Triple somersault On marine yachts, the ballast is low, which does not allow them to heel too much and generally capsize. However, it happens that the yacht still flies somersaults, like a ballastless iol, and this happens only here - in the great Southern Ocean. I know

Problems solved with the help of proportions are traditionally studied in the course of arithmetic in grades 5–6. It is believed that it is at this age that students should learn to solve proportions, get acquainted with two practically important dependencies - direct and inverse proportions, learn to distinguish them and solve the corresponding problems. The study of proportions and the indicated dependencies has little to do with the needs of the arithmetic course itself or with the needs of teaching problem solving in grade 6 - there are no direct and inverse proportional problems in textbooks that could not be solved without proportions. However, the use of proportions is of great importance for the subsequent study of mathematics. In grade 6 textbooks, it is often proposed to solve percentage problems using proportions. Although, in our opinion, the solution of problems for percentages does not require the use of proportions.

Let's consider a technique for solving problems on proportions, which, apparently, belongs to chemistry teachers, tired of students' poor knowledge of percentage calculations. It boils down to advice: on the record

400 gsolution - 100%

20 g salt - x%

separate two lines of numeric data in two lines, bring the two lines together until you get a sign
"=" and solve the resulting proportion:

400 / 20 = 100 /X.

Sometimes in the process of solving the proportion is not fixed explicitly. For example, in the student manual "500 problems in chemistry" (Prosveshchenie, 1981), a brief record of the solution is given:

b) 32 gsulfurs combine with 32 g of oxygen, and

x g » 8 g »

x = 32 8/ 32 = 8(d).

in) 32 gsulfurs combine with 48 g of oxygen, and

x g » 8 g »

x = 32 8/ 48 = 5.33(g).

As you can see, here the proportions are left “behind the scenes”, students can multiply and divide numbers “crosswise”. There is nothing reprehensible in this way of designing a solution, it is quite possible to use it when solving a large number similar tasks in chemistry lessons. True, we would not apply the cumbersome general trick in the obvious case "b" and use the sign "=" instead of "≈" in the case "c". But we are sure that if the student does not understand proportions and cannot explain the meaning of his actions, then solving problems according to the model will be of little use for his development.

Good for chemists! They deal with direct proportionality. And students in grade 6 (especially those who missed the teacher’s explanation) sometimes bring from home this way of solving the first problem without proportion: “we multiply the numbers crosswise: 20 times 100, x- by 400, we equate the results obtained and find x". It is difficult for such students to teach the use of proportions, since they consider their own method to be simpler, but this difficulty is easily removed after trying to solve inverse proportionality problems using the “crosswise” method.

Note that the rule "multiply and divide crosswise" is akin to the rules that were used in the old days when solving arithmetic problems. Let us take advantage of this circumstance and return once more to the history of the question. But first, let's clarify the terminology.

AT old times to solve many types of problems, there were special rules for solving them. Problems familiar to us on direct and inverse proportionality, in which it is necessary to find the fourth by three values ​​of two quantities, were called problems on the triple rule (simple triple rule). If five values ​​were given for three quantities and it was required to find the sixth, then the rule was called five. Similarly, for the four quantities there was a "septenary" rule. These rules were also called tasks for the complex triple rule.

In the introductory article to the first paragraph of our book, we cited a fragment from the book by I. Beshenshtein (1514), which reflects the almost mystical attitude of teachers to triple rule, and the presentation of the material itself has a pronounced prescription character. Training according to the rules was widespread in Russia as well. Wishing to describe the methodology of teaching solving problems of the times of L.F. Magnitsky, we will refer to S.I. Shokhor-Trotsky, who in his “Methods of Arithmetic for Teachers of Secondary Educational Institutions” wrote: “How much the books on arithmetic abounded in the old days with rules can be judged by the work of Leonty Magnitsky, very respectable for its time ... In the first book ... in addition to many rules about integer and fractional numbers, the rules are set out, which the author calls “similar” (now called triple) ... the author distinguishes: the rule is triple in integers, the rule is triple in fractions, the rule is triple contractile, the rule is “reactive” (inversely proportional), the rule is five, the rule is “septenary” ..., and then, in the form of applying these rules, he proposes a number of "articles": a triple trade article ("in whole" and "in shares"), a triple trade article on purchases and sales, a triple trade article in marketable vegetables and "with a sign" ( that is, about the calculation of the container of goods), about "buy-in" and "overheads", "question" about the triple rule, "question from the time", "business in the triple rule", trade "exchange in the triple rule" ...

Further S.I. Shokhor-Trotsky cites a fragment from "Arithmetic" by L.F. Magnitsky, from which it is clearly seen that the prescription style of presentation of the material, characteristic of earlier European sources, has not yet been overcome in the first Russian arithmetic textbook. In this fragment on the application of the five rule, first the definition of the rule and an example of its application are given (the text of the problem is in italics here), then the recipe for obtaining an answer; in other cases it is recommended to do the same.

“There is a five-fold rule, when such estimates happen, they cannot be understood by any other rank or rule, only through this five-point or five-point rule, the three-fold one is also spoken ... because five lists [numbers] are supplied in the rule, and the sixth is invented ...: someone who has a hundred rubles in the merchants for one year, and if they acquire only 7 rubles, and he gives the packs 1000 rubles to the merchants for 5 years, how much he will acquire with them, and you do shite, putting the beginning of the triple rule:

year year

100 –––––– 1 –––––– 7 –––––– 1000 –––––– 5

And multiply the two lists from the left hand among themselves, also the other three like right hand, so multiply among yourselves in order, and divide their product by that product from the first two, having produced: as it is here. [ibid.]

We will talk about the possibility of using tasks of this kind in the learning process, but for now, following the rule, we will get the correct answer:

(7 1000 5):(100 1) = 350 ( R.).

At the time of S.I. Shokhor-Trotsky still preserved the tradition of solving problems according to the rules. The most famous arithmetic textbook of that time was A.P. Kiseleva (first edition in 1884). In order for the reader to get an idea of ​​the method of presenting the material related to tasks on the triple rule in this textbook, to be able to imagine the practice of teaching schoolchildren to solve problems on direct and inverse proportionality at that time, we will give a few excerpts from the 9th edition of this textbook (1896). .). Our comments in the text are in italics.

A simple rule of three.

Tasks for this rule are solved by the method of proportions or reduction to unity.

A task. 8 arshins of cloth cost 30 rubles; how much is 15 arshins of this cloth?

With p about with about about about about and y. Let's denote by a letter x price
15 ars. cloth and arrange the numbers like this:

The number of arshins. Their cost.

8 arsh. . . . . . . . 30 rub.

fifteen " . . . . . . . x »

Since the cost of cloth is proportional to the number of arshins, then

x : 30 = 15: 8.

Where: x\u003d 30 15 / 8 \u003d 56 1 / 4 rubles.

Reduction to unity. To solve the problem in this way, we first find out how many rubles cost 1 arshin (from this the very method is called reduction to unity). For clarity, we will arrange the solution in lines:

8 arsh. cost 30 rubles.

1 arsh. costs 30/8 rubles.

8 arsh. cost 30 / 8 · 15 \u003d 56 1/4 rub.

Note that the presentation of the material in the textbook could be simpler. After all, the second way to solve the problem is just another record of the decision by actions:

1) 30: 8 = 30 / 8 (rub.); 2) 30 / 8 15 = 56 (rub.)

In this way, but with the cost of the cloth expressed in kopecks, the students had to be able to solve the problem even before learning the operations with fractions. The method of reduction to unity with the intentional preservation of reducible fractions was necessary for presenting the solution of the problem for a complex triple rule, for the “final formula”, for teaching schoolchildren to sequentially change first one value (as here), and then several quantities (as in solving problems on complex triple rule).

Also, in two ways (first with the help of proportion, then reduction to unity), the problem of inverse proportionality was also solved.

A way to solve such problems in which one corresponding value of two quantities is given, directly or inversely proportional, but it is required find, what value will one of them take if the other receives a new given value, called simple triple rule.

The following is a task for a complex triple rule, the complexity of which exceeds the needs of the initial training - here it would be enough to take three values, not four (that is, take the task for the five rule, as in L.F. Magnitsky, and not for the “septenary”) .

Complicated triple rule.

A task. For lighting 18 rooms in 48 days, 120 pounds were spent. kerosene, with 4 lamps burning in each room. How many days will 125 lb. kerosene, if 20 rooms are illuminated and 3 lamps are lit in each room?

With p about with about about about about and y. Let's arrange the data of this task in two lines:

twenty " - X» – 125 » – 3 »

If we leave the number of pounds and lamps unchanged (these quantities are in brackets), then we can find x 1 is the number of days corresponding to 20 rooms, solving the simple triple rule problem.

18 rooms – 48 days - 120 lbs. – 4 lamps

twenty " - X 1" - 120" - 4"

X 1 \u003d 48 18 / 20 \u003d 216 / 5 (days).

20 rooms – 216 / 5 days – 120 pounds – 4 lamps

twenty " - X 2" - 125" - 4"

X 2 = 216 125 / 5 120 = 45 (days).

Now let's replace 4 lamps with 3 lamps:

20 rooms – 45 days - 125 lb. – 4 lamps

twenty " - X» – 125 » – 3 »

X= 45 4 / 3 = 60 (days).

A way to solve such problems when there are more than two given quantities, called. complex triple rule.

A n d e r t i o n to unity. ... Let us arrange, for convenience, the data and the desired number so that x stood in the last column on the right:

20 » 125 » 3 » x »

Now we will find out what the number of days will be if it is illuminated 1 room, kerosene will 1 pound and every room will have 1 lamp. This we learn by leading to 1 gradually one condition after another.

18 rooms 120 lb. 4 lamps 48 days

1" 120" 4" 48 18"

1 » 1 » 4 » 48 18 / 120 »

1 » 1 » 1 » 48 18 4 / 120 »

Now we will gradually replace the units with the numbers given in the question of the problem:

1 room 1 lb. 1 lam. 48 18 4 / 120 days.

20 » 1 » 1 » 48 18 4 / 120 20 »

20 » 125 » 1 » 48 18 4 125 / 120 20 »

20 » 125 » 3 » 48 18 4 125 / 120 20 3 »

It remains to reduce the resulting formula and calculate.

F u n t for m u l a l . With sufficient skill in solving problems for a complex triple rule, you can immediately write final formula for x. Let's show how it's done. Let's take the problem above:

18 rooms – 48 days - 120 lbs. – 4 lamps

twenty " - X» – 125 » – 3 »

The number of days would be 48 if 18 rooms were illuminated; if only one room was illuminated, then there would be 48 days · 18, and when lighting 20 rooms, the days should be 48 18 / 20 (under the same other conditions). Such a number of days would be subject to 120 pounds of kerosene; if there were 1 pound of kerosene, then the number of days would be 48 18 / 20 120, and with 125 pounds of kerosene it should be 48 18 125 / 20 120. Such a number of days would be subject to 4 lamps; with 1 lamp it was 48 18 125 4 / 20 120, and with 3 lamps it should be:

x = 48 18 125 4 / 120 20 3 , or x= 48 18/20 125/120 4/3.

Rule. To get the desired number, it is enough to multiply the given value of the same magnitude sequentially by the ratio of the given values ​​of the other magnitudes, taking the ratio of the new value to the previous one, if the value is directly proportional to the one whose value is being sought, and the old value to the new one, when the value is inversely proportional to the value which is being sought.

To memorize and accurately apply this rule was apparently not so easy. Let us pay attention to the fact that it was supposed to pass to the final formula "with sufficient skill in solving problems on a complex triple rule" in the first two ways. Is it any wonder that such training was difficult and of little use for students, and aroused objections from teachers and methodologists. So, for example, in the program for the I and II stages of the seven-year school of the unified labor school of 1921, it is quite clearly written: “All the rest of the “rules” are remnants of the past and nonsense, not even natural, but artificial.” And further: "The complex triple rule covers the collection artificial tasks which should be thrown out of school life long ago due to their meaninglessness.

Such a sharp categoricalness of the authors of the program, apparently, was connected not so much with the tasks themselves (their conditions could well be brought closer to the experience of the child), but with the little useful method of teaching schoolchildren to solve problems "according to the rules." The above fragments of the text from the textbook by A.P. Kiseleva give an idea of ​​the method of presenting the material of interest to us in pre-revolutionary textbooks. Note that in the revised version of the textbook in 1938, the problems for the complex triple rule were still preserved, and a little more than a page of the textbook is devoted to the analysis of one such problem - immediately for the "septenary" rule. However, only the "final formula" is considered here and the rule is not formulated. Obviously, this change did not solve the problem of using tasks of the type in question.

Only by simplifying the methodology for using this type of problem can one usefully preserve a whole class of traditional problems in school practice. As we will see later, many of them may have content that is quite close to practice, and the implementation preparatory work when learning to solve problems on a simple triple rule and building a chain of problems from simple to complex, they will increase the availability of problems of this type. True, the question remains unresolved: should all students be trained to solve such problems? The answer to it depends on what we see practical value learning to solve text problems - only in learning to solve problems encountered in practice or, in addition, in the development of schoolchildren's thinking in the process of solving a wide variety of problems, including artificial ones. The achievement of the second goal may well be facilitated by the use of tasks for a complex triple rule in the educational process. Of course, the requirement to be able to solve such problems cannot be mandatory for all students, but participation in the analysis of their solution, training in distinguishing between direct and inverse proportions will be useful to each of them.

As for the use of direct and inverse proportionality problems in modern textbooks, then in the textbook N.Ya. Vilenkin and others. direct and reverse proportional dependencies item 22 is allocated. It contains 18 tasks. Moreover, starting from the samples in the educational text, the corresponding values ​​of the quantities are expressed in decimal fractions or natural numbers, whose ratios are not expressed as integers. This makes learning difficult. In addition, a third of the tasks are tasks for percentages. When initially learning the use of proportions, it is better to separate the difficulties: study proportions separately from decimal fractions and percent. AT the following paragraphs In the textbook, from time to time there are tasks “for proportion”, but there are not many of them and most of them are also easy to solve without proportions.

Thus, the proportions themselves do not greatly enrich the arsenal of methods for solving problems used by schoolchildren in the process of studying the entire course of mathematics in grades 5–6, and without increasing the complexity of the problem, direct and inverse proportionality do not have the desired effect on the development of schoolchildren. On a small number of simple tasks of the same type, it is not always possible to achieve another important goal - to teach schoolchildren to distinguish well between direct and inverse proportionality.

We do not claim that in the old days, direct and inverse proportionality problems were used much more effectively. But still more diverse tasks, including tasks “for a complex triple rule”, left the teacher with the opportunity to develop the strongest students. That is why we recommend teachers to use in their work with all students, especially with the most prepared of them, these now almost forgotten tasks. Of course, we will simplify their inclusion in studying proccess and make the necessary adjustments to the method of teaching them to solve them. We do not at all propose to teach all schoolchildren to solve such problems as the problem about kerosene lamps, and in exactly the same way that was shown above. Perhaps this task should be made the last in the chain of tasks, solving which the student will be able not only to understand the solutions offered by the teacher, but also independently move forward from simple to complex. Such work would be more useful than marking time when solving problems of the same type of the same complexity, it would allow students to get a good training in distinguishing between direct and inverse proportionality. Where should one start?

First, we need to teach students how to solve proportions. The main way to solve them should be based on the main property of proportions. When this goal is achieved, then you can show the use of proportion properties to simplify their solution. For example, to solve the proportion
X/ 5 \u003d 1 / 10, you can multiply the right and left sides of the equality by 5 or swap the middle members of the proportion.

Secondly, it is necessary to teach schoolchildren to single out two quantities in the conditions of problems, to establish the type of dependence between them.

Thirdly, you need to teach them to make a proportion according to the condition of the problem.

Thus, students will master the minimum range of skills provided for by the current program in mathematics. Only after that, in order to prepare for the solution of more challenging tasks to proportional values ​​(a complex triple rule), you need to show students a way to solve the studied problems without any proportions at all. Let's solve the problem:

- At a speed of 80 km / h, a freight train traveled 720 km. What distance will be covered then the same time as a passenger train with a speed of 60 km/h?

The path is proportional to the speed at a constant time of movement, which means that with a decrease in speed by 80 / 60 times, the path will decrease by 80 / 60 times.

720: 80 / 60 = 540 (km).

The problem is solved in the same way if the speed did not decrease, but increased, if the values ​​are not directly, but inversely proportional. Of course, the first application of this technique should be preceded by questions asked in solving previous problems: how many times has this value increased (decreased)? The first answers to them should be expressed in whole numbers, and then in fractions, always obtained by division greater value values ​​to a smaller one. Only after students learn how to determine how the value of the second quantity will change with a corresponding change in the first, can they proceed to solving problems first with two quantities (triple rule), then with three and four quantities (complex triple rule).

There is no expression strong enough that the compilers of medieval arithmetic would be stingy with to praise the triple rule. “That line is triple laudable and the best line of all other lines.” "Philosophers call it the golden line." In German textbooks, he was referred to as one that is “above all praise”, it is the “key of merchants”. In the same way, among the French, it was known under the name of règle dorée - the golden rule. It was opposed to the whole science of algebra.

Why, then, are such immoderate praises given to a department, which in our time is accustomed to occupy a more modest place? It is very interesting to find out, and we allow ourselves to go back a little and give brief description goals pursued by arithmetic since ancient times.

Any science in the initial stage of its development is caused by practical needs and strives, in turn, to satisfy them. Then, depending on the conditions under which it develops, science sometimes rather quickly, sometimes more slowly takes on a theoretical coloration and acts educationally on those who study it, i.e., improves their spiritual abilities: mind, feeling and will: with slow growth, science remains for a long time the leader of skill, imparts only skill, gives a person mechanical skills and gives him the features of mechanicalness. Both directions were tested by arithmetic. On the one hand, the Greek scholars saw in arithmetic, most of all, an educational element; they constantly asked “why?” questions. and “why?”, always looking for reasons and conclusions; the students of the Greek schools delved into the essence of science, thought about it, and therefore the study acted on them in an educational and developing way. On the other hand, the Hindus looked at arithetics rather from the side of art, they did not like the question "why?", but their main question was always: "how to do it?" The direction of the Hindus passed to the Arabs, and from there to medieval Europe. In it, it met with an extremely cordial reception, and the soil for it turned out to be quite grateful: after the great migration of peoples and with incessantly ongoing wars, there was nothing to even think about the development of an exact, frequent, abstract science, and at the time it was necessary to confine oneself to its applied part, it was enough only teach "how to do" rather than "why to do it." And so the practical coloring remained behind arithmetic on for a long time, almost to the present day, at the same time, its study was narrowly mechanical: without conclusions, explanations, without deepening into the foundations; everywhere in the textbooks there was “do this”, “you must do this”, and the student had only to confirm and apply to the case; our Magnitsky also has a number of characteristic expressions “see the see”, “see the invention”; suppose that among these expressions he has "think and come", but how exactly to think, very few hints are given. In accordance with the practical significance of arithmetic, everything that could bring direct benefit, deliver earnings, was especially distinguished and valued in it.

“Whoever knows this wisdom,” says the Russian arithmetic of the 17th century, “can be with the sovereign in great honor and in salary; according to this wisdom, guests trade in states, and in all sorts of goods and trades, they know strength and in all sorts of weights and measures, and in earthly layout and in the sea current, they are evilly skilled, and they know the account from any number of the list.

But what part of arithmetic can give more practical, directly applicable skills than problem solving? Therefore, all the efforts of medieval authors were directed towards collecting as many problems as possible and, at the same time, the most diverse everyday content. Here were problems a about sale and purchase, about bills of exchange and about interest, about mixing, about exchange; the diversity was terrible and there was no way to sort out the whole mass of problems. In order to group at least a little and introduce some system and order, they tried to distribute all tasks by departments or types. This idea, of course, is a good one, but it was usually carried out very unsuccessfully, and tasks were distributed not according to the methods of their solution, as it should, but according to their content, that is, according to their appearance; for example, there was a special kind of problem about dogs chasing a hare, about trees, about girls, etc.

The solution of problems with division according to their content did not bring almost any benefit, because it did not help in the least to better understand the solution. And, in the opinion of ancient authors, it was hardly necessary to understand.

“That’s nothing,” the mentor used to console his pupils: “that you don’t understand anything, you won’t understand much ahead either.”

Instead of understanding, it was recommended not to be carried away, but to memorize everything that was asked, and then try to apply it to the case, that is, to examples, and all the power of understanding was concentrated not on understanding the conclusion of the rule, but on a more modest one, on that how to apply general rule to examples.

And so the triple rule was outstanding and worthy of special attention in many respects. Firstly, the range of his tasks is quite extensive, secondly, the rule itself is expressed quite simply and clearly, and thirdly, it was relatively easy to apply this rule. For all these merits, he was given the name "gold", "the key of merchants", etc.

The triple rule originated with the Hindus, where its tasks were solved for the most part by reduction to unity. The Arab scholar Alkhvarizmi (9th century A.D.) attributed it to algebra. Leonardo Fibonacci, 13th century Italian according to R. X., devotes a special section to the triple rule under the title: ad majorem guisam, where tasks are given for calculating the value of goods. Example: 100 rotuli (Pisan weight) cost 40 lire, what does 5 rotuli cost? The condition was written like this:

The rule prescribed to solve this problem in the following order: the product of 40 by 5 divided by 100.

Particular attention has been paid to the triple rule since the 16th century, that is, since the time when European trade and industry immediately moved forward, thanks to important inventions and the discovery of new countries. But this did not prevent us from developing this chapter in a completely unsatisfactory way, at least from our point of view. First of all, the rule was determined purely externally: “the problem consists of three numbers and gives itself a fourth number, just as if you put three corners of a house, then this will determine the 4th corner; the second number must be multiplied by the 3rd, and what happens, then divided by the 1st number. Such a definition could not but lead to inconsistency, and above all, the question was: what should be considered the first number, and can any problems with three given numbers be solved by the triple rule? The textbooks did not consider it necessary to clarify this misunderstanding. In addition, problems were solved not only with integers, but also with fractions, and in other arithmetic they were arranged so inconsistently that problems with fractional numbers on the triple rule, the chapters on fractions were placed earlier, because the entire triple rule went before the arithmetic of fractional numbers.

After the triple rule with integers and fractions, special rule"reducing", in which it was explained how it is possible to reduce some given numbers, and then the "reflexive" rule already went; it was a very confused department, to which questions with inverse proportion belonged, and the authors of the textbooks could not distinguish in any way which problems belonged to this group; the disciples had to rely on their own hunches and content themselves with ingenuity. In the XV and XXII centuries. the explanation was given as follows: “If a measure of grain costs 1½ marks, then two poods of bread are given for 1 mark; how many poods of bread will be given per mark if a measure of grain costs 1¾ marks; solve with the triple rule, it turns out

but the understanding one will realize that when grain rises in price, then they will give less bread, not more, so the question must be turned around, it will be

Magnitsky (1703) interprets in a similar spirit

“There is a return rule, when it is necessary in the assignment to put the third list instead of the first: it is necessary in civil frequent cases, as if speaking on the butt: a certain gentleman called for a carpenter and ordered the yard to be built, giving him twenty workers: and asked how many days then he will build his courtyard, he answered, in thirty days; but the master needs to build the whole thing in 5 days, and for this he asked the carpenter's packs, how many people are worth having, so that you can build a courtyard with them in 5 days, and that carpenter, perplexed, asks you arithmetically: how many people he deserves to have in order to build him that courtyard in 5 days, and if you start to create according to the order of the triple rule simply; then truly err; but it’s not appropriate for you: 30-20-5, but turning it into a sit: 5-20-30; 30X20=600; 600: 5=120".

The triple rule was followed by the five, followed by the seven. It is easy to guess that these are special cases of a complex triple rule, precisely when, according to 5 or 7 data, which are proportionally dependent on each other, the 6th or 8th, the corresponding number is found, in other words: the five-fold rule requires 2 proportions, and the seventh is three. The rule of five was explained in the eighteenth century as follows:

they make such calculations that cannot be made according to another rule; 5 numbers are given in it, and the sixth desired number is found from them; for example, someone put a hundred rubles into circulation, and they brought him a profit of 7 rubles, the question is how much profit he would receive with 100 rubles. for 5 years;

solved like this: 100-1-7-1000-5, multiply the two left numbers, and also multiply the 3 right numbers and divide the last product by the first, the answer will be 350, so many rubles of profit will give 1000 rubles. within 5 years.

A simple and complex triple rule was usually distributed in the 16th-18th centuries. into a mass of small departments, which bore very intricate names, depending on the content of the tasks. Here are these names according to Magnitsky: a “triple trading rule”, i.e., the calculation of the cost of the purchased goods; b “triple trading about purchases and sales”, - the same as the previous one, but only more complicated; c “triple trading in marketable vegetables and with a sign”, when you have to make a deduction for dishes and casings in general; d “on profit and loss”; e “a question article in the triple rule”, in it the tasks of a very diverse content, for the most part with inverse proportion; f “a questionable article with time”, where it is asked to calculate the duration of work, paths, etc.

At the beginning of the 19th century, Bazedov proposed another change in the triple rule and again in the same direction of mechanical, unconscious habit. This German teacher set himself the goal of simplifying the solution of problems on the triple rule even more, by further reducing the reasoning in solving them and replacing it with writing a ready-made formula. He advises to arrange the given numbers in 2 columns: in the left one is written an unknown amount and all those numbers that should be included in the numerators of the formula, and in the right one - all the factors that make up the denominator. Example: for the food of 1200 people for 4 months, 2400 centners of flour are required; how many people will 4000 centners come out in 3 months? We write 2 columns:

and get the answer formula

Why are the numbers 1200, 4000 and 4 included in the numerator, and 2400 and 3 in the denominator? This can be answered with the following rule: the numerator includes a number that is homogeneous with the desired one, that is, in our case, the number 1200; in addition, it also includes all those numbers of the second condition (4000 4), which are directly proportional to the desired one; if they are inversely proportional, as in our example 3, then they are replaced by the corresponding numbers of the 1st condition (4th).

That is all we can say about the historical development of the triple rule. From all that has been said, one can draw a conclusion that is suitable for our time. Medieval arithmetic, with its tendency to give only rules and omit conclusions, with its mechanical solution of questions, had too much influence on the whole subsequent school life, and so large that traces of it appear at every step even in our time. No matter how hard we try to shake off tradition, to free ourselves from habit, but they seize us too closely and are attracted to us too strongly to be thrown away without a trace. Our school is still guilty of rote learning of arithmetic, without sufficient participation of consciousness. The triple rule is a good proof of this. Often forgets our average and lower school that it is intended to provide a general education, and not to train accountants, clerks, accountants, etc. Meanwhile, the craft methods of Italians and Germans, who sought not to develop a person, but to make him calculating machine, are often used even now. Why all these rules: triple, mixtures, etc.? What purpose are they supposed to serve? They should be a conclusion from the solved problems, and not precede the solution of problems; it is harmful to solve problems according to a previously learned rule, but one must try to arrive at an answer by free personal consideration. In a word, the rule should not be understood in the form of a recipe, which is enough to memorize in order to prepare various intricate solutions according to it; but they should be valued only as a conclusion to which the student comes: if the student cannot draw this conclusion, then this means that the problems are taken little, or they are not arranged systematically, and this error must be corrected by a more systematic arrangement of problems; if the student does not draw such a complete and detailed conclusion as the teacher would like, then it is better to be satisfied with him than to force him to learn the rule imposed by the textbook: it will soon be forgotten and will not have a developing effect, since the necessary quality of a mathematical conclusion should be independence, but a necessary condition of consciousness there must be a close connection of all parts of the course, which is why there can be no place for the mechanical insertion into the head of separate pieces assimilated by memory.

Part Three

RELATIONS AND PROPORTIONS.

TASKS SOLVED WITH THE HELP OF PROPORTIONS AND
BY THE METHOD OF REDUCTION TO ONE.

SECTION VIII..

§ 50. Complicated triple rule.

2661. 45 masons were paid 216 rubles for six days of work; How much should 30 masons work for 8 days?

2662. 5 pumps pumped out 1800 buckets of water within 3 hours. How much water will be pumped out by 4 similar pumps in 4 hours?

2663. 25 workers dug a canal in 12 days, 36 fathoms long. What length of canal could be dug by 15 similar workers in 10 days?

2664. A capital of 100 rubles in 12 months brings 6 rubles of profit. How much profit will the capital of 8600 rubles bring in 4 months?

2665. From a rectangular field, 40 sazhens long and 30 sazhens wide, 6 quarters 2 quarters of oats were harvested. How many oats were harvested from another field, which is 96 fathoms long and 50 fathoms wide, if the sowing and harvest conditions for both fields were the same?

2666. For 15 pairs of dresses, 45 arshins of cloth 1 arshin wide were used. 14 inches. What width was the other cloth, if it went 60 arshins for 10 of the same pairs of dresses?

2667 .18 workers, working 7 hours a day, completed some work in 30 days and received 201 rubles for this. 60 kop. 14 employees, working daily for 4 hours, received 67.2 rubles for performing other work. Assuming that the hourly pay for the worker of both parties was the same, determine how many days the second party of workers worked.

2668. For the transportation of 420 poods of goods by rail over a distance of 24 versts, 2 rubles were paid. 52 kopecks. According to this calculation, for the transportation of 50 pounds of goods along the Nikolaev railway, from St. Petersburg to Moscow, 7 rubles should have been paid. 61 1/4 kop. Find the length of this road.

2669. 155 passenger tickets of the second class, taken by rail from Paris to Rouen, cost 1488 francs. Knowing that the price of 10 second-class tickets taken for a journey of 4 kilometers is equal to 3 francs, and that 16 kilometers are 15 versts, express in versts the length railway between Paris and Rouen.

2670. If the wheel of a machine that makes iron wire rotates at 60 revolutions per minute, then this machine will produce 240 arsh. wire for 3 hours 20 minutes. How long will it take her to make 33 1/8 fathoms of wire if the wheel makes 41 2/3 revolutions per minute?

2671. From a rectangular field, which is 125 sazhens long and 0.08 versts wide, 12 1/2 quarters of wheat were harvested; thus, the calculation showed a yield of self-six. From another rectangular field, whose length is 0.3 (9) versts, 8 1/3 quarters of wheat were harvested, which amounted to a crop of five. Assuming that the sowing conditions of both fields were the same, determine the width of the second field.

2672. The stone slab, 5.3 feet long, 0.8 feet wide, and 2 5/8 inches thick, weighs 4.2 pounds. Another slab of the same stone as the first weighs 7 poods 35 pounds and is 15 inches wide and 2 inches thick. How long is the second plate?

2673 . An iron strip, 2 arshins long, 1 1/2 inches wide and 2/3 inches thick, weighs 0.4375 pounds. How much will an iron strip weigh, which is 2 feet long, 1 3/7 inches wide and 0.16666 .... feet thick?

2674. 36 workers, working daily for 12 hours and 30 minutes, built wooden house in 30 days. How many hours a day must 27 workers work to build the same house in 50 days?

2675. The length of the corridor is 6 sazhens. 2 arsh. 9 1/7 inch, width 1.4(9) sazhens. and height 5, (3) yards (yard-English measure of length). Atmospheric air contained in the corridor weighs 17 pounds. 34 lbs. The air filling the room adjacent to the corridor weighs 11.9 pounds. Knowing that 0.58 (3) yards = 0.75 ars., and that the height of the room is 5 5/7 ars., and its width is 0.945 of the height, calculate the length of this room.

2676. For lighting the stairs of the house with 6 gas jets that burned for 40 evenings, for 6 hours and 12 minutes every evening, 22 rubles were paid to the gas company. 32 kopecks. On another staircase, 5 similar horns burned for 60 evenings, for which 27 rubles were paid. How many hours each evening did the gas burn on the second staircase?

2677 . For 4 lamps, which were lit every evening for 7 1/2 hours, 2.25 poods of kerosene were consumed during 30 evenings. In how many evenings will 1.8 poods of kerosene be used up if 5 such lamps are lit every evening for 4 hours and 30 minutes?

2678 . 32 masons, working daily for 8 1/2 hours, in 42 days laid down a brick wall 10 sazhens long, 7 1/2 inches thick and 1 sazhen 3.5 feet high. In how many days will 40 masons, of the same strength as the first, working daily for 6.8 hours, lay a brick wall 15 sazhens long, 0.9375 arshins thick and 2 1/2 arshins high?

2679. Length mail road between Vitebsk and Orel is 483 versts; one traveler covered this distance in 7 days, being in the city for 10 hours every day and traveling the same number of miles per hour. Another traveler left Vitebsk for Mogilev and, being on the road every day for 12 hours, made his way in 4 days. How many versts from Witsbsk to Mogilev, if it is known that the second traveler traveled 10 versts at the same time as the first traveler traveled 23 versts?

2680. Brick (clinker), 0.375 arshins long, 3 inches wide and 1 1/2 inches thick, weighs 10 pounds 38.4 spools. How much will it weigh rectangular shape a piece of marble, which is 8.75 inches long, 2 1/4 inches wide, and 2 inches thick, whereby marble is known to be 1 1/2 times heavier than brick?

2681. 25 weavers, working 8 1/3 hours a day, wove in 32 days 120 arshins of linen, 1 arshin wide. 5 1/3 inches. In how many days will 40 weavers, working daily for 4 hours and 10 minutes, weave 320 arshins of linen with a width of 0.75 arshins?

2682. The capital of 1200 rubles in 8 months brought 40 rubles of profit; what time 100 rub. will bring 5 rubles. arrived?

2683. A capital of 30,000 rubles in 7 1/2 months brought 1,125 rubles of profit. How much profit is brought by each 100 rubles of this capital within 1 year?

2684. The capital of 24,400 rubles for 10 months brought 1,525 rubles of profit. What kind of capital must one have in order for it, being in circulation under the same conditions as the first, to make 1,250 rubles of profit within 2 1/2 months?

2685. 54 diggers, working 10 hours a day, made a mound in 33 days, 124 fathoms long, 1 fathoms wide 2 1/2 arshins and 6 3/4 feet high. How many diggers need to be hired so that, working daily for 7 1 / 2 hours, they make in 30 days an embankment, 0.31 versts long, 7 1 / 3 arsh sprin. and a height of 3 6/7 arshins?

2686. 48 diggers, working daily for 9 hours and 20 minutes, made in 55 days Earthworks, 40 1/3 fathoms long, 4 1/2 arshins wide, and 7 arshins high. What height will 40 diggers make in 64 days, working daily for 6 hours and 45 minutes, if the length of the shaft is 44 fathoms and the width is 1 fathom?

2687 . 14 sazhens of pine firewood were spent on heating the apartment with 6 stoves for 2 months and 10 days. How long will it take 10 sazhens of birch firewood to heat an apartment with 8 stoves, if the amount of heat emitted by each stove should be the same as for the first apartment, and if 9 sazhens of pine firewood give as much heat as 7 1/2 fathoms of birch?

2688. From a rectangular field, having a length of 2 versts and a width of 1 1/2 versts, with a crop of sam-27, so much sugar beet was harvested that 937 1/2 poods of sugar were extracted from it at the factory. From another field, which had a width of 400 sazhens, with a harvest of 18 sam, sugar beet was harvested, from which 250 pounds of sugar were extracted. Assuming that the sowing conditions and the quality of the beet for both fields were the same, find the length of the second field.

2689. 4 scribes, working daily for 7 1/2 hours, copied 225 sheets in 15 days, with an average of 32 lines on each page. How many scribes need to be hired so that, working daily for 5 hours and 20 minutes, they could copy 64 leaves in 9 days, placing an average of 36 lines on each page?

2690. 3 pipes in the course of 4 1 / 2 hours filled the reservoir, 1 soot in length. 2 arshins, 1.5 arshins wide and 3 2/3 feet deep. To what depth will 4 pipes fill another reservoir within 5.4 hours, if the length of this reservoir is 1 soot. 2 5/8 feet, 1.2 ars wide, and if each of the first pipes pours 16 buckets of water at the same time, into which one of the last pipes pours 9 buckets?

2691 . 22 weavers, working 10 hours a day, prepared 120 pieces of linen in 30 days. How many such weavers need to be hired so that, working a day for 7 1/2 hours, in 40 days they can prepare 300 pieces of linen, and the length of each of these pieces should be 1 1/10 times the length of the first, and the width should be 0.8(3) the width of the first?

2692. For food for a certain number of soldiers, a supply of grain for 60 days will be obtained if each soldier is given 2 1 / 2 pounds daily. How many days will 3/4 of this supply last if the number of soldiers is reduced by 3/8 of the previous number, and the daily ration of each is increased by 1.25 pounds.?

2693. Fifteen workers and 12 workers, working daily for 10 hours and 30 minutes, removed bread from the field in 12 days. How many days will 21 workers and 8 workers, working 8.4 hours a day, remove bread from the field, the length of which is related to the length of the first, as 0.3: 1 / 5, and whose width is related to the width of the first, as 0, 51: 0.5(6) - if it is known that the strength of a man is related to the strength of a woman, how 0.2(6) : 0.1(9)?

2694. To pump out the water from the pool, 3 large and 5 small pumps were supplied, which, acting together, could pour out all the water in 6 hours. After 2 1/2 hours of their combined action, two large pumps deteriorated and were immediately replaced by 5 small ones. Knowing that the strength of each small pump is related to the strength of each large one, how 2 1 / 2: 4 1 / 6 determine how many hours it took to pump water out of the pool.

2695. 4215 bricks were used to build the wall of the house, each of which was 10 1/2 inches long and 5.25 inches wide. and 2 5/8 inches thick. In order to build another wall, bricks were used, each of which was 5 1/2 inches long, 3 1/3 inches wide and 1 1/4 inches thick. How many of these bricks will be used to build the second wall, if its length is 0.8 (3) the length of the first, the thickness is 1.1 times the thickness of the first, and the height is 0. (5) the height of the first wall?

2696. Twenty-five people, working every day for 5 hours, managed to do 0.27 of some work in 15 days. How many more people need to be hired, so that they, studying together with the first for 8 1 / 3 hours a day, can complete the rest of the same work in 20 days?

Shvetsov K.I., BEVZ G.P.
HANDBOOK OF ELEMENTARY MATH
ARITHMETIC, ALGEBRA, 1965


1. Simple triple rule. Of the problems on proportional quantities, the most common are problems on the so-called simple triple rule. In these tasks, three numbers are given and it is required to determine the fourth, proportional to them.

Problem 1. 10 bolts weigh 4 kg. How much do 25 of these bolts weigh? Such tasks can be solved in several ways.

Solution I (by reduction to unity).

1) How much does one bolt weigh?

4 kg: 10 = 0.4 kg.

2) How much do 25 bolts weigh?

0.4 kg 25 = 10 kg.

Solution II (method of proportions). Since the weight of the bolts is directly proportional to their number, the ratio of the weights is equal to the ratio of the pieces (bolts). Denoting the desired weight with the letter x, we get the proportion:

X : 4 = 25: 10,

(kg)

You can argue like this: 25 bolts are 2.5 times more than 10 bolts. Therefore, they are also 2.5 times heavier than 4 kg:

4 kg 2.5 = 10 kg.

Answer. 25 bolts weigh 10 kg.

Problem 2. The first gear makes 50 rpm. The second gear, meshed with the first, makes 75 rpm. Find the number of teeth of the second wheel if the number of teeth of the first is 30.

Solution (by reduction to unity). Both meshed gears will move in a minute by the same number of teeth, so the number of revolutions of the wheels is inversely proportional to the number of their teeth.

50 rev. - 30 teeth

75 rev. - X tooth.

X : 30 = 50: 75; (teeth).

You can also argue like this: the second wheel makes revolutions 1.5 times more than the first (75: 50 \u003d 1.5). Therefore, it has teeth 1.5 times smaller than the first:

30: 1.5 = 20 (teeth).

Answer. 20 teeth.

2. Complicated triple rule. Tasks in which, for a given series of corresponding values ​​of several (more than two) proportional quantities, it is required to find the value of one of them corresponding to another series of given values ​​of the remaining quantities, they are called tasks for a complex triple rule.

A task. 5 pumps pumped out 1800 buckets of water within 3 hours. How much water will be pumped out by 4 such pumps in 4 hours?

5 us. 3 hours - 1800 ved.

4 us. 4 h - X ved.

1) How many buckets of water did 1 pump pump out in 3 hours?

1800: 5 = 360 (buckets).

2) How many buckets of water did 1 pump pump out in 1 hour?

360: 3 = 120 (buckets).

3) How much water will be pumped out by 4 pumps in 1 hour?

120 4 = 480 (buckets).

4) How much water will be pumped out by 4 pumps in 4 hours?

480 4 = 1920 (buckets).

Answer. 1920 buckets

Shortcut solution for numerical formula:

(buckets).

A task. Divide the number 100 into two parts in direct proportion to the numbers 2 and 3,

This task should be understood as follows: divide 100 into two parts so that the first relates to the second as 2 to 3. If we denote the desired numbers by letters X 1 and X 2, this problem can be formulated as follows. Find X 1 and X 2 such that

X 1 + X 2 = 100,

X 1: X 2 = 2: 3.