Open lesson on mathematics in grade 5b.
Teacher: Bambutova M.I.
Topic: How to find a part of a whole and a whole from its part.
Goal: learn to solve problems of finding a part from a whole and a whole from its part.
Educational: derive a rule for finding a part from a whole and a whole from its part,
solve problems of finding a part from a whole and a whole from its part.
Educational: develop memory and mathematical speech
Educational: develop communication skills.
Lesson plan:
1).Introductory and motivational stage.
1. Org. Moment
2. Updating basic knowledge
Answer the questions (slide)
1) What does a fraction mean?
2) What does a fraction mean? 
?
3) 
Statement of the problem:
1 task:
2 tasks per slide
1) draw a rectangle with sides 2 cm and 5 cm. What is its area?
Solve the problem
1) The area of the rectangle is 10 cm 2. Parts of the rectangle's area are shaded. What is the area of the shaded part of the rectangle?
2) The shaded part of the rectangle is equal to 4 cm 2, which is part of the entire rectangle. What is the area of the rectangle?
Answer the questions: ( )
part of the whole , and in which the whole according to its part ?
What do we find in task 1 (the whole by its part), what do we find in task 2 (part of the whole)
Task 2: Read the tasks and answer the questions:
1) Field area – 50 hectares. During the day, a team of tractor drivers plowed the fields. How many hectares did the team plow in a day?
2) During the day, the team plowed 20 hectares, which was the area of the entire field. What is the area of the field?
Answer the questions: ( distribute tasks in the form of cards)
What quantity is taken as an integer in each problem?
In which of the problems is this quantity known and in which is it not?
Which problem requires finding part of the whole , and in which the whole according to its part ?
What are these tasks? (reciprocal)
What do these tasks have in common? What were we looking for in these tasks?
-Part of the whole And whole according to its part.
So what is our topic today? ?
Topic: How to find a part of a whole and a whole from its part .(slide)
The correct solution to the last two problems is found in the textbook on page 95.
Now we have solved 4 problems, generalize all the problems and derive a rule for finding a part from a whole and a whole from its part.
Students try, to help them, random word combinations need to be assembled into a logically correct sentence, which will be the rule.
which expresses this part.
corresponding to the whole,
To find a part of the whole,
divide by the denominator
and multiply the result by the numerator of the fraction
I need a number
To find a part of a whole, you need to divide the number corresponding to the whole by the denominator and multiply the result by the numerator of the fraction that expresses this part.
and multiply the result by the denominator of the fraction,
I need a number
divide by the numerator
which expresses this part.
To find the whole from its part,
corresponding to this part,
To find a whole from its part, you need to divide the number corresponding to this part by the numerator and multiply the result by the denominator of the fraction that expresses this part.
Collect this rule on the board.
Students recite this rule to each other.
3. Primary consolidation. Game “Sorting tasks”.
Problem solving workshop. Option 1 solves problems of finding a part of a whole, option 2 solves problems of finding a whole from its part.
1. There are 80 students in the choir, ¼ of them are boys. How many boys are there in the choir?
2. There are 20 boys in the choir, which is ¼ of all students in the choir. How many students are there in the choir?
3. A small deciduous forest purifies the air from 70 tons of dust per year. And coniferous forest is ½ of this amount. How much dust does a coniferous forest filter out per year?
4. 7/12 of the kerosene that was there was poured out of the barrel. How many liters of kerosene were in the barrel if 84 liters were poured out of it?
5. The girl skied 300 m, which was 3/8 of the entire distance. What is the distance?
6. Cleared snow from 2/5 of the skating rink, which is 200 sq.m. Find the area of the entire skating rink?
7. The girl read ¾ of the book, which is 120 pages. How many pages are in the book?
8. The squirrel prepared a total of 600 nuts. In the first week she collected 20% of all nuts. How much did the squirrel collect in the first week?
9. Find the number X, 1/8 of which is equal to 1/24.
10. The girl collected 40 plums, which amounted to 1/3 of all plums. How many plums were collected in total?
11. Mom bought 6 kg of sweets. Vitya immediately ate 2/3 of all the candies and felt sick. After how many sweets did Vitya have a stomach ache?
12. The boy collected 80 nuts, which is 2/3 of all the collected nuts. How many nuts were collected?
13. There were 40 chickens in the chicken coop. In a week, the fox carried away 3/8 of all the chickens. How many chickens did the fox take?
14. Alice fell into a fairy well and flew 90 m in 1 minute. What is the depth of the well if Alice flew ¾ of the entire distance in 1 minute?
15. Before the ball, the stepmother gave Cinderella a lot of work. It took Cinderella 6 hours to complete 3/5 of this work. How long will it take Cinderella to complete all the work?
4. Reflection. The rule is to speak it out.
5. Homework: learn the rule, make a card with tasks for finding a part of a whole and a whole from its part (3 tasks for each rule).
§ 20. Finding a part of a whole and a whole but its part - Textbook on Mathematics, grade 5 (Zubareva, Mordkovich)
Brief description:
It happens that we need to find some part of a number, for example, from a certain number of potatoes we need to peel only a third of it. Or vice versa, when we are told that only a quarter of the class came on an excursion, we need to find out what the total number of students in the class is. Knowing the whole, you can find some given part of it, and in the same way, knowing the part, you can determine what the whole was like. You will learn about this today from this paragraph of the textbook. Determining a part of a whole, and vice versa, is directly related to the simple fractions that you have already studied. In this case, actions occur not 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 a whole number from 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 to divide the known number that corresponds to its part by the numerator and multiply by the denominator of the fraction that denotes this part (which is the action of dividing a 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 is equal to 12, you need 12: 3/4 = 12 4/3 = 48/3 = 16. Or method No. 2, which removes unnecessary mathematical operations - number x, 2/5 from which they are equal to 20: x = 20: 2 5 = 50.
Test yourself when completing 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.”
Objective of the lesson:
- Learn to find a fraction from a number and a number from its fraction.
- Generalize the concept of a common fraction and operations with common fractions.
Equipment: Multimedia projector, Power Point presentation ( Application ).
PROGRESS OF THE LESSON
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 identifies the “character” of his mood.
II. Updating knowledge
We are already familiar with the concept of a common fraction.
– What does the numerator of a fraction show? (How many parts is the whole divided into?)
– What does the denominator of a fraction show? (How many parts did they take).
– Look at the picture and answer the questions:
Students are asked to reproduce it.
III. Oral counting. (Best counter)
Each team is given a task on the screen. Teams take turns completing the task.
1st team
2nd team
3rd team
4th team
The bottom line is which team is the best counter.
IV. Dictation
The dictation is carried out followed by self-test. It is possible to make a carbon copy; students submit one copy to the teacher for checking.
1. Instead of x, insert the missing number:
2. Reduce a fraction:
3. Arrange the fractions in descending order:
4. Follow these steps:
5. Giant turtles live on the islands of the Pacific Ocean. They are so big 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 their answers.
V. New material
The teacher offers to solve problems (5 – 7 minutes are given to think about them)
1. 12 birds were sitting on a branch. Then it flew away from them. How many birds flew away?
2. In your math class, 6 people received a grade of “5” in the third quarter. This is the number of all students in the class. How many students are in the class?
Then the solution is checked and shown on the slide.
Method 1: 12: 3 2 = 8 (birds)
Method 2: 12 = 8 (birds)
Task 2. 6: = 6 = 34 (persons)
The teacher draws attention to the fact that two types of tasks can be distinguished:
1. To find part of the number, expressed as a fraction, you need this number multiply for this fraction.
2. To find number according to its frequency and, expressed as a fraction, you need divide for this fraction the number corresponding to it.
Students are asked to memorize this rule in class and retell it to each other in pairs.
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 . These prepositions are found in problems of finding numbers by their fraction.
VI. Consolidating new material
On the slide there are six problems 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 were carp. How many kg of carp did the store receive?
2. We carried out 18 experiments, this amounted to 2/9 of the entire series of experiments. How many experiments should be carried out?
3. The teacher checked 20 notebooks. This amounted to 4/5 of all notebooks. How many notebooks does a teacher need to check?
4. Of the 72 fifth-graders, 3/8 are involved in athletics. How many students play this sport?
5. 30 paintings were selected for the exhibition. This amounted to 2/3 of the paintings available in the museum. How many paintings were taken to the exhibition?
6. From a rope 18 m long, 3/4 of its length was cut off. How many meters of rope are left?
VII. Lesson Summary
The teacher returns students to the purpose of the lesson and suggests identifying two types of fraction problems and algorithms for solving them. Leaflets with mood diagnostics are collected.
VIII. Homework: P. 9.6, No. 1050, 1058, 1060.
§ 20. Finding a part of a whole and a whole but its part - Textbook on Mathematics, grade 5 (Zubareva, Mordkovich)
Brief description:
It happens that we need to find some part of a number, for example, from a certain number of potatoes we need to peel only a third of it. Or vice versa, when we are told that only a quarter of the class came on an excursion, we need to find out what the total number of students in the class is. Knowing the whole, you can find some given part of it, and in the same way, knowing the part, you can determine what the whole was like. You will learn about this today from this paragraph of the textbook.
Determining a part of a whole, and vice versa, is directly related to the simple fractions that you have already studied. In this case, actions occur not 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 a whole number from 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 to divide the known number that corresponds to its part by the numerator and multiply by the denominator of the fraction that denotes this part (which is the action of dividing a 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 is equal to 12, you need 12: 3/4 = 12 4/3 = 48/3 = 16. Or method No. 2, which removes unnecessary mathematical operations - number x, 2/5 from which they are equal to 20: x = 20: 2 5 = 50.
Test yourself when completing tasks from the textbook and do not forget to review the material to better master and remember it!
So, let us be given some integer a. We need to find half of this number. This can be done using ordinary fractions:
- Let us denote the whole as one, then half of one 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 that 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 a whole number, then the result will be less than the original number.
There may be different tasks on finding a part of a whole: if you need to find, for example, a quarter of the number a, then you need a * 1/4 = a/4. If you need 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 that needs to be found.
Let's look at 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 a whole by a part: 75 * 1/3 = 25. This means that the third part of the number 75 is the number 25. We 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.
