Topic: Finding a part from a whole and a whole from its part

Target: Systematize, expand, generalize and consolidate the acquired knowledge on the topic “Finding a part of a whole and a whole by its part. Computer science among us"
Tasks:
Activate students' knowledge of fraction concepts and solving fraction problems.
Teach students to solve problems on a topic, be able to distinguish between ways to solve problems.
Application of acquired theoretical knowledge in solving practical problems.
Expand students' horizons in the field of computer science.
Stages of conducting a lesson.

Goal setting - 2 min.
Updating basic knowledge – 8 min.
Consolidation and generalization of material. – 23 min.
Summing up the lesson and setting homework. - 5 minutes.

Expected results: Students must learn to apply the necessary solution methods to a particular problem, must be able to solve problems, and be able to calculate fractions.

During the classes:

Organizing time. - 2 minutes.
Greetings students.
Goal setting – 2 min.
Guess the rebus.

What word is encrypted here? That's right, the Internet.
What topic are we studying now? (correct, “Finding a part from a whole and a whole from its part”)
How will the Internet be connected to this topic? (we will solve problems on this topic on knowledge of the Internet0
Who can formulate the topic of today's lesson? (The Internet is among us)
Do you know what the Internet is? (They express their versions)
Internet - (from the Latin inter - between and net - network), a global computer network that connects both users of computer networks and users of individual (including home) computers.
Updating of reference knowledge– 8 min.
Do orally:
A) Find the part of the number:
3/4 of 16;
2/5 of 80;
7/10 from 120;
3/5 of 150;
6/11 from 121;
5/6 from 108

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

Consolidation and generalization of material. – 23 min.
Where and when do you think the Internet appeared? (express opinions)
In 1957, after the Soviet Union launched the first artificial Earth satellite, the US Department of Defense considered that in case of war the United States needed a reliable information transmission system. The US Defense Advanced Research Projects Agency proposed developing a computer network for this purpose.

Now we will solve several problems.

Alena has 140 photos uploaded to her personal page on the Odnoklassniki website. 2/7 of all photos are uploaded to the “Personal Photos” album, 1/4 to the “Hobby” album, 3/35 to the “Recreation” album, 5/28 to the “Family” album, and the rest to the “On photos of friends." How many photos does Alena have in each album?
140: 7 * 2 = 40 (f) “Personal photos”
140: 4 * 1 =35 (f) “Hobby”
140: 35 * 3 =12 (f) “Rest”
140: 28 * 5 = 25 (f) “Family”
140 – 40 – 35 – 12 – 25=28 (f) “In the photo of friends”

Misha has 276 letters in his email, which is 3/5 of the number of letters in Kolya’s email. How many more letters does Kolya have than Misha?
276: 3 * 5 = 460
460 – 276 = 184.

On a flash card designed for 4G bytes (1G byte = 1024 M bytes) there are various files. Photos occupy 3/16 of the total memory, movies - 1/8 of the total memory) more than photos, text documents - 5/64 of the total memory more than photos. How many M bytes are there for each file?
4 * 1024 = 4096
4096: 16 *3 =768(M bytes) in the photo
4096: 8 * 1=512
768 + 512 = 1280 (M bytes) for movies
4096: 64 *5 = 320
320 +768 = 1088 (M bytes) for text documents.

Guys, why do you need the Internet?
Communication;
Information;
Games.
What social networks do you know? (express their opinion)
Let's name the pros and cons of social networks:
"Pros":
Communication;
Information.
"Minuses":
Negative impact on health;
Internet is an addiction;
Immersion in the virtual world;
Danger from strangers.

Let's solve the following problem.

A survey was taken among 5th grade students at one of the schools on the topic “Social networks and children.” To the question “How much time a day do you spend on the Internet”, 3/10 of all schoolchildren surveyed answered “5 – 6 hours”. How many schoolchildren spend this time on the Internet every day if 150 children participated in the survey?
150: 10 * 3 =45 (children).
45 children! This is a very large number! After all, every day they waste so much time sitting at the computer.
Guys, what do you think can be harmful to health from spending a long time on the Internet?
Possible student answers:
Deterioration of vision;
Decreased physical activity;
Psychological stress;
The person loses the ability to communicate;
Rachiocampsis;
Headache;
Sleep disturbance.

You see how much negative stuff you can earn by sitting on the Internet for several hours!

5. Summing up the lesson and setting homework. - 5 minutes.
What new did you learn in class today?
What do you think is the optimal time to spend on the Internet every day?
What will you mainly use the Internet for?
Do you think that 5–6 hours on the Internet every day is the norm?
Homework: prepare a message on the topic “History of the Internet”
Announcement of grades.
Thank you for the lesson!

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.

The rule for finding a number by its fraction:

To find a number from a given value of its fraction, you need to divide this value by the fraction.

Let's look at how to find a number by its fraction, using specific examples.

Examples.

1) Find a number whose 3/4 are equal to 12.

To find a number by its fraction, divide the number by that fraction. To do this, you need to multiply this number by the inverse of the fraction (that is, by an inverted fraction). To do this, you need to multiply the numerator by this number and leave the denominator unchanged. 12 and 3 by 3. Since we got one in the denominator, the answer is an integer.

2) Find a number if 9/10 of it equals 3/5.

To find a number given the value of its fraction, divide this value by this fraction. To divide a fraction by a fraction, multiply the first fraction by the inverse of the second (inverted). To multiply a fraction by a fraction, multiply the numerator by the numerator, and the denominator by the denominator. We reduce 10 and 5 by 5, 3 and 9 by 3. As a result, we get the correct irreducible fraction, which means this is the final result.

3) Find a number whose 9/7 are equal

To find a number by the value of its fraction, divide that value by that fraction. Mixed number and multiply it by the inverse of the second number (inverted fraction). We reduce 99 and 9 by 9, 7 and 14 by 7. Since we received an improper fraction, we need to separate the whole part from it.

BASIC TYPES OF SOLVING PERCENTAGE PROBLEMS

I. FINDING A PART OF THE WHOLE

To find a part (%) of a whole, you need to multiply the number by the part (percent converted to a decimal fraction).

EXAMPLE: There are 32 students in the class. During the test, 12.5% ​​of students were absent. Find how many students were absent?
SOLUTION 1: The integer in this problem is the total number of students (32).
12,5% = 0,125
32 · 0.125 = 4
SOLUTION 2: Let x students be absent, which is 12.5%. If 32 students –
total number of students (100%), then
32 students – 100%
x students – 12.5%

ANSWER: There were 4 students missing from the class.

II. FINDING THE WHOLE BY ITS PART

To find a whole from its part (%), you need to divide the number by the part (percents converted to a decimal fraction).

EXAMPLE: Kolya spent 120 crowns in the amusement park, which amounted to 75% of all his pocket money. How much pocket money did Kolya have before coming to the amusement park?
SOLUTION 1: In this problem you need to find the whole if the given part and value are known
this part.
75% = 0,75
120: 0,75 = 160

SOLUTION 2: Let Kolya have x crowns, which is a whole, i.e. 100%. If he spent 120 crowns, which was 75%, then
120 CZK – 75%
x CZK – 100%

ANSWER: Kolya had 160 crowns.

III. EXPRESSION AS A PERCENTAGE OF THE RATIO OF TWO NUMBERS

SAMPLE QUESTION:
WHAT % IS ONE VALUE FROM ANOTHER?

EXAMPLE: The width of the rectangle is 20m and the length is 32m. What % is the width of the length? (Length is the basis for comparison)
SOLUTION 1:

SOLUTION 2: In this problem, the length of a rectangle of 32m is 100%, then the width of 20m is x%. Let's compose and solve the proportion:
20 meters – x%
32 meters – 100%

ANSWER: The width is 62.5% of the length.

NB! Notice how the solution changes as the question changes.

EXAMPLE: The width of the rectangle is 20m and the length is 32m. What % is the length of the width? (Width is the basis for comparison)
SOLUTION 1:

SOLUTION 2: In this problem, the width of a rectangle of 20m is 100%, then the length of 32m is x%. Let's compose and solve the proportion:
20 meters – 100%
32 meters – x%

ANSWER: The length is 160% of the width.

IV. EXPRESSION AS PERCENTAGE OF CHANGE IN QUALITY

SAMPLE QUESTION:
BY HOW MUCH % DID THE INITIAL VALUE CHANGE (INCREASED, DECREASED)?

To find the change in value in % you need to:
1) find how much the value has changed (without %)
2) divide the resulting value from step 1) by the value that is the basis for comparison
3) convert the result to % (by multiplying by 100%)

EXAMPLE: The price of the dress has decreased from 1250 CZK to 1000 CZK. Find by what percentage the price of the dress has decreased?
SOLUTION 1:


2) The basis for comparison here is 1250 CZK (i.e. what it was originally)
3)

ANSWER: The price of the dress has decreased by 20%.

NB! Notice how the solution changes as the question changes.

EXAMPLE: The price of the dress increased from 1000 CZK to 1250 CZK. Find by what percentage the price of the dress has increased?
SOLUTION 1:

1) 1250 –1000= 250 (kr) how much the price has changed
2) The basis for comparison here is 1000 CZK (i.e. what it was originally)
3)
Solving a problem in one step:

SOLUTION 2:
1250 –1000= 250 (cr) how much the price has changed
In this problem, the initial price of 1000 kroner is 100%, then the change in price of 250 kroner is x%. Let's compose and solve the proportion:
1000 CZK – 100%
250 CZK – x%

x =
ANSWER: The price of the dress has increased by 25%.

V. CONSEQUENTIAL CHANGE OF QUANTITY (NUMBER)

EXAMPLE: The number was reduced by 15% and then increased by 20%. Find by what percentage the number has changed?

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

SOLUTION 1:
1) Although the original number is not given, for ease of solution it can be taken as 100 (i.e. one integer or 1)
2) If the number is decreased by 15%, then the resulting number will be 85%, or from 100 it would be 85.
3) Now the result obtained must be increased by 20%, i.e.
85 – 100%
and the new number x is 120% (since it has increased by 20%)

x =
4) Thus, as a result of the changes, the number 100 (original) changed and became 102, which means that the original number increased by 2%

SOLUTION 2:
1) Let the initial number X
2) If the number decreased by 15%, then the resulting number will be 85% of X, i.e. 0.85X.
3) Now the resulting number must be increased by 20%, i.e.
0.85Х – 100%
what about the new number? – 120% (since increased by 20%)

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

ANSWER: The number increased by 2%.

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)

Formulation 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 parts ?

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 parts ?

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 the 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 600 nuts in total. 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 was 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).