actions in this case do not occur with two numbers, which are denoted by a fraction, but with one fraction and one integer. For example, finding 1/2 of 16 would mean multiplying 16 by 1/2, in which case the denominator of 16 = 1 and the expression can be written as: 1/2 16/1 = 16/2 = 8.
To find an integer by its part, we use the reverse method, and multiply the known number by the inverted fraction (that is, divide by it). In another way, this can be explained as follows: in order to find a whole from its part, you need the known number that corresponds to its part, divide by the numerator and multiply by the denominator of the fraction that denotes this part (which is the action of dividing the fraction, or multiplying to an inverted fraction - you can remember the most convenient way for you to solve such problems). Thus, to find an integer whose 3/4 are equal to 12, you need 12: 3/4 = 12 4/3 = 48/3 = 16. Or method number 2, which removes unnecessary mathematical operations - the number x, 2/5 from which are 20: x = 20: 2 5 = 50.
Test yourself with the tasks from the textbook and do not forget to review the material to better master and remember it!
§ 1 Rules for finding a part from a whole and a whole from its part
In this lesson, we will formulate the rules for finding a part from a whole and a whole by its part, and also consider solving problems using these rules.
Consider two tasks:
How many kilometers did the tourists walk on the first day if the entire tourist route is 20 km long?
Find the length of the entire journey of the tourists.
Let's compare these tasks - in both the whole path is taken as a whole. In the first problem, the integer 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 whole by its part. The value of 20 km known in the first problem is unknown in the second problem, and vice versa, the known value of 8 km in the second problem must be found in the first problem. Such problems are called mutually inverse, since in them the known and sought values are interchanged.
Consider the first task:
The denominator 5 shows how many parts the whole was divided into, i.e. if the whole 20 is divided by 5, we find out how many kilometers is one part, 20: 5 \u003d 4 km. The numerator 2 shows that the tourists went 2 parts of the way, so 4 must be multiplied by 2, it will be 8 km. On the first day, the hikers walked 8 km.
It turned out the expression 20: 5 ∙ 2 = 8.
Let's move on to the second task.
Therefore, one part will be equal to the quotient 8 and 2, it will turn out 4, the denominator is 5, which means there are 5 parts in total.
Multiply 4 by 5, you get 20. The answer is 20 km, the length of the entire journey.
Let's write the expression: 8: 2 ∙ 5 = 20
Using the meaning of multiplying and dividing a number by a fraction, the rules for finding a part of a whole and a whole by its part can be formulated as follows:
To find a part of a whole, you need to multiply the number corresponding to the whole by the fraction corresponding to this part;
to find a whole by its part, you need to divide the number corresponding to this part into the corresponding part of the fraction.
Accordingly, the solution of problems can now be written in a different way:
for the first task 20 ∙ 2/5 = 8 (km),
for the second task 8: 2/5 = 20 (km).
In order to avoid difficulties, we write the solution of such problems as follows:
Whole: all the way, known - 20 km.
Answer: 8 km.
Whole: all the way - unknown.
Answer: 20 km.
§ 2 Algorithm for solving problems for finding the whole by its part and part of the whole
Let us compose an algorithm for solving such problems.
First, let's analyze the condition and the question of the problem: find out what is the whole, whether it is known or not, then find out how a part of the whole is represented and what needs to be found.
If it is necessary to find a part of the whole, then we multiply the whole by the fraction corresponding to this part, if it is necessary to find the whole by its part, then the number corresponding to the part is divided by the fraction corresponding to this part. As a result, we get an expression. Next, we find the value of the expression and write down the answer, after reading the question of the problem again before that.
So, before solving such problems, it is necessary to answer the following questions:
What value is taken as an integer?
Is this value known?
What is required to find: a part of a whole or a whole in its part?
Let's summarize: in this lesson you got acquainted with the rules for finding a part from a whole and a whole from its part, and also learned how to solve problems according to these rules.
List of used literature:
- Maths. Grade 6: lesson plans for the textbook by I.I. Zubareva, A.G. Mordkovich // author-compiler L.A. Topilin. Mnemosyne, 2009.
- Maths. Grade 6: a textbook for students of educational institutions. I.I. Zubareva, A.G. Mordkovich.- M.: Mnemozina, 2013.
- Maths. Grade 6: textbook for educational institutions / G.V. Dorofeev, I.F. Sharygin, S.B. Suvorov and others/ edited by G.V. Dorofeeva, I.F. Sharygin; Russian Academy of Sciences, Russian Academy of Education, Moscow: Education, 2010.
- Maths. Grade 6: textbook. for general education institutions /N.Ya. Vilenkin, V.I. Zhokhov, A.S. Chesnokov, S.I. Schwarzburd. - M.: Mnemosyne, 2013.
- Maths. Grade 6: textbook / G.K. Muravin, O.V. Ant. – M.: Bustard, 2014.
