Tasks for fractions with different denominators. Addition and subtraction of fractions with different denominators. Adding fractions with different denominators
Lesson Objectives:
- To develop skills in comparing fractions,
- Addition and subtraction of fractions with different denominators,
- To consolidate knowledge of finding the least common multiple of numbers.
Today at the lesson we continue to work on the topic “Addition and subtraction of fractions with different denominators”.
This is already the second lesson of the topic, you will face goal:
If in the first lesson we dealt with fractions, in which the denominators are coprime or multiples of each other, then today our task becomes more complicated, for some cases we will have to find a common denominator by expanding the denominators into prime factors according to the rule for finding the LCM.
At the end of the lesson, you should know the rule well:
how to add fractions with different denominators and be able to apply this rule when solving problems.
After 3 lessons, a test will take place, in which there will be tasks that check how you have learned the topic. On the test there will be 2 tasks on our topic: the third task is to perform the addition and subtraction of fractions with different denominators and the fourth task: solving the problem of applying the rule. So, today we are working on tasks for the standard.
1. a) Let's work orally.
42 | 48 | 6 |
36 | 54 | 12 |
30 | 24 | 18 |
Look carefully at this rectangle and try to remember the location of the numbers, maybe you will notice some pattern.
Now try to restore these numbers in the draft.
Who remembers the numbers?
How could you remember the location of these numbers well?
(Numbers that are multiples of 6 are in ascending order clockwise, starting from the top right rectangle)
Let's repeat the comparison of fractions with different denominators and with equal numerators.
Compare the following fractions: ; .
Arrange them in ascending order.
b) Look carefully at the following row of numbers:
16, 10, 8, , 2007, 1961.
How many numbers are written?
How many even numbers? Name them.
Name the third number.
Second number from the end.
Three digit number.
A number that is a multiple of 5.
Multiple of 10
Multiple 3.
Multiple of 9. Why is the number 1961 famous?
What number is different from the rest, that is, does not fit into a series of numbers?
Is this fraction correct or incorrect?
Reducible or irreducible?
Reduce this fraction.
2. Checking homework.
How do you compare two fractions with different denominators?
How do you add fractions with different denominators?
How do you subtract fractions with different denominators?
Do you have questions about homework? Line check by teacher.
3. Working with the rule according to the textbook after inaccurate answers of students.
In mathematics, you cannot skip a single word in some rules. The common denominator and the lowest common denominator are not always the same.
Listen to the parable of a certain mayor.
When there was no electricity yet, the mayor of one city liked to walk along the city streets in the evening. Once he ran into a city dweller, a bump popped up on his forehead. the next day he issued a decree: "In the dark, go out into the street with a lantern." And in the evening the same citizen ran into him. The mayor demanded a lantern from him.
Here, said the passer-by.
Where is the candle? the mayor asked.
And the decree does not say that there should be a candle in the lantern, - he answered.
The mayor issued a second decree: “In the dark, go out into the street with a lantern with a candle.”
On the third day, history repeated itself.
The mayor has already lost his temper.
Do you think that the passer-by answered the mayor?
The order does not say that the candle of the lantern must be lit.
The mayor had to issue a decree a third time, only after that a passer-by left him alone.
Our task is to know the rule well and be able to apply it. Once again, we are working on a standard.
4. Exercise.
Solve the following examples on the board as desired.
You solved examples where the denominators are relatively prime numbers and when the larger denominator is a multiple of the smaller one.
In this lesson, we will solve more complex tasks for adding and subtracting fractions with different denominators.
Write down the task:
If the student decides the way we decided, then he knows well how to find the LCM of two numbers and knows how to isolate an integer part from an improper fraction, knows that the denominators are not coprime numbers.
And if the student finds a common denominator by multiplying the denominators, he shows ignorance of finding the LCM, that is, the rules: how to add fractions with different denominators. Therefore, first of all, if the denominators are not coprime numbers and are not multiples of each other, it is necessary to find the LCM of the denominators.
No. No. are written on the board, which must be solved in class: 309 d - and, 328, 340 (repetition)
e) ; perform on the board
e) ; repeated the reduction of the fraction, this task is on the test, it checks the assimilation of the standard.
g) (on one's own)
h) ; we find LCM(21,15) = 3*7*5 =105.
6. Solve problem No. 327 on your own.
7. Repetition of previously studied material. No. 340.
Reduce fractions:
There is also a reduction in fractions on the test, this is a task for the standard.
8. The result of the lesson.
a) How do you add and subtract fractions with different denominators?
b) Marking.
c) Homework: item 11,
In a real educational process, not so many tasks for adding and subtracting fractions with the same denominators are required - there will be enough tasks from the textbook. We will pay more attention to problems in which the entire quantity is taken as a unit. Moreover, at first it is better to represent it as 2/2, 3/3, etc. quantities.
163 . The girl read 2/5, then another 1/5 of the book. What part of the book did she read?
164 . Tourists have passed 1/7, then another 3/7 of the whole route. What part of the route do they have left to go?
165 . Two tractor drivers mowed 5/9 of the meadow, and the first tractor driver mowed 2/9 of the meadow. What part of the meadow was mowed by the second tractor driver?
166 . The first tractor driver plowed 2/7 of the field, the second - 3/7 of the field. Together they plowed 10 ha. Determine the area of the field.
167 . Solve problems 150 (a-c) using fraction subtraction.
168 . Solve problems 154 (1-2) using fraction subtraction.
169 . 1) Sparrows were sitting on a branch. When the third part of the sparrows flew away, there were 6 of them left. How many sparrows were on the branch initially?
2) Someone spent 3/4 of his money and had 200 left R. How much money did he have?
3) On the first day, tourists covered 2/5 of the planned route, and on the second day, the remaining 15 km. What is the length of the route?
4) Vasya has 200 stamps in his collection. Over the past year the number of stamps in the collection has increased by 1 / 4 . How many stamps were in the collection a year ago?
170 . Before lunch, the turner completed 2/8 of the task, after lunch - 3/8 of the task, after which he had 24 parts to turn. How many parts did he have to grind?
171 . From « Arithmetic » L.N. Tolstoy. Husband and wife took money from the same chest, and nothing was left. Husband took 7/10 of all money and wife 690 R. How much was all the money?
172 . Solve problems from Egyptian papyri in two ways.
1) The quantity and its fourth part give together 15. Find
amount.
2) The number and half of it is 9. Find the number.
173 . Make up a problem similar to the Egyptian problems and solve it in two ways.
Starting from the next problem, addition and subtraction of fractions with different denominators occur in the solutions. If this material was not studied in the 5th grade, then the remaining tasks related to fractions should be postponed until the 6th grade.
174 . a) At every hour, the first pipe fills 1/2 of the pool, and the second - 1/3 of the pool. What part of the pool is filled by both pipes in 1 h collaboration?
b) The first brigade can complete 1/12 of the task per day, and the second - 1/8 of the task. What part of the task will two teams complete in 1 day of joint work?
c) A passenger car travels 1/10 of the distance between cities per hour, and a truck - 1/12 of this distance. How much of this distance do they converge in 1 h cars moving towards each other?
175 . a) Two tractor drivers plowed 2/3 of the field in 1 day of joint work. The first tractor driver plowed 1/2 of the field. What part of the field was plowed by the second tractor driver?
b) Two cars traveling towards each other approached within 1 h 1/3 of the distance between two cities. The first car traveled 1/8 of that distance. What fraction of the total distance was traveled by the second car?
c) Through two pipes, 1/3 of the pool is filled every hour. Through the first pipe for 1 h 1/10 of the pool is filled. What part of the pool is filled in 1 h through the second pipe?
176 . First, 1/2 of the water in it was poured out of the barrel, then 1/3, 1/15 and 1/10. What part of the water was poured out?
177 .* I drank half a cup of black coffee and topped it up with milk. Then I drank 1/3 cup and topped it up with milk. Then I drank 1/6 cup and topped it up with milk. Finally, I drank the contents of the cup to the end. What did I drink more: coffee or milk?
178 . Vintage problems. 1) Two pedestrians came out at the same time towards each other from two villages. The first one can walk the distance between two villages in 8 h, and the second for 6 h. What part of the distance do they approach in 1 h?
2) Three carpenters were hired to build the bath; the first did on the day 2/33 of all the work, the second 1/11, the third 7/55. What fraction of the total work did they all do in a day?
3) 4 scribes were hired to copy the essay; the first one could rewrite the essay alone in 24 days, the second in 36 days, the third in 20 and the fourth in 18 days. What part of the essay will they rewrite in one day if they work together?
179 . 1) The typist retyped the third part of the manuscript, then another 10 pages. As a result, she retyped half of the entire manuscript. How many pages are in the manuscript?
2) Old problem. A passerby, catching up with another, asked: « How far is it to the village we have ahead of us? » Another passerby replied: « The distance from the village from which you are walking is equal to one third of the entire distance between the villages, and if you walk another 2 versts, then you will be exactly in the middle between the villages » . How many versts are left for the first passer-by to walk?
180 . Problem of Adam Riese (XVI century). Three won some amount of money. The first accounted for 1/4 of this amount, the second 1/7, and the third 17 florins. How big is all the winnings?
To express a part as a fraction of the whole, you need to divide the part by the whole.
Task 1. There are 30 students in the class, four are missing. What proportion of students are missing?
Decision:
Answer: there are no students in the class.
Finding a fraction from a number
To solve problems in which it is required to find a part of a whole, the following rule is true:
If a part of the whole is expressed as a fraction, then to find this part, you can divide the whole by the denominator of the fraction and multiply the result by its numerator.
Task 1. There were 600 rubles, this amount was spent. How much money have you spent?
Decision: to find from 600 rubles, you need to divide this amount into 4 parts, thereby we will find out how much money is one fourth:
600: 4 = 150 (p.)
Answer: spent 150 rubles.
Task 2. It was 1000 rubles, this amount was spent. How much money has been spent?
Decision: From the condition of the problem, we know that 1000 rubles consists of five equal parts. First we find how many rubles are one fifth of 1000, and then we find out how many rubles are two fifths:
1) 1000: 5 = 200 (p.) - one fifth.
2) 200 2 \u003d 400 (p.) - two fifths.
These two actions can be combined: 1000: 5 2 = 400 (p.).
Answer: 400 rubles were spent.
The second way to find a part of a whole:
To find a part of a whole, you can multiply the whole by a fraction expressing that part of the whole.
Task 3. According to the charter of the cooperative, for the validity of the reporting meeting, it must be attended by at least members of the organization. The cooperative has 120 members. With what composition can the reporting meeting be held?
Decision:
Answer: the reporting meeting can be held if there are 80 members of the organization.
Finding a number by its fraction
To solve problems in which it is required to find the whole by its part, the following rule is true:
If a part of the desired integer is expressed as a fraction, then to find this integer, you can divide this part by the numerator of the fraction and multiply the result by its denominator.
Task 1. We spent 50 rubles, this amounted to the original amount. Find the original amount of money.
Decision: from the description of the problem, we see that 50 rubles is 6 times less than the initial amount, i.e., the initial amount is 6 times more than 50 rubles. To find this amount, you need to multiply 50 by 6:
50 6 = 300 (r.)
Answer: the initial amount is 300 rubles.
Task 2. We spent 600 rubles, this amounted to the initial amount of money. Find the original amount.
Decision: we will assume that the desired number consists of three thirds. By condition, two-thirds of the number are equal to 600 rubles. First, we find one third of the initial amount, and then how many rubles are three-thirds (initial amount):
1) 600: 2 3 = 900 (p.)
Answer: the initial amount is 900 rubles.
The second way to find the whole by its part:
To find a whole by the value of its part, you can divide this value by a fraction that expresses this part.
Task 3. Line segment AB, equal to 42 cm, is the length of the segment CD. Find the length of a segment CD.
Decision:
Answer: segment length CD 70 cm
Task 4. Watermelons were brought to the store. Before lunch, the store sold, after lunch - brought watermelons, and it remains to sell 80 watermelons. How many watermelons were brought to the store in total?
Decision: first, we find out what part of the imported watermelons is the number 80. To do this, we take the total number of imported watermelons as a unit and subtract from it the number of watermelons that we managed to sell (sell):
And so, we learned that 80 watermelons are from the total number of watermelons brought. Now we will find out how many watermelons of the total amount is, and then how many watermelons are (the number of watermelons brought):
2) 80: 4 15 = 300 (watermelons)
Answer: in total, 300 watermelons were brought to the store.
Actions with fractions.
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
So, what are fractions, types of fractions, transformations - we remembered. Let's tackle the main question.
What can you do with fractions? Yes, everything is the same as with ordinary numbers. Add, subtract, multiply, divide.
All these actions with decimal operations with fractions are no different from operations with integers. Actually, this is what they are good for, decimal. The only thing is that you need to put the comma correctly.
mixed numbers, as I said, are of little use for most actions. They still need to be converted to ordinary fractions.
And here are the actions with ordinary fractions will be smarter. And much more important! Let me remind you: all actions with fractional expressions with letters, sines, unknowns, and so on and so forth are no different from actions with ordinary fractions! Operations with ordinary fractions are the basis for all algebra. It is for this reason that we will analyze all this arithmetic in great detail here.
Addition and subtraction of fractions.
Everyone can add (subtract) fractions with the same denominators (I really hope!). Well, I’ll remind you completely forgetful: when adding (subtracting), the denominator does not change. The numerators are added (subtracted) to give the numerator of the result. Type:
In short, in general terms:
What if the denominators are different? Then, using the main property of the fraction (here it came in handy again!), We make the denominators the same! For example:
Here we had to make the fraction 4/10 from the fraction 2/5. Solely for the purpose of making the denominators the same. I note, just in case, that 2/5 and 4/10 are the same fraction! Only 2/5 is uncomfortable for us, and 4/10 is even nothing.
By the way, this is the essence of solving any tasks in mathematics. When we're out uncomfortable expressions do the same, but more convenient to solve.
Another example:
The situation is similar. Here we make 48 out of 16. By simple multiplication by 3. This is all clear. But here we come across something like:
How to be?! It's hard to make a nine out of a seven! But we are smart, we know the rules! Let's transform every fraction so that the denominators are the same. This is called "reduce to a common denominator":
How! How did I know about 63? Very simple! 63 is a number that is evenly divisible by 7 and 9 at the same time. Such a number can always be obtained by multiplying the denominators. If we multiply some number by 7, for example, then the result will certainly be divided by 7!
If you need to add (subtract) several fractions, there is no need to do it in pairs, step by step. You just need to find the denominator that is common to all fractions, and bring each fraction to this same denominator. For example:
And what will be the common denominator? You can, of course, multiply 2, 4, 8, and 16. We get 1024. Nightmare. It is easier to estimate that the number 16 is perfectly divisible by 2, 4, and 8. Therefore, it is easy to get 16 from these numbers. This number will be the common denominator. Let's turn 1/2 into 8/16, 3/4 into 12/16, and so on.
By the way, if we take 1024 as a common denominator, everything will work out too, in the end everything will be reduced. Only not everyone will get to this end, because of the calculations ...
Solve the example yourself. Not a logarithm... It should be 29/16.
So, with the addition (subtraction) of fractions is clear, I hope? Of course, it is easier to work in a shortened version, with additional multipliers. But this pleasure is available to those who honestly worked in the lower grades ... And did not forget anything.
And now we will do the same actions, but not with fractions, but with fractional expressions. New rakes will be found here, yes ...
So, we need to add two fractional expressions:
We need to make the denominators the same. And only with the help multiplication! So the main property of the fraction says. Therefore, I cannot add one to x in the first fraction in the denominator. (But that would be nice!). But if you multiply the denominators, you see, everything will grow together! So we write down, the line of the fraction, leave an empty space on top, then add it, and write the product of the denominators below, so as not to forget:
And, of course, we don’t multiply anything on the right side, we don’t open brackets! And now, looking at the common denominator of the right side, we think: in order to get the denominator x (x + 1) in the first fraction, we need to multiply the numerator and denominator of this fraction by (x + 1). And in the second fraction - x. You get this:
Note! Parentheses are here! This is the rake that many step on. Not brackets, of course, but their absence. Parentheses appear because we multiply the whole numerator and the whole denominator! And not their individual pieces ...
In the numerator of the right side, we write the sum of the numerators, everything is as in numerical fractions, then we open the brackets in the numerator of the right side, i.e. multiply everything and give like. You don't need to open the brackets in the denominators, you don't need to multiply something! In general, in denominators (any) the product is always more pleasant! We get:
Here we got the answer. The process seems long and difficult, but it depends on practice. Solve examples, get used to it, everything will become simple. Those who have mastered the fractions in the allotted time, do all these operations with one hand, on the machine!
And one more note. Many famously deal with fractions, but hang on examples with whole numbers. Type: 2 + 1/2 + 3/4= ? Where to fasten a deuce? No need to fasten anywhere, you need to make a fraction out of a deuce. It's not easy, it's very simple! 2=2/1. Like this. Any whole number can be written as a fraction. The numerator is the number itself, the denominator is one. 7 is 7/1, 3 is 3/1 and so on. It's the same with letters. (a + b) \u003d (a + b) / 1, x \u003d x / 1, etc. And then we work with these fractions according to all the rules.
Well, on addition - subtraction of fractions, knowledge was refreshed. Transformations of fractions from one type to another - repeated. You can also check. Shall we settle a little?)
Calculate:
Answers (in disarray):
71/20; 3/5; 17/12; -5/4; 11/6
Multiplication / division of fractions - in the next lesson. There are also tasks for all actions with fractions.
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Lesson contentProblems for fractions
Task 1. In the class of schoolchildren are excellent students. What part is the rest? Make a graphic description of the task. The drawing can be anything.
Decision
If they make up excellent students, then they make up the rest
Task 2. In the class of schoolchildren, there are excellent students, good students, and three students. Make a graphic description of the task. The drawing can be anything.
Task 3. There are 24 students in the class. schoolchildren are excellent students, good students, three students. How many excellent students, good students and three students are in the class?
Decision
24: 6 × 1 = 4 × 1 = 4 (excellent)
24: 6 × 3 = 4 × 3 = 12 (good students)
24: 6 × 2 = 4 × 2 = 8 (triples)
Examination
4 + 12 + 8 = 24 (student)
24 = 24
Task 4. In the class of schoolchildren there are excellent students, there are good students. What part are triplets?
Decision
The students are divided into 6 parts. One of the parts has excellent students, three parts have good students. It is not difficult to guess that the other two parts are triples. So schoolchildren are triplets
Without drawing figures, you can add the fractions and, and subtract the result from the fraction, which expresses the entire part of the students. In other words, add up the A's and B's, then subtract those A's and B's from the total number of students.
Task 5. There are 16 students in the class. Of them are excellent students, are good students. How many excellent students and good students are in the class? Make a graphic description of the task. The drawing can be anything.
Decision
16: 4 × 1 = 4 × 1 = 4 (excellent)
16: 16 × 12 = 1 × 12 = 12 (good guys)
Task 6. There are 16 students in the class. Of them are excellent students, good students, three students. How many A's, B's and C's are in the class? Make a graphic description of the task. The drawing can be anything.
Decision
16: 8 × 1 = 2 × 1 = 2 (excellent)
16: 16 × 10 = 1 × 10 = 10 (good students)
16:4 = 4 (triples)
Task 7. Poltava groats are produced from wheat grains, the mass of which is the mass of wheat grains, and the rest is feed waste. How much Poltava groats and feed waste can be obtained from 500 centners of wheat
Decision
We find from 500 centners:
Now let's find a lot of feed waste. To do this, we subtract the mass of Poltava groats from 500 centners:
This means that from 500 centners of wheat grains you can get 320 centners of Poltava groats and 180 centners of feed waste.
Task 8. A kilogram of sugar costs 88 rubles. How much is a kg of sugar? kg? kg? kg?
Decision
1) A kg is half of one kilogram. If one kilogram costs 88 rubles, then half a kilogram will cost half of 88, that is, 44 rubles. If we find half of 88 rubles, we get 44 rubles
88: 2 = 44
44 × 1 = 44 rubles
2) kg is a quarter of a kilogram. If one kilogram costs 88 rubles, then a quarter of a kilogram will cost a quarter of 88 rubles, that is, 22 rubles. If we find from 88 rubles, we will get 22 rubles
88: 4 = 22
22 × 1 = 22 rubles
3) A fraction means that a kilogram is divided into eight parts, and three parts are taken from there. If one kilogram costs 88 rubles, then the cost of three eight kilograms will cost from 88 rubles. If we find from 88 rubles, we will get 33 rubles.
4) A fraction means that a kilogram is divided into eight parts, and eleven parts are taken from there. But it is impossible to take eleven parts if there are only eight. We are dealing with an improper fraction. First, let's select the whole part in it:
Eleven-eighths is one whole kilogram and kilogram. Now we can separately find the cost of one whole kilogram and the cost of three-eighths of a kilogram. One kilogram, as mentioned above, costs 88 rubles. We also found the cost per kg and got 33 rubles. This means that a kg of sugar will cost 88 + 33 rubles, that is, 121 rubles.
The cost can be found without highlighting the whole part. To do this, it is enough to find from 88.
88: 8 = 11
11 x 11 = 121
But having singled out the whole part, one can understand well how the price per kg of sugar was formed.
Task 9. Dates contain sugar and mineral salts. How many grams of each substance are contained in 4 kg of dates?
Decision
Find out how many grams of sugar are contained in one kilogram of dates. One kilogram is one thousand grams. Find from 1000 grams:
1000: 25 = 40
40 × 18 = 720 g
One kilogram of dates contains 720 grams of sugar. To find out how many grams of sugar are in four kilograms, you need to multiply 720 by 4
720 × 4 = 2880 g
Now let's find out how many mineral salts are contained in 4 kilograms of dates. But first, let's find out how many mineral salts are contained in one kilogram. One kilogram is one thousand grams. Find from 1000 grams:
1000: 200 = 5
5 × 3 = 15 g
One kilogram of dates contains 15 grams of mineral salts. To find out how many grams of mineral salts are contained in four kilograms, you need to multiply 15 by 4
15 × 4 = 60 g
This means that 4 kg of dates contain 2880 grams of sugar and 60 grams of mineral salts.
The solution for this problem can be written much shorter, in two expressions:
The bottom line is that from 4 kilograms they found and the resulting 2.88 were converted to grams, multiplying by 1000. The same was done for mineral salts - from 4 kg they found and the resulting kilograms were converted to grams, multiplied by 1000. Also note that that a fraction of a number is found in a simplified way - by directly multiplying a number by a fraction.
Task 10. The train traveled 840 km, which is its path. How far does he have to go? What is the distance of the entire journey?
Decision
The problem says that 840 km is from his path. The denominator of the fraction indicates that the entire path is divided into seven equal parts, and the numerator indicates that four parts of this path have already been covered and amount to 840 km. Therefore, dividing 840 km by 4, we find out how many kilometers fall on one part:
840: 4 = 210 km.
And since the whole path consists of seven parts, the distance of the entire path can be found by multiplying 210 by 7:
210 × 7 = 1470 km.
Now let's answer the second question of the problem - what distance is left for the train to travel? If the length of the path is 1470 km, and 840 are covered, then the remaining path is 1470−840, that is, 630
1470 − 840 = 630
Task 11. One of the groups that conquered the mountain peak of Everest consisted of athletes, guides and porters. There were 25 athletes in the group, the number of guides was the number of athletes, and the number of athletes and guides together was only 9/140 of the number of porters. How many porters were on this expedition?
Decision
Athletes group 25. Conductors is the number of athletes. Let's find from 25 and find out how many conductors are in the group:
25: 5 × 4 = 20
Athletes and guides together - 45 people. This number is based on the number of porters. Knowing that the number of porters is 45 people, we can find the total number of porters. To do this, find the number by fraction:
45: 9 x 140 = 5 x 140 = 700
Task 12. 900 new textbooks were brought to the school, of which mathematics textbooks made up all books, Russian language textbooks of all books, and the rest of the books were literature. How many books on literature were brought
Find out how many math textbooks are:
900: 25 × 8 = 288 (math books)
How many Russian language textbooks:
900: 100 × 33 = 297 (Russian language books)
Find out how many textbooks on literature. To do this, subtract textbooks in mathematics and Russian from the total number of books:
900 – (288+297) = 900 – 585 = 315
Examination
288 + 297 + 315 = 900
900 = 900
Task 13. On the first day they sold, and on the second day the grapes arrived at the store. What fraction of the grapes were sold in two days?
Decision
The grapes were sold in two days. This part is obtained by adding fractions and
You can imagine the grapes that arrived at the store in the form of six bunches. Then grapes are two bunches, grapes are three bunches, and grapes are five out of six bunches sold in two days. Well, it is not difficult to see that there is one cluster left, a expressed fraction (one cluster out of six)
Task 14. Vera read books on the first day, and less on the second day. What part of the book did Vera read on the second day? Did she manage to read the book in two days?
Decision
Determine the part of the book read on the second day. It is said that less was read on the second day than on the first day. Therefore, you need to subtract from
On the second day, Vera read books. Now let's answer the second question of the problem - did Vera have time to read the book in two days? Let's sum up what Vera read on the first and second days:
Vera read the books in two days, but there are still books left. So Vera did not have time to read the whole book in two days.
Let's do a check. Suppose that the book that Vera was reading had 180 pages. On the first day, she read books. We will find from 180 pages
180: 9 × 5 = 100 (pages)
On the second day, Vera read less than on the first. Find from 180 pages, and subtract the result from 100 sheets read on the first day
180: 6 × 1 = 30 × 1 = 30 (pages)
100 − 30 = 70 (pages on the second day)
Let's check if 70 pages are part of a book:
180: 18 × 7 = 10 × 7 = 70 (pages)
Now let's answer the second question of the problem - did Vera manage to read all 180 pages in two days. The answer is that she did not have time, because in two days she read only 170 pages.
100 + 70 = 170 (pages)
There are 10 more pages to read. In the problem, we had a fraction as a remainder. Let's check if 10 pages are part of the book?
180: 18 × 1 = 10 × 1 = 10 (pages)
Task 15. In one package kg, and in the other kg less. How many kilograms of sweets are in two packages together?
Decision
Let us determine the mass of the second package. It is kg less than the mass of the first package. Therefore, from the mass of the first package, subtract the mass of the second:
Weight of the second package kg. Let's determine the mass of both packages. Add the mass of the first and the mass of the second:
Weight of both packages kg. A kilogram is 800 grams. You can solve this problem by working with fractions, adding and subtracting them. You can also first find the number according to the fractions given in the problem and proceed with the solution. So a kilogram is 500 grams and a kg is 200 grams.
1000: 2 × 1 = 500 × 1 = 500 g
1000: 5 × 1 = 200 × 1 = 200 g
The second package has 200 grams less, so to determine the mass of the second package, you need to subtract 200 g from 500 g
500 − 200 = 300 g
And finally, add the masses of both packages:
500 + 300 = 800 g
Task 16. Tourists went from the camp site to the lake in 4 days. On the first day they traveled the whole way, on the second day they traveled the rest of the way, and on the third and fourth days they walked 12 km each. What is the length of the entire journey from the camp site to the lake?
Decision
The problem says that on the second day the tourists passed the rest of the way . The fraction means that the remaining path is divided into 7 equal parts, of which the tourists went through three parts, but it remains to go through the rest. These account for the distance that the tourists traveled on the third and fourth days, that is, 24 km (12 km each day). Let's draw a visual diagram illustrating the second, third and fourth days:
On the third and fourth days, the hikers walked 24 km, which is equal to the distance covered on the second, third and fourth days. Knowing what is 24 km, we can find the whole distance traveled on the second, third and fourth day:
24: 4 × 7 = 6 × 7 = 42 km
On the second, third and fourth day, the tourists walked 42 km. Now let's find from this path. So we find out how many kilometers the tourists traveled on the second day:
42: 7 × 3 = 6 × 3 = 18 km
Now back to the beginning of the task. It is said that on the first day the tourists went all the way. The whole path is divided into four parts, and the first part is the path covered on the first day. And we have already found the path that falls on the other three parts - these are 42 kilometers traveled on the second, third and fourth days. Let's draw a visual diagram illustrating the first and the remaining three days:
Knowing that the paths are 42 kilometers, we can find the length of the entire path:
42: 3 × 4 = 56 km
This means that the length of the path from the camp site to the lake is 56 kilometers. Let's do a check. To do this, we add up all the paths traveled by tourists on each of the four days.
First, find the path traveled on the first day:
56: 4 × 1 = 14 (first day)
14 + 18 + 12 + 12 = 56
56 = 56
A problem from the arithmetic of the famous Central Asian mathematician Muhammad ibn Musa al-Khwarizmi (9th century AD)
“Find a number knowing that if you subtract one third and one quarter from it, you get 10”
Let's depict the number we want to find as a segment divided into three parts. In the first part of the segment, we mark a third, in the second - a quarter, the remaining third part will depict the number 10.
Let's add a third and a quarter:
Now let's draw a segment divided into 12 parts. We mark a fraction on it, the remaining five parts will go to the number 10:
Knowing that five twelfths of a number make up the number 10, we can find the whole number:
10: 5 x 12 = 2 x 12 = 24
We found the whole number - it is 24.
This problem can be solved without drawing pictures. To do this, you first need to add a third and a quarter. Then, from the unit, which plays the role of an unknown number, subtract the result of adding a third and a quarter. Then, using the resulting fraction, determine the whole number:
Problem 17. A family of four earns 80,000 rubles a month. The budget is planned as follows: for food, for utilities, for the Internet and TV, for treatment and trips to doctors, for donations to an orphanage, for living in a rented apartment, in a piggy bank. How much money is allocated for food, utilities, Internet and TV, treatment and visits to doctors, a donation to an orphanage, living in a rented apartment, and a piggy bank?
Decision
80: 40 × 7 = 14 (thousand food)
80: 20 × 1 = 4 × 1 = 4 thousand (for utilities)
80: 20 × 1 = 4 × 1 = 4 thousand (on the Internet and TV)
80: 20 × 3 = 4 × 3 = 12 thousand (for treatment and trips to doctors)
80: 10 × 1 = 8 × 1 = 8 thousand (for a donation to an orphanage)
80: 20 × 3 = 4 × 3 = 12 thousand (for living in a rented apartment)
80: 40 × 13 = 2 × 13 = 26 thousand (to the piggy bank)
Examination
14 + 4 + 4 + 12 + 8 + 12 + 26 = 80
80 = 80
Problem 18. Tourists during the trip covered km in the first hour, and more in the second hour. How many kilometers did the tourists walk in two hours?
Decision
Let's find numbers by fractions. this is three whole kilometers and seven tenths of a kilometer, and seven tenths of a kilometer is 700 meters:
This is one whole kilometer and one fifth of a kilometer, and one fifth of a kilometer is 200 meters.
Determine the length of the path traveled by tourists in the second hour. To do this, add 1 km 200 m to 3 km 700 m
3 km 700 m + 1 km 200 m = 3700 m + 1200 m = 4900 m = 4 km 900 m
Determine the length of the path traveled by tourists in two hours:
3 km 700 m + 4 km 900 = 3700 m + 4900 m = 8600 m = 8 km 600 m
This means that in two hours the tourists walked 8 kilometers and another 600 meters. Let's solve this problem using fractions. So it can be shortened considerably.
We received a kilometer answer. This is eight whole kilometers and six tenths of a kilometer, and six tenths of a kilometer is six hundred meters.
Problem 19. Geologists passed the valley, located between the mountains, in three days. On the first day they passed, on the second the whole way and on the third the remaining 28 km. Calculate the length of the path through the valley.
Decision
Let's depict the path as a segment divided into three parts. In the first part we mark the paths, in the second part of the path, in the third part the remaining 28 kilometers:
Let's add the parts of the path traveled on the first and second days:
During the first and second days, the geologists went all the way. The remaining paths account for 28 kilometers covered by geologists on the third day. Knowing that 28 kilometers is the entire path, we can find the length of the path passing through the valley:
28: 4 × 9 = 7 × 9 = 63 km
Examination
63: 9 × 5 = 7 × 5 = 35
63: 9 × 4 = 7 × 4 = 28
35 + 28 = 63
63 = 63
Problem 20. To prepare the cream, cream, sour cream and powdered sugar were used. Sour cream and cream are 844.76 kg, and powdered sugar and cream are 739.1 kg. How much cream, sour cream and powdered sugar are contained in 1020.85 kg of cream?
Decision
sour cream and cream - 844.76 kg
powdered sugar and cream - 739.1 kg
We will take out sour cream and cream (844.76 kg) from 1020.85 kg of cream. So we find the mass of powdered sugar:
1020.85 kg - 844.76 kg = 176.09 (kg of powdered sugar)
We will extract powdered sugar (176.09 kg) from powdered sugar and cream. So we find a lot of cream:
739.1 kg - 176.09 kg = 563.01 (kg of cream)
Take the cream out of the sour cream and cream. So we find the mass of sour cream:
844.76 kg - 563.01 kg = 281.75 (kg of sour cream)
176.09 (kg powdered sugar)
563.01 (kg cream)
281.75 (kg sour cream)
Examination
176.09 kg + 563.01 kg + 281.75 kg = 1020.85 kg
1020.85 kg = 1020.85 kg
Problem 21. The mass of a can filled with milk is 34 kg. The mass of a half-filled can is 17.75 kg. What is the mass of the empty can?
Decision
Subtract from the mass of the can filled with milk, the mass of the can filled halfway. So we get the mass of the contents of the can, half filled, but without taking into account the mass of the can:
34 kg − 17.75 kg = 16.25 kg
16.25 is the weight of the contents of a half-filled can. Multiply this mass by 2, we get the mass of the can filled completely:
16.25 kg × 2 = 32.5 kg
32.5 kg is the mass of the contents of the can. To calculate the mass of an empty can, you need to subtract the mass of its contents from 34 kg, that is, 32.5 kg
34 kg − 32.5 kg = 1.5 kg
Answer: the weight of the empty can is 1.5 kg.
Problem 22. Cream is 0.1 mass of milk and butter is 0.3 mass of cream. How much butter can be obtained from a cow's daily milk yield of 15 kg of milk?
Decision
Let's determine how many kilograms of cream can be obtained from 15 kg of milk. To do this, we find 0.1 part of 15 kg.
15 × 0.1 = 1.5 (kg of cream)
Now let's determine how much butter can be obtained from 1.5 kg of cream. To do this, we find 0.3 part of 1.5 kg
1.5 kg × 0.3 = 0.45 (kg of butter)
Answer: from 15 kg of milk you can get 0.45 kg of butter.
Problem 23. 100 kg of linoleum adhesive contains 55 kg of asphalt, 15 kg of rosin, 5 kg of drying oil and 25 kg of gasoline. What part of this glue is formed by each of its constituents?
Decision
Imagine that 100 kg of glue is like 100 parts. Then asphalt is 55 parts, rosin is 15 parts, drying oil is 5 parts, gasoline is 25 parts. We write these parts in the form of fractions, and, if possible, reduce the resulting fractions:
Answer: glue makes up asphalt, makes up rosin, makes up drying oil, makes up gasoline.
Tasks for independent solution
Task 3. In the first hour, the skier covered the entire distance that he must cover, in the second, the entire path, and in the third, the rest of the path. What part of the total distance did the skier cover in the third hour?
Decision
Let us determine the part of the path covered by the skier in two hours of movement. To do this, add the fractions expressing the paths traveled in the first and second hours:
Let us determine the part of the path covered by the skier in the third hour. To do this, from all parts we subtract the part of the path traveled in the first and second hours of movement:
Answer: in the third hour the skier covered the entire distance.
Task 4. All the boys of the class took part in school competitions: some of them entered the football team, some of the basketball team, some of them competed in long jumps, the rest of the students of the class were on the run. What fraction of runners had more (or fewer) than footballers? basketball players?