Triple somersault sea ​​yachts the ballast is placed low, which does not allow them to heel strongly 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

Among the tasks in two actions, there is a group of tasks that are solved unity. Solving such problems, children should practically learn the properties of quantities that are in direct proportion.

Let's take for example the problem: A steamboat traveled 40 km in 2 hours. How many kilometers will the ship travel in 4 hours at the same speed? In this problem, two time values ​​and one distance value are known, corresponding to the first time value; it is known that the speed of movement does not change, it is required to find another value of the distance.

Let's consider various ways of solving this problem, writing down the solution on the left and its justification on the right.

I method of solution - a method of direct reduction to unity

oral solution

2 hours - 40 km
1 hour – 20 km
4 hours - 80 km

Written decision

1) 40km: 2 = 20km
2)20km x 4 = 80km

The numerical value of time, two values ​​of which are known, is reduced to one.

At constant speed if the time is reduced by 2 times, the distance will decrease by 2 times, if it is then increased by 4 times, the distance will increase by 4 times.

The second method of solution is the method of inverse reduction to unity.

oral solution

40 km - 2 hours = 120 min.
1 km - 3 min.
4 hours (240 min.) – 80 km

Written decision

1) 120 min. : 40 = 3 min.
2) 240 min. : 3 min. = 80 (km)

The numerical value of the distance is reduced to one, one value of which is known, and the other is unknown.

At a constant speed, the passage of 1 km of the path will take 40 times less time than the passage of 40 km of the path, that is, 3 minutes, and in 4 hours (240 minutes) the steamer will cover as many times as many kilometers as 240 minutes . more than 3 min.

The third way of solving is the way of finding the relation.

A brief record of the task condition:

2 hours - 40 km
4 hours - x

1) 4 hours: 2 hours = 2
2) 40 km x 2 = 80 km

At a constant speed of movement, how many times the time increases, the distance traveled increases by the same amount

IV method of solution - the method of finding numerical value constant value.

Brief statement of the task condition

2 hours - 40 km
4 hours -?

1) 40 km: 2 = 20 km
2) 20 km x 4 = 80 km

When solving this problem, method IV coincides with method I.

To find the distance traveled in 4 hours, you need to multiply the speed, which is found by dividing the distance by the corresponding time value, by the new time value.

Let's apply the method of finding the numerical value of a constant value to another problem:

The ship traveled 40 km at a speed of 20 km per hour. How many kilometers will the ship travel in the same time at a speed of 30 km per hour?

Solution. According to the condition of this problem, time is a constant value.

1) How many hours did the ship take to travel 40 km?

40 km: 20 km = 2 (hours)

2) How many kilometers will the steamer travel in 2 hours at the new speed?

30 km x 2 – 60 km

Answer: 60 km.

When solving this problem, the method of finding the numerical value of a constant differs from the method of direct reduction to unity. This can be seen from a comparison of the above method with the method direct reduction to unity.

The possibility of applying one or another method of solving problems on a simple triple rule in the framework of operations with integers depends on the characteristics of numerical data. So, for example, the method of finding a ratio can only be applied if the numbers expressing two different meanings of the same magnitude, are multiples of one another.

Method of back reduction to unity it is convenient to use when solving problems in which it is required to find an unknown value of quantity or time. Therefore, in arithmetic textbooks for elementary grades, tasks for a simple triple rule are selected in groups according to the methods for solving them. At the same time, according to the current program, problems solved by methods of direct and inverse reduction to unity are assigned to class II, and problems solved by the method of finding the ratio are assigned to class IV.

There is reason to believe that the easier of the tasks, solved by the method of finding the ratio, can be introduced in grade II, where students are already solving simple tasks for multiple comparison. There are no problems solved by the method of finding the numerical value of a constant value in existing arithmetic textbooks, and it is useful to offer them for solution already in grade II.

When teaching the solution of these problems, one should rely on the ability previously acquired by students to solve simple problems of multiplication and division, in which it is required to find out the value of one of the three quantities related to each other, for example, find out the cost by price and quantity of items, quantity by price and value, price in terms of cost and quantity.

Children's good knowledge of the relationship between quantities serves as the basis, based on which they master the solution of problems by the method of reduction to unity.

To explain to students how to find a relationship, you can use visual aids(Fig. 22). Let it be necessary to solve the problem: 2 envelopes with stamps cost 9 kopecks. How much are 6 of these envelopes?

Looking at the image of these envelopes grouped in pairs will help students understand that increasing the number of pairs of envelopes by several times entails an increase in their value by the same amount.

rice. 22

Students ask the question: how many times 6 envelopes are more than 2 envelopes? - They find the answer, which is 3 times more, and find out the cost of 6 envelopes, multiplying 9 kopecks. on 3.

Joint consideration of tasks and independent work children to convert direct tasks into inverse ones contribute to a better understanding of how to solve them.

For example, the task of 3 cups costs 6 rubles. How much do 5 of these cups cost? by replacing the desired number with the found number, and one of the data with the desired number, can be transformed into the following inverse problems:

1. 5 cups cost 10 rubles. How much do 3 of these cups cost?
2. 3 cups cost 6 rubles. How many of these cups can you buy for 10 rubles?
3. 5 cups cost 10 rubles. How many of these cups can you buy for 6 rubles?

The solution of the original problem and the first of the transformed ones is performed method of direct reduction to unity, the solution of the second and third - back to unity.

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 of the 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 a 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 a square-shaped piece of marble weigh, which is 8.75 inches long, 2 1/4 inches wide, and 2 inches thick, and 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 an earthen rampart, 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 7 1/2 hours a day, 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. In 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?

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 this out, and we allow ourselves to go back a little and give a brief description of the goals that arithmetic has pursued 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 quite 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 Indians 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 the 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 coloration remained behind arithmetic 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 the 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 in any way distinguish 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 people 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 completely discarded. 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 give a general education, and not to train accountants, clerks, accountants, etc. Meanwhile, the craft methods of Italians and Germans, who did not seek to develop a person, but to make a calculating machine out of him, 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 independence should be a necessary quality of mathematical derivation, 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.

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

Abbreviated solution by 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.