Adding and subtracting fractions with different denominators. Operations with fractions Adding and subtracting fractions assignments
Fraction problems
Task 1. The class of schoolchildren consists of excellent students. What part is the rest? Make a graphic description of the task. The drawing can be anything.
Solution
If the excellent students make up the rest, then the rest make up
Problem 2. In a class of schoolchildren, there are excellent students, some good students, and some C students. Make a graphic description of the task. The drawing can be anything.
Task 3. There are 24 students in the class. schoolchildren are made up of excellent students, made up of good students, and made up of C grade students. How many excellent students, good students and C students are there in the class?
Solution
24: 6 × 1 = 4 × 1 = 4 (excellent students)
24: 6 × 3 = 4 × 3 = 12 (good players)
24: 6 × 2 = 4 × 2 = 8 (C grades)
Examination
4 + 12 + 8 = 24 (schoolchildren)
24 = 24
Task 4. In a class of schoolchildren, there are excellent students and good students. What part are C students?
Solution
Schoolchildren 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 remaining two parts are filled with C students. So the schoolchildren are made up of C students
Without giving pictures, you can add the fractions and , and subtract the resulting result from the fraction , which expresses the entire part of the schoolchildren. In other words, add up the excellent and good students, then subtract these excellent and good students from the total number of schoolchildren
Problem 5. There are 16 students in the class. Some of them are excellent and some are good. How many excellent and good students are there in the class? Make a graphic description of the task. The drawing can be anything.
Solution
16: 4 × 1 = 4 × 1 = 4 (excellent students)
16: 16 × 12 = 1 × 12 = 12 (good)
Problem 6. There are 16 students in the class. Of these, there are excellent students, some good students, and some C students. How many excellent, good and C students are there in the class? Make a graphic description of the task. The drawing can be anything.
Solution
16: 8 × 1 = 2 × 1 = 2 (excellent students)
16: 16 × 10 = 1 × 10 = 10 (good)
16: 4 = 4 (C grades)
Task 7. Poltava grains are produced from wheat grains, the mass of which is the mass of wheat grains, and the rest is feed waste. How much Poltava grain and feed waste can be obtained from 500 centners of wheat
Solution
Let's find from 500 centners:
Now let's find a lot of feed waste. To do this, subtract the mass of Poltava cereal from 500 c:
This means that from 500 centners of wheat grains you can get 320 centners of Poltava grain and 180 centners of feed waste.
Task 8. A kilogram of sugar costs 88 rubles. How much does a kg of sugar cost? kg? kg? kg?
Solution
1) 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 highlight the whole part of 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 stated above, costs 88 rubles. We also found the cost of kg and received 33 rubles. This means that a kg of sugar will cost 88+33 rubles, that is, 121 rubles.
The cost can be found without isolating the whole part. To do this, just find from 88.
88: 8 = 11
11 × 11 = 121
But by highlighting the whole part, you can clearly understand 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?
Solution
Let's find out how many grams of sugar are contained in one kilogram of dates. One kilogram is a thousand grams. Let's 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 contained in four kilograms, you need to multiply 720 by 4
720 × 4 = 2880 g
Now we will 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 a thousand grams. Let's 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 to this problem can be written much more briefly, in two expressions:
The point is that they found 4 kilograms and converted the resulting 2.88 into grams, multiplying by 1000. The same thing was done for mineral salts - they found 4 kg and converted the resulting kilograms into grams, multiplying by 1000. Also note that that the fraction of a number was found in a simplified way - by directly multiplying the number by the fraction.
Problem 10. The train traveled 840 km, which is its journey. How far does he have to go? What is the distance of the entire journey?
Solution
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 completed and amount to 840 km. Therefore, dividing 840 km by 4, we find out how many kilometers are in one part:
840: 4 = 210 km.
And since the entire 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 - how much distance does the train have left to travel? If the length of the path is 1470 km, and 840 have been covered, then the remaining path is 1470−840, that is, 630
1470 − 840 = 630
Problem 11. One of the groups that conquered Mount 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 there on this expedition?
Solution
There are 25 athletes in the group. The guides make up the number of athletes. Let's find from 25 and find out how many conductors are in the group:
25: 5 × 4 = 20
There are 45 athletes and guides together. 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 × 140 = 5 × 140 = 700
Problem 12. 900 new textbooks were brought to the school, of which all books were mathematics textbooks, Russian language textbooks were all books, and the rest were literature books. How many books on literature were brought?
Let's find out how much mathematics textbooks consist of:
900: 25 × 8 = 288 (math books)
Let's find out how many textbooks on the Russian language:
900: 100 × 33 = 297 (books on the Russian language)
Let's find out how many literature textbooks there are. To do this, we 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
Problem 13. On the first day they sold, and on the second day the grapes that arrived at the store. How many grapes were sold in two days?
Solution
They sold the grapes in two days. This part is obtained by adding fractions and
You can imagine the grapes arriving at the store in the form of six bunches. Then the grapes are two bunches, the grapes are three bunches, and the grapes are five bunches out of six, sold in two days. Well, it’s not difficult to see that there is only one bunch left, an expressed fraction (one bunch out of six)
Problem 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?
Solution
Let's determine the part of the book read on the second day. It is said that on the second day less was read than on the first day. Therefore, we need to subtract from
On the second day, Vera read books. Now let’s answer the second question of the problem - did Vera manage to read the book in two days? Let's add up what Vera read on the first and second days:
In two days, Vera read the books, but there were still books left. This means Vera did not have time to read the entire book in two days.
Let's do a check. Let's assume that the book 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. Let's find 180 pages or more, 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 the 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 didn’t have time, because in two days she read only 170 pages
100 + 70 = 170 (pages)
There are still 10 pages left to read. In the problem, we had a fraction as the remainder. Let's check if 10 pages are part of the book?
180: 18 × 1 = 10 × 1 = 10 (pages)
Problem 15. One package contains kg, and the other contains kg less. How many kilograms of candy are in two bags together?
Solution
Let's 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. Let's 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 using the fractions given in the problem and start solving it. 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 bag contains 200 grams less, so to determine the mass of the second bag, you need to subtract 200 g from 500 g
500 − 200 = 300 g
And finally, add up the masses of both packages:
500 + 300 = 800 g
Problem 16. Tourists walked from the camp site to the lake in 4 days. On the first day they walked the entire distance, on the second the remaining distance, and on the third and fourth days they walked 12 km each. What is the length of the entire path from the camp site to the lake?
Solution
The problem says that on the second day the tourists walked the rest of the way . The fraction means that the remaining path is divided into 7 equal parts, of which tourists have completed three parts, but the rest remains to be completed. These account for the distance that tourists walked on the third and fourth days, that is, 24 km (12 km on each day). Let's draw a visual diagram illustrating the second, third and fourth days:
On the third and fourth days, tourists walked 24 km and this is equal to the distance covered on the second, third and fourth days. Knowing what 24 km is, we can find the entire distance covered on the second, third and fourth days:
24: 4 × 7 = 6 × 7 = 42 km
On the second, third and fourth days, tourists walked 42 km. Now let's find a path from this. This is how we find out how many kilometers the tourists walked on the second day:
42: 7 × 3 = 6 × 3 = 18 km
Now let's go back to the beginning of the task. It is said that on the first day the tourists walked the entire distance. The entire path is divided into four parts, and the first part accounts for the path covered on the first day. And we have already found the path that falls on the other three parts - it is 42 kilometers covered on the second, third and fourth days. Let's draw a visual diagram illustrating the first and remaining three days:
Knowing that the paths are 42 kilometers long, we can find the length of the entire path:
42: 3 × 4 = 56 km
This means 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 taken by tourists on each of the four days.
First, let's find the path taken on the first day:
56: 4 × 1 = 14 (on the 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 will mark a third, in the second - a quarter, the remaining third part will represent the number 10.
Let's add a third and a quarter:
Now let's draw a segment divided into 12 parts. Let's mark the fraction on it, the remaining five parts will go to the number 10:
Knowing that five twelfths of a number makes up the number 10, we can find the whole number:
10: 5 × 12 = 2 × 12 = 24
We found the whole number - it is 24.
This problem can be solved without providing drawings. To do this, you first need to fold 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 entire number:
Problem 17. A family of four earns 80 thousand rubles a month. The budget is planned as follows: for food, for utilities, for the Internet and TV, for treatment and visits to doctors, for a donation to an orphanage, for living in a rented apartment, for a piggy bank. How much money is allocated for food, utilities, Internet and TV, for treatment and visits to doctors, a donation for an orphanage, for living in a rented apartment, and for a piggy bank?
Solution
80: 40 × 7 = 14 (thousand for food)
80: 20 × 1 = 4 × 1 = 4 thousand (for utilities)
80: 20 × 1 = 4 × 1 = 4 thousand (on Internet and TV)
80: 20 × 3 = 4 × 3 = 12 thousand (for treatment and visits 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. During the hike, tourists walked a kilometer in the first hour, and a kilometer more in the second. How many kilometers did the tourists walk in two hours?
Solution
Let's find numbers using 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
Let us determine the length of the path traveled by tourists in the second hour. To do this, you need to 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
Let us 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 significantly shortened
We received an answer of a kilometer. 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 walked, on the second the entire journey and on the third the remaining 28 km. Calculate the length of the path passing through the valley.
Solution
Let us 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 up the parts of the path covered on the first and second days:
During the first and second days, geologists covered the entire route. The remaining routes 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. Cream, sour cream and powdered sugar were used to prepare the cream. Sour cream and cream are 844.76 kg, and powdered sugar and cream are 739.1 kg. How much individual cream, sour cream and powdered sugar are contained in 1020.85 kg of cream?
Solution
sour cream and cream - 844.76 kg
powdered sugar and cream - 739.1 kg
Let's take out sour cream and cream from 1020.85 kg of cream (844.76 kg). This is how we find the mass of powdered sugar:
1020.85 kg - 844.76 kg = 176.09 (kg of powdered sugar)
Take out the powdered sugar and cream (176.09 kg). So we will find a lot of cream:
739.1 kg - 176.09 kg = 563.01 (kg cream)
Remove the cream from the sour cream and cream. This is how we find the mass of sour cream:
844.76 kg - 563.01 kg = 281.75 (kg 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?
Solution
Let us subtract from the mass of the can filled with milk the mass of the can half filled. So we get the mass of the contents of a half-filled can, but without taking into account the mass of the can:
34 kg − 17.75 kg = 16.25 kg
16.25 is the mass of the contents of the can half filled. Let's multiply this mass by 2, we get the mass of a completely filled can:
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 mass of the empty can is 1.5 kg.
Problem 22. Cream makes up 0.1 weight of milk, and butter makes up 0.3 weight of cream. How much butter can be obtained from a cow's daily milk yield of 15 kg of milk?
Solution
Let's determine how many kilograms of cream can be obtained from 15 kg of milk. To do this, find 0.1 part of 15 kg.
15 × 0.1 = 1.5 (kg cream)
Now let’s determine how much butter can be obtained from 1.5 kg of cream. To do this, find 0.3 part of 1.5 kg
1.5 kg × 0.3 = 0.45 (kg butter)
Answer: from 15 kg of milk you can get 0.45 kg of butter.
Problem 23. 100 kg of linoleum glue contains 55 kg of asphalt, 15 kg of rosin, 5 kg of drying oil and 25 kg of gasoline. What part of this glue does each of its constituents form?
Solution
Let's imagine that 100 kg of glue is 100 parts. Then 55 parts are asphalt, 15 parts are rosin, 5 parts are drying oil, and 25 parts are gasoline. Let's write these parts as fractions, and, if possible, reduce the resulting fractions:
Answer: glue makes up asphalt, makes up rosin, makes up drying oil, makes up gasoline.
Problems to solve independently
Problem 3. In the first hour the skier covered the entire distance that he had to cover, in the second the entire distance, and in the third the remaining part of the path. What part of the total distance did the skier cover in the third hour?
Solution
Let us determine the part of the path covered by the skier in two hours of movement. To do this, we 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 part of the path traveled during the first and second hours of movement:
Answer: in the third hour the skier covered the entire distance.
Task 4. All the boys in the class took part in school competitions: some were on the football team, some on the basketball team, some competed in the long jump, and the rest of the class competed in running. What percentage of runners were more (or fewer) than football players? basketball players?
Actions with fractions.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
So, what are fractions, types of fractions, transformations - we remembered. Let's get to the main issue.
What can you do with fractions? Yes, everything is the same as with ordinary numbers. Add, subtract, multiply, divide.
All these actions with decimal working with fractions is no different from working with whole numbers. Actually, that’s what’s good about them, decimal ones. The only thing is that you need to put the comma correctly.
Mixed numbers, as I already said, are of little use for most actions. They still need to be converted to ordinary fractions.
But the actions with ordinary fractions they will be more cunning. 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.
Adding and subtracting fractions.
Everyone can add (subtract) fractions with the same denominators (I really hope!). Well, let me remind those who are 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 basic property of a fraction (here it comes 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. For the sole purpose of making the denominators the same. Let me note, just in case, that 2/5 and 4/10 are the same fraction! Only 2/5 are uncomfortable for us, and 4/10 are really okay.
By the way, this is the essence of solving any math problems. When we from uncomfortable we do expressions the same thing, but more convenient for solving.
Another example:
The situation is similar. Here we make 48 from 16. By simple multiplication by 3. This is all clear. But we came 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”:
Wow! How did I know about 63? Very simple! 63 is a number that is divisible by 7 and 9 at the same time. Such a number can always be obtained by multiplying the denominators. If we multiply a number by 7, for example, then the result will certainly be divisible 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 common to all fractions and reduce 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’s easier to estimate that the number 16 is perfectly divisible by 2, 4, and 8. Therefore, from these numbers it’s easy to get 16. 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 you take 1024 as the common denominator, everything will work out, in the end everything will be reduced. But not everyone will get to this end, because of the calculations...
Complete the example yourself. Not some kind of logarithm... It should be 29/16.
So, 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 worked honestly 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 rake will be discovered here, yes...
So, we need to add two fractional expressions:
We need to make the denominators the same. And only with the help multiplication! This is what the main property of a fraction dictates. Therefore, I cannot add one to X in the first fraction in the denominator. (that would be nice!). But if you multiply the denominators, you see, everything grows together! So we write down the line of the fraction, leave an empty space at the 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 the parentheses! And now, looking at the common denominator on the right side, we realize: in order to get the denominator x(x+1) in the first fraction, you need to multiply the numerator and denominator of this fraction by (x+1). And in the second fraction - to x. This is what you get:
Note! Here are the parentheses! This is the rake that many people step on. Not parentheses, of course, but their absence. The parentheses appear because we are multiplying all numerator and all 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. We multiply everything and give similar ones. There is no need to open the parentheses in the denominators or multiply anything! In general, in denominators (any) the product is always more pleasant! We get:
So we got the answer. The process seems long and difficult, but it depends on practice. Once you solve the examples, get used to it, everything will become simple. Those who have mastered fractions in due time do all these operations with one left hand, automatically!
And one more note. Many smartly deal with fractions, but get stuck on examples with whole numbers. Like: 2 + 1/2 + 3/4= ? Where to fasten the two-piece? You don’t need to fasten it anywhere, you need to make a fraction out of two. It's not easy, but 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) = (a+b)/1, x=x/1, etc. And then we work with these fractions according to all the rules.
Well, the knowledge of addition and subtraction of fractions was refreshed. Converting fractions from one type to another was repeated. You can also get checked. Shall we settle it 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 operations with fractions.
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You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
To express a part as a fraction of the whole, you need to divide the part into the whole.
Task 1. There are 30 students in the class, four are absent. What proportion of students are absent?
Solution:
Answer: There are no students in the class.
Finding a fraction from a number
To solve problems in which you need to find a part of a whole, the following rule applies:
If a part of a 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 did you spend?
Solution: to find 600 rubles or more, we need to divide this amount into 4 parts, thereby we will find out how much money one fourth part is:
600: 4 = 150 (r.)
Answer: spent 150 rubles.
Task 2. There were 1000 rubles, this amount was spent. How much money was spent?
Solution: from the problem statement we know that 1000 rubles consists of five equal parts. First, let’s find how many rubles are one-fifth of 1000, and then we’ll find out how many rubles are two-fifths:
1) 1000: 5 = 200 (r.) - one fifth.
2) 200 · 2 = 400 (r.) - two fifths.
These two actions can be combined: 1000: 5 · 2 = 400 (r.).
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 the fraction expressing that part of the whole.
Task 3. According to the charter of the cooperative, for the reporting meeting to be valid, at least at least members of the organization must be present. The cooperative has 120 members. What composition can a reporting meeting take place?
Solution: 
Answer: the reporting meeting can take place if there are 80 members of the organization.
Finding a number by its fraction
To solve problems in which you need to find a whole from its part, the following rule applies:
If part of the desired whole is expressed as a fraction, then to find this whole, you can divide this part by the numerator of the fraction and multiply the result by its denominator.
Task 1. We spent 50 rubles, which was less than the original amount. Find the original amount of money.
Solution: from the description of the problem we see that 50 rubles is 6 times less than the original amount, i.e. the original 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, which was less than the original amount of money. Find the original amount.
Solution: We will assume that the required number consists of three thirds. According to the condition, two-thirds of the number equals 600 rubles. First, let's find one third of the original amount, and then how many rubles are three thirds (the original amount):
1) 600: 2 3 = 900 (r.)
Answer: the initial amount is 900 rubles.
The second way to find a whole from its part:
To find a whole by the value expressing its part, you can divide this value by the fraction expressing this part.
Task 3. Line segment AB, equal to 42 cm, is the length of the segment CD. Find the length of the segment CD.
Solution: 
Answer: segment length CD 70 cm.
Task 4. Watermelons were brought to the store. Before lunch, the store sold the watermelons it brought, and after lunch, there were 80 watermelons left to sell. How many watermelons did you bring to the store?
Solution: First, let’s find out what part of the brought watermelons is the number 80. To do this, let’s take the total number of watermelons brought as one and subtract from it the number of watermelons that were sold (sold):
And so, we learned that 80 watermelons make up the total number of watermelons brought. Now we find out how many watermelons from the total amount make up, and then how many watermelons make up (the number of watermelons brought):
2) 80: 4 15 = 300 (watermelons)
Answer: In total, 300 watermelons were brought to the store.
In the real educational process, not many problems on adding and subtracting fractions with the same denominators are required - here there will be enough problems from the textbook. We will pay more attention to problems in which the entire quantity is taken as one. Moreover, at first it is better to imagine it as 2/2, 3/3, etc. quantities.
163 . The girl read 2/5, then another 1/5 of the book. How much of the book did she read?
164 . Tourists walked 1/7, then another 3/7 of the entire route. How much of the route do they have left to cover?
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 there 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 walked 2/5 of the planned route, and on the second day the remaining 15 km. How long is 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 tasks, after lunch - 3/8 tasks, after which he had 24 parts left to turn. How many parts did he have to grind?
171 . From « Arithmetic » L.N. Tolstoy. The husband and wife took money from the same chest, and there was nothing left. The husband took 7/10 of all the money, and the 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 together give 15. Find
quantity.
2) The number and its half make 9. Find the number.
173 . Compose a problem similar to the Egyptian problems and solve it in two ways.
Starting with the next problem, the solutions involve addition and subtraction of fractions with different denominators. If this material was not studied in 5th grade, then the remaining problems related to fractions should be postponed until 6th grade.
174 . a) At every hour, the first pipe fills 1/2 of the pool, and the second pipe fills 1/3 of the pool. What part of the pool can both pipes fill in 1 h working together?
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 travels 1/12 of this distance. To what fraction of this distance do they approach each other in 1 h cars driving towards each other?
175 . a) Two tractor drivers plowed 2/3 of the fields in 1 day of joint work. The first tractor driver plowed 1/2 of the field. What part of the field did the second tractor driver plow?
b) Two cars driving towards each other approached in 1 h at 1/3 of the distance between the two cities. The first car traveled 1/8 of this distance. What part of the total distance did the second car travel?
c) Through two pipes, 1/3 of the pool is filled every hour. Through the first pipe in 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 finished the contents of the cup. 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 cover the distance between two villages in 8 h, and the second for 6 h. What fraction of the distance do they approach in 1 h?
2) Three carpenters were hired to build the bathhouse; the first did 2/33 of the entire work on the day, the second 1/11, the third 7/55. How much of the total work did they all do in a day?
3) 4 scribes were hired to copy the work; the first 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) An ancient problem. A passerby who caught up with the other asked: « How far is it to the village ahead of us? » Another passer-by answered: « The distance from the village from which you are walking is equal to a third of the total distance between the villages, and if you walk another 2 miles, then you will be exactly in the middle between the villages » . How many miles does the first passerby still have to go?
180 . Adam Riese's problem (XVI century). The three won some money. The first accounted for 1/4 of this amount, the second 1/7, and the third 17 florins. How big is the total winnings?
Solving problems from the problem book Vilenkin, Zhokhov, Chesnokov, Shvartsburd for grade 5 on the topic:
26. Adding and subtracting fractions with like denominators
1005 A salad was made from tomatoes weighing 5/16 kg and cucumbers weighing 9/16 kg. What is the mass of the salad?
SOLUTION
1006 The mass of the machine is 73/100 t, and the mass of its packaging is 23/100 t. Find the mass of the machine including the packaging.
SOLUTION
1007 On the first day, potatoes were planted on 2/7 of the plot, and on the second day on 3/7 of the plot. What part of the plot was planted with potatoes during these two days?
SOLUTION
1008 One brigade received 7/10 tons of nails, and the second 3/10 tons less. How many nails did the second brigade receive?
SOLUTION
1009 In two days, 10/11 fields were sown. On the first day, 4/11 fields were sown. What part of the field was sown on the second day?
SOLUTION
1010 The tank is 3/5 filled with gasoline, 1/5 of the tank has been poured into a barrel. What part of the tank remains filled with gasoline?
SOLUTION
1012 Find the value of the expression
SOLUTION
1013 Of the 11 greenhouses of the vegetable farm, 4 are planted with tomatoes, and 2 with cucumbers. What part of the greenhouses is occupied by cucumbers and tomatoes? Solve the problem in two ways.
SOLUTION
1014 An area of 300 hectares was allocated for forest planting. Spruce was planted on 3/10 of the plot, and pine on 4/10 of the plot. How many hectares are occupied by spruce and pine together?
SOLUTION
1015 The team decided to produce 175 items above plan. On the first day she produced 9/25 of this quantity, on the second day 13/25 of this quantity. How many products did the team produce in these two days? How many items does she have left to make?
SOLUTION
1016 11/17 fields of the vegetable farm were planted with potatoes. 1/17 more fields are sown with cucumbers than carrots, and 8/17 fields less than potatoes. What part of the field is sown with cucumbers and what part with carrots? What part of the field is occupied by potatoes, cucumbers and carrots together?
SOLUTION
1019 There were 2 quintals of 70 kg of fruit in the tent. Apples made up 5/9 of all fruits, and pears made up 1/9 of all fruits. How much is the mass of apples greater than the mass of pears? Solve the problem in two ways.
SOLUTION
1020 On the first day the tourist walked 5/14 of the entire route, and on the second day 7/14. It is known that during these two days the tourist walked 36 km. How many kilometers is the entire tourist route?
SOLUTION
1021 The first story took up 5/13 of the book, and the second story took up 2/13 of the book. It is known that the first story took up 12 pages more than the second. How many pages are in the whole book?
SOLUTION
1022 Using the equality 4/25 + 12/25= 16/25, find the values of the expression and solve the equations
SOLUTION
1024 260 people go on an excursion. How many buses should be ordered if each bus should carry no more than 30 passengers?
SOLUTION
1025 Draw a line segment. Then draw a line segment whose length is equal to
SOLUTION
1026 Find the coordinates of points A, B, C, D, E, M, K (Fig. 128) and compare these coordinates with 1.
SOLUTION
1027 Calculate the perimeter and area of triangle ABC (Fig. 129)
SOLUTION
1030 Find all values of x for which the fraction x/15 is a regular fraction and the fraction 8/x is an improper fraction.
SOLUTION
1031 Name 3 proper fractions whose numerator is greater than 100. Name 3 improper fractions whose denominator is greater than 200.
SOLUTION
1033 The length of a rectangular parallelepiped is 8 m, width is 6 m and height is 12 m. Find the sum of the areas of the largest and smallest faces of this parallelepiped.
SOLUTION
1034 To produce 750 m of viscose fabric, 10 kg of cellulose is required. From 1 m3 of wood you can get 200 kg of cellulose. How many meters of viscose fabric can be obtained from 20 m3 of wood?
SOLUTION
1035 The combination lock has six buttons. To open it, you need to press the buttons in a certain sequence and enter a code. How many code options are there for this lock?
SOLUTION
1036 Solve the equation: a) (x - 111) · 59 = 11,918; b) 975(x - 615) = 12,675; c) (30,901 - a) : 605 = 51; d) 39,765: (b - 893) = 1205.
SOLUTION
1037 Solve the problem: 1) Out of 30 planted seeds, 23 germinated. What part of the planted seeds germinated? 2) 40 swans swam on the pond. Of these, 30 were white. What proportion of all swans were white swans?
SOLUTION
1038 Find the value of the expression: 1) 76 · (3569 + 2795) - (24,078 + 30,785); 2) (43 512-43 006) 805 - (48 987 + 297 305)
SOLUTION
1039 In the first hour, 5/17 of the entire road was cleared of snow, and in the second hour, 9/17 of the entire road. How much of the road was cleared of snow during these two hours? Which part of the road was cleared less in the first hour than in the second?
SOLUTION
1040 6/25 m of fabric was used for the dress for the first doll, and 9/25 m of fabric for the dress for the second doll. How much fabric did you use for both dresses? How much more fabric was used on the second doll's dress than on the first doll's dress?