Analyzing these problems, observing what the problems have in common from the point of view of mathematics, what are the differences, find an extraordinary way to solve problems, create a piggy bank of problem solving techniques, learn how to solve one problem different ways.A simulator of problems grouped under the same theme “Arithmetic methods for solving problems”, tasks for working in a group and for individual work.
“tasks for the simulator manual”
Trainer: “Arithmetic methods for solving problems”
“Comparing numbers by sum and difference.”
There are 80 boletus mushrooms in two baskets. The first basket contains 10 less boletus than the second. How many boletus mushrooms are there in each basket?
The sewing studio received 480 m of denim and drape. Denim fabric was supplied 140 m more than drapery. How many meters of denim did the studio receive?
The TV tower model consists of two blocks. The lower block is 130 cm shorter than the upper one. What are the heights of the upper and lower blocks if the height of the tower is 4 m 70 cm?
Two boxes contain 16 kg of cookies. Find the mass of cookies in each box if one of them contains 4 kg more cookies.
Problem from “Arithmetic” by L. N. Tolstoy.
a) Two men have 35 sheep. One has 9 more sheep than the other. How many sheep does each person have?
b) Two men have 40 sheep, and one has 6 sheep less than the other. How many sheep does each man have?
There were 23 cars and motorcycles with sidecars in the garage. Cars and motorcycles have 87 wheels. How many motorcycles are there in the garage if each sidecar has a spare wheel?
"Eulerian Circles".
The house has 120 residents, some of whom have dogs and cats. There is a circle in the picture WITH depicts residents with dogs, circle TO – residents with cats. How many tenants have both dogs and cats? How many tenants have only dogs? How many tenants have only cats? How many tenants have neither dogs nor cats?
Of the 52 schoolchildren, 23 play volleyball and 35 basketball, and 16 play both volleyball and basketball. The rest do not play any of these sports. How many schoolchildren do not play any of these sports?
There is a circle in the picture A depicts all university employees who know English language, circle N – who know German and circle F - French. How many university employees know: a) 3 languages; b) English and German; c) French? How many university employees are there? How many of them don't speak French?
The international conference was attended by 120 people. Of these, 60 speak Russian, 48 speak English, 32 speak German, 21 speak Russian and German, 19 speak English and German, 15 speak Russian and English, and 10 people spoke all three languages. How many conference participants do not speak any of these languages?
82 students sing in the choir and practice dancing. rhythmic gymnastics There are 32 students, and 78 students sing in the choir and do rhythmic gymnastics. How many students sing in a choir, do dance and do rhythmic gymnastics separately, if it is known that each student does only one thing?
Every family living in our house subscribes to either a newspaper or a magazine, or both. 75 families subscribe to a newspaper, and 27 families subscribe to a magazine, and only 13 families subscribe to both a magazine and a newspaper. How many families live in our house?
"Method of data adjustment".
There are 29 flowers in 3 small and 4 large bouquets, and 35 flowers in 5 small and 4 large bouquets. How many flowers are in each bouquet individually?
The mass of 2 chocolate bars - large and small - is 120 g, and 3 large and 2 small - 320 g. What is the mass of each bar?
5 apples and 3 pears weigh 810 g, and 3 apples and 5 pears weigh 870 g. How much does one apple weigh? One pear?
Four ducklings and five goslings weigh 4 kg 100 g, five ducklings and four goslings weigh 4 kg. How much does one duckling weigh?
For one horse and two cows, 34 kg of hay is given daily, and for two horses and one cow - 35 kg of hay. How much hay is given to one horse and how much to one cow?
3 red cubes and 6 blue cubes cost 165 tenge rubles. Moreover, five red ones are 95 tenge more expensive than two blue ones. How much does each cube cost?
2 sketchbooks and 3 stamp albums together cost 160 rubles, and 3 sketchbooks cost 45 rubles. more expensive than two stamp albums.
"Counts".
Seryozha decided to give his mother a bouquet of flowers (roses, tulips or carnations) for her birthday and put them either in a vase or in a jug. In how many ways can he do this?
How many three-digit numbers can be made from the digits 0, 1, 3, 5 if the digits in the number are not repeated?
On Wednesday in 5th grade there are five lessons: mathematics, physical education, history, Russian and science. How many various options Can you make a schedule for Wednesday?
“An ancient way to solve problems involving mixing substances.”
How to mix oils? A certain person had two types of oil for sale: one at a price of 10 hryvnia per bucket, the other at 6 hryvnia per bucket. He wanted to make oil from these two oils, mixing them, costing 7 hryvnia per bucket. What parts of these two oils do you need to take to get a bucket of oil worth 7 hryvnia?
How much caramel do you need to take at a price of 260 tenge per 1 kg and at a price of 190 tenge per 1 kg to make 21 kg of the mixture at a price of 210 tenge per kilogram?
Someone has three varieties of tea - Ceylon for 5 hryvnia per pound, Indian for 8 hryvnia per pound and Chinese for 12 hryvnia per pound. In what proportions should these three varieties be mixed to get tea worth 6 hryvnia per pound?
Someone has silver of different standards: one is 12th standard, another is 10th standard, the third is 6th standard. How much silver should you take to get 1 pound of 9th standard silver?
The merchant bought 138 arshins of black and blue cloth for 540 rubles. The question is, how many arshins did he buy for both, if the blue one cost 5 rubles? for an arshin, and black - 3 rubles?
Various tasks.
For New Year's gifts, we bought 87 kg of fruit, and there were 17 kg more apples than oranges. How many apples and how many oranges did you buy?
At the New Year's tree, there were 3 times more snowflakes for children in carnival costumes than in Parsley costumes. How many children were there in Parsley costumes if there were 12 fewer of them?
Masha received 2 times less New Year's greetings than Kolya. How many congratulations did each person receive if there were 27 in total? (9 and 18).
28 kg of sweets were purchased for New Year's prizes. Candies “Swallow” made up 2 parts, “Muse” - 3 parts, “Romashka” - 2 parts. How many sweets of each type did you buy? (8, 8, 12).
There are 2004 kg of flour in the warehouse. Can it be put into bags weighing 9 kg and weighing 18 kg?
There are 5 different cups and 3 different saucers in the "Everything for Tea" store. In how many ways can you buy a cup and saucer?
A horse eats a haystack in 2 days, a cow in 3, a sheep in 6. How many days will it take them to eat the haystack if they eat it together?
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"lesson summary arif sp"
"Arithmetic methods for solving word problems."
For a student of mathematics, it is often more useful to solve the same problem in three different ways than to solve three or four different problems. By solving one problem in different ways, you can find out by comparison which one is shorter and more efficient. This is how experience is developed.
W.W. Sawyer
The purpose of the lesson: use the knowledge acquired in previous lessons, show imagination, intuition, imagination, and ingenuity to solve test problems in various ways.
Lesson objectives: educational: by analyzing these problems, observing what the problems have in common from the point of view of a mathematician, what are the differences, finding an extraordinary way to solve problems, creating a piggy bank of techniques for solving problems, learning to solve one problem in different ways.
Developmental: feel the need for self-realization when finding yourself in a certain role situation.
Educational: develop personal qualities, form a communicative culture.
Means of education: a simulator of problems grouped under the same theme “Arithmetic methods for solving problems”, tasks for working in a group and for individual work.
DURING THE CLASSES.
Hello guys. Sit down. Today we have a lesson on the topic “Arithmetic methods for solving word problems.”
II. Updating knowledge.
Mathematics is one of the ancient and important sciences. People used a lot of mathematical knowledge in ancient times - thousands of years ago. They were necessary for merchants and builders, warriors and land surveyors, priests and travelers.
And nowadays, not a single person can get by in life without a good knowledge of mathematics. The basis good understanding mathematics - the ability to count, think, reason, find successful solutions to problems.
Today we will look at arithmetic methods for solving word problems, we will analyze ancient problems that have come down to us from different countries and times, tasks on equalization, comparison by sum and difference, and others.
The purpose of the lesson is to involve you in amazing world beauty, richness and diversity - the world interesting tasks. And, therefore, introduce you to some arithmetic methods that lead to very elegant and instructive solutions.
A task is almost always a search, the discovery of some properties and relationships, and the means of solving it are intuition and conjecture, erudition and mastery of mathematical methods.
The main ones in mathematics are arithmetic and algebraic methods of solving problems.
Solving a problem using the arithmetic method means finding the answer to the requirement of the problem by performing arithmetic operations on numbers.
With the algebraic method, the answer to the question of the problem is found as a result of composing and solving the equation.
It is no secret that a person who owns different tools and uses them depending on the nature of the work being performed achieves significantly better results than a person who owns only one universal tool.
There are many arithmetic methods and non-standard techniques for solving problems. Today I want to introduce you to some of them.
1.Method of solving word problems “Comparing numbers by sum and difference.”
Task : Grandma collected 51 kg of carrots and cabbage from her summer cottage in the fall. There was 15 kg more cabbage than carrots. How many kilograms of carrots and how many kilograms of cabbage did grandma collect?
Questions that correspond to the points of the problem solving algorithm of this class.
1. Find out what quantities are being discussed in the problem
About the number of carrots and cabbage that grandma collected, together and separately.
2. Indicate the values of which quantities need to be found in the problem.
How many kilograms of carrots and how many kilograms of cabbage did grandma collect?
3. Name the relationship between the quantities in the problem.
The problem talks about the sum and difference of quantities.
4. Name the sum and difference of the values of quantities.
Sum – 51 kg, difference – 15 kg.
5. By equalizing the quantities, find the double value of the smaller quantity (subtract the difference of the quantities from the sum of the quantities).
51 – 15 = 36 (kg) – double the amount of carrots.
6. Knowing the doubled value, find the smaller value (divide the doubled value by two).
36: 2 = 18 (kg) – carrots.
7. Using the difference between the quantities and the value of the smaller quantity, find the value of the larger quantity.
18 + 15 = 33 (kg) – cabbage. Answer: 18 kg, 33 kg. Task.There are pheasants and rabbits in the cage. There are 6 heads and 20 legs in total. How many rabbits and how many pheasants are in a cage
?
Method 1. Selection method:
2 pheasants, 4 rabbits.
Check: 2 + 4 = 6 (goals); 4 4 + 2 2 = 20 (feet).
This is a selection method (from the word “to select”). Advantages and disadvantages of this solution method (difficult to select if the numbers are large) Thus, there is an incentive to search for more convenient solution methods.
Discussion results: the selection method is convenient when working with small numbers; when the values increase, it becomes irrational and labor-intensive.
Method 2. Complete search of options.
A table is compiled:
Answer: 4 rabbits, 2 pheasants.
The name of this method is “full”. Discussion results: the exhaustive search method is convenient, but for large values it is quite labor-intensive.
Method 3. Guessing method.
Let's take an old Chinese problem:
The cage contains an unknown number of pheasants and rabbits. It is known that the entire cell contains 35 heads and 94 legs. Find out the number of pheasants and the number of rabbits.(Problem from the Chinese mathematical book “Kiu-Chang”, compiled 2600 BC).
Here is a dialogue found in the old masters of mathematics. - Let’s imagine that we put a carrot on the cage in which the pheasants and rabbits are sitting. All rabbits will stand on their hind legs to reach the carrot. How many feet will be on the ground at this moment?
But in the problem statement, 94 legs are given, where are the rest?
The remaining legs are not counted - these are the front legs of the rabbits.
How many are there?
24 (94 – 70 = 24)
How many rabbits are there?
12 (24: 2 = 12)
What about pheasants?
23 (35- 12 = 23)
The name of this method is “deficiency guessing method.” Try to explain this name yourself (those sitting in a cage have 2 or 4 legs, and we assumed that everyone has the smallest of these numbers - 2 legs).
Another way to solve the same problem. - Let’s try to solve this problem using the “surplus assumption method”: Let’s imagine that pheasants now have two more legs, then there will be all legs 35 × 4 =140.
But according to the conditions of the problem, there are only 94 legs, i.e. 140 – 94= 46 extra legs, whose are they? These are the legs of pheasants, they have an extra pair of legs. Means, pheasants will 46: 2 = 23,
then rabbits 35 -23 = 12.
Discussion results: the assumption method has two options- By deficiency and excess; Compared to previous methods, it is more convenient because it is less labor-intensive.
Task. A caravan of camels is slowly walking through the desert, there are 40 of them in total. If you count all the humps on these camels, you will get 57 humps. How many dromedary camels are there in this caravan?1 way. Solve using equation.
Number of humps per person Number of camels Total humps
2 x 2 x
1 40 - X 40 - X 57
2 x + 40 - X = 57
x + 40 = 57
X = 57 -40
X = 17
Method 2.
- How many humps can camels have?
(there may be two or one)
Let's attach a flower to each camel's hump.
- How many flowers will you need? (40 camels – 40 flowers)
- How many humps will be left without flowers?
(There will be such 57-40=17 . This second humps Bactrian camels).
How many Bactrian camels? (17)
How many dromedary camels? (40-17=23)
What is the answer to the problem? ( 17 and 23 camels).
Task.In the garage there were cars and motorcycles with sidecars, 18 of them all together. The cars and motorcycles had 65 wheels. How many motorcycles with sidecars were in the garage, if cars have 4 wheels and motorcycles have 3 wheels?
1 way. Using the equation:
Number of wheels for 1 Number of total wheels
Mash. 4x 4 x
Mot. 3 18 -X 3(18 - X ) 65
4 x + 3(18 - X ) = 65
4 x + 5 4 -3 X =65
X = 65 - 54
X = 11, 18 – 11 = 7.
Let's reformulate the problem : The robbers, who came to the garage where 18 cars and motorcycles with sidecars were parked, removed three wheels from each car and motorcycle and took them away. How many wheels are left in the garage if there were 65 of them? Do they belong to a car or a motorcycle?
3×18=54 – that’s how many wheels the robbers took away,
65- 54 = 11 – so many wheels left (cars in the garage),
18 - 11 = 7 motorcycles.
Answer: 7 motorcycles.
On one's own:
There were 23 cars and motorcycles with sidecars in the garage. Cars and motorcycles have 87 wheels. How many motorcycles are there in the garage if each sidecar has a spare wheel?
- How many wheels do cars and motorcycles have together? (4×23=92)
- How many spare wheels did you put in each stroller? (92 - 87= 5)
- How many cars are in the garage? (23 - 5=18).
Task.In our class you can study English or French languages(optionally). It is known that 20 schoolchildren study English, and 17 study French. In total, there are 32 students in the class. How many students study both English and French?
Let's draw two circles. In one we will record the number of schoolchildren studying English, in the other - schoolchildren studying French. Since according to the conditions of the problem there are students studyingboth languages: English and French, then the circles will have a common part. The conditions of this problem are not so easy to understand. If you add 20 and 17, you get more than 32. This is explained by the fact that we counted some schoolchildren here twice - namely those who study both languages: English and French. So, (20 + 17) – 32 = 5 Students learn both languages: English and French.
English Fran.
20 lessons 17 school
(20 + 17) – 32 = 5 (students).
Schemes similar to the one we used to solve the problem are called in mathematics Euler circles (or diagrams). Leonhard Euler (1736) born in Switzerland. But long years lived and worked in Russia.
Task.Every family living in our house subscribes to either a newspaper or a magazine, or both. 75 families subscribe to a newspaper, and 27 families subscribe to a magazine, and only 13 families subscribe to both a magazine and a newspaper. How many families live in our house?
Newspapers magazines
The picture shows that 89 families live in the house.
Task.The international conference was attended by 120 people. Of these, 60 speak Russian, 48 speak English, 32 speak German, 21 speak Russian and German, 19 speak English and German, 15 speak Russian and English, and 10 people spoke all three languages. How many conference participants do not speak any of these languages?
Russian 15 English
21 10 19
German
Solution: 120 – (60 + 48 + 32 -21 – 19 – 15 + 10) = 25 (persons).
Task. Three kittens and two puppies weigh 2 kg 600 g, and two kittens and three puppies weigh 2 kg 900 g. How much does the puppy weigh?
3 kittens and 2 puppies – 2kg 600 g
2 kittens and 3 puppies – 2 kg 900 g.
It follows from the condition that 5 kittens and 5 puppies weigh 5 kg 500 g. This means that 1 kitten and 1 puppy weigh 1 kg 100 g
2 cats and 2 puppies. weigh 2 kg 200 g
Let's compare the conditions -
2 kittens + 3 puppies = 2kg 900 g
2 kittens + 2 puppies = 2 kg 200 g, we see that the puppy weighs 700 g.
Task.For one horse and two cows, 34 kg of hay is given daily, and for two horses and one cow - 35 kg of hay. How much hay is given to one horse and how much to one cow?
Let's write it down short condition tasks:
1 horse and 2 cows -34kg.
2 horses and 1 cow -35kg.
Is it possible to know how much hay is needed for 3 horses and 3 cows?
(for 3 horses and 3 cows – 34+35=69 kg)
Is it possible to find out how much hay is needed for one horse and one cow? (69: 3 – 23kg)
How much hay does one horse need? (35-23=12kg)
How much hay does one cow need? (23 -13 =11kg)
Answer: 12kg and 11kg.
Task.Madina decided to have breakfast at the school cafeteria. Study the menu and answer, in how many ways can she choose a drink and a confectionery item?
| Confectionery |
|
| Cheesecake |
|
Let's assume that Madina chooses tea as a drink. What confectionery product can she choose for tea? (tea - cheesecake, tea - cookies, tea - bun)
How many ways? (3)
What if it's compote? (also 3)
How can you find out how many ways Madina can use to choose her lunch? (3+3+3=9)
Yes you are right. But to make it easier for us to solve this problem, we will use graphs. The word "graph" in mathematics means a picture with several points drawn, some of which are connected by lines. Let's designate the drinks and confectionery dots and connect the pairs of those dishes that Madina chooses.
tea milk compote
cheesecake cookies bun
Now let's count the number of lines. There are 9 of them. This means that there are 9 ways to choose dishes.
Task.Seryozha decided to give his mother a bouquet of flowers (roses, tulips or carnations) for her birthday and put them either in a vase or in a jug. In how many ways can he do this?
How many ways do you think? (3)
Why? (3 colors)
Yes. But there is still different dishes: either a vase or a jug. Let's try to complete the task graphically.
vase jug
roses tulips carnations
Count the lines. How many are there? (6)
So, how many ways does Seryozha have to choose? (6)
Lesson summary.
Today we solved a number of problems. But the work is not completed, there is a desire to continue it, and I hope that this will help you successfully solve word problems.
We know that problem solving is a practical art, like swimming or playing the piano. You can learn it only by imitating good examples and constantly practicing.
These are only the simplest of problems; complex ones remain a subject for future study. But there are still many more of them than we could solve. And if at the end of the lesson you can solve problems “behind the pages” educational material", then we can assume that I have completed my task.
Knowledge of mathematics helps to solve a certain life problem. In life, you will have to regularly resolve certain issues, for this you need to develop intellectual abilities, thanks to which internal potential develops, the ability to foresee the situation, make predictions, and make non-standard decisions develops.
I want to end the lesson with the words: “Every well-solved mathematical problem gives mental pleasure.” (G. Hesse).
Do you agree with this?
Homework .
The following assignment will be given at home: using the texts of solved problems as a sample, solve problems No. 8, 17, 26 using the methods that we have studied.
Based on the similarity in mathematical meaning and the interchangeability of different solution methods, all arithmetic methods can be combined into the following groups:
- 1) method of reduction to unity, reduction to a general measure, inverse reduction to unity, method of relations;
- 2) a way to solve problems from the “end”;
- 3) a method for eliminating unknowns (replacing one unknown with another, comparing unknowns, comparing data, comparing two conditions by subtraction, combining two conditions into one); way of guessing;
- 4) proportional division, similarity or finding of parts;
- 5) a method of transforming one problem into another (decomposing a complex problem into simple, preparatory ones; bringing unknowns to such values for which their relationship becomes known; the method of determining an arbitrary number for one of the unknown quantities).
In addition to the above methods, it is advisable to consider also the arithmetic mean method, the surplus method, the method of rearranging the known and the unknown, and the method of “false” rules.
Since it is usually impossible to determine in advance which of the methods is rational, to foresee which of them will lead to the simplest and most understandable solution for the student, then students should be introduced to different ways and give them the opportunity to choose which one to use when solving a specific problem.
Method for excluding unknowns
This method is used when there are several unknowns in the problem. This problem can be solved using one of five techniques: 1) replacing one unknown with another; 2) comparison of unknowns; 3) comparison of two conditions by subtraction; 4) comparison of data; 5) combining several conditions into one.
As a result of using one of the listed techniques, instead of several unknowns, there remains one that can be found. Having calculated it, they use the data in the dependence condition to find other unknowns.
Let's take a closer look at some of the techniques.
1. Replacing one unknown with another
The name of the technique reveals its idea: based on the dependencies (multiple or difference) that are given according to the conditions of the problem, it is necessary to express all the unknowns through one of them.
Task. Sergei and Andrey have only 126 stamps. Sergei has 14 marks more than Andrey. How many stamps did each boy have?
Brief description of the condition:
Sergey -- ? marks, 14 marks more
Andrey -- ? stamps
Total -- 126 stamps
Solution 1.
- (replacing a larger unknown with a smaller one)
- 1) Let Sergei have as many stamps as Andrey. Then total there would be 126 marks - 14 = 112 (marks).
- 2) Since the boys now have the same number of marks, we will find how many marks Andrei had at the beginning: 112: 2 = 56 (stamps).
- 3) Considering that Sergei has 14 marks more than Andrey, we get: 56 + 14 = 70 (marks).
Solution 2.
- (replacing a smaller unknown with a larger one)
- 1) Let Andrei have the same number of stamps as Sergei. Then the total number of stamps would be 126 + 14 = 140 (stamps).
- 2) Since the boys now have the same number of marks, let’s find how many marks Sergei had at first: 140: 2 = 70 (marks).
- 3) Considering that Andrey had 14 marks less than Sergei, we get: 70 - 14 = 56 (marks).
Answer: Sergei had 70 marks, and Andrey had 56 marks.
For best absorption students of the method of replacing a smaller unknown with a larger one, before considering it, it is necessary to find out with the students the following fact: if the number A is greater than the number B by C units, then in order to compare the numbers A and B it is necessary:
- a) subtract number C from number A (then both numbers are equal to number B);
- b) add number C to number B (then both numbers are equal to number A).
The ability of students to replace a larger unknown with a smaller one, and vice versa, further contributes to the development of the ability to choose an unknown and express other quantities through it when composing an equation.
2. Comparison of unknowns
Task. There were 188 books on four shelves. On the second shelf there were 16 fewer books than on the first, on the third - 8 more than on the second, and on the fourth - 12 fewer than on the third shelf. How many books are on each shelf?
Task Analysis
For better awareness dependencies between four unknown quantities (the number of books on each shelf) we use the following diagram:
I_________________________________
II___________________________
III______________________________
IV_______________________ _ _ _ _ _
Comparing the segments that schematically depict the number of books on each shelf, we come to the following conclusions: there are 16 more books on the first shelf than on the second; on the third there are 8 more than on the second; on the fourth - 12 - 8 = 4 (books) less than on the second. Therefore, the problem can be solved by comparing the number of books on each shelf. To do this, remove 16 books from the first shelf, 8 books from the third, and put 4 books on the fourth shelf. Then there will be the same number of books on all shelves, namely, as there were on the second one at first.
- 1) How many books are there on all shelves after the operations described in the problem analysis?
- 188 -- 16 -- 8 + 4 = 168 (books)
- 2) How many books were on the second shelf?
- 168: 4 = 42 (books)
- 3) How many books were on the first shelf?
- 42 + 16 = 58 (books)
- 4) How many books were on the third shelf?
- 42 + 8 = 50 (books)
- 5) How many books were on the fourth shelf?
- 50 -- 12 = 38 (books)
Answer: There were 58, 42, 50 and 38 books on each of the four shelves.
Comment. You can invite students to solve this problem in other ways by comparing the unknown number of books that were on the first, or on the second, or on the fourth shelf.
3. Comparison of two conditions by subtraction
The plot of the problem that is solved by this technique often includes two proportional quantities (the quantity of goods and its cost, the number of workers and the work they performed, etc.). The condition gives two values of one quantity and the difference of two proportional to them numerical values of a different size.
Task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time for 4 kg of oranges and 3 kg of bananas bought at the same prices they paid 500 rubles. How much does 1kg of oranges and 1kg of bananas cost?
Brief description of the condition:
- 4kg app. and 5kg ban. - 620 rubles,
- 4kg app. and 3kg ban. - 500 rub.
- 1) Let's compare the cost of two purchases. Both the first and second time they bought the same number of oranges at the same price. The first time we paid more because we bought more bananas. Let's find how many more kilograms of bananas were purchased the first time: 5 -- 3 = 2 (kg).
- 2) Let's find out how much more we paid the first time than the second time (that is, we find out how much 2 kg of bananas cost): 620 - 500 = 120 (rub.).
- 3) Find the price of 1 kg of bananas: 120: 2 = 60 (rub.).
- 4) Knowing the cost of the first and second purchases, we can find the price of 1 kg of oranges. To do this, first find the cost of purchased bananas, then the cost of oranges, and then the price of 1 kg. We have: (620 -- 60*5) : 4 = 80 (rub).
Answer: the price of 1 kg of oranges is 80 rubles, and the price of 1 kg of bananas is 60 rubles.
4. Data comparison
Application this technique makes it possible to compare data and apply the subtraction method. You can compare data values:
- 1) using multiplication (comparing them with the least common multiple);
- 2) using division (comparing them with the largest common divisor).
Let's show this with an example.
Task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time for 6 kg of oranges and 3 kg of bananas bought at the same prices they paid 660 rubles. How much does 1kg of oranges and 1kg of bananas cost?
Brief description of the condition:
- 4kg app. and 5kg ban. - 620 rubles,
- 6kg app. and 3kg ban. - 660 rub.
Let's equalize the number of oranges and bananas by comparing them with the least common multiple: LCM(4;6) = 12.
Solution1.
- 1) Let us increase the number of fruits purchased and their cost in the first case by 3 times, and in the second - by 2 times. We get the following short statement of the condition:
- 12kg app. and 15kg ban. - 1860 rubles,
- 12kg app. and 6kg ban. - 1320 rub.
- 2) Find out how many more bananas you bought the first time: 15-6 = 9 (kg).
- 3) How much does 9kg of bananas cost? 1860 -- 1320 = 540 (rub).
- 4) Find the price of 1 kg of bananas: 540: 9 = 60 (rub).
- 5) Find the cost of 3 kg of bananas: 60 * 3 = 180 (rub).
- 6) Find the cost of 6 kg of oranges: 660 -- 180 = 480 (rub).
- 7) Find the price of 1 kg of oranges: 480: 6 = 80 (rub).
Solution2.
Let's equalize the number of oranges and bananas by comparing them with the greatest common divisor: GCD (4; 6) = 2.
- 1) To equalize the number of oranges purchased the first time and the second time, we reduce the quantity of the purchased product and its cost in the first case by 2 times, in the second - by 3 times. Let us obtain a problem that has the following short form of condition:
- 2kg app. and 2.5 kg ban. - 310 rubles,
- 2kg app. and 1kg ban. - 220 rub.
- 2) How many more bananas are they buying now: 2.5 -- 1 = 1.5 (kg).
- 3) Let's find how much 1.5 kg of bananas costs: 310 -- 220 = 90 (rub).
- 4) Find the price of 1 kg of bananas: 90: 1.5 = 60 (rub).
- 5) Find the price of 1 kg of oranges: (660 -- 60*3) : 6 = 80 (rub).
Answer: the price of 1 kg of oranges is 80 rubles, 1 kg of bananas is 60 rubles.
When solving problems using the technique of comparing data, you can not do such detailed analysis and recordings, but only record the changes that were made for comparison and write them down in the form of a table.
5. Combining several conditions into one
Sometimes you can get rid of unnecessary unknowns by combining several conditions into one.
Task. The tourists left the camp and first walked for 4 hours, and then rode bicycles for another 4 hours at a certain constant speed and moved 60 km away from the camp. The second time they left the camp and first rode bicycles at the same speed for 7 hours, and then turned in the opposite direction and, walking for 4 hours, found themselves at a distance of 50 km from the camp. How fast did the tourists ride their bikes?
There are two unknowns in the problem: the speed at which the tourists rode their bicycles, and the speed at which they walked. In order to exclude one of them, you can combine two conditions into one. Then the distance that tourists will cover in 4 hours, moving forward on foot the first time, is equal to the distance they covered in 4 hours, moving back the second time. Therefore, we do not pay attention to these distances. This means that the distance that tourists will cover in 4 + 7 = 11 (hours) on bicycles will be equal to 50 + 60 = 110 (km).
Then the speed of tourists on bicycles is: 110: 11 = 10 (km/h).
Answer: The speed of bicycles is 10 km/h.
6. Method of assumption
Using the assumption method when solving problems does not cause difficulties for most students. Therefore, to avoid students mechanically memorizing the diagram of steps of this method and misunderstanding the essence of the actions performed on each of them, students should first be shown the trial method (“false rule” and “rule of the ancient Babylonians”).
When using the sampling method, in particular the “false rule,” one of the unknown quantities is given (“allowed”) a certain value. Then, using all the conditions, they find the value of another quantity. The resulting value is checked against the one specified in the condition. If the resulting value is different from the one given in the condition, then the first value specified is not correct and it must be increased or decreased by 1, and again the value of another value must be found. This must be done until we obtain the value of another quantity such as in the problem statement.
Task. The cashier has 50 coins of 50 kopecks and 10 kopecks, totaling 21 rubles. Find how many separate 50k coins the cashier had. and 10k each.
Solution1. (sampling method)
Let's use the rule of the “ancient” Babylonians. Let's assume that the cashier has an equal number of coins of each denomination, that is, 25 pieces each. Then the amount of money will be 50*25 + 10*25 = 1250+250=1500 (k.), or 15 rubles. But in the condition 21 rubles, that is, 21 UAH more than received - 15 rubles = 6 rubles. This means that it is necessary to increase the number of 50-kopeck coins and decrease the number of 10-kopeck coins until we get a total of 21 rubles. We will record the change in the number of coins and the total amount in the table.
|
Number of coins |
Number of coins |
Amount of money |
Amount of money |
total amount |
Less or more than in the condition |
|
Less by 6 rubles. |
|||||
|
Less by 5rub60k |
|||||
|
As in condition |
As can be seen from the table, the cashier had 40 coins of 50 kopecks and 10 coins of 10 kopecks.
As it turned out in solution 1, if the cashier had an equal number of 50k coins. and 10k each, then in total he had 15 rubles of money. It is easy to see that each coin replacement is 10k. per coin 50k. increases the total amount by 40k. This means we need to find how many such replacements need to be made. To do this, let’s first find how much money we need to increase the total amount by:
21 rubles -- 15 rubles. = 6 rub. = 600 k.
Let's find how many times such a replacement needs to be made: 600 k. : 40 k. = 15.
Then 50 kopecks will be 25 +15 = 40 (coins), and 10 kopeck coins will remain 25 -- 15 = 10.
The check confirms that the total amount of money in this case is 21 rubles.
Answer: The cashier had 40 coins of 50 kopecks and 10 coins of 10 kopecks.
By asking students to choose for themselves different meanings number of coins of 50 kopecks, it is necessary to bring them to the idea that the best from the point of view of rationality is the assumption that the cashier had only coins of one denomination (for example, all 50 coins of 50 kopecks or all 50 coins of 10 kopecks each). Due to this, one of the unknowns is excluded and replaced by another unknown.
7. Residue method
This method has some similarities with thinking when solving problems using trial and guesswork methods. We use the method of remainders when solving problems involving movement in one direction, namely, when it is necessary to find the time during which the first object, which is moving behind at a higher speed, will catch up with the second object, which has a lower speed. In 1 hour, the first object approaches the second at a distance that is equal to the difference in their speeds, that is, equal to the “remainder” of the speed that it has in comparison with the speed of the second. To find the time it takes for the first object to cover the distance that was between it and the second at the beginning of the movement, you should determine how many times the “remainder” is placed in this distance.
If we abstract from the plot and consider only the mathematical structure of the problem, then it talks about two factors (the speed of movement of both objects) or the difference between these factors and two products (the distances they travel) or their difference. The unknown factors (time) are the same and need to be found. From a mathematical point of view, the unknown factor shows how many times the difference of the known factors is contained in the difference of the products. Therefore, problems that are solved using the method of remainders are called problems of finding numbers by two differences.
Task. The students decided to paste photos from the holiday into an album. If they stick 4 photos on each page, there won't be enough space in the album for 20 photos. If you paste 6 photos onto each page, then 5 pages will remain free. How many photos are students going to put in the album?
Task Analysis
The number of photos remains the same for the first and second gluing options. According to the conditions of the problem, it is unknown, but it can be found if the number of photographs that are placed on one page and the number of pages in the album are known.
The number of photographs that are pasted onto one page is known (the first multiplier). The number of pages in the album is unknown and remains unchanged (second multiplier). Since it is known that 5 pages of the album remain free for the second time, you can find how many more photographs could be pasted into the album: 6 * 5 = 30 (photos).
This means that by increasing the number of photos on one page by 6 - 4 = 2, the number of pasted photos increases by 20 + 30 = 50.
Since the second time they pasted two more photographs onto each page and in total they pasted 50 more photographs, we will find the number of pages in the album: 50: 2 = 25 (pages).
Therefore, there were 4*25 + 20 = 120 (photos) in total.
Answer: The album had 25 pages and 120 photographs.
General notes on solving problems using the arithmetic method.
Problems of finding unknowns based on the results of actions.
Proportional division problems.
Problems involving percentages and parts.
Problems solved in reverse.
1. The arithmetic method is the main method for solving word problems in primary school. It also finds its application in the middle level of secondary schools. This method allows you to better understand and appreciate the importance and significance of each stage of working on a task.
In some cases, solving a problem using the arithmetic method is much simpler than using other methods.
While captivating with its simplicity and accessibility, the arithmetic method is at the same time quite complex, and mastering the techniques for solving problems using this method requires serious and painstaking work. The wide variety of types of problems does not allow us to form a universal approach to analyzing problems and finding ways to solve them: problems, even combined into one group, have completely different ways of solving them.
2 . To the tasks on finding unknowns by their difference and ratio These include problems in which, using the known difference and quotient of two values of a certain quantity, it is required to find these values.
Algebraic model:
The answer is found using the formulas: X= ak/(k – 1), y = a/(k – 1).
Example. In the reserved seat carriages of the fast train there are 432 more passengers than in the compartment carriages. How many passengers are there in reserved seat and compartment carriages separately, if there are 4 times fewer passengers in compartment carriages than in reserved seat carriages?
Solution. A graphical model of the problem is presented in Fig. 4.
Rice. 4
We will take the number of passengers in compartment cars as 1 part. Then you can find how many parts there are per number of passengers in reserved seat cars, and then how many parts there are per 432 passengers. After this, you can determine the number of passengers making up 1 part (located in compartment cars). Knowing that there are 4 times more passengers in reserved seat carriages, we can find their number.
1 4 = 4 (hours) – accounts for passengers in reserved seat carriages;
4 – 1 = 3 (h.) – accounts for the difference between the number of passengers in reserved seat and compartment carriages;
432: 3 = 144 (p.) – in compartment cars;
144 4 = 576 (p.) – in reserved seat carriages.
This problem can be verified by solving it in another way, namely:
1 4 = 4(h);
4 – 1 = 3 (h);
432: 3 = 144 (p.);
144 + 432 = 576 (p.).
Answer: there are 144 passengers in compartment carriages, and 576 in reserved seat carriages.
To the tasks on finding unknowns from two residues or two differences, include problems in which two directly or inversely proportional quantities are considered, such that two values of one quantity and the difference of the corresponding values of another quantity are known, and it is required to find the values of this quantity themselves.
Algebraic model:
The answers are found using the formulas:
Example. Two trains traveled at the same speed - one 837 km, the other 248 km, and the first was on the road 19 hours longer than the second. How many hours did each train travel?
Solution. A graphical model of the problem is presented in Figure 5.
Rice. 5
To answer the question of the problem, how many hours was this or that train on the way, you need to know the distance it traveled and the speed. The distance is given in the condition. To find out the speed, you need to know the distance and the time during which this distance was covered. The condition says that the first train took 19 hours longer, and the distance it covered during this time can be found. He walked for an extra 19 hours - obviously, during this time he also covered an extra distance.
837 – 248 = 589 (km) – the first train traveled so many kilometers more;
589: 19 = 31 (km/h) – speed of the first train;
837: 31 = 27 (hours) – the first train was on its way;
4) 248: 31 = 8 (hours) – the second train was on its way.
Let's check the solution to the problem by establishing a correspondence between the data and the numbers obtained when solving the problem.
Having found out how long each train was on the road, we will find how many hours more the first train was on the road than the second: 27 – 8 = 19 (hours). This number matches the one in the condition. Therefore, the problem was solved correctly.
This problem can be verified by solving it in another way. All four questions and the first three actions remain the same.
4) 27 –19 = 8 (hours).
Answer: the first train took 31 hours to travel, the second train took 8 hours.
Problems to find three unknowns from three sums of these unknowns, taken in pairs:
Algebraic model:
The answer is found using the formulas:
x =(A -b + c)/2, y = (a +b – c)/2, z = (b + With -a)/ 2.
Example. English and German languages 116 schoolchildren study German and spanish languages 46 students are studying, and 90 students are studying English and Spanish. How many students study English, German and Spanish separately if it is known that each student studies only one language?
Solution. A graphical model of the problem is presented in Figure 6.
How many students study each language?
The graphical model of the problem shows: if we add up the number of schoolchildren given in the condition (116 + 90 + 46), we get double the number of schoolchildren studying English, German and Spanish. Dividing it by two, we find the total number of schoolchildren. To find the number of schoolchildren studying English, it is enough to subtract from this number the number of schoolchildren studying German and Spanish. Similarly, we find the remaining required numbers.
Let's write down the decision on actions with explanations:
116 + 90 + 46 = 252 (schoolchildren) – double the number of schoolchildren studying languages;
252: 2 = 126 (school) – study languages;
126 – 46 = 80 (school) – study English;
126 – 90 = 36 (school) – study German;
126 – 116 = 10 (school) – learn Spanish.
This problem can be verified by solving it in another way.
116 – 46 = 70 (schoolchildren) – so many more schoolchildren study English than Spanish;
90 + 70 = 160 (schoolchildren) – double the number of schoolchildren studying English;
160: 2 = 80 (school) – learn English;
90 – 80 = 10 (school) – learn Spanish;
116 – 80 = 36 (school) – study German.
Answer: 80 schoolchildren study English, 36 schoolchildren study German, and 10 schoolchildren study Spanish.
3. Proportional division problems include problems in which a given value of a certain quantity needs to be divided into parts proportional to given numbers. In some of them, the parts are presented explicitly, while in others, these parts must be distinguished by taking one of the values of this quantity as one part and determining how many such parts are accounted for by its other values.
There are five types of proportional division problems.
1) Problems involving dividing a number into parts, directlyproportional to a series of whole or fractional numbers
To the tasks of this type include tasks in which the number A X 1, X 2 , x 3 , ..., X n directly proportional to the numbers A 1 , A 2 , A 3 , ..., A n .
Algebraic model:
The answer is found using the formulas:
Example. The travel company has four recreation centers, which have buildings of the same capacity. On the territory of the 1st recreation center there are 6 buildings, the 2nd - 4 buildings, the 3rd - 5 buildings, the 4th - 7 buildings. How many campers can each base accommodate if all 4 bases can accommodate 2,112 people?
Solution. A summary of the task is shown in Figure 7.
Rice. 7
To answer the question of the problem, how many vacationers can be accommodated at each base, you need to know how many vacationers can be accommodated in one building and how many buildings are located on the territory of each base. The number of buildings on each base is given in the condition. To find out how many vacationers can be accommodated in one building, you need to know how many vacationers can be accommodated at all 4 bases (this is given in the condition) and how many buildings are located on the territory of all 4 bases. The latter can be determined by knowing from the condition how many buildings are located on the territory of each base.
Let's write down the decision on actions with explanations:
6 + 4 + 5 + 7 = 22 (k.) – located on the territory of 4 bases;
2112: 22 = 96 (hours) – can be placed in one building;
96 6 = 576 (h) – can be placed on first base;
96 4 = 384 (h) – can be placed on second base;
96 5 = 480 (h) – can be placed on third base;
96 7 = 672 (h) – can be placed on fourth base.
Examination. We calculate how many vacationers can be accommodated at 4 bases: 576 + 384 + 480 + 672 = 2,112 (hours). There is no discrepancy with the task conditions. The problem was solved correctly.
Answer: the first base can accommodate 576 vacationers, the second - 384 vacationers, the third - 480 vacationers, the fourth - 672 vacationers.
2) Problems involving dividing a number into parts inversely proportional to a series of integers or fractions
These include tasks in which the number A(the value of a certain quantity) needs to be divided into parts x 1 i , x 2 , x 3 i , ..., X" inversely proportional to numbers A 1b A 2 , A 3 ,..., A n .
Algebraic model:
or
x 1 : x 2 :X 3 :...:х„ = a 2 a 3 ...A n :A 1 A 3 ...A P :A 1 A 2 A 4 ...A n :...:A 1 A 2 ...A n -1
The answer is found using the formulas:
Where S = A 2 A 3 ...a„ +a l a i ... a n + a ] A 2 A 4 ...A n + ... + a 1 A 2 ...A n -1.
Example. For four months, the fur farm’s income from the sale of furs amounted to 1,925,000 rubles, and by month the money received was distributed in inverse proportion to the numbers 2, 3, 5, 4. What is the farm’s income in each month separately?
Solution. To determine the income mentioned in the condition, the total income for four months is given, that is, the sum of the four required numbers, as well as the relationship between the required numbers. The required income is inversely proportional to the numbers 2, 3, 5, 4.
Let's denote 
the required incomes respectively through x, X 2
, X 3
, X 4
.
Then the problem can be briefly written as shown in Figure 8.
Rice. 8
Knowing the number of parts per each of the required numbers, we will find the number of parts contained in their sum. Based on the given total income for four months, that is, based on the sum of the required numbers and the number of parts contained in this amount, we find out the value of one part, and then the required income.
Let's write down the decision on actions with explanations:
1. The required incomes are inversely proportional to the numbers 2, 3, 5, 4, which means they are directly proportional to the inverse numbers, that is, there are relations 
. Let us replace these ratios in fractional numbers with ratios of integers:
2. Knowing that X contains 30 equal parts, X 2 – 20, X 3 – 12, X 4 – 15, let’s find how many parts are contained in their sum:
30 + 20 + 12+ 15 = 77 (hours).
3. How many rubles are there for one part?
1,925,000: 77 = 25,000 (r.).
4. What is the farm's income in the first month?
25,000 30 = 750,000 (r.).
5. What is the farm income in the second month?
25,000 20 = 500,000 (r.).
6. What is the farm income in the third month?
25,000–12 = 300,000 (r.).
7. What is the farm income in the fourth month?
25,000–15 = 375,000 (r.).
Answer: in the first month, the farm’s income was 750,000 rubles, in the second – 500,000 rubles, in the third – 300,000 rubles, in the fourth – 375,000 rubles.
3) Problems involving dividing a number into parts, when separate ratios are given for each pair of required numbers
Problems of this type include those tasks in which the number A(the value of a certain quantity) must be divided into parts x 1, X 2 , x 3, ..., X", when a series of relations is given for the required numbers, taken in pairs. Algebraic model:
x 1: X 2 = a 1 : b 1, X 2 : X 3 = a 2 : b 2, x 3 : X 4 = a 3 : b 3 , ..., X n-1 : X n = a n -1 : b n-1 .
n = 4. Algebraic model:
X X :X 2 = a 1 : b 1, X 2 :X 3= A 2 : b 2, X 3 : X 4 = a 3: b 3 .
So, X 1: X 2 : x 3: X 4 = A 1 A 2 A 3 : b 1 A 2 A 3 : b 1 b 2 A 3 : b 1 b 2 b 3 .
Where S = A 1 A 2 A 3 + b 1 A G A 3 + b 1 b 2 A 3 + b 1 b 2 b 3
Example. The three cities have 168,000 inhabitants. The numbers of residents of the first and second cities are in the ratio 
, and the second and third cities – in relation to . How many inhabitants are there in each city?
Solution. Let us denote the required population numbers accordingly by X 1 , X 2 , X 3 . Then the problem can be briefly written as shown in Figure 9.
Rice. 9
To determine the number of inhabitants, the numbers of inhabitants in three cities are given, that is, the sum of the three required numbers, as well as individual relationships between the required numbers. Replacing these relations with a series of relations, we express the number of inhabitants of the three cities in equal parts. Knowing the number of parts per each of the required numbers, we will find the number of parts contained in their sum. From the given total number of inhabitants in three cities, that is, from the sum of the required numbers and from the number of parts contained in this sum, we find out the size of one part, and then the required numbers of inhabitants.
Let's write down the decision on the actions with explanations.
1. Replace the ratio of fractional numbers with the ratio of integers:
We match the number of residents of the second city with the number 15 (the least common multiple of the numbers 3 and 5).
We change the resulting relationships accordingly:
X 1: X 2 = 4: 3 = (4-5): (3-5) = 20: 15, x 2: x 3 = 5: 7 = (5-3): (7-3) = 15: 21.
From individual relations we create a series of relations:
X 1: X 2 : X 3 = 20: 15: 21.
2. 20 + 15 + 21 = 56 (h) – the number 168,000 corresponds to so many equal parts;
3. 168,000: 56 = 3,000 (f.) – per part;
4. 3,000 20 = 60,000 (f.) – in the first city;
5. 3,000 15 = 45,000 (f.) – in the second city;
3,000 21 = 63,000 (f.) - in the third city.
Answer: 60,000 inhabitants; 45,000 inhabitants; 63,000 inhabitants.
4) Problems involving dividing a number into parts proportional to two, three, and so on rows of numbers
Problems of this type include problems in which the number A(the value of a certain quantity) needs to be divided into parts X 1, X 2 , X 3 ,..., X n proportional to two, three, ..., N rows of numbers.
Due to the cumbersomeness of the formulas for solving the problem in general view Let's consider a special case when n = 3 and N = 2. Let X 1 X 2 , X 3 directly proportional to numbers A 1 , A 2 , A 3 and inversely proportional to the numbers b 1 , b 2 , b 3 .
Algebraic model:
(see paragraph 1 of this paragraph),
Example. Two workers received 1,800 rubles. One worked 3 days for 8 hours, the other 6 days for 6 hours. How much did each earn if for 1 hour of work they received equally?
Solution. A summary of the task is shown in Figure 10.
Rice.10
To find out how much each worker received, you need to know how many rubles were paid for 1 hour of work and how many hours each worker worked. To find out how many rubles they paid for 1 hour of work, you need to know how much they paid for the entire work (given in the condition) and how many hours both workers worked together. To find out the total number of hours worked, you need to know how many hours each person worked, and for this you need to know how many days each person worked and how many hours a day. This data is included in the condition.
Let's write down the decision on actions with explanations:
8 3 = 24 (hours) – the first worker worked;
6 6 = 36 (hours) – the second worker worked;
24 + 36 = 60 (hours) – both workers worked together;
1800: 60 = 30 (r.) – workers received for 1 hour of work;
30 24 = 720 (r.) – earned by the first worker;
30 36 = 1080 (r.) - earned by the second worker. Answer: 720 rub.; 1080 rub.
5) Problems on finding several numbersaccording to their relations and the sum or difference (the sum or difference of some of them)
Example. The school administration spent 49,000 rubles on equipment for the playground, greenhouse and gym. Equipment for the playground cost half as much as greenhouses, and greenhouses cost 3 times less than gym and playground together. How much money was spent on equipment for each of these facilities?
Solution. A summary of the task is shown in Figure 11.
Rice. elevenTo find out the amount of money spent on the equipment of each object, you need to know how many parts of all the money spent were on the equipment of each object and how many rubles were for each part. The number of parts of money spent on the equipment of each object is determined from the conditions of the problem. Having determined the number of parts for the equipment of each object separately, and then finding their sum, we calculate the value of one part (in rubles).
Let's write down the decision on the actions with explanations.
We take as 1 part the amount of money spent on equipment for the playground. According to the condition, 2 times more was spent on greenhouse equipment, that is, 1 2 = 2 (h); 3 times more was spent on the equipment of the playground and sports hall than on the greenhouse, that is, 2 3 = 6 (hours), therefore, 6 – 1 = 5 (hours) were spent on the equipment of the sports hall.
1 part was spent on equipment for the playground, 2 parts for greenhouses, and 5 parts for the gym. The entire consumption was 1 + 2 + + 5 = 8 (h).
8 parts equal 49,000 rubles, one part is 8 times less than this amount: 49,000: 8 = 6,125 (rub.). Consequently, 6,125 rubles were spent on equipment for the playground.
Twice as much was spent on greenhouse equipment: 6,125 2 = 12,250 (r.).
5 parts were spent on equipment for the gym: 6,125 5 = 30,625 (r.).
Answer: 6,125 rubles; RUB 12,250; RUR 30,625
6) Problems to exclude one of the unknowns
Problems in this group include problems in which the sums of two products that have two repeating factors are given, and it is required to find the values of these factors. Algebraic model
The answer is found using the formulas:
These problems are solved by the method of equalizing data, the method of equalizing data and the required ones, the method of replacing data, as well as the so-called “assumption” method.
Example. At a garment factory, 24 coats and 45 suits used 204 m of fabric, and 24 coats and 30 suits used 162 m. How much fabric is used for one suit and how much for one coat?
Solution. Let's solve the problem using data adjustment method. Brief description of the task.
Kowtowed Maria, Lyudmila Bryantseva
The work shows ways to solve word problems.
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Municipal educational institution average comprehensive school No. 64 Volgograd
City competition of educational and research works
"Me and the Earth" named after. IN AND. Vernadsky
(district stage)
ARITHMETIC METHOD OF SOLUTION
TEXT PROBLEMS IN MATHEMATICS
Section "Mathematics"
Completed by: Lyudmila Bryantseva,
Student of class 9 A, Municipal Educational Institution Secondary School No. 64,
Lowly Mary,
Student of class 9 A, Municipal Educational Institution Secondary School No. 64.
Head: Noskova Irina Anatolyevna,
Mathematics teacher, Municipal Educational Institution Secondary School No. 64
Volgograd 2014
Introduction ……………………………………………………………………………… 3
Chapter 1. Non-standard methods problem solving
- Tasks on the topic " Integers" ………………….. 5
- . Problems “in parts and percentages” …………………………... 8
- Movement problems……………………………………...... 11
- Collaboration tasks…………………………… 14
Conclusion ………………………………………………………. 16
Literature………………………………………………………. 16
Introduction.
It is known that historically for a long time mathematical knowledge was passed down from generation to generation in the form of a list of practical problems along with their solutions. Initially, mathematics was taught using models. The students, imitating the teacher, solved problems based on a certain “rule”. Thus, in ancient times, someone who knew how to solve certain types of problems encountered in practice (in trade calculations, etc.) was considered trained.
One reason for this is that historically, for a long time, the goal of teaching children arithmetic was to teach them a specific set of computational skills related to practical calculations. At the same time, the line of arithmetic - the line of numbers - had not yet been developed, and teaching calculations was carried out through tasks. In “Arithmetic” L.F. Magnitsky, for example, fractions were considered as named numbers (not just, A ruble, pood, etc.), and actions with fractions were studied in the process of solving problems. This tradition continued for quite a long time. Even much later, problems with implausible numerical data were encountered, for example: “ Sold kg of sugar per ruble per kilogram...",which were brought to life not by the needs of practice, but by the needs of learning to calculate.
The second reason increased attention to the use of word problems in Russia is that in Russia they not only adopted and developed the ancient method of transmitting mathematical knowledge and reasoning techniques using word problems. With the help of problems, we learned to form important general educational skills related to text analysis, identifying the conditions of the problem and the main question, drawing up a solution plan, searching for conditions from which one can get an answer to the question. main question, checking the result obtained. An important role was also played by teaching schoolchildren to translate text into the language of arithmetic operations, equations, inequalities, and graphic images.
Another point that cannot be ignored when we talk about solving problems. Training and development are in many ways reminiscent of the development of humanity, therefore the use of ancient problems and various arithmetic methods for solving them allows you to go to historical context, which develops creativity. In addition, a variety of solutions awaken children’s imagination and allow them to organize the search for a solution in a new way each time, which creates a favorable emotional background for learning.
Thus, the relevance of this work can be summarized in several points:
Word problems are important means teaching mathematics. With their help, students gain experience working with quantities, comprehend the relationships between them, and gain experience in applying mathematics to solving practical problems;
The use of arithmetic methods for solving problems develops ingenuity and intelligence, the ability to pose questions and answer them, that is, develops natural language;
Arithmetic methods for solving word problems allow you to develop the ability to analyze problem situations, build a solution plan taking into account the relationships between known and unknown quantities, interpret the result of each action, check the correctness of the solution by composing and solving the inverse problem;
Arithmetic methods for solving word problems accustom one to abstractions, allow one to cultivate a logical culture, can contribute to the creation of a favorable emotional background for learning, the development of an aesthetic sense in relation to problem solving and the study of mathematics, arousing interest in the process of finding a solution, and then in the subject itself;
The use of historical problems and a variety of ancient (arithmetic) methods for solving them not only enriches the experience mental activity, but also allows us to master an important cultural and historical layer of human history associated with the search for solutions to problems. This is an important internal incentive to find solutions to problems and study mathematics.
From all of the above, we draw the following conclusions:
subject of researchis a block of text problems in mathematics for grades 5-6;
object of studyis an arithmetic way of solving problems.
purpose of the studyis consideration sufficient quantity text problems of a school mathematics course and the application of an arithmetic method of solving them;
tasks to achieve the research goalare the analysis and solution of word problems in the main sections of the course “Natural numbers”, “Rational numbers”, “Proportions and percentages”, “Motion problems”;
research methodis a practical search engine.
Chapter 1. Non-standard ways of solving problems.
- Problems on the topic “Natural numbers”.
At this stage of working with numbers, arithmetic methods for solving problems have an advantage over algebraic ones already because the result of each individual step in solving actions has a completely clear and specific interpretation that does not go beyond the scope of life experience. Therefore they are absorbed faster and better various techniques reasoning based on imaginary actions with known quantities, rather than a single solution method for problems with different arithmetic situations, based on the use of an equation.
1. We thought of a number, increased it by 45 and got 66. Find the number you thought of.
To solve the problem, you can use a schematic drawing to help you visualize the relationship between the operations of addition and subtraction. Especially effective assistance the drawing will be at more actions with unknown magnitude.We thought of the number 21.
2. In the summer, my window was open all day. In the first hour, 1 mosquito flew in, in the second - 2 mosquitoes, in the third - 3, etc. How many mosquitoes flew in per day?
Here we use the method of dividing all terms into pairs (the first with the last; the second with the penultimate, etc.), find the sum of each pair of terms and multiply by the number of pairs.
1 + 2 + 3 + … + 23 + 24 = (1 + 24) + (2 + 23) + …. + (12 + 13) = 25 12 = 300.
300 mosquitoes flew in.
3. The guests asked: how old were each of the sisters? Vera replied that she and Nadya have been together for 28 years; Nadya and Lyuba are 23 years old together, and all three are 38 years old. How old is each sister?
1. 38 – 28 = 10 (years) – Lyuba;
2. 23 – 10 = 13 (years) – Nadya;
3.28 – 13 = 15 (years) – Vera.
Lyuba is 10 years old, Nadya is 13 years old, Vera is 15 years old.
4. There are 30 students in our class. 23 people went on an excursion to the museum, 21 went to the cinema, and 5 people did not go on either the excursion or the cinema. How many people went on both the excursion and the cinema?
Let's consider solving the problem; the figure shows the stages of reasoning.
- 30 – 5 = 25 (persons) – went to the cinema, or to
Excursion;
- 25 – 23 = 2 (persons) – went only to the cinema;
- 21 – 2 = 19 (persons) – went to the cinema and to
Excursion.
19 people went to both the cinema and an excursion.
5. Someone has 24 bills of two types - 100 and 500 rubles each for a total of 4,000 rubles. How many 500 ruble bills does he have?
Since the resulting amount is a “round” number, it follows that the number of 100 ruble bills is a multiple of 1000. Thus, the number of 500 ruble bills is also a multiple of 1000. Hence we have - 100 ruble bills are 20; 500 rubles - 4 bills.
Someone has 4 bills of 500 rubles.
6. The summer resident came from his dacha to the station 12 minutes after the train left. If he had spent 3 minutes less on each kilometer, he would have arrived just in time for the train to leave. How far does the summer resident live from the station?
Spending 3 minutes less per kilometer, a summer resident could save 12 minutes at a distance of 12: 3 = 4 km.
The summer resident lives 4 km from the station.
7. The spring gives a barrel of water in 24 minutes. How many barrels of water does the spring produce per day?
Since we need to work around fractions, we don’t need to find which part of the barrel is filled in 1 minute. Let's find out how many minutes it will take to fill 5 barrels: 24 · 5 = 120 minutes, or 2 hours. Then in a day 24: 2 = 12 times more barrels will be filled than in 2 hours, that is, 5·12 = 60 barrels.
The spring produces 60 barrels per day.
8. In some areareplace old rails 8 m long with new ones 12 m long. How many new rails are needed instead of 240 old ones?
On a section 24 m long, instead of 3 old rails, 2 new ones will be installed. The rails will be replaced in 240: 3 = 80 such sections, and 80 · 2 = 160 new rails will be placed on them.
160 new rails will be required.
9. The bakery had 654 kg of black and white bread. After 215 kg of black and 287 kg of white bread were sold, there was an equal amount of both types of bread left. How many kilograms of black and white bread were there in the bakery separately?
1) 215 + 287 = 502 (kg) – sold bread;
2) 654 – 502 = 152 (kg) – bread left to sell;
3) 152: 2 = 76 (kg) of white (and black) bread left to sell;
4) 215 + 76 = 291 (kg) – there was initially black bread;
5) 287 + 76 = 363 (kg) – there was white bread initially.
There were 291 kg of black bread initially and 363 kg of white bread initially.
- Problems “in parts and percentages”.
As a result of working with tasks this section it is necessary to take a suitable value for 1 part, determine how many such parts fall on another value, their sum (difference), then get an answer to the question of the problem.
10. The first brigade can complete the task in 20 hours, and the second in 30 hours. First, the teams completed ¾ of the task while working together, and the rest of the task was completed by the first team alone. How many hours did the task take to complete?
Work performance tasks are less clear than movement tasks. Therefore, a detailed analysis of each step is necessary here.
1) If the first team works alone, then it will complete the task in 20 hours - this means that every hour it completes the entire task.
2) Arguing in a similar way, we obtain labor productivity for the second team - the entire task.
3) First, working together, the teams completedthe entire task. How much time did they spend?. That is, in one hour of joint work, both teams complete the twelfth part of the task.
4)Then they will complete the task in 9 hours, since(according to the main property of a fraction).
5) All that remains is to completetasks, but only to the first team, which completes in 1 hourthe entire task. So the first brigade needs to work 5 o'clock to bring the matter to an end, since.
6) Finally, we have 5 + 9 = 14 hours.
The task will be completed in 14 hours.
eleven . Volumes annual production from the first, second, and third wells are ratioed as 7: 5: 13. It is planned to reduce the annual oil production from the first well by 5% and from the second by 6%. By what percentage should the annual oil production from the third well be increased so that the total volume of oil produced per year does not change??
Problems with parts and percentages are an even more time-consuming and incomprehensible area of problems. Therefore, the most concrete way for us to understand them was through numerical examples. Example 1. Let annual oil production be 1000 barrels. Then, knowing that this production is divided into 25 parts (7+5+13=25, i.e. one part is 40 barrels) we have: the first tower pumps 280 barrels, the second – 200 barrels, the third – 520 barrels per year. If production decreases by 5%, the first rig loses 14 barrels (280·0.05 = 14), that is, its production will be 266 barrels. If production decreases by 6%, the second rig loses 12 barrels (200·0.06 = 12), that is, its production will be 188 barrels.
In just a year, they together will pump 454 barrels of oil, then the third tower will need to produce 546 barrels instead of 520 barrels.
Example 2. Let annual oil production be 1500 barrels. Then, knowing that this production is divided into 25 parts (7+5+13=25, i.e. one part is 60 barrels) we have: the first tower pumps 420 barrels, the second - 300 barrels, the third - 780 barrels per year. If production decreases by 5%, the first rig loses 21 barrels (420·0.05 = 21), that is, its production will be 399 barrels. With a 6% decline in production, the second rig loses 18 barrels(300·0.06 = 18), that is, its production will be 282 barrels.
In total, in a year they will pump together 681 barrels of oil, then the third tower will need to produce 819 barrels instead of 780 barrels.
This is 5% more than previous production, since.
It is necessary to increase the annual oil production from the third well by 5% so that the total volume of oil produced per year does not change.
We can consider another version of a similar problem. Here we introduce some variable, which is just a “symbol” of volume units.
12. The volume of annual oil production from the first, second and third wells is ratioed as 6:7:10. It is planned to reduce annual oil production from the first well by 10% and from the second by 10%. By what percentage should the annual oil production from the third well be increased so that the total volume of oil produced does not change?
Let the volumes of annual oil production from the first, second and third wells be equal to 6x, 7x, 10x of some volume units, respectively.
1) 0.1 ·6x = 0.6x (units) – reduction in production at the first well;
2)0.1 ·7x = 0.7x (units) – reduction in production at the second well;
3) 0.6x + 0.7x = 1.3x (units) – should amount to an increase in the volume of oil production at the third well;
The annual oil production from the third well must be increased by this percentage.
Annual oil production from the third well needs to be increased by 13%.
13. We bought 60 notebooks - there were 2 times more squared notebooks than lined ones. How many parts are there in a lined notebook? on a squared notebook; for all notebooks? How many lined notebooks did you buy? How many per cage?
When solving a problem, it is better to rely on a schematic drawing, which can be easily reproduced in a notebook and supplemented with the necessary notes as the solution progresses. Let the lined notebooks make up 1 part, then the squared notebooks make up 2 parts.
1) 1 + 2 = 3 (parts) – covers all notebooks;
2) 60: 3 = 20 (notebooks) – accounts for 1 part;
3) 20 · 2 = 40 (notebooks) – squared notebooks;
4) 60 – 40 = 20 (notebooks) – lined.
We bought 20 lined notebooks and 40 squared notebooks.
14. In 1892, someone thinks of spending as many minutes in St. Petersburg as he will spend hours in the village. How long will someone spend in St. Petersburg?
Since 1 hour is equal to 60 minutes and the number of minutes is equal to the number of hours, then someone in the village will spend 60 times more time than in St. Petersburg (travel time is not taken into account here). If the number of days spent in St. Petersburg is 1 part, then the number of days spent in the village is 60 parts. Since we are talking about a leap year, there are 366 per part: (60 + 1) = 6 (days).
Someone will spend 6 days in St. Petersburg.
15. Apples contain 78% water. They were dried a little and now contain 45% water. What percentage of their mass did the apples lose during drying?
Let x kg be the mass of apples, then it contains 0.78x kg of water and x – 0.78x = 0.22x (kg) of dry matter. After drying, the dry matter is 100 - 45 = 55 (%) of the mass of dry apples, so the mass of dry apples is 0.22x: 0.55 = 0.46x (kg).
So, during drying, the apples lost x - 0.46x = 0.54x, that is, 54%.
During drying, the apples lost 54% of their mass.
16. Grass contains 82% water. It was dried a little, and now it contains 55% water. How much mass did the grass lose during drying?
At initial conditions the live weight of the grass was 100% - 82% = 18%.
After drying, this value increased to 45%, but the total mass of the grass decreased by 40% (45: 18 · 10% = 40%).
The grass lost 40% of its mass during drying.
- Movement tasks.
These tasks are considered traditionally difficult. Therefore, there is a need to analyze in more detail the arithmetic method for solving this type of problem.
17. Two cyclists travel from point A to point B at the same time. The speed of one of them is 2 km/h less than the other. The cyclist who arrived at B first immediately turned back and met another cyclist 1 hour 30 minutes later. after leaving A. At what distance from point B did the meeting take place?
This problem is also solved using the example of subject images and associations.
After a number of examples have been considered, and no one doubts the number - the distance is 1.5 km, it is necessary to justify its finding from the data of the presented problem. Namely, 1.5 km is the difference in the lag of 2 from the 1st cyclist in half: in 1.5 hours the second will lag behind the first by 3 km, since 1 returns, then both cyclists get closer to each other by half the difference in the distance traveled, that is, by 1 .5 km. This implies the answer to the problem and the method for solving this kind of word problems.
The meeting took place at a distance of 1.5 km from point B.
18. Two trains left Moscow for Tver at the same time. The first passed at the hour 39 versts and arrived in Tver two hours earlier than the second, which passed at the hour 26 versts. How many miles from Moscow to Tver?
1) 26 · 2 = 52 (versts) – how much is the second train behind the first;
2) 39 – 26 = 13 (versts) – this is how much the second train lagged behind the first in 1 hour;
3) 52: 13 = 4 (h) - this is how long the first train was on the way;
4) 39 · 4 = 156 (versts) – the distance from Moscow to Tver.
From Moscow to Tver 156 versts.
- Collaboration tasks.
19. One team can complete the task in 9 days, and the second in 12 days. The first team worked on this task for 3 days, then the second team finished the job. In how many days was the task completed?
1) 1: 9 = (tasks) – will be completed by the first team in one day;
2 ) 3 = (tasks) - completed by the first brigade in three days;
3) 1 - = (tasks) – completed by the second brigade;
4) 1: 12 = (tasks) – will be completed by the second team in one day;
5) 8 (days) – the second team worked;
6) 3 + 8 = 11 (days) – spent on completing the task.
The task was completed in 11 days.
20. A horse eats a load of hay in a month, a goat in two months, a sheep in three months. How long will it take a horse, goat and sheep to eat the same load of hay together?
Let the horse, goat and sheep eat hay for 6 months. Then the horse will eat 6 carts, the goat – 3 carts, the sheep – 2 carts. There are only 11 carts, which means they arecart, and one cart will be eaten for 1:= (months).
A horse, goat, sheep will eat a cartload of hay for months.
21. Four carpenters want to build a house. The first carpenter can build a house in 1 year, the second in 2 years, the third in 3 years, the fourth in 4 years. How long will it take them to build a house if they work together?
In 12 years, each individual carpenter can build: the first - 12 houses; second – 6 houses; third – 4 houses; fourth – 3 houses. Thus, in 12 years they can build 25 houses. Therefore, working together, they will be able to build one yard in 175.2 days.
Carpenters will be able to build a house by working together in 175.2 days.
Conclusion.
In conclusion, it should be said that the problems presented in the study are only a small example of the use of arithmetic methods in solving word problems. One thing must be said important point– choosing the plot of the tasks. The fact is that it is impossible to foresee all the difficulties when solving problems. But nevertheless, at the moment of initial mastery of a method for solving any type of problem, their plot should be as simple as possible.
The given samples represent a special case, but they reflect the direction - bringing the school closer to life.
Literature
1. Vileitner G. Reader on the history of mathematics. – Issue I. Arithmetic and algebra / trans. with him. P.S. Yushkevich. – M.-L.: 1932.
2.Toom A.L. Text problems: applications or mental manipulatives // Mathematics, 2004.
3.Shevkin A.V. Text problems in a school mathematics course. M, 2006.
Solving problems using arithmetic methods
Math lesson in 5th grade.
“If you want to learn to swim, then boldly enter the water, and if you want to learn to solve problems, then solve them.”.
D. Polya
Goals and objectives of the lesson:
developing the ability to solve problems using an arithmetic method;
development creativity, cognitive interest;
development logical thinking;
nurturing love for the subject;
fostering a culture of mathematical thinking.
Equipment: signal cards with numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
During the classes
I. Organizational moment (1 min.)
The lesson is devoted to solving problems using an arithmetic method. Today we will solve problems different types, but all of them will be solved without the help of equations.
II. Historical reference (1 min.)
Historically, for a long time, mathematical knowledge was passed down from generation to generation in the form of a list of practical problems along with their solutions. In ancient times, someone who knew how to solve certain types of problems encountered in practice was considered trained.
III. Warm-up
(solving problems orally - 6 min.)
a) Problems on cards.
Each student is given a card with a problem, which he solves orally and gives an answer. All tasks for action 3 - 1 = 2.
(Students solve the problems correctly, and some not. All orally. They raise the cards and the teacher sees who solved the problem; the cards should contain the number 2.)
b) Problems in verse and logic problems. (The teacher reads the problem aloud, the students raise the card with the correct answer.
The hedgehog gave the ducklings
Which one of the guys will answer?
Eight leather boots
How many ducklings were there?
(Four.)
Two nimble piglets
They were so cold, they were shaking.
Count and say:
How many felt boots should I buy them?
(Eight.)
I entered a pine forest
And I saw a fly agaric
Two honey mushrooms,
Two morels.
Three oil cans,
Two lines...
Who has the answer ready:
How many mushrooms did I find?
(Ten.)
4. Chickens and dogs were walking in the yard. The boy counted their paws. It turned out to be ten. How many chickens and how many dogs could there be? (Two dogs and one chicken, one dog and three chickens.)
5. According to a doctor’s prescription, we bought 10 tablets at the pharmacy. The doctor prescribed me to take 3 tablets a day. How many days will this medicine last? (Full days.)
6. Brother is 7 years old, and sister is 5. How old will sister be when brother is 10 years old?
7. Given numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. which is greater: their product or sum?
8. When building the fence, the carpenters placed 5 pillars in a straight line. The distance between the posts is 2 m. What is the length of the fence?
IV. Problem solving
(Tasks for children are given on cards - 15 minutes. Children solve problems at the board)
Tasks a) and b) are aimed at repeating the connection between the relations “by... more” and “by... less” with the operations of addition and subtraction.
a) A turner's apprentice turned 120 parts per shift, and the turner turned 36 parts more. How many parts did the turner and his apprentice turn together?
b) The first team collected 52 devices during the shift, the second? - 9 devices less than the first, and the third - 12 devices more than the second. How many devices did the three teams collect during the shift?
Using problem c), students can be shown the solution to the problem “in reverse.”
c) There are 44 girls in three classes - this is 8 less than boys. How many boys are there in three classes?
In problem d) students can propose several solutions.
d) Three sisters were asked: “How old is each of the sisters?” Vera replied that she and Nadya were 28 years old together, Nadya and Lyuba were 23 years old together, and all three were 38 years old. How old are each of the sisters?
Task e) is intended to repeat the connection between “more in...” and “less in...”.
e) Vasya had 46 marks. Over the course of a year, his collection increased by 230 stamps. How many times has his collection increased?
V. Physical education minute (2 minutes.)
Stand on one leg
It's like you're a steadfast soldier.
Raise your left leg.
Look, don't fall.
Now stand on the left,
If you are a brave soldier.
VI. Ancient, historical problems. Problems with fairy tale content (10 min.)
Problem e) to find two numbers by their sum and difference.
e)(from “Arithmetic” by L.N. Tolstoy)
Two men have 35 sheep. One has 9 more than the other. How many sheep does each person have?
Movement task.
and)(An old problem.)Two trains left Moscow for Tver at the same time. The first passed at 39 versts per hour and arrived in Tver two hours earlier than the second, which traveled 26 versts per hour. How many miles from Moscow to Tver?
(It's easier to get to the answer using an equation. But students are encouraged to look arithmetic solution tasks.)
1) 26 * 2 = 52 (versts) - the second train was so many miles behind the first;
2) 39 - 26 = 13 (versts) - by so many miles the second train was 1 hour behind the first;
3) 52: 13 = 4 (h) - that’s how long it took the first train to travel;
4) 39 * 4 = 156 (versts) - the distance from Moscow to Tver.
You can look in reference books to find the distance in kilometers.
1 verst = 1 km 69 m.
The task is divided into parts.
h)Kikimora's task.The merman decided to marry the kikimore Ha-Ha. He planted several leeches on his kikimore veil, and twice as many on his cape. During the holiday, 15 leeches fell off, and only 435 remained. How many leeches were on the kikimora’s veil?
(The problem is given to be solved using an equation, but we solve it in an arithmetic way)
VII. Live numbers (unloading pause - 4 min.)
The teacher calls 10 students to the board and gives them numbers from 1 to 10. The students receive different tasks;
a) the teacher calls the numbers; those named take a step forward (eg: 5, 8, 1, 7);
b) only the neighbors of the named number come out (for example: the number 6, 5 and 7 come out);
c) the teacher comes up with examples, and only the one who has the answer to this example or problem comes out (for example: 2 ´ 4; 160: 80; etc.);
d) the teacher makes several claps and also shows a number (one or two); a student must come out whose number is the sum of all heard and seen numbers (for example: 3 claps, number 5 and number 1.);
what number is 4 greater than four?
I thought of a number, subtracted 3 from it, I got 7. What number did I think of?
if you add 2 to the intended number, you get 8. What is the intended number?
We must try to select tasks so that the same numbers are not repeated in the answers, so that everyone can actively participate in the game.
VIII. Summing up the lesson (2 minutes.)
- What did we do in class today?
- What does it mean to solve a problem using arithmetic?
- We must remember that the solution found to the problem must satisfy the conditions of the problem.
IX. Homework assignment. Grading (2 minutes.)
№ 387 (solve problems using arithmetic method), for weak students. For average and strong students, homework assignments are given on cards.
1. The bakery had 645 kg of black and white bread. After selling 215 kg of black and 287 kg of white bread, there was an equal amount of both types of bread left. How many kilograms of black and white bread were there in the bakery separately?
Brother and sister found 25 porcini mushrooms in the forest. The brother found 7 more mushrooms than his sister. How many porcini mushrooms did your brother find?
For the compote, we took 6 parts of apples, 5 parts of pears and 3 parts of words. It turned out that the pears and plums together took 2 kg 400 g. Determine the mass of the apples taken; mass of all fruits.
Literature
Vilenkin N., Zhokhov V., Chesnokov A.Mathematics. 5th grade. - M., “Mnemosyne”, 2002.
Shevkin A.V.Text problems in a school mathematics course. - M.: Pedagogical University “First of September”, 2006.
Volina V.Holiday of numbers. - M.: Knowledge, 1994.
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