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

Target: To systematize, expand, generalize and consolidate the acquired knowledge on the topic “Finding a part from a whole and a whole by its part. Informatics among us»
Tasks:
Activate students' knowledge about the concepts of fractions, solving problems on fractions.
To teach students to solve problems on a topic, to be able to distinguish between ways to solve problems.
Application of the obtained theoretical knowledge in solving practical problems.
Expand students' horizons in the field of computer science.
Stages of the lesson.

Goal setting - 2 min.
Updating of basic knowledge - 8 min.
Consolidation 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 of solving a particular problem, must be able to solve problems, be able to calculate fractions.

During the classes:

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

What word is encoded 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 related to this topic? (we will solve problems on this topic on knowledge of the Internet0
Who can formulate the topic of today's lesson? (Internet among us)
Do you know what the Internet is? (Give their version)
Internet - (from 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 basic knowledge– 8 min.
Perform 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 are 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 are equal to 42;
2/11 of it is equal to 99.

Consolidation and generalization of the material. – 23 min.
Where and when do you think the Internet appeared? (express opinions)
In 1957, after the launch of the first artificial Earth satellite by the Soviet Union, the US Department of Defense decided that in case of war, the US needed a reliable information transmission system. The US Defense Advanced Research Projects Agency has offered to develop a computer network for this.

Now we will solve several problems.

Alena has 140 photos uploaded on her personal page on the Odnoklassniki website. 2/7 of the total number of photos uploaded to the Personal Photos album, 1/4 to the Hobby album, 3/35 to the Rest album, 5/28 to the Family album, and the rest to Na photo 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 \u003d 12 (f) "Rest"
140: 28 * 5 = 25 (f) "Family"
140 - 40 - 35 - 12 - 25 \u003d 28 (f) "On the photo of friends"

Misha has 276 emails, which is 3/5 of the number of emails 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 bytes = 1024 M bytes), there are various files. Photos take up 3/16th of the total memory, movies - 1/8th (of the total memory) more than photos, text documents - 5/64th (of the total memory) more than photos. How many M bytes are in 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, what do you need the Internet for?
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 - addiction;
Immersion in the virtual world;
Danger from strangers.

Let's solve the following problem.

Among the 5th grade students of one of the schools, a survey was conducted on the topic “Social networks and children”. To the question "How much time per day do you spend on the Internet", 3/10 of all surveyed schoolchildren 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 spend so much time wasted sitting at the computer.
Guys, what do you think, what harm to health can cause a long pastime on the Internet?
Possible student responses:
visual impairment;
Decreased motor activity;
Psychological overstrain;
The person loses the ability to communicate;
Rachiocampsis;
Headaches;
Sleep disturbance.

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

5. Summing up the lesson and setting homework. - 5 minutes.
What new did you learn at the lesson today?
What do you think is the best time to spend on the Internet every day?
What do you mainly use the internet for?
Do you think that 5-6 hours on the Internet every day is the norm?
Homework: prepare a report on the topic "The History of the Internet"
Announcement of grades.
Thank you for the lesson!

Lesson topic: Finding the whole from its part.

Target: develop the skill of oral counting, develop logical thinking,

develop the ability to work independently and in a group,

cultivate interest in mathematics, cultivate a sense of friendship and

mutual understanding, to cultivate love for the native land.

During the classes.

1. Organizational moment. (Slide number 1, 2)

Long awaited call

The lesson starts.

2. Oral account.

Let's think!

a) Lyuda and Nadya bought a bun each at the buffet, but Lena forgot to take the money with her. Then Lyuda and Nadya gave Lena 1/2 loaf each. Who got more buns? (Lena got a whole bun, and Luda and Nadia got half each) (Slide #3)

b) The hedgehog has 3 whole apples, 10 halves, 8 quarters. How many apples does the hedgehog have? (The hedgehog has 10 apples) (Slide number 4)

c) A snail moves along a vertical column 6 m high. During the day it rises by 4 m, and during the night it falls by 3 m. How many days does it take for a snail to reach the top? (3 days) (Slide number 5)

d) How many centimeters:

1/4 m, 3/5 m, 6/10 m. (25 cm, 60 cm, 60 cm)

How many meters:

1/5 km, 4/5 km, 7/10 km. (200m, 800m, 700m) (Slide number 6)

e) What part of the segment AB is the segment SD. Find the length of segment AB if segment CD is 5 cm (A

(Slide number 7)

3. Work with a new theme.

a) 1/8 segment AB - 8 mm. Draw line segment AB.

8*8=64mm=6cm 4mm (Slide number 8)

e) The cake costs 160 rubles. It was cut into 4 pieces. How much will 1/4 of the cost. You and two of your friends came to the cafe. How much money will you pay if everyone eats one piece of cake?

Solution (160:4=40(r.) costs 1 piece, 40*3=120(r.) must be paid (Slide number 9, 10)

Fizminutka(Slide number 11)

c) M.d. 1/2 hour, 1/3 hour, 1/4 hour, 1/10 hour. (30min, 20min, 15min, 6min) (Slide number 12)

d) Problem solving

The length of the Don River in the Voronezh region is 530 km. This is 1/3 of the entire length of the Don River. Find the length of the Don River.

Solution: (530*3=1590 (km) the length of the Don River) (Slide number 13, 14)

Birch lives 240 years. This is 1/5 of the life of a blue spruce. How many years does a blue spruce live.

240 * 5 \u003d 1200 (l) w - blue spruce lives (Slide number 15, 16, 17 )

Fizminutka (Slide number 18)

4. Consolidation of what has been learned.

Task number 227. (Slide number 19)

We bought 5 coils of electrical wire, 56 meters each. We used up 2/7 of the entire wire. How many meters of wire are left?

Solution: (56*5=280m - total wire, 280:7*2=80m - used up, 280-80= 200(m) - wire left)

5.Repetition of the past

a) Task No. 231. (independent work) (Slide number 20)

Lemons were laid out in baskets, 100 pieces in each. How many lemons were there if 15 baskets were filled and 30 more lemons were left?

Solution: (100 * 15 + 30 \u003d 1530 (l) - was)

b) Division with remainder. No. 229 (verification) (Slide number 21)

76:8=9 (rest.4) 8*9+4=76,

54:11=4 (rest 10) 4*11+10=54

612:7=87 (rest.3) 87 *7+3=612

793:6= 132 (rest 1) 132*6+1=793

939:4 =234 (rest.3) 234 *4+3=939

c) Problem number 228. (Slide number 22)

For 3 hours of work, the bulldozer leveled 234 square meters of the road. How many square meters of road will the bulldozer level in 10 hours if it works with the same productivity?

Solution: (234:3=78 - in 1 hour, 78* 10=780 - in 10 hours)

6. Group work in rows

Problem solving (by cards)

6 candies is 1/7 of all candies. How many candies?

8 candies is 1/3 of all candies. How many candies?

3 candies make up 1/8 of all candies. How many candies

Share all the sweets with all the students in our class. How many candies will each get?

Solution (6*7=42, 8*3=24, 3*8=24, 42+24+24=90, 90:18=5)

7. Summary of the lesson (Slide number 23)

By what action do we find the whole from its part? (multiplication)

What is the action to find a part of a whole number (division)

8. Homework: pp. 48. No. 229, 228. (Slide number 24)

The lesson was prepared by a primary school teacher MOU secondary school No. 21

MAIN TYPES OF SOLVING PROBLEMS FOR INTEREST

I. FINDING THE PART OF THE WHOLE

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

EXAMPLE: There are 32 students in the class. During the control work, 12.5% ​​of students were absent. Find how many students were missing?
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 -
the 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 IN ITS PART

To find a whole by its part (%-s), you need to divide the number by a part (percentages 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 joining 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 an integer, i.e. 100%. If he spent 120 crowns, which was 75%, then
120 CZK - 75%
x kroons - 100%

ANSWER: Kolya had 160 crowns.

III. EXPRESSION IN PERCENTAGE OF THE RATIO OF TWO NUMBERS

MODEL 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 the 32m rectangle is 100%, then the width of 20m is x%. 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 20m rectangle is 100%, so the length of 32m is x%. Compose and solve the proportion:
20 meters - 100%
32 meters - x%

ANSWER: The length is 160% of the width.

IV. EXPRESSION IN PERCENTAGE OF VALUE CHANGE

MODEL QUESTION:
WHAT % HAS THE INITIAL VALUE CHANGED (INCREASED, DECREASED)?

To find the change in % you need:
1) find how much the value has changed (without %)
2) divide the value obtained from point 1) by the value that is the basis for comparison
3) translate the result into% (by multiplying by 100%)

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


2) The basis for comparison here is 1250 crowns (i.e. what 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 has increased from 1000 crowns to 1250 crowns. Find the percentage increase in the price of the dress?
SOLUTION 1:

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

SOLUTION 2:
1250 –1000= 250 (cr) how much the price has changed
In this problem, the initial price of 1000 crowns is 100%, then the price change of 250 crowns is x%. Compose and solve the proportion:
1000 CZK - 100%
250 CZK - x %

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

V. CONSECUTIVE CHANGE OF VALUE (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 simplicity of the solution, you can take it as 100 (i.e. one integer or 1)
2) If the number has 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% (because it 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 original number X
2) If the number has 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.
0.85X - 100%
what about the new number? – 120% (because it 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%.

The rule for finding a number by its fraction:

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

Consider how to find a number by its fraction, using specific examples.

Examples.

1) Find a number whose 3/4 equals 12.

To find a number by its fraction, this number is divided by this fraction. To, you need to multiply this number by the reciprocal of the fraction (that is, by the inverted fraction). To , 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, this value is divided by this fraction. To divide a fraction by a fraction, multiply the first fraction by the reciprocal 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 got 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, this value is divided by this fraction. Mixed number and multiply it by the reciprocal of the second (inverted fraction). We reduce 99 and 9 by 9, 7 and 14 - by 7. Since we got an improper fraction, it is necessary to select an integer part from it.

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.