Elementary school program "Perspective": feedback from teachers. Elementary school program "Perspective": teachers' reviews Delimoe
Here we are, finally, and finished our studies safely up to the second grade. The lessons started again and again they are homework. To make doing homework with your child and checking answers much easier, you can use our ready-made math homework assignments in the form of a solution book.
GDZ in this section of the site 7gurus to the textbook of mathematics for grade 2, to its first part. Textbook of the current year of publication. Authors G.V. Dorofeev, T.N. Mirakova, T.B. Buka. Program Perspective.
The answers, as usual on our website, are approved by the primary school teacher. The most difficult to understand tasks and tasks, as well as tasks from the category of increased complexity, we will analyze in more detail.
Select the pages you need from the list to view the GDZ.
Answers to assignments for the textbook of mathematics, part 1 for grade 2 Dorofeev
Analysis of answers and explanations for the tasks of the textbook
By themselves, the tasks in this textbook are quite simple, but there are tricky questions on logic and non-standard thinking. Everything else is simple. Before you start homework, we recommend that you review how to design problems of different types, since teachers sometimes lower grades for processing, and the design rules may vary in different schools. At the beginning of the textbook, the given tasks are solved by actions, but closer to the second part, the teacher may ask you to solve the problem with an expression.
GDZ to the topic Numbers from 1 to 20. Addition and subtraction
Repetition
5 page of the textbook, task 8. Guess how to read the text and read it.
If you attach a mirror to the text, it can be easily read in the reflection.
Task 9. Masha is standing in a round dance of girls. The fourth girl on the left of Masha is the same as the fifth on the right. How many girls are in the round dance.
Solution. We have 1 Masha, 1 "the same girl", and between them there are 3 people on one side and 4 on the other. It is not difficult to count: 1 + 1 + 3 + 4 = 9 people in a round dance. If you draw, marking the girls in a round dance with circles, it will be even easier for the child to figure out the answer.
GDZ to page 7, task 9 (increased complexity). Each bike needs one large wheel and 2 small ones. Made 8 small wheels and 5 large ones. How many bikes can be built using these wheels?
The problem is solved only by the selection method. Draw schematically bicycles, everything will become clear and clear. 5 big wheels are enough for 5 bikes. But eight small wheels are only enough for 4 bicycles (8=2+2+2+2), so we won't be able to make 5 big ones. You will get 4 bicycles and 1 large wheel will remain in stock.
Directions and beams
8 page, 3 task. Think about whether it is still possible to draw rays starting at point O. How many such rays can be drawn.
Answer: an unlimited number of rays can be drawn from any point, that is, an infinite number.
9 page of the textbook, 8 task. The Magnificent Seven.
Since pictures from the application cannot be cut out from the library textbook, you can download them from us, print them on a printer and cut them out. Click on the picture to enlarge the image. Actually
9 page 1 task. Explain for each picture the direction of movement to the objects indicated on it.
The first drawing - there are no difficulties: gas station to the left, first-aid post straight ahead, canteen to the right. But with the second picture, the trick is that the arrows are not from the reader, but towards us. We turn the textbook over and the answer becomes obvious: the village of Solnechny straight ahead, the village of Novinki to the left.
11 page, 9 task. Several numbers are written in a series according to a certain rule. Determine what this rule is and write down the last two numbers in this row. 3 8 5 10 7 12 9
Solution. The pattern is such that 5 is added to one number, and 3 is subtracted from the next.
number beam
15 page, task 9. The Magnificent Seven...
In no case do not cut out a square from a library textbook, but use a scan.
16 page, 4 task. Great staircase.
First we solve the right side of the stairs, it's simple: subtract 2 from 7 and write the answer 5 on the step. Now we add 4 from the next step to this answer, and so on.
The left side is more difficult. They took 2 from a certain number and got 7, so this number is 9. We write it on a step. Further, from to an unknown number, 5 was added and received 9, this number is 4. We write. Next, we fill the stairs by analogy.
19 page, 8 task. There were 4 varieties of cakes in the buffet: puff, shortbread, biscuit and custard. How many different sets of 2 cakes of different varieties can be made from them?
Answer. Puff and sand, puff and biscuit, puff and custard, sand and biscuit, sand and custard, biscuit and custard - only 6 different sets.
9 task. There are 2 numbers in a strip of 11 cells: in the first cell, the number is 6, and in the ninth cell, the number is 4. Is it possible to arrange the numbers in the remaining cells so that the sum of the numbers in any three cells in a row is equal to 15.
Solution. If the amount should be the same, then 3 numbers should alternate with us. We have numbers 6 and 4.
15 - (6 + 4) \u003d 5, that is, the third number is 5. We write down in turn, alternating, 6 5 4.
Beam designation
GDZ 22 page, task 10 of increased complexity. The gnomes of a mountain cave decided to help the giant pick apples. On the first day they worked 6 hours, and on the second - 1 hour more than on the third. How many hours did the gnomes work on the second day and how many on the third, if they did not work 15 hours in just three days?
Solution. And again, the authors of the workbook, without explanation, throw the children a task for the 4th grade, and this is not the first time. Children have not yet passed the division that is used here! Well, okay, if there is this problem in your homework, we will figure it out. So...
We know how much time the gnomes worked in total (15 hours) and how much they worked on the first day (6 hours), we can find out how much they worked on the second and third days together: 15-6=9 hours.
That is, they worked 9 hours in 2 days. But it is impossible to divide in half, because on the second day they worked 1 hour more than the third. That is, you need to divide the days so that the difference is 1 hour. This is 5 hours (on the second day) and 4 hours (on the third).
We check: 6+5+4=15 hours. All right.
If you want to explain to your child how to solve this type of problem correctly, see the article finding terms by sum and difference >>. It will come in handy again and again.
23 page of the textbook, task 8. There are 3 red balls and 2 yellow ones in the bag. 3 balls were drawn at random. What color balloons could you get?
We just sort through all the options in order, there are only 3 of them.
Injection
Page 25, 9 problem. Pineapples of the same weight and melons of the same weight lie on the scales. Find the mass of one pineapple. Is it possible to find the mass of a melon if it is known that the mass of all fruits lying on the scales is 17 kg?
Let's look at the picture. If the same part is removed from each scale (and these are 2 melons and 2 pineapples), then the balance will not be disturbed. A 5 kg weight will remain on the left bowl, and a pineapple and a 4 kg weight will remain on the right bowl. They are balanced, so the pineapple weighs 1kg.
From the mass of all fruits on, for example, the left bowl, we subtract the mass of pineapples (there are 2 pieces of 1 kg each, which means 2 kg) and the mass of weights: 17-2-5 \u003d 10 kg - the remaining 2 melons weigh. So the mass of one melon is 10:2 = 5 kg.
Answers to the lesson Angle designation
Page 27, task 8. The bag contains 3 red balls and 3 blue ones. 3 balls were drawn at random. What color balloons could you get?...
Answer: there are only 4 options, we go through in turn and draw a diagram.
The sum of the same terms
29 page, 10 task. Several numbers are written in a series according to a certain rule. Determine what this rule is and write down the last two numbers in this row. 0 1 1 2 3 5
The rule is simple, two adjacent numbers are added and the next number in the series is obtained. 3+5=8 5+8=13
GDZ on the topic of multiplication and division
Multiplication. Number multiplication 2
32 page, task 8. There are 15 balls in a box: black, white and red. There are 12 fewer red balls than white ones. How many black marbles are in the box?
Solution. If there are 12 fewer red balls than white ones, then there are exactly 12 white balls + some more. If we remove 12 white marbles from the box, only 3 marbles will remain in the box (15-12=3). We have just three colors of balls, so there will be 1 ball of each color in the box. Therefore, there is 1 black ball in the box.
Broken line. Polyline notation
37 page, task 8. Can a triangle and a polyline have only 2 points in common? 3 common points? Make drawings.
Answer: a triangle and a polyline can have 2 common points - these are the corners of the triangle and the vertices of the polyline, a triangle and a polyline can have 3 common points if the polyline is closed (links and sides coincide) or if 2 of its links coincide with the sides of the triangle.
Answers to the lesson Polygon
39 page, task 10. Can a quadrilateral and an angle have 2 common points? 3 common points? Make drawings.
Answer: A quadrilateral and a corner can have 2 common points if one of the vertices of the corner and the quadrilateral coincides and one side of the corner coincides with the side of the quadrilateral. 3 common points - if at the same time 2 sides of the angle coincide with the sides of the quadrilateral.
Multiplication of the number 3
Page 42, task 8. How to divide the figure in the figure into 6 identical triangles with a broken line of three links
The answer is on the scan gdz. The links of the polyline will pass through the opposite corners of the quadrilaterals.
Page 43, task 9. From a bag containing 2 blue and 2 red balls, the girl randomly chooses 2 balls in turn. Are all possible choices of balls shown in the schematic drawing? What option is missing?
Answer: not all, there is not enough option with two identical red balls.
Cube
45 page of the textbook, task 9. Pen, eraser, ruler and bookmark cost 20 rubles together. Pen, ruler and eraser together cost 17 rubles. A bookmark, an eraser and a ruler cost 12 rubles together. The eraser is 1 ruble more expensive than the ruler. How much does each item cost?
Solution. We know that the whole purchase costs 20 r, and we know that the same items without a bookmark cost 17 r, so we can find out how much the bookmark costs: 20-17= 3 p worth bookmark. An eraser, a ruler and a bookmark cost 12 r, so 12-3=9 r cost an eraser and a ruler. And since the eraser is 1 p more expensive than the ruler, then eraser costs 5 r, a ruler 4 p. Find out how much a pen costs: 17-9= 8 r is worth a pen.
We check: the entire purchase should be 20 rubles. 8+5+4+3=20. The answer is correct.
GDZ to page 47, 6 task.
We carefully consider, taking into account those cubes that are hidden behind the front rows. If the child does not quite imagine this figure figuratively, lay it out from real cubes and count how many cubes are used. You should get 14 cubes.
Task 7 on page 47 of the textbook. From a bag containing 2 blue and 2 red balls, the girl chooses 3 balls in turn. Depict all possible options for choosing balls using a schematic drawing. Write down the resulting options using the letters C and K.
The task is similar to the one solved on page 43, task 9, only we draw 1 more ball. For clarity, in order to explain the task to the child, we cut out 2 blue circles and 2 red ones from colored paper, put it in a hat and let the child pull it out in turn one by one. After that, you can draw up a diagram and write down the options. KKS, KSS, KSK, SSK, SKS, SCM
8 task. The guide needs to choose a route through the halls of the museum so as to go around all the halls without entering any of them twice. Where should you start and end your inspection? Find one of the possible routes. Write down the numbers of the halls in the order in which the guide will walk around them.
Solutions: 1 2 3 6 5 4 7 8 9
1 2 3 6 9 8 5 4 7
5 2 1 4 7 8 9 6 3
and many more similar options, starting either from the corner halls or from the middle one.
Multiplying the number 4
Page 49, task 9. How many angles do you see in the drawing? Write down their designations.
The catch is that any 2 rays coming out of the same point form an angle. That is, in the first figure there are 3 corners - AOK, COD and AOD, in the second there are 6 corners - RNS, RNL, RNV, SNL, START, LNV.
Page 51, task 8. Alyosha, Borya, Vasya and Gena are the best mathematicians in the class. You need to submit a team of three people to the school Olympiad. In how many ways can this be done?
The solution is easy to find if you exclude each boy from the team in turn and write down the rest. There will be 4 ways in total: by the first letters of the names ABV, ABG, AVG, BVG.
Page 52, task 2. The mass of one package with flour is 2 kg. 4 such packages were placed on the first pan of the scales, and 3 weights of 2 kg each were placed on the second. How many weights of mass 2 kg must be added to the second scale to bring them into equilibrium?
Here you need to find a lot of things on the first bowl and on the second. We see that the difference is 2 kg, and this is just 1 weight. One weight must be added to balance the scales.
Task 3. The mass of one melon is 2 kg. 3 such melons were placed on the first scale, and 2 weights of 5 kg each on the second. How to balance the scales? Try to find some options.
Let's calculate how much things weigh on 1 and 2 bowls. 4 kg difference. That is, you need
to report melons or 2 more melons,
or 2 weights of 2 kg.
Or replace 1 weight of 5 kg with a weight of 1 kg.
Or put another 3 melons to the melons, and another 2 kg weight to the weights.
Page 53, assignment 10. Cancer, folded from matches, creeps up. Move 3 matches so that he crawls down.
A common mistake is that you start to shift symmetrically and change the top and bottom, and in such puzzles with matches, as a rule, the matches are shifted diagonally or perpendicular to the way you wanted. The solution is in the picture.
GDZ on the topic Multiplication of the number 5
Page 56, task 4. Masha marked 5 points in her notebook and connected them with segments, drawing one segment every two points. How many segments did Masha get in total? Draw this figure in your notebook. Write down the designations of the segments drawn.
In order not to be mistaken, we first draw segments connecting point A with the rest of the points, then point B with all but A, point B with all but A and B, and so on. You should get 10 segments (a star in a hexagon).
Multiplication of the number 6
GDZ to page 57, task 9 increased complexity. Three friends met in the cafe: Belov, Chernov and Ryzhov. "It's amazing that one of us is blond, the other is brunette, the third is red, and at the same time, none of us has a hair color that matches the last name," the black-haired man remarked. "You're right," Belov said. Determine the color of Ryzhov's hair.
GDZ to the textbook section Multiplication of numbers 0 and 1. Multiplication of numbers 7,8,9 and 10. Multiplication table within 20
Division tasks. Division. Division by 2
Pyramid
Page 80. Pyramid. Cut out from the application a figure consisting of 4 triangles...
You can't cut it out of the textbook, so print out the template and cut it out. Click on the image to view the full size template and print it. In fact, the template is quite primitive and will not make a stable pyramid out of it. There are not enough allowances for gluing, we advise you to finish them before cutting.
Page 82, task 9. Borya and Olya played school. “I thought of a number,” Olya said. “If you subtract 10 from it, and then multiply the result by 5, you get 10. What number did I think of?”
We find the solution by carrying out the same actions exactly the opposite: first, divide what happened by 5, and then add 10.
Olya thought of the number 12.
Division by 3
Page 86, task 7. Put + or - signs instead of circles to get the correct entries.
We solve by selection method. Answer: 12-6+9=15 8-5+14=17 9+7-8=8
Page 88, task 8. Vanya laid out a row of pebbles on the table at a distance of 2 cm from one another. How many pebbles did he arrange on a segment 16 cm long?
Solution. The first thing that comes to mind is 16:2=8. But do not jump to conclusions that these are 8 pebbles. By this action, we get 8 segments of 2 cm, which are located between the pebbles. And since the segment has a beginning and an end, then 1 pebble, the very first one, must be taken into account here. Vanya laid out 8+1=9 pebbles.
Dividend. Divider. Private
Page 90, task 9. Can a pentagon and a polyline have 2 points in common? 3 common points? 4 common points? Make drawings.
A pentagon and a polyline can have at least all 5 points in common. Scan answer.
GDZ for a math lesson Division by 4
Page 92, task 9. Fill in the gaps with numbers from 0 to 9 so that you get three correct addition examples. Numbers cannot be repeated. Find two ways.
Solution. In the first example, 2 cells are left for the answer, which means there will be a two-digit number. If any number is added to 0, then we get the same number, and according to the assignment, the numbers should not be repeated. So there is only one place for zero - in the answer of the first example. And since 20, when added using the two listed digits, will not work, then this answer is 10. One and zero were used. We select other numbers by the selection method.
6+4=10 7+2=9 5+3=8
7+3=10 5+4=9 6+2=8
Page 93, task 10. There are 3 keys for 3 suitcases with different locks. Are three samples enough to pick up the keys to the suitcases?
Start talking like this. Let's take some key. If he went to the first suitcase, then the other two keys - from the rest of the suitcases. With one sample, we select the keys to them.
If the first key did not fit the first suitcase, then it is from one of the other suitcases. We take the second key (2nd test). Trying to open the first suitcase. If it was possible, we select the key to the next suitcase with the third breakdown.
If the second key does not fit the first suitcase, then the third one will definitely fit. The other two are from the second and third suitcases. We also select the key of the third breakdown.
Answer: three samples are enough to pick up the keys to three suitcases.
Division by 5
Page 96, task 6. Think of two different problems from the pictures, which are solved like this: 12:3. Sign the names in the answers.
GDZ. a) Mom baked 12 pancakes and divided them into 3 plates equally. How many pancakes are on each plate? 12:3=4 (b.)
b) Ira placed 12 flowers in vases, 3 in each. How many vases did Ira need? 12:3=4 (c.)
Page 96, task 9. How can you release 17 kg of nails from the warehouse in boxes of 3 kg and 2 kg without breaking the packaging? Try to find three options.
Solution. To find out how many whole boxes of 3 kg we can release, we find the nearest number, which is divisible by 3. This is 15. 15:3=5 boxes of 3 kg. 17-15=2 kg of nails left. This is one box of 2kg.
Second option. If you take 4 boxes of 2 kg. 2*4=8 kg Then there will be 17-8=9 kg of nails. 9:3=3 boxes of 3 kg
Third option. We will find out how many whole boxes of 2 kg can be released. The nearest number that is divisible by 2 is 16. But then 1 kg remains, and this is not a whole package. The second number is 14. 14:2=7 boxes of 2 kg. 17-14 \u003d 3 kg, and this is 1 box of 3 kg.
Order of actions
Page 100, task 4. Try to place signs + -, * or: between the numbers so that you get the correct entries.
We decide by choice. 9:3+3=6 12:4+7=10 2*8:4+1=5
Task 7 increased complexity. The boy wrote the number 6 on paper and said to his friend: "Without making any notes, increase this number by 3 and show me the answer." Without thinking twice, the comrade showed the answer. How did he do it?
6+3=9. Nine is an inverted six. You just need to turn over the sheet with the number.
Division by 6
Page 102, task 9. The doctor prescribed the patient 3 injections, one every 2 hours. How long will it take to make all these injections?
Solution. The doctor gave the first injection immediately, then we wait 2 hours and give the second injection, wait another 2 hours and give the third injection. 2+2=4 (h) will be required to make 3 injections.
Page 103, task 7. How can you place action signs between these numbers so that you get the correct record? 1 2 3 4 5 =5
Solution. 1+2+3+4-5=5 or 1*2*3+4-5=5
Task 9. The game "The Third Extra". Try to group the figures in two so that the third is superfluous. Explain why it is redundant.
1 polyline is not closed, the rest are closed.
2 figures are red, the rest are green.
3 figure consists of 5 links, the rest of the four.
Task 10. Yura, Misha, Volodya, Sasha and Oleg took part in ski competitions. Yura came to the finish line before Misha, but later than Oleg. Volodya and Oleg did not come one after another, and Sasha did not come next to either Oleg, or Yura, or Volodya. In what order did the boys finish?
You need to draw a number line and mark points on it - guys, it will be easier to solve the problem. Yura arrived earlier than Misha, but later than Oleg. So Oleg 1st, then Yura, then Misha. We put 3 points: O Yu M
Volodya and Oleg did not come one after another, so Volodya came either after Yura or after Misha.
Sasha did not come next to either Oleg, or Yura, or Volodya, which means he came after Misha - at the very end, which means Volodya came after Yura.
The answer looks like this: O Yu V M S
Division by 7,8,9 and 10
Page 105, task 8. Try to draw up a plan for building a wireframe model of a quadrangular pyramid shown in the figure. Build a pyramid model according to this plan.
A similar plan is on page 103 of the textbook, where it was proposed to build a wireframe cube and on page 87 (build a wireframe model of a triangular pyramid). We draw up a plan by analogy.
1. Roll 5 pea-sized balls from plasticine (for the tops of the pyramid).
2. Prepare 8 matches or counting sticks (for the edges of the pyramid).
3. Build the base of the pyramid. To do this, connect 4 matches with plasticine balls in the form of a square.
4. Take 1 more ball and connect it with a match with each of the balls.
Page 106, task 8. Masha gave Vita a sheet of paper with a square and a triangle drawn on it. Vitya put 3 dots inside the square, and 2 dots inside the triangle. There were 4 points in total, and none of them were located on the sides of a square or triangle. Show how Vitya put the dots.
The fact is that the square and the triangle overlap each other and have a common area. In this general area, we put one point, it will be both inside the square and inside the triangle. We place the rest of the points outside this area.
GDZ on the topic Numbers from 1 to 100. Numbering
Counting in tens. round numbers
Page 112, task 9. Five points A, B, C, D and E were connected by segments and got the figure shown in the figure. Try to draw this figure in one stroke, without lifting the pencil from the sheet of paper and without drawing the same line twice.
We draw, alternately connecting the dots: DBGAWDABVGD
Page 113, task 6. How many cubes are used to build the figure shown in the drawing?
We have 4 layers of 4 cubes each + 3 more cubes. 4*4+3=19 (c.) used.
Page 114, task 9. The chocolate bar has the shape of a square and consists of 9 slices. How many breaks do you need to make to separate the tile into separate slices?
With the first two faults, we divide the tile into 3 parts of 3 slices. Now each of the three parts needs to be broken 2 times to divide into slices. 2+2*3=8 breaks must be made to divide the tile into slices.
Page 115, task 6. How many rays are in the drawing? Write down their designations. What rays intersect?
There are 4 beams in the drawing: OD, VK, IG, TE. If the rays continue along the ruler, it becomes obvious that the rays OD and VC intersect.
Formation of numbers that are greater than 20
Page 117, task 11. Brothers Sasha, Vanya and Dima put on new yellow, lilac and orange jackets and hats of the same colors. Sasha's jacket and cap turned out to be the same color. Vanya never wears yellow clothes. Dima put on a lilac hat and a jacket of a different color. How were the guys dressed?
GDZ for this task. Dima put on a lilac hat, then Vanya gets an orange one (he does not wear yellow), and Sasha a yellow one. Then Sasha's jacket is also yellow. Since Dima has a lilac hat and a jacket of a different color, she is orange. Vanya was left with a lilac jacket.
Answer: Sasha in yellow, Vanya in an orange hat and lilac jacket, Dima in a lilac hat and orange jacket.
Page 118, task 9. From 12 roses, 5 carnations and 6 chrysanthemums made up a bouquet of 15 flowers. Are there roses in this bouquet?
6 + 5 = 11 carnations with chrysanthemums, that is, they will not be enough for a bouquet of 15 flowers and in any case you will have to add roses.
The answer is YES.
Task 10. Vanya placed 16 points on eight lines so that each line had 4 points. Try to guess how he did it.
If there were 4 points each on 8 separate lines, then there would be 32 points in total. We have 16 of them - 2 times less. This means that each point stands at the intersection of lines and belongs to two lines at the same time.
Let's go straight. We mark 4 points. We draw a straight line through each of them. On each line we mark 4 points and so on. We get a quadrilateral, divided by two segments vertically and two horizontally, at the intersection of the lines we mark the points.
Page 120, task 8. Draw any rectangle on the cells of the notebook. A broken line consisting of three links, divide it into 4 identical polygons.
The middle link of the polyline will divide the rectangle in half (either horizontally or vertically), and the 1st and 3rd links will divide the resulting two identical quadrangles diagonally, dividing them into 2 identical triangles each.
Write in the comments what pages you are currently going through.
How often do we teachers hear from the parents of our students that when doing homework, children get distracted, do it carelessly, often ask for help. Here is an excerpt from the letter: “I analyze the task with him, he understands everything, and I see that he can do it himself, without my help. Because of this, I get angry at him, annoyed. The result - he is in tears, I am in a scream.
Very often a child, especially in grades 1 and 2, experiences difficulties with self-organization and self-control. First of all, parents need to understand the real problems of the child, and not label him as lazy and incompetent. There can be many reasons for such behavior of a child.
And they manifest themselves especially clearly precisely in the inability to navigate in the performance of a task, to single out the main, essential in it. He is hardly included in any work that requires stress, it is difficult to switch to the next task, while doing only part of the necessary work or making a large number of mistakes. And this is not laziness or unwillingness to work, but quite objective difficulties that he experiences in such a difficult educational activity.
A lot depends on the parents themselves. After all, the so-called organizing assistance can be a good help in the work for our student. This is not a hint, but a benevolent parental indication of what the child needs to pay attention to in their work.
It is impossible not to say that, of course, it is necessary to start with the help of the child in organizing his workplace. Of course, parents should think about this in advance. Not everyone can allocate a separate room for preparing lessons for a child. In this case, it is especially important to organize two separate zones in the room where the child lives, “playing” and “working”, and visually separate them from each other, so that nothing distracts the first grader from classes. To do this, you can use a movable partition - a screen, a rack, or suspended fabric blinds will do. You can also visually divide the space by sticking wallpaper in a neutral color in the “working” area, unlike the rest of the room.
When designing a child's workplace, parents should remember that the atmosphere created should be conducive to work, study (toys, TV, etc. are best placed in the "play" part of the room). For example, you can hang on the wall developed by the parents (together with the child!) The daily routine or lesson schedule, any educational tables, geographical maps. On a free wall, parents can place a shelf or special fabric pockets in which the child will put some important things for him.
Now about doing the homework itself. Sometimes it is advisable to break the entire amount of the child's work into separate small parts and work with each step by step, while helping him switch from one task to another.
Situations may arise when a first-grader should be reminded which textbook to take out of the portfolio, along with him to find the right page, exercise number. This will save your child time and effort.
And most importantly - do not rush it! Let him work at his "natural" pace at home. After all, forcing the pace of work can quickly exhaust a novice student, increase his nervousness. Under pressure from an adult, a child can write faster, but it is unlikely that he will learn to think faster. However, I repeat once again that thoughtful external control on the part of the parent (or teacher, if the child does homework in the classroom), as a rule, increases the student's work efficiency.
In addition, calm and friendly help will help the baby not only save energy, but also give him the opportunity to believe in himself and his success. And there is no need to be afraid that the child will never become independent - after all, with accompanying, supportive help, we do not deprive him of the initiative, we do not rigidly impose our own way of action, but simply help.
If you cannot overcome difficulties on your own, then you can always turn to a specialist: a teacher, a psychologist, a defectologist, a neurologist. It is they who will allow you to understand the objective root causes of learning difficulties. Professionally competently advise how to help the child.
Of course, these are general recommendations: each situation is individual, like a child. It is important that a growing person believe that his parents will love him regardless of all the difficulties and difficulties, that they are happy with his very desire to do something, his cognitive activity at least on the simplest tasks.
Our dear schoolchildren have a great variety of various tasks and aspirations. This also applies to their school life, when, having come home, it is necessary to do homework and do homework and other preferences and desires ... So, in order to somehow help them save time and effort, so that there is more of the latter for exactly what they wanted, we have created a page of our site.
Here you can find answers to homework in mathematics for grade 3, part 1, according to the Perspective program, author Dorofeev and others. In the people, such homework is also called GDZ. It looks like homework is done. We wanted to add that you should not abuse such tasks, blindly rewrite everything without thinking and without learning. First of all, the information given here is intended for reconciliation, verification, and not for cheating. If you study the topic, do the work, and then check in, then you are doing everything right!
So, let's look at our GDZ.
Answers to homework Grade 3, part 1, Dorofeeva, textbook on the program "Perspective"
Mathematics grade 3, part 1, Dorofeev, textbook, page 3
Numbers from 0 to 100
1. Orally. Answer the questions.
1) The number twenty-five is followed by the number twenty-six. Forty-eight is forty-nine. For eighty-one, the number is eighty-two. Ninety-nine is one hundred.
2) The number thirty-six, preceded by the number thirty-five. Before the number forty, the number thirty-nine. Before the number fifty nine, the number fifty eight. Before the number sixty-one, the number sixty is exactly.
3) Between twenty-six and thirty-two there are five numbers: 27, 28, 29, 30, 31. Between the numbers sixty-nine and seventy-three there are three numbers: 70, 71, 72.
4) Yes, this is the number nine (9). Yes, two-digit number ninety-nine (99).
5) Yes, a small two-digit is ten (10).
2. Calculate: 20 + 4 = 24; 3 + 50 = 53; 61 - 1 = 60;
65 – 1 = 64; 1 + 72 = 73; 9 + 80 = 89;
30 + 8 = 38; 94 – 4 = 90; 50 – 1 = 49;
27 – 7 = 20; 84 – 80 = 4; 35 – 35 = 0;
49 + 1 + 1 = 51; 22 – 1 – 1 = 20; 60 – 1 + 1 = 60.
3. From two boxes of pencils:
1) In the second box: 4 + 16 = 20 pencils;
2) Colored pencils: 12 - 3 = 9 in the first box;
3) Total: 20 + 12 = 32 pencils;
4) Total: 3 + 4 = 7 simple pencils;
5) In the second, there are 16 more colored pencils - 9 = 7 pencils;
+ Question: How many colored pencils are in 2 boxes? 9 + 16 = 25;
+ Question: How many more colored pencils than ordinary ones? 25 - 7 = 18.
Mathematics grade 3, part 1, Dorofeev, page 4
4. To find out, you need to divide this number by 4:
8 / 4 = 2; 12 / 4 = 3; 16 / 4 = 4; 40 / 4 = 10; 80 / 4 = 20.
5. The squirrel will be:
a) at point six (6)
b) at point nine (9)
c) at point fifteen (15)
To get to point 12, you need to make four jumps. When jumping, the squirrel will not end up at point 16.
6. In the first table, the product: 3 * 2 = 6, 5 * 3 = 15; 6 * 2 = 12; 4 * 5 = 20; 8 * 2 = 16; 2 * 7 = 14.
In the second table, the quotient: 8 / 4 = 2; 12 / 6 = 2; 14 / 7 = 2; 15 / 3 = 5; 18 / 9 = 2; 20 / 5 = 4.
7. In total, a segment with a length of 24 cells (12 cm) will be obtained; it will consist of three segments of 8 cells (4 cm) We mark the segments with points C and D. We get segments A - C, C - D, G - B equal to 4 cm.
8. Petya has the most stamps, 15 more than Zhenya and 35 more than Igor.
Mathematics grade 3, part 1, Dorofeev, page 5
1. The first digit in a two-digit number is tens, the second is units.
2. Calculate the meanings of expressions with oral explanation:
43 + 5 = 48 (three plus five equals eight);
24 + 3 = 27 (four plus three equals seven);
55 + 4 = 59 (five plus four equals nine);
69 - 4 \u003d 65 (nine minus four equals five);
56 - 2 = 54 (six minus two equals four);
35 - 3 \u003d 32 (five minus three equals two);
34 + 20 = 54 (three plus two equals five);
65 + 30 = 95 (six plus three equals nine);
47 + 40 = 87 (four plus four equals eight);
78 - 40 \u003d 38 (seven minus four equals three);
53 - 20 = 33 (five minus two equals three);
96 - 50 = 46 (nine minus five equals four).
3. In total, 35 + 40 = 75 seedlings were brought, from lindens 35 - 20 = 15 seedlings remained to be planted.
1) (35 + 40) - 20 \u003d 75 - 20 \u003d 55 subtract those that were planted from the total number of seedlings;
2) add oak seedlings to the remaining linden seedlings: 40 + (35 - 20) = 40 + 15 = 55.
4. Right angle is 90*
1) Right angles in figures: AOB, VDE, STF, TFR.
2) What is the name of a quadrilateral, in which:
a) all right angles of a rectangle;
b) all sides are equal and the angles are right at the square.
Mathematics grade 3, part 1, Dorofeev, page 6
5. Fill in the gaps in the tables:
First table 32 + 2 = 34; 32 + 3 = 35; 32 + 4 = 36; 32 + 5 = 37; 32 + 6 = 38; 32 + 7 = 39.
Second table 78 - 50 = 38; 79 - 50 = 29; 80 - 50 = 30; 81 - 50 = 31; 82 - 50 = 32; 83 - 50 = 33.
1) The sum has increased by one because the term has also increased by one;
2) The difference increased by one because the minuend also increased by one.
6. There are sixty minutes in one hour (60 minutes); there are ten centimeters (10 cm) in one decimeter; one meter is one hundred centimeters (100 cm); There are ten decimeters (10 dm) in one meter
7. Compare.
2 m. 6 dm. less than 32 dm; 7 dm. 4 cm less than 1 m; 2m more than 97cm;
1 hour 10 minutes more than 50 min; 1 hour 35 minutes equals 95 min; 1 hour 2 minutes less than 67 min.
8. 1) 1 hour 12 minutes \u003d 72 min, it took the pedestrian to go; 2) 72 - 24 = 48 minutes, the pedestrian spent so much more time.
9. There are more students in the class who completed the task, because Among them are girls who completed the task. The numbers are the difference and the quotient of which are equal: 4 - 2 \u003d 4 / 2.
Mathematics grade 3, part 1, Dorofeev, page 7
1. Using the diagram, answer the questions:
1) One division 3/6 = 2 fish. In total, 21 * 2 = 42 fish swim, 4 * 2 = 8 - barbs, 9 * 2 = 18 - neon, 5 * 2 = 10 - guppies, 6 - limia.
2) 18 - 10 = 8, so much less guppies than neon.
+ Question: How many fish are there in the aquarium in total, except for the Limes? 8 + 18 + 10 = 36 or 42 - 6 = 36.
2. Gaps in tables.
First table: 2 * 4 = 8; 2 * 5 = 10; 2 * 6 = 12; 2 * 7 = 14; 2 * 8 = 16; 2 * 9 = 18.
Second table: 20 / 2 = 10; 18 / 2 = 9; 16 / 2 = 8; 14 / 2 = 7; 12 / 2 = 6; 10 / 2 = 5.
1) The product has increased by 2 because the multiplier has increased by 1;
2) The quotient has decreased by 1 because the dividend has decreased by 2.
3. 1) 2 * 5 = 10 m, pine height. 2) 5 + 2 = 7 m, pine height. In the first problem, the conditions for the height of a pine tree are given in multiples of 2, and in the second, the difference is by 2. Different operations, multiplication and addition.
Mathematics grade 3, part 1, Dorofeev, page 8
4. 1) 4 * 2 = 8 pages Vanya wrote in his math notebook. 2) 4 + 8 = 12 pages he wrote in both notebooks. If he wrote 2 more pages: 1) 4 + 2 = 6 pages, 2) 4 + 6 = 10 pages.
5. 1) 6 + 14 = 20 grades five and four were given by the teacher; 2) 20 / 4 = 5 marks three points.
6. 1) Square ABCD, perimeter AB * 4; Pentagon DESIQUE, perimeter sum of sides; Triangle LMN, perimeter is the sum of the sides. Right angles (90 *) A, B, C, D, D, K, M.
7. A total of 14 pies, M - with meat, K - with cabbage, G - with mushrooms. 2 * M \u003d K, there are half as many pies with meat as with cabbage. M is less than G, there are fewer meat pies than with mushrooms:
2 * M + M + G = 14; take M = 3, then G = 5, K = 6.
Let's check: 6 + 5 + 3 = 14.
Mathematics grade 3, part 1, Dorofeev, page 9
1. 4 + 6 = 10, 10 / 2 = 5;
14 + 6 = 20, 20 / 2 = 10;
34 + 6 = 40, 40 / 2 = 20;
54 + 6 = 60, 60 / 2 = 30;
94 + 6 = 100, 100 / 2 = 50.
2. (Oral) 1) Kolya learned twice as many lines as Masha, this is six (6) times two, twelve (12) 2) Four (4) times fewer cheesecakes were baked in a pan, sixteen (16) should be divided by four (4), we get four cheesecakes (4) 3) All paint 40 kg. divide by the number in one class 20, 40 / 20 = 2 classes can be painted. 4) All the money is 60 rubles, divided by the cost of one notebook 30, 60 / 30 = 2 notebooks can be bought.
3. I will add the first term or first subtract all individual units from the reduced one.
1) I will add the first term to 30: 30 + 5 = 35 then subtract 4, we get 31;
2) First, subtract from the reduced all individual units 40 - 18 \u003d 22, add 5, we get 27;
3) First, subtract from the reduced all individual units 40 + 47 = 87; add 3, get 90;
4) First subtract from the subtrahend all the individual units 60 - 10 = 50; then subtract 4, we get 46;
5) I will add the first term to 50. 50 + 47 \u003d 97, then subtract 3, we get 94.
4. Perform calculations with oral explanation.
8 + 6 = 14, add 8 + 8 = 16, subtract 2, get 14;
5 + 9 = 14, add 5 + 10 = 15, subtract 1, get 14;
45 + 9 = 54, add 45 + 10 = 55, subtract 1, get 54;
56 + 7 = 63, add 6 + 7 = 13, add 50, get 63;
24 - 7 \u003d 17, subtract 14 - 7 \u003d 7, add 10, we get 17;
43 - 9 \u003d 34, subtract 10 - 9 \u003d 1, add 33, we get 34;
60 - 12 \u003d 48, subtract 60 - 10 \u003d 50, subtract 2, we get 48;
70 - 26 \u003d 44, subtract 30 - 26 \u003d 4, add 40, we get 44;
63 + 17 = 80, add 3 + 7 = 10, add 60 + 10 = 70, total 80;
39 + 31 = 70, add 9 + 1 = 10, add 30 + 30 = 60, total 70.
5. The price of one ball is 20 rubles, one doll - 48 rubles. How much does a model cost if the total amount of toys is 90 rubles. 1) (20 + 48) - 90 = 22 rubles. the model costs.
Return data: The ball costs 20 rubles, the doll is 48 rubles, the model is 22 rubles. How much are all the toys together? 20 + 48 + 22 = 90 rubles
Mathematics grade 3, part 1, Dorofeev, page 10
6. Convenient scale, use one cell in a notebook for one student, depict 4 vertical columns with a common base, but different in height: 24, 27, 18 and 24 cells.
7. Specify the order of actions in expressions. Comput.
2 * 8 + 30 = 16 + 30 = 46, first multiplication, then addition;
53 - 24 / 6 \u003d 53 - 4 \u003d 49, first division, then subtraction;
80 - (30 + 7) \u003d 80 - 37 \u003d 43, first action in brackets, then subtraction;
(21 - 15) / 3 \u003d 14 / 3 \u003d 4, first the action in brackets, then the division.
8. Compare the expressions in each column. Comput.
3 * 6 + 20 = 18 + 20 = 38; 3 * 6 + 2 = 18 + 2 = 20. 38 is greater than 20;
5 * 3 + 7 = 15 + 7 = 22; 5 * 3 + 70 = 15 + 70 = 85. 22 is less than 85;
80 / 2 - 30 = 40 - 30 = 10; 80 / 2 - 3 = 40 - 3 = 37. 10 is less than 37;
60 / 2 - 2 = 30 - 2 = 28; 60 / 2 - 20 = 30 - 20 = 10. 28 is greater than 10.
9. From point A to point B, observing the conditions of the problem, you can go in 6 ways:
1) A-3-6-7-B; 2) A-3-4-7-B; 3) A-3-4-5-B; 4) A-1-4-7-B; 5) A-1-4-5-B 6) A-1-2-5-B.
1. Calculate.
1) Private. 12 / 3 = 4;
2) The work. 8 * 2 = 16;
3) Amount. 27 + 40 = 67;
4) Difference. 70 - 15 = 55.
2. Calculate with oral explanation.
52 + 16 = 68, units 2 + 6 = 8, tens 5 + 1 = 6, result 68;
39 - 24 = 15, units 9 - 4 = 5, tens 3 - 2 = 1, result 15;
47 + 35 = 82, units 7 + 5 = 12 (+ 1 tens), tens 4 + 3 + 1 = 8, result 82;
70 - 46 \u003d 24, ones 10 - 6 \u003d 4 (- 1 ten), tens 7 - 4 - 1 \u003d 2, result 24;
22 + 68 = 90, units 8 + 2 = 10 (+ 1 tens), tens 2 + 6 + 1 = 9, result 90.
Mathematics grade 3, part 1, Dorofeev, page 1 1
3. Calculations in a column. 65 + 24 = 89; 78 - 43 = 35; 36 + 12 = 48; 52 - 24 = 28; 90 - 17 = 73.
4. The boy has 32 rubles left.
1) 100 - (50 + 18) = 32, add up all costs and subtract from the total;
2) (100 - 50) - 18 = 32, subtracts all costs in turn from the total.
5. We consider how much a towel costs: 97 - 17 \u003d 80 rubles. A napkin costs 80 / 2 = 40 rubles. To get 8, 10, 20 in the answer, we change the cost of a napkin 10, 8, 4 times cheaper than a towel.
6. The first and second diagrams are equal in value and ratio, but in the first, the height of the trees is represented by divisions, and in the second, the scale is 5 meters.
1) Pine is 10 meters higher than birch;
2) Below all the rowan trees;
3) The oak is 5 meters below the spruce.
Mathematics grade 3, part 1, Dorofeev, page 12
7. First cut 4 cm; segment, a) 4 + 3 = 7 cm; segment b) 4 * 3 = 12 cm.
8. If Yura takes 7 pencils from the box, he may have 5 blue and 2 red, if he takes 8 pencils, he may have 5 blue and 3 red.
1. 1) 38 + 20 = 58;
2) 15 / 3 = 5;
3) 14 / 7 + 20 = 2 + 20 = 22;
4) 16 + 4 – 5 = 20 – 5 = 15.
2. 1) How many pies were with blueberries 25 - 11 = 14;
2) How many pies did mom bake in total? 25 + (25 - 11) = 39;
3) How many fewer pies were there with blueberries? 25 - (25 - 11) = 11.
3 * 4 / 2 = 6; 3 * 6 / 9 = 2; 3 * 5 / 3 = 5;
(12 + 8) / 4 = 5; (35 + 45) / 8 = 10; (46 + 14) / 6 = 10;
(57 - 42) / 5 = 3; (72 – 60) / 6 = 2; (90 - 30) / 3 = 20;
74 – (43 – 23) * 3 = 74 – 20 * 3 = 14; 8 * 2 + 90 / 90 = 16 + 1 = 17; (70 / 7 + 40) / 5 = (10 + 40) / 5 = 10.
Mathematics grade 3, part 1, Dorofeev, page 13
4. 1 hour 20 minutes = 80 min. more than 75 min.;
1 hour 5 minutes = 65 min. more than 55 min.;
1 hour 13 minutes = 73 min. less than 80 min.;
2 dm. 3 cm = 23 cm. less than 16 cm + 8 cm = 24 cm;
3 m. 6 dm. = 36 dm. more than 42 dm - 7 dm. = 35 dm.;
6 dm. 1 cm = 61 cm less than 1 m – 35 cm = 65 cm
5. Total 7 geese and 9 ducks 7 + 9 = 16 birds. If there are 16 birds in total, of which 7 are geese, how many ducks will there be?
Answer: 16 - 7 = 9 ducks.
If there are 16 birds in total, 9 of which are ducks, how many geese will there be? 16 - 9 = 7 geese.
6. The cake costs: 18 / 2 = 9 rubles. For the price of one cake, you can buy 90 / 9 = 10 cakes.
7. Closed line - hexagon. Line length 15 * 6 = 90 cm.
8. Fluff caught the most - 4 mice, Basilio - B, Vaska - C, Leopold - L:
B + C \u003d L + 4; Using the selection method, we get the equality: 2 + 3 = 1 + 4.
Basilio - 2 mice, Vaska - 3 mice, Leopold - 1 mouse.
Mathematics grade 3, part 1, Dorofeev, page 14
Addition and subtraction.
The sum of several terms.
6 + 9 + 4 = 19. 1st method, add the sum of red and yellow dots, then add green, (6 + 9) + 4 = 19;
2nd method, add the sum of red and green dots, then add yellow,
(6 + 4) + 9 = 19;
3rd method, add yellow and green, then add red, (9 + 4) + 6 = 19.
Conclusion: From a change in the places of the terms, the sum does not change.
(7 + 9) + 3 = 19; (7 + 3) + 9 = 19; (9 + 3) + 7 = 19.
(12 + 8) + 7 = 27; (12 + 7) + 8 = 27; (8 + 7) + 12 = 27.
(16 + 5) + 25 = 46; (25 + 5) + 16 = 46; (16 + 25) + 5 = 46.
2. Calculate in a convenient way.
(28 + 2) + 14 = 44; (16 + 4) + 35 = 55; (17 + 3) + 52 = 72.
3. The perimeter of a triangle is 21 cm + 16 cm + 34 cm = (16 + 34) + 21 = 71 cm.
Mathematics grade 3, part 1, Dorofeev, page 15
4. The price of a pen is 25 rubles, the price of an album is 42 rubles. How much does a notebook cost if everything together costs 100 rubles. 25 + 42 = 67 rubles pen and album price. 100 - 67 \u003d 33 rubles. notebook price. Inverse problem 1) Unknown, pen price: 100 - (33 + 42) = 100 - 75 = 25 rubles, 2) unknown, album price: 100 - (25 + 33) = 100 - 58 = 42 rubles.
5. Compare: 5 dm. = 50 cm more than 48 cm;
1 m. = 100 cm. more than 20 cm;
8 dm. = 80 cm. less than 94 cm;
7 dm. = 70 cm more than 63 cm;
83 cm. more than 3 dm. 8cm=38cm;
6 m. 2 dm. less than 72 dm. \u003d 7 m. 2 dm;
1 dm. 8 cm = 18 cm less than 81 dm. = 810 cm;
3 m. 9 dm. = 39 dm. less than 40 dm;
1 hour 28 minutes = 88 min. more than 78 min;
1 hour 40 minutes equals 100 min;
1 hour 35 minutes = 95 min. more than 85 min;
2 hours 5 minutes = 125 min. more than 1 hour 55 minutes = 115 min.
6. 23 - 6 = 17 kg. cucumbers in a box. 17 + 15 = 32 kg. cucumbers in a bag.
7. AB - ray, VGD - angle, KIL - triangle, MNOP - square, ZE - segment, RSTUF - pentagon, TsCh - line.
8. All numbers from 20 to 29, also 12, 32, 42, 52, 62, 72, 82, 92 - only eighteen numbers (18)
Mathematics grade 3, part 1, Dorofeev, page 16
1. Find the meaning of each expression in three ways, underline the most convenient one.
(6 + 4) + 11 = 21, (4 + 11) + 6 = 21, (11 + 6) + 4 = 21;
(16 + 4) + 8 = 28, (16 + 8) + 4 = 28, (8 + 4) + 16 = 28;
(37 + 13) + 6 = 56, (6 + 37) + 13 = 56, (6 + 13) + 37 = 56.
2. Calculate in a convenient way.
42 + 19 + 18 = (42 + 18) + 19 = 79;
59 + 17 + 11 = (59 + 11) + 17 = 87;
37 + 45 + 3 = (37 + 3) + 45 = 85.
3. Put all the paper clips together. (17 + 43) + 25 = 85 staples in three boxes.
4. 69 - (28 + 15) \u003d 69 - 43 \u003d 26, the length of the third side. To get 30 in the answer, the sum of the lengths of the first and second sides must be 39. For example, 25 and 14.
5. Three faces are visible. Figure 2: Three faces missing, purple (left), green (back), and brown (bottom). Figure 3: Three faces missing, purple (right), yellow (back), and brown (bottom). Figure 4: Three faces missing, green (top), blue (back), and brown (left). Figure 5: Three edges missing, purple (right), green (back), and brown (top)
6. Compare.
68 min. more than 1 h. 05 min. = 65 min;
90 min. equals 1 hour 30 minutes;
84 min. more than 1 h. 20 min. = 80 min;
4 dm. = 40 cm. less than 22 dm + 18 cm = 238 cm;
92 dm. - 6 dm. = 86 dm. more than 8 m = 80 dm;
9 dm. \u003d 90 cm. less than 1 m. - 5 cm. \u003d 95 cm;
2 dm. + 15cm = 35cm less than 1m = 100cm;
50 cm + 5 dm. = 10 dm. less than 5 m. = 50 dm.
Mathematics grade 3, part 1, Dorofeev, page 17
7. The weight of the dog is 18 kg, the cat weighs 5 kg. How much does a pig weigh if the mass of all animals is 63 kg? 63 - (18 + 5) \u003d 63 - 23 \u003d 40 kg, piglet weight. Inverse problem 1) unknown, dog weight: 63 - (40 + 5) = 63 - 45 = 18 kg. 2) unknown, weight of the cat: 63 - (40 + 18) = 63 - 58 = 5 kg.
8. 60 / 2 = 30 tickets sold on the second day. 30 + 37 = 67 tickets sold on the third day.
9. Calculate the sum of all numbers from 1 to 9.
(1 + 2 + 3 + 4) + 5 + 6 + 7 + 8 + 9 = 10 + (5 + 6) + (7 + 8) + 9 = 10 + 11 + 15 + 9 = 25 + 20 = 45.
1. Find the meaning of each expression in three ways, underline the most convenient one.
(15 + 5) + 8 = 20 + 8 = 28, (8 + 15) + 5 = 23 + 5 = 28, (8 + 5) + 15 = 13 + 15 = 28;
(12 + 8) + 13 = 20 + 13 = 33, (13 + 12) + 8 = 25 + 8 = 33, (13 + 8) + 12 = 21 + 12 = 23;
(29 + 11) + 7 = 40 + 7 = 47, (7 + 29) + 11 = 36 + 11 = 47, (7 + 11) + 29 = 18 + 29 = 47.
2. In ascending order: 17 + 5 = 22; 17 + 14 = 31; 28 + 14 = 42; 35 + 14 = 49; 35 + 23 = 58.
3. 25 - 7 = 18 kg, collected red currants. 25 + 18 = 43 kg. currants, all collected by summer residents.
4. (17 + 23) + 11 \u003d 40 + 11 \u003d 51 cm, the perimeter of the triangle. It is necessary to reduce the length of the first and second sides by 11 cm, for example, (12 + 17) + 11 = 29 + 11 = 40.
Mathematics grade 3, part 1, Dorofeev, page 18
5. 76 - (24 + 15) = 76 - 39 = 37 is intended.
6. 1) A, B, O, N, D, C, K, M - 8 vertices of this cube; 2) ABSD, BOX, DSCM - 3 visible edges of the cube. ABON, OKMN, ANMD - 3 invisible edges of the cube; 3) AB, BO, BS, OK, SK, SD, DA, DM, MK - 9 visible faces of the cube. AN, NO, NM - 3 invisible faces of the cube.
7. Fish weighs 12 kg, meat 25 kg. How much does cheese weigh if the mass of all products is 60 kg? 60 - (25 + 12) = 60 - 37 = 23 kg. cheese weight. Inverse problem 1) unknown, fish weight, 60 - (25 + 23) = 60 - 48 = 12 kg, 2) unknown, meat weight: 60 - (23 + 12) = 60 - 35 = 25 kg.
8. Compare.
58 min. less than 1 hour 8 minutes = 68 min;
80 min. more than 1 h. 10 min. = 70 min;
72 min. equals 1 hour 12 minutes;
82 cm + 18 cm = 100 cm equals 10 dm;
5 m = 50 dm. less than 57 dm. - 5 dm. = 52 dm;
1 m. - 2 dm. = 8 dm. more than 7 dm.
9. 12 + 3 = 15 books, in the second pack. 15 / 5 = 3 books in the third pack. 12 + 15 + 3 = 30 books in total.
10. Sum: (1 + 3 + 5 + 7) = 16; (9 + 11 + 13) = 33; (16 + 15) + (33 + 17) + 19 = 31 + 19 + 50 = 100.
Mathematics grade 3, part 1, Dorofeev, page 19
Price. Quantity. Price.
Task about 3 albums: 20 * 3 = 60 rubles. worth the whole purchase.
Mathematics grade 3, part 1, Dorofeev, page 20
1. Make up tasks according to the table and solve them.
1) The price of one pen is 5 rubles. How much will 4 pieces cost? 5 * 4 = 20 rubles
2) The price of one eraser is 2 rubles. How much will the cost of 7 pieces? 2 * 7 = 14 rubles.
3) The price of one notebook is 6 rubles. How much will 3 pieces cost? 6 * 3 = 18 rubles
2. 1) 3 buns, 5 rubles each. will cost 3 * 5 = 15 rubles. 2) One bun will cost 15 / 3 = 5 rubles. 3) For 15 rubles. you can buy 15 / 5 \u003d 3 buns for 5 rubles. At a price of 10 rubles. 3 buns will cost 10 * 3 = 30 rubles. With the same price, 4 buns 10 * 4 = 40 rubles. To find the price, you need to know the quantity and amount of goods. To find the quantity, you need to know the price and the amount of the goods.
3. Calculate in a convenient way.
(41 + 19) + 28 = 60 + 28 = 88; (26 + 34) + 25 = 60 + 25 = 85;
(25 + 45) + 29 = 70 + 29 = 99; (47 + 13) + 16 = 60 + 16 = 76;
(45 + 25) + 22 = 70 + 22 = 92; (27 + 53) + 18 = 80 + 18 = 98.
4. Draw a rectangle, calculate the perimeter: (7 * 2) + (5 * 2) = 14 + 10 = 24 cm.
Mathematics grade 3, part 1, Dorofeev, page 21
5. Express in decimeters or in decimeters and centimeters: 60 cm = 6 dm; 95 cm = 9 dm. and 5 cm; 33 cm = 3 dm. and 3 cm; 1 m. \u003d 10 dm.; 10 cm = 1 dm; 28 cm = 2 dm. and 8 cm.
6. 1) 04:10 exact 03:50; 2) 07:55 exact 07:35; 3) 11:30 is the exact 11:10.
7. Compare: 5 * 4 / 2 = 10, equals 10; 16 / 8 * 5 = 10, less than 20; 20 / 4 + 20 = 25, less than 20 * 5 = 100; 20 * 4 - 20 \u003d 60, equals 20 * 3 \u003d 60; 12 / (6 / 2) = 4, greater than 1; 15 - 7 * 2 = 1, equals 1.
8. 60 - 8 = 52 m left for the first time. 52 - 8 * 2 = 36 m left in the piece.
9. Out of 2 flowers, one each is tulips and carnations, out of 4 - 1 \u003d 3 rose flowers. 3 + 1 + 1 = 5 flowers in total in a bouquet.
1. Decrease the numbers by 30, and reduce the result by 3 times:
45 – 30 = 15, 15 / 3 = 5;
39 – 30 = 9, 9 / 3 = 3;
60 – 30 = 30, 30 / 3 = 10;
48 – 30 = 18, 18 / 3 = 6.
2. Write down: 74 - 24 = 50;
56 + 39 = 95 (+ 1 ten);
81 - 35 \u003d 46 (- 1 ten);
60 - 19 \u003d 41 (- 1 ten);
72 - 27 \u003d 45 (- 1 ten).
3. Calculate: 54 - (47 - 9) \u003d 54 - 38 \u003d 16; 70 - (28 + 27) = 70 - 55 = 15; 81 - (8 + 59) = 81 - 67 = 14;
12 / 3 * 4 = 24; 20 / 4 * 3 = 15; 2 * (14 / 2) = 2 * 7 = 14;
2 * (72 - 64) = 2 * 8 = 16; 3 * (100 / 20) = 3 * 5 = 15; 7 * (60 / 30) = 7 * 2 = 14;
9 + 70 / 10 = 9 + 7 = 16; 30 – 3 * 5 = 30 – 15 = 15; 18 / 3 + 8 = 6 + 8 = 14.
Mathematics grade 3, part 1, Dorofeev, page 22
4. For 18 rubles, at the price of 6 rubles. you can buy 18 / 6 = 3 pencils.
1) How much will 3 pencils of 6 rubles each cost? 3 * 6 = 18 rubles.
2) How much will 1 pencil cost if you can buy 3 pencils for 18 rubles? 18 / 3 \u003d 6 rubles.
Answers: 1) Multiply the price by the quantity; 2) Divide the cost by the quantity; 3) Divide the cost by the price.
5. 1) Visible - 9 ribs, not visible - 3 ribs; 2) No, 1 vertex is not visible, there are 8 in total.
6. Expressions: 18 / 3 = 6, the number of cakes in each box; (18 / 3) / 2 = 3, half of all cakes in one box; 18 - 18 / 3 = 12, the cakes were laid out in a vase.
7. They brought: 9 * 10 \u003d 90 kg., Cabbage. 90 - 47 \u003d 43 kg., Cabbage left.
8. There were 37 fish in total. Perch \u003d bream * 5, and ruffs \u003d bream + 9. We get: L * 5 + L + 9 + L \u003d 37. (L - bream) Using the selection, we find L \u003d 4, then ruffs 4 + 9 \u003d 13, and perches 4 * 5 = 20.
Mathematics grade 3, part 1, Dorofeev, page 23
Addition check.
1. Write down the amounts in a column. Do a check.
14 + 29 = 43, 43 – 14 = 29, 43 – 29 = 14;
34 + 58 = 92, 92 – 34 = 58, 92 – 58 = 34;
56 + 27 = 83, 83 – 56 = 27, 83 – 27 = 56;
42 + 18 = 60, 60 – 42 = 18, 60 – 18 = 42.
2. Make up a problem: It was 19 kg. and 26 kg. honey. 1) We spent 14 kg. How much honey is left? 19 + 26 = 45, 45 - 14 = 31 kg. left. 2) Added 14 kg. How much honey was there? 45 + 14 = 59 kg. became. The tasks are similar to the source data, the difference is only in the action of addition or subtraction.
Mathematics grade 3, part 1, Dorofeev, page 24
3. Compare.
16 cm - 1 dm. \u003d 6 cm is less than 16 cm. - 1 cm. \u003d 15 cm;
1 m. - 5 dm. \u003d 50 cm is more than 1 dm. - 5 cm = 5 cm;
1 m. - 2 dm. = 80 cm is less than 25 cm + 75 cm = 100 cm;
4 dm. + 60 cm = 10 dm. it is more than 1 m. - 1 dm. = 9 dm.
4. Calculate.
2 * 6 / 4 = 3; 4 * 3 / 6 = 2;
16 / 4 / 2 = 2; 18 / 2 / 3 = 3;
(36 - 18) / 6 = 3; (45 - 29) / 8 = 2. These expressions can be divided - without brackets and with brackets. The operations of multiplication and division are performed in order from left to right. Actions in parentheses are executed first.
5. From the problem: (30 - 12) / 2 = 9 buckets were in the second barrel. 1) 9 buckets left in each barrel; 2) 9 + 12 = 21 buckets were in the first barrel.
6. Square ABCD 3 * 4 = 12 cm; rectangle EZhZD (2 + 4) * 2 = 12 cm; rectangle KLMI (1 + 5) * 2 = 12 cm.
7. Granddaughter + 53 years old = Father + 28 years old = Grandfather. 53 - 28 = 25 years difference between father and daughter.
Mathematics grade 3, part 1, Dorofeev, page 25
1. Descending. 9 * 5 = 45; 9 * 3 = 27; 3 * 8 = 24; 7 * 3 = 21; 5 * 3 = 15; 2 * 5 = 10.
2. Calculate. (17 + 3) + 59 = 20 + 59 = 79; (15 + 5) + 26 = 20 + 26 = 46; (36 + 4) + 48 = 40 + 48 = 88.
3. Check. 52 + 37 = 89, 89 - 52 = 37, 89 - 37 = 52; 64 + 18 = 82, 82 - 64 = 18, 82 - 18 = 64; 39 + 25 = 64, 64 - 39 = 25, 64 - 25 = 39; 41 + 19 = 60, 60 - 41 = 19, 60 - 19 = 41.
4. Mom + daughter = 38 years old.; Mom: 38 - 9 = 29 years; Grandmother: 90 - 38 = 52 years old.
5. Compare 15 + 28 less than 15 + 30; 60 - 19 more than 59 - 19; 20/5 is less than 20/4; 83 - 40 more than 83 - 45; 22 + 77 is equal to 77 + 22; 0 * 10 is less than 1 * 9.
6. Subtract the cost of products from the total amount: 100 - 52 - 23 \u003d 25 rubles. cheese cost.
7. Calculate. 4 * 5 - 17 = 37; 9 / 3 + 28 = 31; (52 - 32) / 5 = 4; (89 - 75) / 7 = 2; 18 / (18 - 12) = 3; 28 - (36 - 8) = 0; 97 - (56 - 7 * 2) = 55; 61 + 20 / 5 * 3 = 73.
8. In total, 6 * 3 = 18 pieces were collected from three bushes. tomatoes. 18 / 9 = 2 packs required.
9. A book and a magazine together cost 100 rubles. Book for 50 rubles. more than a magazine. Book - K, magazine - J.
K \u003d W + 50, we get the equality: W + W + 50 \u003d 100;
2F = 100 - 50;
2W = 50;
F = 25.
The magazine costs 25 rubles, and the book: K = 25 + 50 = 75 rubles.
Mathematics grade 3, part 1, Dorofeev, page 26
1. Zoom in 3 times. eighteen; 6; 90; thirty; 12; 60. Zoom in 2 times. 12; 4; 60; twenty; eight; 40.
Spring. 3 * 4 \u003d 12 cm. length of the stretched spring. Two segments OM - 3 cm and OT - 12 cm.
2. Segment AB 2 * 7 \u003d 14 cm. \u003d 1.4 dm.
Mathematics grade 3, part 1, Dorofeev, page 27
3. Add and check.
28 + 36 = 64, 64 – 28 = 36, 64 – 36 = 28;
35 + 45 = 80, 80 – 35 = 45, 80 – 45 = 35;
16 + 69 = 85, 85 – 16 = 69, 85 – 69 = 16;
38 + 38 = 76, 76 – 38 = 38;
47 + 26 = 73, 73 – 47 = 26, 73 – 26 = 47.
4. How much did 4 postcards cost if the price for one piece is 5 rubles? 4 * 5 = 20 rubles.
1) How many postcards were bought for 20 rubles, if the price for one was 5 rubles? 20 / 5 = 4 pieces;
2) How much does a postcard cost if for 20 rubles. did you buy 4 pieces? 20 / 4 \u003d 5 rubles.
5. Do the calculations. 14 / 7 * 4 = 2 * 4 = 8; 3 * 6 / 9 = 18 / 9 = 2; 15 / (12 - 7) + 29 = 15 / 5 + 29 = 32; 7 - 20 / (10 / 2) = 7 - 20 / 5 = 3; 4 * 4 - 2 * 8 = 16 - 16 = 0; 6 * 2 + 9 / 3 = 12 + 3 = 15; 40 / 4 + 20 * 4 = 10 + 80 = 90; 30 * 3 + 30 / 3 = 90 + 10 = 100.
6. 1 dm. 4 cm = 14 cm; 14 / 7 \u003d 2 cm. length of the other side. 14 * 2 + 2 * 2 \u003d 28 + 4 \u003d 32 cm. perimeter.
7. 12 / 6 = 2 rubles. there is one clothespin. 9 * 2 = 18 rubles there are 9 such clothespins.
8. Count the number of whole rows on the figure - 3, multiply by the number of cubes - 5 pcs. and add 2 pcs. 3 * 5 + 2 = 15 + 2 = 17 dice in the blueprint.
9. Fill out. 7 m. \u003d 70 dm.; 4 dm. = 40 cm; 2 m. 6 dm. = 26 dm.; 1 dm. 9 cm = 19 cm; 8 m. + 3 dm. \u003d 1 m. 1 dm.; 5 dm - 9 cm = 4 dm. 1 cm
10. Working with the application.
Mathematics grade 3, part 1, Dorofeev, page 28
1. Calculate in a convenient way.
15 + 28 + 7 = 22 + 28 = 50; 23 + 41 + 7 = 30 + 41 = 71;
42 + 36 + 8 = 50 + 36 = 86; 35 + 2 + 18 = 35 + 20 = 55;
27 + 3 + 54 = 30 + 54 = 84; 84 + 6 + 10 = 90 + 10 = 100.
2. Working with the application. Rice. 2, 2 faces are visible, 2 faces are not visible, red and green. Below is the red line. Behind the green line. Rice. 3, 2 faces are visible, 2 faces are not visible, yellow and green. Below is the blue line. Behind yellow and green.
3. At 9:25 a.m. the students were in the museum. At 9:25 a.m. + 1 hour = 10 hours 25 minutes the tour ended. At 10:25 a.m. + 30 min. = 10 hours 55 minutes The students have returned from the field trip.
4. Task 1. There were 46 liters in a barrel. water. First, 12 liters were added, and then another 8 liters. water. How much water was in the barrel? 46 + 12 + 8 = 66 liters.
Task 2. There were 46 meters of wire in the bay. First they cut off 12 meters, and then another 8 meters. How much wire was left in the bay. 46 - (12 + 8) = 26 m.
These tasks are similar in terms of condition, because in both tasks the source data is changed twice. They differ in that in the first case it is addition, in the second case it is subtraction.
5. In the drawing there are 5 rays, OA, OB, VI, DM, E -.
Mathematics grade 3, part 1, Dorofeev, page 29
6. The length of the second side of the triangle is 24 + 15 = 39 cm. The length of the third side is 39 - 6 = 33 cm. The perimeter of the triangle: 24 + 39 + 33 = 96 cm.
7. Fill in the blanks. 20 + 16 + 10 = 46; 34 + 6 + 12 = 52; 5 + 60 + 15 = 80; 18 + 4 + 32 = 54.
8. Fill in the blanks. 87 cm = 8 dm. 7 cm; 93 cm. = 9 dm. 3 cm; 70 cm = 7 dm; 4 dm. 7 cm = 47 cm; 5 m. 6 dm. = 56 dm.; 9 m. = 90 dm.
9. Diagram. 1) The largest mass is in a pig (100 kg.), The smallest mass is in a goose (10 kg.); 2) For 50 - 10 = 40 kg. the mass of a goose is less than the mass of a sheep; 3) For 100 - 40 = 60 kg. The mass of a pig is greater than that of a goat. Question: 1) How much is the mass of a sheep greater than that of a goat? 50 - 40 = 10 kg. 2) What is the mass of all animals together? 50 + 10 + 40 + 100 = 200 kg.
Mathematics grade 3, part 1, Dorofeev, page 30
1. Calculate in a convenient way.
33 + 17 + 9 = 50 + 9 = 59; 37 + 15 + 13 = 50 + 15 = 65; 16 + 9 + 41 = 16 + 50 = 66;
37 + 8 + 13 = 50 + 8 = 58; 18 + 63 + 7 = 18 + 70 = 88; 51 + 9 + 18 = 60 + 18 = 78;
18 + 9 + 21 = 18 + 30 = 48; 36 + 8 + 14 = 50 + 8 = 58; 65 + 14 + 5 = 70 + 14 = 84;
42 + 11 + 29 = 42 + 40 = 82; 22 + 17 + 18 = 40 + 17 = 57; 45 + 5 + 11 = 50 + 11 = 61.
2. On the first day there were 46 + 27 = 73 bags. On the second day it became 73 + 27 = 100 bags.
3. Task 1. On the road section of 84 m, 41 m were asphalted on the first day, and 23 m on the second day. How many meters of the road remained to be asphalted? 84 - (41 + 23) = 20 m.
Task 2. From 46 kg. potatoes were sold 12 kg., then another 5 kg. How many potatoes are left? 46 - (12 + 5) = 29 kg.
These tasks are similar in terms of condition, because in both tasks the source data is changed twice.
4. Pyramid. 1) Vertex O; 2) Visible edges OA, OD, OS; YES, DS; 3) Invisible edges ABOUT; BA, BS; 4) Visible faces of AOD, DOS, invisible faces of AOB, BOS.
Mathematics grade 3, part 1, Dorofeev, page 31
5. Decide and check. 1) Wires 36 + 40 = 76 m in two pieces. Check 76 - 36 = 40, 76 - 40 = 36.
2) Total 58 + 26 = 84 rubles. the boy had. Check 84 - 58 = 26, 84 - 26 = 58.
6. Fill out.
33 + 24 + 20 = 77; 26 + 26 + 20 + 53 = 99; 10 + 58 + 21 = 89; 27 + 5 + 43 = 75.
7. Length of side BV 40 - 17 = 23 cm. Length of side VD 23 - 5 = 18 cm. Length of side AD 100 - (40 + 23 + 18) = 19 cm.
8. Andryusha weighs toys: car = 2 cubes + 1 ball; car + 1 die = 2 balls. Then 1 ball + 3 dice = 2 balls, remove one ball each, we get 3 dice = 1 ball.
Answer: 5 cubes will balance the car.
Mathematics grade 3, part 1, Dorofeev, page 32
1. Locomotive and wagons. 50 + 30 = 80; 80 + 15 = 95; 95 - 25 = 70; 70 - 18 = 52; 52 + 38 = 90; 90 - 75 = 15; 15 + 5 = 20.
2. Remaining wire: 1) (27 + 27) - 7 = 54 - 7 = 47 m;
2) (27 - 7) + 27 = 20 + 27 = 47 m. The second method is calculated in a more convenient way.
3. Fill in the blanks. 5 + 5 + 5 = 15; 10 + 5 + 5 = 20; 25 + 7 + 3 = 35; 44 + 12 + 10 = 66; 12 + 15 + 33 = 60.
4. The length of the second side of the triangle: 10 + 2 = 12 cm. The sum of the first and second lengths: 10 + 12 = 22 cm. The length of the third side: 22 - 9 = 13 cm.
5. Calculate.
45 + 17 + 15 = 60 + 17 = 77;
29 + 22 + 38 = 29 + 60 = 89;
37 + 13 + 48 = 50 + 48 = 98.
Mathematics grade 3, part 1, Dorofeev, page 33
6. Fairy tale. 70 - 54 = 16 (P); 56 + 33 = 89 (U); 50 / 10 = 5 (C); 9 * 2 = 18 (A); 5 * 3 = 15 (L); 35 - 0 = 35 (O); 18 + 24 = 42 (H); 20 / 5 = 4 (K); 100 - 100 = 0 (A); THE LITTLE MERMAID.
7. Solve problems. 1) 28 + 12 = 40 trees had to be planted; 2) 35 + 40 = 75 pages in a book.
8. Pyramid. 1) 5 ribs are visible, 1 rib is not visible; 2) Yes; 3) At the base of the pyramid is a triangle.
9. Extra number 32, because it is not divisible by 9.
Mathematics grade 3, part 1, Dorofeev, page 34
Designation of geometric shapes.
1. Points “O”, “JI”, “ASh”, angle “KA”, “ES”, “EN”, angle “PI”, “ER”, “EF”.
Mathematics grade 3, part 1, Dorofeev, page 35
2. Shapes. Segment “AB”, ray “PQ”, polygon “KLMNF”, line “RS”, triangle “CDE”.
3. Calculate. 3 * 5 + 10 = 25; 2 * 4 + 30 = 38; 5 * 4 + 40 = 60; 60 - 2 * 6 = 48; 80 - 4 * 5 = 60; 50 - 6 * 2 = 38; 2 * 9 + 12 = 30; 7 * 2 + 36 = 50; 6 * 3 + 52 = 70; 40 / (12 - 8) = 10; 60 / (22 - 19) = 20; 80 / (11 - 7) = 20.
4. Fill in the tables.
1) 37 + 5 = 42; 37 + 4 = 41; 37 + 3 = 40; 37 + 2 = 39; 37 + 1 = 38; 37 + 0 = 37;
2) 59 – 28 = 31; 58 – 28 = 30; 57 – 28 = 29; 56 – 28 = 28; 55 – 28 = 28; 54 – 28 = 26.
The sum has decreased by one because the term has decreased by one. The difference has decreased by one because the minuend has decreased by one.
Mathematics grade 3, part 1, Dorofeev, page 36
5. 30 - 18 = 12 boxes left for the carpenter to make on the second day. 12 / 3 = 4 hours for a carpenter.
6. In buses 20 * 2 = 40 people, in cars 5 * 3 = 15 people. Total 40 + 15 = 55 people.
7. Do the calculations.
1 dm. 2 cm + 5 dm. 7 cm = 12 cm + 57 cm = 68 cm;
8 m. 8 dm. - 3 m. 7 dm. = 88 dm. - 37 dm. = 51 dm.;
6 dm. 8 cm + 2 dm. 2 cm = 68 cm + 22 cm = 90 cm;
4 m. 7 dm. - 37 dm. = 47 dm. - 37 dm. = 10 dm.;
9 dm. 3 cm - 93 cm = 93 cm - 93 cm = 0;
2 dm. 7 cm + 53 cm = 27 cm + 53 cm = 80 cm;
5 dm. 6 cm + 44 cm = 56 cm + 44 cm = 100 cm;
7 m. 4 dm. + 8 dm. = 74 dm. + 8 dm. = 82 dm.
8. Make two-digit numbers. 55, 50, 56, 57, 65, 60, 66, 67, 75, 70, 76, 77.
1. Calculate.
37 + (20 + 7) = 37 + 27 = 64; (40 + 19) – 30 = 59 – 30 = 29; 38 + (2 + 15) = 38 + 17 = 55;
(43 + 19) – 3 = 62 – 3 = 59; 36 – (6 + 18) = 36 – 24 = 12; 57 + (14 + 3) = 57 + 17 = 74;
29 – (10 + 19) = 29 – 29 = 0; (81 + 12) – 31 = 93 – 31 = 62; 57 + (29 + 13) = 57 + 42 = 99.
2. There were 10 + 8 = 18 participants in total. Each team had 18 / 3 = 6 people.
3. Compare.
1 dm. 3 cm is equal to 13 dm.;
1 dm. 5 cm less than 11 dm;
70 dm. less than 7 m. 1 cm;
1 m. more than 9 dm. 4 cm;
7 dm. equal to 70 cm;
2 m. more than 2 dm. 4cm
Mathematics grade 3, part 1, Dorofeev, page 37
4. Write down the designations of the corners. “MAC”, “NKT”, “EFS”, “BOD” – right angle.
5. In a wallet. 3 * 5 = 15 rubles. 5 rub. 6 * 10 = 60 rubles 10 rub. 15 + 60 = 75 rubles was in the wallet.
6. Fill in the blanks. 24 - 13 = 11; 47 - 13 = 34; 62 - 37 = 25; 53 - 26 = 27; 61 - 54 = 7; 32 - 14 = 18.
7. Recordings. 6 + 24 8. The width of the rectangle is 17 - 5 = 12 cm. The perimeter is 17 * 2 + 12 * 2 = 34 + 24 = 58 cm.
9. Write it down. a) 5, 10, 15, 20, 25, 30, 35, 40, 45, 50; b) 20, 18, 16, 14, 12, 10, 8, 6, 4, 2.
Mathematics grade 3, part 1, Dorofeev, page 38
Subtracting a number from a sum.
1. Find the meaning of each expression in three ways, underline the most convenient one.
(47 + 26) - 7 = 73 - 7 = 66 or (47 - 7) + 26 = 40 + 26 = 66, (26 - 7) + 47 = 19 + 47 = 66.
(31 + 29) - 20 = 60 - 20 = 40 or (29 - 20) + 31 = 9 + 31 = 40, (31 - 20) + 29 = 11 + 29 = 40.
(70 + 24) - 14 = 94 - 14 = 80 or (24 - 14) + 70 = 10 + 70 = 80, (70 - 14) + 24 = 56 + 24 = 80.
2. Calculate in a convenient way.
(15 + 26) – 6 = (26 – 6) + 15 = 20 + 15 = 35;
(40 + 54) – 34 = (54 – 34) + 40 = 20 + 40 = 60;
(63 + 9) – 13 = (63 – 13) + 9 = 50 + 9 = 59.
3. Calculate a number from the sum.
(36 + 8) – 5 = 44 – 5 = 39;
(19 + 50) – 30 = 69 – 30 = 39;
(18 + 29) – 8 = 47 – 8 = 39;
The difference of all expressions is 39.
Mathematics grade 3, part 1, Dorofeev, page 39
4. In two boxes left: (23 + 19) - 15 = 42 - 15 = 27 kg .; (23 - 15) + 19 \u003d 8 + 19 \u003d 27 kg .; (19 - 15) + 23 = 4 + 23 = 27 kg.
5. 1) The length of the third side of the triangle (34 + 29) - 30 \u003d (34 - 30) + 29 \u003d 4 + 29 \u003d 33 cm; 2) The perimeter of the triangle is 34 + 29 + 33 = 96 cm.
6. Compare.
10 * 7 is equal to 7 * 10;
16/4 is less than 16/2;
18/6 is less than 18 - 6;
20*4 is more than 20/4;
15/3*4 is more than 15/5*4;
30 * 2 * 0 is less than 30 * 2 * 1.
7. Lyosha and Masha together 12 + 8 = 20 years old. Grandpa 20 * 3 = 60 years old.
8. Fill in the blanks.
a) 2, 12, 22, 32, 42, 52, 62, 72;
b) 85, 79, 73, 67, 61, 55, 49;
c) 1, 4, 5, 9, 14, 23, 37, 60, 97.
9. The cube will be depicted in the figure with the MNPK face. 1) front; 2) behind.
Mathematics grade 3, part 1, Dorofeev, page 40
1. Find the meaning of each expression in three ways, underline the most convenient one.
(56 + 35) – 11 = 91 – 11 = 80; (56 – 11) + 35 = 45 + 35 = 80; 56 + (35 – 11) = 56 + 24 = 80;
(65 + 19) – 24 = 84 – 24 = 60; (65 – 24) + 19 = 41 + 19 = 60; 65 + (19 – 24) = 65 – 5 = 60;
(68 + 34) – 28 = 102 – 28 = 74; (68 – 28) + 34 = 40 + 34 = 74; 68 + (34 – 28) = 68 + 6 = 74.
2. Calculate. (47 + 29) - 17 = (47 - 17) + 29 = 30 + 29 = 59; (50 + 37) - 27 = 50 + (37 - 27) = 50 + 10 = 60; (78 + 9) - 48 = (78 - 48) + 9 = 30 + 9 = 39.
3. Left in the tent: 1) (20 + 35) - (13 + 29) \u003d 55 - 42 \u003d 13 kg .; 2) (20 - 13) + (35 - 29) = 7 + 6 = 13 kg.
4. There are: (15 + 12) - 13 = 27 - 13 = 14 participants left in the competition.
5. Answer: (67 + 8) - 27 = 75 - 27 = 48; (49 + 40) - 20 = 89 - 20 = 69; (78 + 9) - 8 = 87 - 8 = 79; 48 6. Remaining: (15 + 10) - 7 = 25 - 7 = 18 kg. fresh cucumbers.
7. The length of the second side of the triangle: 18 + 4 = 22 cm;
1) The length of the third side of the triangle: (18 + 22) - 5 = 40 - 5 = 35 cm;
2) Triangle perimeter: 18 + 22 + 35 = 75 cm.
8. Fill in the gaps: a) 18, 20, 24, 30, 38, 48; b) 78, 73, 67, 60, 52, 43; c) 10, 16, 15, 21, 20, 26, 25.
Mathematics grade 3, part 1, Dorofeev, page 41
9. Face ASB in the pyramid will be 1) Front; 2) Behind.
1. Find the meaning of each expression in three ways, underline the most convenient one.
(47 + 38) – 15 = 85 – 15 = 70; (47 – 15) + 38 = 32 + 38 = 70; 47 + (38 – 15) = 47 + 23 = 70;
(53 + 38) – 33 = 91 – 33 = 58; (53 – 33) + 38 = 20 + 38 = 58; 53 + (38 – 33) = 53 + 5 = 58;
(57 + 32) – 27 = 89 – 27 = 62; (57 – 27) + 32 = 30 + 32 = 62; 57 + (32 – 27) = 57 + 5 = 62.
2. Calculate. (52 + 29) - 12 = (52 - 12) + 29 = 40 + 29 = 69; (48 + 34) - 24 = 82 - 24 = 58;
(85 + 9) – 35 = (85 – 35) + 9 = 50 + 9 = 59.
3. Shapes. ABC - angle; DE - segment; MN - beam; LK - beam. The rays intersect. Ray MN intersects segment DE.
4. The teacher had: (25 + 25) - 18 = 50 - 18 = 32 notebooks.
Mathematics grade 3, part 1, Dorofeev, page 42
5. Find out how many fruits were brought in total (40 - 15) + 40 = 65 boxes. Remaining: 65 - 23 = 42 boxes.
6. Extra figure No. 4, its shape represents the rotated letter “Z”
7. On segment AD, 72 cm long, put two points B and C, provided that the distance between points C and D is 18 cm, and between B and C is 25 cm. What is the distance between points A and B?
Answer: 72 - (25 + 18) = 72 - 43 = 29 cm.
8. Compare: 12/3 is less than 5; 20 / 4 is more than 3; 16 / 8 is less than 7;
2 * 8 + 30 is less than 50; 20 - 3 * 5 is equal to 5; 0 * 6 + 48 is greater than 46;
3 * 4 / 2 is equal to 6; 9 * 2 / 6 is less than 4; 2 * 7 / 2 is greater than 1.
9. Masha conceived the number X. From the conditions we get the equation: (X + 25) - 15 = 75;
X + 25 = 75 + 15;
X + 25 = 90;
X \u003d 90 - 25;
X = 65.
10. Fill out. a) 5, 9, 12, 16, 19, 23, 26; b) 1, 0, 6, 5, 11, 10, 16, 15, 21, 20; c) 3, 8, 18, 33, 53, 78.
Mathematics grade 3, part 1, Dorofeev, page 43
Check subtraction.
1. Write in a column. 67 - 24 = 43, check 24 + 43 = 67, or 67 - 43 = 24;
80 - 36 = 44, check 36 + 44 = 80, or 80 - 44 = 36;
53 - 18 = 35, check 18 + 35 = 53, or 53 - 35 = 18;
71 - 45 = 26, check 45 + 26 = 71, or 71 - 26 = 45.
2. Task 1. They poured out of the barrel: 30 - 14 = 16 buckets, check 14 + 16 = 30, 30 - 16 = 14.
Task 2. In the class library there were: 52 - 28 = 24 books, check 28 + 24 = 52, 52 - 24 = 28.
3. Calculate. 4 * 3 / 6 = 2; 9 * 2 / 3 = 6; 16 / 4 * 5 = 20; 12 / 3 / 4 = 1; (21 - 9) / 2 = 6; (7 + 53) / 3 = 20; 45 - 20 / 4 = 40; 98 - 9 * 2 = 80.
Mathematics grade 3, part 1, Dorofeev, page 44
4. 1st column 15 + (26 + 8) = 15 + 34 = 49; 32 + (40 + 24) = 32 + 64 = 96; 30 + (47 + 20) = 30 + 67 = 97.
2nd column. (26 + 8) - 15 = 34 - 15 = 19; (40 + 24) - 32 = 64 - 32 = 32; (47 + 20) - 30 = 67 - 30 = 37.
5. Task 1. Out of 40 meters of wire, 15 meters were used up, and then another 9 meters. How much wire is left?
Answer: 40 - (15 + 9) \u003d 40 - 24 \u003d 16 m. Task 2. At 40 l. 15 liters of gasoline were added, then 9 liters were used up. How much gasoline has become? Answer: (40 + 15) - 9 \u003d 55 - 9 \u003d 46 liters. Tasks are similar in condition, but differ in action - added.
6. For a triangle: measure and fold the sides KL + LM + MK.; square: (PO + PR) * 2.
7. Total collected: 4 * 3 = 12 kg. currants. It took: 12 / 2 = 6 packages.
8. Let's write: 2 * 7 = 14. The product 14 / 7 = 2, more than one twice and 14 / 2 = 7, more than the other 7 times.
Mathematics grade 3, part 1, Dorofeev, page 45
1. Write in a column.
52 - 17 = 35, check 17 + 35 = 52, or 52 - 35 = 17;
70 - 28 = 42, check 28 + 42 = 70, or 70 - 42 = 28;
45 - 16 = 29, check 16 + 29 = 45, or 45 - 29 = 16;
84 - 39 = 45, check 39 + 45 = 84, or 84 - 45 = 39.
2. Calculate.
1st column: 35 - 19 = 16; 19 + 16 = 35; 35 - 16 = 19;
2nd column: 27 + 46 = 73; 73 - 27 = 46; 73 - 46 = 27;
3rd column: 50 - 24 = 26; 24 + 26 = 50; 50 - 26 = 24;
4th column: 32 + 18 = 50; 50 - 32 = 18; 50 - 18 = 32.
3. Task 1. Performed: 27 + 28 = 55 vocalists. Check: 55 - 27 = 28, 55 - 28 = 27;
Task 2. Marshmallow costs: 90 - 18 = 72 rubles. Check: 18 + 72 = 90, 90 - 72 = 18.
4. Calculate. 2 * 8 = 16; 3 * 6 = 18; 4 * 4 = 16; 18 / 9 = 2; 15 / 3 = 5; 14 / 7 = 2; 36 - 12/6 = 34; 27 + 3 * 4 = 39; 70 - 15 / 5 = 67; 50 - 6 * 2 = 38; 16 + 0 * 7 = 16; 32 - 4 * 5 = 12; 12 / 3 = 4; 16 / 4 = 4; 18 / 6 = 3.
5. A loaf of bread costs 25 rubles, and a package of kefir costs 20 rubles. more expensive. How much will one loaf and two packages of kefir cost together? Answer: 1) 25 + 20 = 45 rubles. is kefir; 2) 25 + 45 * 2 = 25 + 90 = 115 rubles.
6. Calculate how many cubes are used, multiply the number of cubes in the perimeter of the well (8) by the number of cubes in height (3) and add the ladder (3): 8 * 3 + 3 = 24 + 3 = 27.
Mathematics grade 3, part 1, Dorofeev, page 46
7. In the bank 12 / 6 \u003d 2 liters. milk; 2 * 9 \u003d 18 liters. in a can.
8. 7 people played in the chess tournament, each in 6 games: 7 * 6 = 42. 2 people participate in each game, so 42 / 2 = 21 games were played in the chess tournament.
Subtract the amount from a number.
Mathematics grade 3, part 1, Dorofeev, page 47
1. Calculate the value of each in different ways. Underline the most convenient.
90 – (16 + 50) = 90 – 66 = 24; 90 – (16 + 50) = (90 – 16) – 50 = 74 – 50 = 24; 90 – (16 + 50) = (90 – 50) – 16 = 40 – 16 = 24;
36 – (6 + 17) = 36 – 23 = 13; 36 – (6 + 17) = (36 – 6) – 17 = 30 – 17 = 13; 36 – (6 + 17) = (36 – 17) – 6 =
19 – 6 = 13;
52 – (2 + 39) = 52 – 41 = 11; 52 – (2 + 39) = (52 – 2) – 39 = 50 – 39 = 11; 52 – (2 + 39) = (52 – 39) – 2 =
13 – 2 = 11.
2. Calculate. 45 - (5 + 30) = (45 - 5) - 30 = 40 - 30 = 10; 72 - (9 + 21) = 72 - 30 = 42; 80 - (50 + 7) = (80 - 50) - 7 = 30 - 7 = 23.
3. Calculate. 16 + 8 + 5 = 29; 29 - (16 + 8) = 5; 7 + 43 + 20 = 70; 70 - (43 + 7) = 20; 24 + 35 + 6 = 65; Write down: 65 - (24 + 35) = 6.
4. 1st method: 52 - (9 + 12) = 52 - 21 = 31; 2nd method: (52 - 9) - 12 = 43 - 12 = 31; 3rd way: (52 - 12) - 9 = 40 - 9 = 31 passengers remained in the train car.
5. The length of the third side: 36 - (12 + 9) \u003d 36 - 21 \u003d 15 m.
6. Fill in the blanks. Table 1: 0 * 3 = 0; 1 * 3 = 3; 2 * 3 = 6; 3 * 3 = 9; 4 * 3 = 12; 5 * 3 = 15.
Table 2: 20 / 4 = 5; 16 / 4 = 4; 12 / 4 = 3; 8 / 4 = 2; 4 / 4 = 1; 0 / 4 \u003d 0. 1) The product has increased by 3, because one of the factors has increased by 1, and the other is 3. 2) The dividend has decreased by 4, because the divisor is 4.
Mathematics grade 3, part 1, Dorofeev, page 48
7. Instead of asterisks, you can put: 2085; 76 = 76; 39>38.
8. Pyramid. 1) FMA face in front; 2) FMA edge at the back.
9. Two-digit numbers: 11, 13, 15, 10, 31, 33, 35, 30, 51, 53, 55, 50.
1. Find the meaning of each expression. Underline the most convenient.
70 – (14 + 30) = 70 – 44 = 26; (70 – 30) – 14 = 40 – 14 = 26; (70 – 14) – 30 = 56 – 30 = 26;
54 – (16 + 4) = 54 – 20 = 34; (54 – 4) – 16 = 50 – 16 = 34; (54 – 16) – 4 = 38 – 4 = 34;
68 – (9 + 28) = 68 – 37 = 31; (68 – 28) – 9 = 40 – 9 = 31; (68 – 9) – 28 = 59 – 28 = 31.
2. Calculate in a convenient way. 36 - (6 + 19) = (36 - 6) - 19 = 30 - 19 = 11; 83 - (6 + 44) = 83 - 50 = 33; 70 - (30 + 5) = 70 - 35 = 35.
3. Calculate the values of the expressions. 34 + 9 + 11 = 54; 54 - (34 + 9) = 11; 5 + 28 + 12 = 45; 45 - (28 + 12) = 5; Write down: 7 + 16 + 4 = 27; 27 - (7 + 16) = 4.
Mathematics grade 3, part 1, Dorofeev, page 49
4. The hotel had: 50 - (16 + 23) = 50 - 39 = 11 triple rooms.
5. The candy box contained: (8 + 12) / 4 = 5 rows.
6. a) are divided by 3: 3, 6, 9, 12, 15, 18; b) are not divisible by 3: 0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20.
7. To build a diagram, select the scale of 1 cell = 1 cm. Pen - 15 cells, pencil - 18 cells, eraser - 3 cells, ruler - 30 cells.
8. Ride on the carousel: 4 * 3 = 12 guys. Rollercoaster ride: 12 / 2 = 6 guys.
9. Put actions: 6 * 3 = 18; 6 / 3 = 2; 6 + 3 = 9; 6 - 3 = 3; 8 + 2 = 10; 8 - 2 = 6; 8 * 2 = 16; 8 / 2 = 4; 10 / 2 = 5; 10 * 2 = 20; 10 - 2 = 8; 10 + 2 = 12.
Mathematics grade 3, part 1, Dorofeev, page 50
1. Calculate the value of each expression.
80 – (27 + 40) = 80 – 67 = 13; (80 – 40) – 27 = 40 – 27 = 13; (80 – 27) – 40 = 53 – 40 = 13;
67 – (7 + 17) = 67 – 24 = 43; (67 – 17) – 7 = 50 – 7 = 43; (67 – 7) – 17 = 60 – 17 = 43;
72 – (22 + 39) = 72 – 61 = 11; (72 – 22) – 39 = 50 – 39 = 11; (72 – 39) – 22 = 33 – 22 = 11.
2. Calculate in a convenient way.
44 – (14 + 30) = (44 – 14) – 30 = 30 – 30 = 0;
83 – (19 + 31) = 83 – 50 = 33;
88 – (50 + 8) = (88 – 8) – 50 = 80 – 50 = 30.
3. In the second month the cow ate: 45 - 5 = 40 kg. hay. Left in the barn: 90 - (45 + 40) = 5 kg. hay.
4. Length of the second: 25 + 17 = 42 cm. Length of the third: (25 + 42) - 12 = (42 - 12) + 25 = 30 + 25 = 55 cm.
5. Fill in the blanks. Table 1: 0 * 5 = 0; 1 * 5 = 5; 2 * 5 = 10; 3 * 5 = 15; 4 * 5 = 20; 5 * 5 = 25.
Table 2: 30 / 6 = 5; 24 / 6 = 4; 18 / 6 = 3; 12 / 6 = 2; 6 / 6 = 1; 0 / 6 = 0. 1) The product has increased by 5 because the first factor has increased by 1 and the second factor is 5; 2) The dividend has decreased by 6 because the divisor is 6.
6. In total, Anya had: 16 + 24 = 40 balloons. Yulia was given: 8 + 7 = 15. Remaining: 40 - 15 = 25 balls.
Mathematics grade 3, part 1, Dorofeev, page 51
7. In the first figure: ABC, ADF, DBE, FEC, DEF - 5 triangles. In the second figure: KLM, KNP, NLO, POM, NOP, NSR, STO, RTP, RST - 9 triangles.
8. Fill in the blanks. 25 + 14 is less than 25 + 16; 13 - 3 is greater than 4 + 5; 2 * 3 is less than 3 * 3; 8 / 2 more than 6 - 3; 12 - 1 * 0 = 12 * 1; 1 * 0 is less than 1 * 1. Other solutions are suitable, no more than the recorded values.
9. 6 ways.
10. Encrypted word: MATHEMATICS, 14 = M; 1 = A; 20 = T; 6 = E; 10 = AND; 12 = K;
Mathematics grade 3, part 1, Dorofeev, page 52
Reception of rounding when adding.
1. Find the meaning of expressions (Oral). 29 + 18 = (29 + 1) + (18 - 1) = 30 + 17 = 47; 46 + 25 \u003d (46 + 4) + (25 - 4) \u003d 50 + 21 \u003d 71; 67 + 15 \u003d (67 + 3) + (15 - 3) \u003d 70 + 12 \u003d 82; 58 + 27 \u003d (58 + 2) + (27 - 2) \u003d 60 + 25 \u003d 85; 36 + 17 + 28 = 40 + 20 + (28 - 4 - 3) = 40 + 20 + 21 = 81; 18 + 45 + 16 = 20 + 50 + (16 - 2 - 5) = 20 + 50 + 9 = 79.
2. Compare. 1 hour = 60 minutes more than 59 minutes; 80 min. less than 2 hours = 120 minutes; 2 dm. = 20 cm less than 22 cm; 1 m = 10 dm. more than 9 inches; 1 m. 5 dm. = 15 dm. equal to 15 dm.; 30 cm is less than 3 dm. 4 cm = 34 cm
Mathematics grade 3, part 1, Dorofeev, page 53
3. Calculate the perimeter of the quadrilateral.
1) 9 + 16 + 23 + 18 = 10 + 20 + 30 + (18 - 1 - 4 - 7) = 10 + 20 + 6 = 36 cm;
2) 3 dm. \u003d 30 cm, 2 dm. \u003d 20 cm, 1 dm. 9 cm = 19 cm; 30 + 27 + 20 + 19 \u003d 30 + 30 + 20 + (19 - 3) \u003d 30 + 30 + 20 + 16 \u003d 96 cm or 9 dm. 6 cm;
3) 17 + 28 + 17 + 28 = 17 * 2 + 28 * 2 = 34 + 56 = 90 m.
4. Left on the board: 32 - (9 + 12) \u003d (32 - 12) - 9 \u003d 20 - 9 \u003d 11 pieces.
5. Make an expression: 1) (28 + 45) + 9 = 73 + 9 = 82; 2) 64 - (17 + 8) = 64 - 25 = 39; 3) (55 + 36) - 20 = 91 - 20 = 71; 4) 14 + (18 + 56) = 14 + 74 = 88; 5) 72 - (3 * 6) = 72 - 18 = 54.
6. Segment AC = BD
7. Mass of dried berries: (9 + 7) / 4 = 16 / 4 = 4 kg.
8. In the second package X, in the second X + 20. From the second they transferred to the first, it became X + 10, in the second X + 10. Now the number of sweets in the packages is the same.
Mathematics grade 3, part 1, Dorofeev, page 54
1. Calculate in a convenient way. 67 + 24 = 70 + (24 - 3) = 70 + 21 = 91; 48 + 15 = 50 + (15 - 2) = 50 + 13 = 63; 26 + 39 + 17 = 30 + 40 + (17 - 4 - 1) = 30 + 40 + 12 = 82; 18 + 68 + 9 = 20 + 70 + (9 - 2 - 2) = 20 + 70 + 5 = 95; 19 + 28 + 17 + 16 + 15 = 35 + 28 + 32 = 67 + 28 = 95; 15 + 28 + 25 + 10 + 12 = 40 + 40 + 10 = 90.
2. There were 26 boxes of juice in the warehouse, then they brought 18 more. How many boxes became: 26 + 18 = 44
1) There were 44 juice boxes, 18 were sold. How many boxes are left: 44 - 18 = 26. 2) Out of 44 boxes, 18 were sold. How many boxes are left: 44 - 18 = 26 pcs.
3. In ninth place: a) 52 + 8 = 60 (under the segment); b) 70 - 8 = 62. (under the segment)
4. Team Fakel conceded: 30 / 3 = 10 goals. The Sirena team scored: 30 - 20 = 10 goals.
1) 10; 2) 10; 3) 30 + 10 = 40; 4) 10 – 10 = 0.
5. Compare. 25 cm is equal to 2 dm. 5 cm = 25 cm; 18 dm. more than 1 m. and 7 dm. = 17 dm.; 80 cm less than 8 m = 800 cm; 1 hour 20 minutes = 80 min. less than 1 hour 40 minutes = 100 min; 2 hours = 120 minutes equals 1 hour 18 minutes. + 42 min. = 120 min.; 63 min. less than 6 hours 4 minutes = 364 min.
Mathematics grade 3, part 1, Dorofeev, page 55
6. Calculate the values of the expressions. 3 * 6 = 18; 7 * 2 = 14; 5 * 3 = 15; 14 / 7 = 2; 16 / 2 = 8; 20 / 2 = 10; 2 * 5 + 8 = 18; 4 * 3 - 9 = 3; 3 * 3 + 6 = 15; (17 - 3) / 2 = 7; (36 + 14) * 2 = 100; (18 - 6) / 4 = 3; 99 - 40 / 2 = 79; 56 + 6 * 3 = 56 + 18 = 74; 80 / (20 / 5) = 20.
7. In the village (27 + 53) / 4 = 80 / 4 = 20 five-story houses.
8. Let's count how many rows of cubes in this figure - 9. Multiply by the number in one row - 3; 9 * 3 = 27 dice used to build the figure.
9. From the first - 3 dishes, from the second - 2 dishes, from the third - 2 drinks. To find out all possible menu options, multiply the first, second and third dishes: 3 * 2 * 2 = 12 options.
Mathematics grade 3, part 1, Dorofeev, page 56
1. Compare. 5 * 4 > 5 * 3 + 4; 4 * 3 > 4 * 2 - 3; 6 * 2 18 / 2 > 18 / 3 + 2; 15 / 5 > 15 / 3 - 5; 14/2 8/(8/4) > 8/(8/2); 12 / 4 * 3 > 12 / (4 * 3); 16 / 8 * 2 > 16 / (2 * 8).
2. The pot contains: (2 * 4) * 5 = 40 cups of water.
3. The length of the blue ribbon: 8 * 2 \u003d 16 m., The length of the green ribbon: 16 - 5 \u003d 11 m. All the ribbons together: 8 + 16 + 11 \u003d 35 m. This is enough to make a strip 30 m long from them.
4. Calculate. 68 - (28 + 7) = (68 - 28) - 7 = 40 - 7 = 33; 35 + (5 + 19) = 35 + 5 + 19 = 40 + 19 = 59;
49 – (5 + 19) = (49 – 19) – 5 = 30 – 5 = 25; 60 – (3 + 27) = 60 – 30 = 30;
18 + 39 + 16 + 7 = 20 + 40 + 20 + (7 – 2 – 1 – 4) = 20 + 40 + 20 = 80; 26 + 19 + 27 + 11 = 30 + 20 + 30 + (11 – 4 – 1 – 3) = 30 + 20 + 30 + 3 = 83.
5. Left in the box: 1) 45 - (18 + 16) = 45 - 34 = 11 tiles; 2) (45 - 18) - 16 = 27 - 16 = 11 tiles; 3) (45 - 16) - 18 = 29 - 18 = 11 tiles.
6. You need to cut the triangle from the middle of the bottom side to the top corner.
7. Products of identical factors. 7 * 7 = 49; 3 * 3 * 3 = 27; 2 * 2 * 2 * 2 * 2 = 32.
8. Find out how much one watermelon weighs: 12/3 = 4 kg. It will take twenty apples. 2 + 2 = 4 kg.
Mathematics grade 3, part 1, Dorofeev, page 57
Reception of rounding when subtracting.
1. (Oral) Find the meaning of the expressions.
43 – 18 = 43 – (18 + 2) + 2 = 43 – 20 + 2 = 25; 56 – 29 = 56 – (29 + 1) + 1 = 56 – 30 + 1 = 27;
64 – 27 = (64 + 6) – 27 – 6 = 70 – 27 – 6 = 37; 87 – 48 = (87 + 3) – 48 – 3 = 90 – 48 – 3 = 39;
56 + 19 – 37 = 60 + 20 – (37 + 4 + 1) = 60 + 20 – 42 = 38; 18 + 45 – 36 = 18 + 45 – (36 + 4) + 4 = 18 + 45 – 40 + 4 = 18 + 5 + 4 = 27.
2. Remaining: 57 - 18 = 57 - (18 + 2) + 2 = 57 - 20 + 2 = 39 m of hose.
3. It remains to paint (19 + 26) - 37 = 45 - 37 = 8 cups.
Mathematics grade 3, part 1, Dorofeev, page 58
4. Write an expression and calculate its value.
1) 43 – (19 + 3) = 43 – 22 = 43 – (22 + 8) + 8 = 43 – 30 + 8 = 21;
2) (32 + 49) + 8 = 32 + 50 + (8 – 1) = 32 + 50 + 7 = 89;
3) (25 + 47) – 5 = (25 – 5) + 47 = 20 + 47 = 67;
4) 35 + (16 + 4) = 35 + 20 = 55;
5) 85 – (100 / 20) = 85 – 5 = 80.
5. If you remove one package from the scales, the weight of one package of flour will be: 10 - 5 - 3 = 2 kg.
6. Calculate the values of the expressions. (8 + 7) / 3 = 5; (10 + 8) / 9 = 2; (5 + 9) / 7 = 2; 15 / 3 = 5; 18 / 9 = 2; 14 / 7 = 2; 3 * 5 / 3 = 5; 6 * 3 / 9 = 2; 7 * 2 / 7 = 2. When calculating the quotient, the same divisor is used, and the dividend is one number obtained as a result of various operations.
7. Cells in the drawing: 1 - 12 cells; 2 - 12 cells; 3 - 12 cells; 4 - 22 cells; 5 - 12 cells; 6 - 12 cells; 7 - 12 cells. The same number of cells: 1, 2, 3, 5, 6, 7. Figures 1 and 6, as well as 3 and 7, are the same, they are turned relative to each other.
Mathematics grade 3, part 1, Dorofeev, page 59
8. From the conditions, we write down how much Masha and Lisa collected: M + L = 4 kg. Masha with Katya: M + K = 5 kg., and Katya with Lisa: K + L = 3 kg. We get: M \u003d 4 - L, K \u003d 3 - L, then (4 - L) + (3 - L) \u003d 5;
4 + 3 - L - L = 5;
7 - L - L = 5;
L + L \u003d 7 - 5;
2L = 2;
L = 1. Lisa collected 1 kg. Katya collected: K + 1 = 3; K = 2 kg. Masha collected: M + 2 = 5; M = 3 kg.
1. Find the values using rounding. 61 - 28 \u003d 60 - (28 - 1) \u003d 60 - 27 \u003d 33; 34 - 19 \u003d 30 - (19 - 4) \u003d 30 - 15 \u003d 15; 82 - 17 \u003d 80 - (17 - 2) \u003d 80 - 15 \u003d 65; 23 + 28 = 25 + (28 - 2) = 25 + 26 = 51; 47 + 29 - 38 = 50 + 30 - (38 + 3 + 1) = 50 + 30 - 42 = 38; 19 + 46 - 27 = 20 + 50 - (27 + 1 + 4) = 20 + 50 - 32 = 38.
2. Solve problems and check. 1) The buyer had: 55 + 38 = 60 + (38 - 5) = 60 + 33 = 93 rubles. check, 93 - 38 = 100 - (38 + 7) = 100 - 45 = 55; 93 - 55 = 90 - (55 - 3) = 90 - 52 = 38.
2) In another class there were: 54 - 29 = 50 - (29 - 4) = 50 - 25 = 25 students, check 54 - 25 = 60 - (25 + 6) = 60 - 31 = 29; 54 - 29 = 55 - (29 + 1) = 55 - 30 = 25.
3. Calculate the values of the expressions. 17 + 6 + 34 = 20 + 10 + (34 - 3 - 4) = 20 + 10 + 27 = 57; 57 - (17 + 6) = 57 - 23 = 60 - (23 + 3) = 60 - 26 = 34; 23 + 7 + 48 = 30 + 48 = 78; 78 - (7 + 48) = 78 - 55 = 80 - (55 + 2) = 80 - 57 = 23; 85 + 9 - 25 = 90 + 10 - (25 + 5 + 1) = 90 + 10 - 31 = 69; (85 + 9) - 69 \u003d 85 + 10 - (69 + 1) \u003d 85 + 10 - 70 \u003d 25. You can see that the sum and difference consists of a summand and a subtrahend.
4. Compare. 12/6 = 18/9; 14/2 > 16/4; 18/3 > 20/5;
5 * 2 3 * 5; 0 * 4
15 / 3 3 * 0.
5. The cut line should pass between the 4th and 5th cells on the bottom side of the figure - vertically.
Mathematics grade 3, part 1, Dorofeev, page 60
6. Remaining: 20 - (6 * 3) \u003d 20 - 18 \u003d 2 m of fabric.
7. Put action signs instead of circles: 18 + 6 = 24; 20 / 2 = 10; 15 / 5 = 3; 18 / 6 = 3; 20 - 2 = 18; 15 + 5 = 20; 18 - 6 = 12; 20 + 2 = 22; 15 - 5 = 10.
8. There were 10 candies in two boxes, 20 in total. If several candies were taken from the first, and as many as were left in the first from the second, then they took 10 in total. 20 - 10 \u003d 10 candies left.
Equal figures. If the figures overlap, then they are equal.
Mathematics grade 3, part 1, Dorofeev, page 61
1. It turned out two rockets. These figures are equal, they are cut according to the same pattern. If you fold the sheet in four, you get four figures. If you attach them to each other, they will be the same.
2. Equal figures in the figure: 1 = 4; 2=6=7.
3. An extra piece of CD, it is of a different length.
Mathematics grade 3, part 1, Dorofeev, page 62
4. Calculate the value of the expressions: 2 * 8 + 6 = 22; 5 * 4 - 11 = 9; 3 * 4 + 30 = 42; 7 * 2 - 5 = 9;
18 / 6 + 39 = 3 + 39 = 42; 15 / 3 + 58 = 5 + 58 = 63; 27 – 12 / 4 = 27 – 3 = 24; 60 + 90 / 3 = 60 + 30 = 90;
(25 + 7) – 5 = (25 – 5) + 7 = 20 + 7 = 27; 87 – (30 + 6) = 87 – 36 = 51; 18 + (2 + 70) = (18 + 2) + 70 = 20 + 70 = 90; (23 + 9) – 17 = 32 – 17 = 15; 63 – (45 – 18) = 63 – 27 = 36; 22 + 80 / 4 = 22 + 20 = 42; 70 / 7 * 10 = 10 * 10 = 100; 70 / (7 * 10) = 70 / 70 = 1.
5. 1st way. Let's find out how many groups of painters there were in total: 18 / 3 = 6 groups. New tasks received: (6 - 2) * 3 = 12 people.
2nd way. Let's find out how many painters remained to work: 2 * 3 = 6 people. Now let's find out how many people received new tasks: 18 - 6 = 12 people.
6. Compare. 46 dm. > 4 dm. 5 cm; 19 cm. 30 dm. - 12 dm.;
35 dm. > 60 cm + 29 dm; 2 m. - 7 dm. > 10 dm.; 3 dm. 2 cm. 7. From the first figure you get a square if you cut off 4 cells, in the form of a square, on the left and attach them to the right in the center. From the second figure you get a square if you cut it in half.
8. From the conditions, write down how many flags Lena made: L \u003d M * 2, and Sveta did: C \u003d (M * 2) * 3. Adding all the flags, we get the equality: M + M * 2 + (M * 2) * 3 = 18;
M + 2M + 6M = 18;
9M = 18;
M = 2. Masha made 2 flags for the garland. Lena made: L = 2 * 2 = 4, and Sveta: S = 4 * 3 = 12 pcs.
Mathematics grade 3, part 1, Dorofeev, page 63
Tasks in 3 steps.
Mathematics grade 3, part 1, Dorofeev, page 64
1. On the first shelf there were 4 cans of juice, each 3 liters. On the second shelf there were 7 cans of 2 liters. How many liters of juice were on the shelves in total?
1) How many liters were on the first shelf: 3 * 4 = 12 liters;
2) How many liters were on the second shelf: 2 * 7 = 14 liters;
3) How much juice was on the shelves: 12 + 14 = 26 liters.
2. 1) All potatoes cost: 3 * 10 = 30 rubles; 2) All apples cost: 5 * 20 = 100 rubles; 3) Apples were more expensive by: 100 - 30 = 70 rubles.
3. The length of the second side of the triangle: 6 * 2 = 12 m;
1) The length of the third side: 12 - 3 = 9 m;
2) Triangle perimeter: 6 + 12 + 9 = 27 m.
4. 1) Edges of a cube with a common vertex N: NB, NR, NF. Visible edges: NB, NR;
2) Faces of a cube with a common edge AD: ABCD, AFTD. Invisible edges: AFTD.
3) Opposite face ABNF, face DCRT.
Mathematics grade 3, part 1, Dorofeev, page 65
5. Compare. 2 * (3 + 5) = 2 * 3 + 2 * 5; 4 * (5 - 2) = 4 * 5 - 4 * 2;
(9 + 6) / 3 = 9 / 3 + 6 / 3; (14 - 8) / 2 = 14 / 2 - 8 / 2. You can see that the result of multiplying and dividing one number does not depend on the sequence of actions with its components.
6. Calculate. 29 + 29 + 29 = 30 + 30 + (30 - 3) = 30 + 30 + 27 = 87; 31 + 31 + 31 = 30 + 30 + 33 = 93;
23 + 23 + 23 + 23 = 46 + 46 = 92; 18 + 18 + 18 + 18 = 20 + 20 + 20 + (18 – 6) = 20 + 20 + 20 + 12 = 72.
7. Let's pick up the numbers. 2 + 3 + 4 \u003d 1 * 9 \u003d 2 + 7, we get: E \u003d 1; O = 2; T = 3; P = 4; K = 7; W = 9.
1. Write down: (28 - 25) * 2 = 6; (34 - 25) * 2 = 18; (27 - 25) * 2 = 4; (35 - 25) * 2 = 20; (30 - 25) * 2 = 10; (32 - 25) * 2 = 14.
2. Calculate. 2 * 7 = 14; 6 * 3 = 18; 8 * 2 = 16; 4 * 5 = 20; 15 / 3 = 5; 16 / 4 = 4; 12 / 6 = 2; 18 / 9 = 2;
20 * 3 – 15 = 45; 80 / 4 + 6 = 26; 15 / 5 + 27 = 30; 3 * 4 + 60 = 72; 48 + 15 / 3 = 53; 57 – 80 / 8 = 47;
90 + 9 * 1 = 99; 16 / (2 * 4) = 2; 65 – (70 – 43) = 65 – 27 = 38; (81 + 9) / 9 = 10; 8 * (55 – 45) = 80;
20 / (76 – 71) = 4.
3. Make a task according to the table. In the buffet, the guys bought 3 pies for 6 rubles, and 2 apples for 5 rubles. each. 1) How much money did you spend in total? 6 * 3 + 5 * 2 \u003d 18 + 10 \u003d 28 rubles;
2) How much more pies cost more than apples? 6 * 3 - 5 * 2 \u003d 18 - 10 \u003d 8 rubles.
4. In each set of numbers, find an extra number.
1) 16 is a two-digit number; 2) 12 is not a round number; 3) 38 - not divisible by 11; 4) 40 - no number 3.
Mathematics grade 3, part 1, Dorofeev, page 66
5. To assemble 7 bikes for adults you will need: 7 * 2 = 14 wheels, 4 bikes for kids: 4 * 3 = 12 wheels.
6. Fill in the blanks. 3 cm + 3 cm + 3 cm + 3 cm = 1 dm. 2 cm;
4 m. - 4 dm. - 4 dm. - 4 dm. - 4 dm. = 24 dm.;
6 dm. + 6 dm. + 6 dm. \u003d 1 m. 8 dm.;
1 m. - 5 cm. - 5 cm. - 5 cm. = 8 dm. 5 cm
7. The perimeter of a figure can be found by adding up all its sides.
8. Build a square in a notebook, 4 X 4 \u003d 16 cells.
9. Weight of fish head: 2 * 4 = 8 kg. Body weight: (2 * 8) + (5 * 4) = 16 + 20 = 36 kg. Weight of the whole fish: 4 + 8 + 36 = 48 kg.
Mathematics grade 3, part 1, Dorofeev, page 67
Material for repetition and self-control.
1. Calculate in a convenient way.
2 + 19 + 8 = 10 + 19 = 29;
80 – (24 + 6) = 80 – 30 = 50;
18 + 7 + 5 = 25 + 5 = 30;
95 – (35 + 8) = (95 – 35) – 8 = 60 – 8 = 52;
(40 + 8) – 20 = (40 – 20) + 8 = 20 + 8 = 28;
3 + 17 + 9 = 20 + 9 = 29;
25 + 6 + 4 = 25 + 10 = 35;
75 – (48 + 12) = 75 – 60 = 15;
26 + 4 + 53 = 30 + 53 = 83;
(34 + 8) – 12 = (34 – 12) + 8 = 22 + 8 = 30;
34 + 6 + 40 = 40 + 40 = 80;
39 – (19 + 11) = 39 – 30 = 9.
2. Solve the problem.
1) Let's solve the problem in two steps. First, find out how many pears are in the basket: 16 / 2 = 8 pcs.
Now let's find out how many plums there are: 16 + 8 = 24 pcs.
2) How many apples, pears and plums are in the basket? We write all the data in one sum:
16 + (16 / 2) + 16 + (16 / 2) = 16 + 8 + 16 + 8 = 32 + 16 = 48 pcs.
3. Write down the names and designations of the figures: BM - segment; EK - beam; OF - beam; AC - beam; DS - beam;
LN is a segment. Ray EK and segment BM intersect. Rays OF and DS.
4. Find out how much the notebooks cost: 3 * 6 = 18 rubles. Then two pencils cost: 28 - 18 = 10 rubles.
One pencil costs: 10 / 2 = 5 rubles. The solution can be written as: 28 - (3 * 6) = 10, 10 / 2 = 5.
5. In the 1st figure: 3 * 5 = 15 cells; In the 2nd figure: 6 * 2 + 3 = 15 cells; In the 3rd figure: 3 * 3 + 3 * 2 = 9 + 6 = 15 cells; In the 4th 6 * 2 + 3 = 12 + 3 = 15 cells. The number of cells in all figures is the same.
Mathematics grade 3, part 1, Dorofeev, page 68
6. If the store opens at 9:00 and closes at 6:00 pm, the opening hours will be: 6 + 3 = 9 hours. During the eight-hour working day, the store will be closed for a break.
7. Write down the expressions and find their meanings.
1) (26 + 15) – 9 = 41 – 9 = 32;
2) (83 – 57) + 40 = 26 + 40 = 66;
3) 63 – (36 + 18) = (63 – 36) – 18 = 27 – 18 = 9;
4) (12 + 47) + 30 = 59 + 30 = 89.
8. Take measurements, find the perimeter: 3 cm + 4 cm + 3 cm + 2 cm 10 cm + 2 cm = 24 cm.
9. From the sum we find out how many brick and wooden houses have become, and from the difference we find out how much more brick houses have become. Write down: (38 + 12) - (43 + 5) \u003d 50 - 48 \u003d 2. 2 houses more.
10. Compare.
3 dm. > 2 dm. 9 cm; 5 m. 7 m. > 60 dm.; 8 dm. > 10 cm;
1 hour 15 minutes = 75 min.; 65 min. 11. Find out how many both workers produced together: (50 - 10) + 50 = 90 parts. Now we divide all the parts by the number of boxes, and find out how many are in each: 90 / 3 = 30 parts.
12. Put the signs of arithmetic operations:
6 * 3 1 * 6; 15 / 5 > 0 * 5.
Mathematics grade 3, part 1, Dorofeev, page 69
13. Make a task according to the table.
1) The mass of one melon is 2 kg, and the mass of a watermelon is 3 kg. How much will 3 melons and 4 watermelons weigh together?
Answer: 2 * 3 + 3 * 4 \u003d 6 + 12 \u003d 18 kg .;
2) The mass of one melon is 2 kg, and the mass of a watermelon is 3 kg. How much more is the mass of 4 watermelons than 3 melons?
Answer: 3 * 4 - 2 * 3 \u003d 12 - 6 \u003d 6 kg.
14. Calculate the values of the expressions.
3 * 4 + 20 = 12 + 20 = 32; 15 / 5 + 29 = 3 + 29 = 32; 80 / 2 – 10 = 50; 53 – 2 * 6 = 53 – 12 = 41;
14 / 7 + 48 = 2 + 48 = 50; 18 / 3 + 15 = 6 + 15 = 21; 48 – 4 * 4 = 48 – 16 = 32; 0 + 9 / 3 = 3;
(64 + 18) – 8 = (18 – 8) + 64 = 10 + 64 = 74; 35 – (20 + 9) = (35 – 20) – 9 = 15 – 9 = 6;
28 – (7 + 10) = 28 – 17 = 11; (83 + 9) – 23 = (83 – 23) + 9 = 60 + 9 = 69;
90 / (63 – 54) = 90 / 9 = 10; 45 – 80 / 2 = 45 – 40 = 5; (92 – 78) / 7 = 14 / 7 = 2; 0 * (55 – 38) = 0.
15. Write down two-digit numbers whose sum of digits is 15: 69, 78, 87, 96.
16. Orally.
1) The number 14 is divided by 7, we get 2;
2) We multiply 5 by 3, we get 15;
3) Subtract the number 17 from 50, we get 33;
4) Find out by subtracting 8 from 34, we get 26.
17. Calculate in a convenient way.
48 – (18 + 9) = (48 – 18) – 9 = 30 – 9 = 21; 56 + (4 + 17) = (56 + 4) + 17 = 60 + 17 = 77;
67 – (5 + 17) = (67 – 17) – 5 = 50 – 5 = 45; 70 – (3 + 37) = 70 – 40 = 30;
28 + 19 + 15 + 6 = 28 + 19 + 21 = 28 + 40 = 68; 37 + 19 + 15 + 6 = 37 + 18 + 20 = 37 + 38 = 75.
18. Find out how many trucks there were: 12 / 4 = 3. Add cars and trucks together and divide by 5: (12 + 3) / 5 = 15 / 5 = 3 motorcycles were in the parking lot.
19. Multiply the number of raincoats by the amount of fabric for one raincoat: 3 * 5 = 15 m. Let's calculate whether 18 m is enough: 18 - 15 \u003d 3 m. Answer: enough and another 3 m of fabric will remain.
Mathematics grade 3, part 1, Dorofeev, page 70
20. Calculate.
2 * 6 = 12; 5 * 4 = 20; 4 * 3 = 12;
18 / 9 = 2; 12 / 4 = 3; 16 / 8 = 2;
14 / 7 + 8 = 2 + 8 = 10; 4 * 4 – 9 = 16 – 9 = 7; 0 * 5 + 27 = 27;
(23 - 9) / 2 = 14 / 2 = 7; (16 + 14) * 3 = 30 * 3 = 90; (57 – 49) / 4 = 8 / 4 = 2.
21. Write down 5 more numbers:
1) 16, 19, 17, 20, 18, 21, 19, 22, 20, 23, 21;
2) 7, 17, 18, 28, 29, 39, 40, 50, 51, 61, 62;
3) 39, 40, 42, 45, 49, 54, 60, 67, 75, 84, 94.
22. Orally. Comput.
1) The difference between 30 and 5 is 25;
2) The quotient of 18 and 9 is 2;
3) The sum of 57 and 9 is 66;
4) The product of 2 and 8 is 16.
23. In total, there are 5 * 4 = 20 cells in the figure. Construct a 5 x 4 rectangle in your notebook.
24. Find by rounding.
52 – 18 = 50 – (18 – 2) = 50 – 16 = 34; 86 – 39 = 87 – (39 + 1) = 87 – 40 = 47;
63 – 27 = 60 – (27 – 3) = 60 – 24 = 36; 44 + 18 = 50 + (18 – 6) = 50 + 12 = 62;
16 + 19 – 17 = 20 + 20 – (17 + 5) = 20 + 20 – 22 = 18; 28 + 28 – 36 = 30 + 30 – (36 + 4) = 30 + 30 – 40 = 20.
25. Find out how much fabric was in two pieces: 6 + 12 = 18 m. Find out how many suits were sewn by dividing the entire fabric by the amount of fabric for 1 suit: 18 / 3 = 6 suits.
26. Compare.
14 / 7 6 + 3; 1 * 8 = 8 / 1; 20 / 2 = 2 * 5; 15 – 3 > 15 / 3.
Mathematics grade 3, part 1, Dorofeev, page 71
Practical work. Cube image.
Draw a cube in your notebook, the edge length of which is 3 cm. (6 cells) Faces of the cube: ABCD, OSET.
Mathematics grade 3, part 1, Dorofeev, page 72
Multiplication and division.
1. The number that is divisible: a) by the number 2 - 10; b) on the number 3 - 15; c) on the number 5 - 25; d) on the number 9 - 36
Mathematics grade 3, part 1, Dorofeev, page 73
2. The number by which it is divided: a) the number 6 - 3; b) the number 8 - 2; c) the number 15 - 3.
3. In order for everyone to get the cake equally, you need to cut the cake into 8 or 12 pieces, Answer: a, c.
4. Correct statements: 1, 2, 4.
5. From the data in the diagram, we will answer the questions:
1) The oldest father (35 years old), the youngest daughter (5 years old);
2) Dad is older than mom by: 35 - 30 = 5 years, Daughter is younger than son by: 10 - 5 = 5 years.
+ Question. How old is the son and daughter together: 10 + 5 = 15 years;
+ Question. How many years is dad older than son: 35 - 10 \u003d 25 years.
Mathematics grade 3, part 1, Dorofeev, page 74
6. The cut line will run from the middle of the upper side, vertically down, to the lower left corner
7. Find out how many cups are in the second set: 6 * 2 = 12, and in the third: 6 - 2 = 4 cups. In total in three services: 6 + 12 + 4 = 22 cups. Can be written as a single sum: 6 + (6 * 2) + (6 - 2) = 22.
1. Odd numbers from 10 to 20: 11, 13, 15, 17, 19.
2. Even numbers that are divisible by 3: 3, 6, 9, 12, 15, 18.
3. Let's calculate the sum of even and odd from 1 to 10, even: 2 + 4 + 6 + 8 + 10 = 30, odd: 1 + 3 + 5 + 7 + 9 = 25; 30 - 25 = 5. The sum of all even numbers is 5 more than odd ones.
2 * 7 + 9 = 23; 6 / 3 + 24 = 26;
43 + 7 + 15 = 50 + 15 = 65; 52 + 9 + 11 = 52 + 20 = 72;
(34 + 6) – 8 = 40 – 8 = 32; 56 – (7 + 29) = 56 – 36 = 20.
5. Length of the rectangle: 12 + 9 = 21 m. Perimeter of the rectangle: 12 * 2 + 21 * 2 = 24 + 42 = 66 m.
6. Find out how many fewer books there are from the difference: (7 * 10) - (4 * 10) = 70 - 40 = 30. Answer: 30 books.
Mathematics grade 3, part 1, Dorofeev, page 75
7. Equal figures in the drawing: 1 = 8, 3 = 5, 4 = 6, mirror 5 and 2.
8. Express.
a) 54 dm. \u003d 5 m. 4 dm.; 12 dm. \u003d 1 m. 2 dm.; 30 dm. = 3 m; 76 dm. \u003d 7 m. 6 dm.;
b) 32 cm = 3 dm. 2 cm; 20 cm = 2 dm; 45 cm. = 4 dm. 5 cm; 11 cm = 1 dm. 1 cm;
c) 1 hour 14 minutes = 74 min.; 1 hour 32 minutes = 92 min.; 1 hour 5 minutes = 65 min.
9. Find out how many kilograms of apples were using the sum: 9 + (9 * 2) \u003d 9 + 18 \u003d 27 kg.
Multiplication of the number 3. Division by 3.
1. Write down the results: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
2. Add: 3 + 3 + 3 + 3 = 3 * 4 = 12; 3 + 3 + 3+ 3 + 3 + 3 = 3 * 6 = 18; 3 + 3+ 3 + 3 + 3 = 3 * 5 = 15.
Mathematics grade 3, part 1, Dorofeev, page 76
3. Factors in products, each has two factors, these are 3 and 7; 3 and 8; 3 and 9.
Sum: (3 * 7) = 7 + 7 + 7 = 21; (3 * 8) = 8 + 8 + 8 = 24; (3 * 9) = 9 + 9 + 9 = 27.
4. Perform calculations according to the sample:
3 * 7 = 3 * 6 + 3 = 18 + 3 = 21; 3 * 8 = 3 * 7 + 3 = 21 + 3 = 24; 3 * 9 = 3 * 8 + 3 = 24 + 3 = 27.
5. Make a table of multiplication and division by 3 in your notebook.
Multiplication: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Division: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
6. Multiply the sugar in one glass by the number of glasses: 3 * 5 = 15, 3 * 7 = 21, 3 * 9 = 27.
Mathematics grade 3, part 1, Dorofeev, page 77
7. Divide all pies by the number of pies in one plate: 24 / 3 = 8. Answer: 8 plates.
8. Find out how much water was poured: 35 - 8 = 27, divide by the number of buckets per garden bed: 27 / 3 = 8.
9. Vasya spent 13 - 3 = 10 minutes on the Russian language task, and 13 + 10 = 23 minutes on reading.
In total, Vasya spent 13 + 10 + 23 = 46 minutes preparing his homework.
10. Write down the numbers under the figures: 58, 35, 23, 38, 81, 51.
1. 3 times less numbers: 4, 8, 6, 9, 7.
2. 3 times more numbers: 15, 21, 12, 24, 27.
3. An even number that is divisible by 5 is 10 or 20.
4. Orally. Do the calculations.
1) We increase 5 by 3 times, three by five we get 15;
2) Reduce 30 by 19, thirty minus nineteen, we get 11;
3) We multiply the quotient of numbers 16 and 2 by 3, divide sixteen by two, we get 8, 8 * 3 = 16;
4) The sum of the numbers 28 and 15 is 43, add 40, we get 83;
5) The difference between the numbers 72 and 45 is 27, divided by 3, we get 9.
Mathematics grade 3, part 1, Dorofeev, page 78
5. Compare.
4 * 2 * 3 = 4 * 3 * 2; 12 / 4 * 3 6. Solve problems.
1) Together a bun and a cutlet cost: 8 + (8 * 3) \u003d 8 + 24 \u003d 32 rubles;
2) In both skeins: 27 + (27 / 3) \u003d 27 + 9 \u003d 36 m;
3) Poured into a trough: (7 * 3) + (9 * 2) = 21 + 18 = 39 liters.
7. Calculate the values of the expressions.
37 + 27 = 64; 63 – 29 = 34; 24 / 3 = 8; 60 / 2 = 30; (41 – 20) / 3 = 21 / 3 = 7; (85 – 76) * 3 = 9 * 3 = 27;
27 / 3 * 2 = 18; 20 * 4 / 8 = 10.
8. Extra figure No. 3 (green) It differs in shape.
9. To find out how old grandfather is, you need to know a two-digit number from which we subtract 90. This number can be from 91 to 99. Let's select a number to write the result in the same numbers.
(95 - 90) * 3 + 73 = 5 * 3 + 73 = 15 + 73 = 88; Answer: Grandpa is 88 years old, two-digit number 95.
Mathematics grade 3, part 1, Dorofeev, page 79
Multiplying a sum by a number.
1. Find the meaning of each expression in two ways, underline the most convenient one.
(2 + 7) * 2 = 9 * 2 = 18, (2 + 7) * 2 = 2 * 2 + 7 * 2 = 4 + 14 = 18;
(4 + 1) * 3 = 5 * 3 = 15, (4 + 1) * 3 = 4 * 3 + 1 * 3 = 12 + 3 = 15;
(3 + 5) * 2 = 8 * 2 = 16; (3 + 5) * 2 = 3 * 2 + 5 * 2 = 6 + 10 = 16.
2. Calculate in a convenient way.
(3 + 7) * 4 = 10 * 4 = 40; (14 + 6) * 2 = 20 * 2 = 40; (3 + 4) * 5 = 3 * 5 + 4 * 5 = 15 + 20 = 35.
Mathematics grade 3, part 1, Dorofeev, page 80
3. 1st figure (blue) 4 * 4 + 3 * 5 = 16 + 15 = 31; 4 * 7 + 3 = 28 + 3 = 31;
2nd figure (yellow) 3 * 6 + 3 * 4 = 18 + 12 = 30; 3 * 9 + 3 = 27 + 3 = 30.
4. Find out how many lessons have passed by adding together the lessons of mathematics and reading for three weeks:
4 * 3 + 4 * 3 = 12 + 12 = 24 lessons.
5. Compare.
(10 + 3) * 2 80; (20 + 30) * 2 = 100.
6. Rectangle ABCD should be divided by a segment from the middle of the upper side (6 cells from corner B) vertically to the center of the lower side, or from the center on the left side (3 cells from corner B) horizontally to the center of the right side. The KLMNOP polygon must be divided into segments between the corners: MP, NK or LO.
7. If you enter the number 3 in the boxes, the entries will be correct:
21 / 3 > 5; 3 * 8 16; 47 – 6 * 3 = 29.
Mathematics grade 3, part 1, Dorofeev, page 81
8. Find out how much it will cost 3 meters of silk braid, 7 rubles each. per meter: 3 * 7 \u003d 21 rubles, now how much does 5 meters of simple braid cost, 4 rubles each. 5 * 4 \u003d 20. Difference: 21 - 20 \u003d 1 rub.
9. From the conditions of the problem, the boy folded a sheet of newspaper 4 times. 1st time it turned out 2 layers, 2nd time 2 more: 2 + 2 = 4, 3rd time add 4: 4 + 4 = 8, 4th time add 8 more layers: 8 + 8 = 16 holes .
1. Decrease the numbers by 30, reduce the result by 3 times.
(48 – 30) / 3 = 18 / 3 = 6; (57 – 30) / 3 = 27 / 3 = 9; (60 – 30) / 3 = 30 / 3 = 10; (54 – 30) / 3 = 24 / 3 = 8.
2. Find out the mass of the lamb: 24/3 = 8 kg. Pig and lamb weight: 24 + 8 = 32 kg.
3. Calculate the values of the expressions.
21 / 3 = 7; 18 / 9 = 2; 27 / 3 = 9; 80 / 4 = 20;
16 + 3 * 8 = 16 + 24 = 40; 72 – 5 * 4 = 72 – 20 = 52; 60 – 3 * 7 = 60 – 21 = 39; 25 + 8 * 2 = 25 + 16 = 41;
(52 – 34) / 3 = 18 / 3 = 6; (8 + 20) / 4 = 28 / 4 = 7; (19 + 21) * 2 = 40 * 2 = 80; (10 + 8) * 3 = 18 * 3 = 54;
82 – (39 + 12) = 82 – 51 = 31; 64 – (50 – 27) = 64 – 23 = 41; 76 – (100 – 87) = 76 – 13 = 63; 18 / (45 – 39) = 18 / 6 = 3.
4. How many meters of fabric are left in the workshop after the jackets have been sewn? 52 - 3 * 9 \u003d 52 - 27 \u003d 25 m.
5. Compile tasks according to the table.
1) The price of one chocolate bar is 20 rubles. How much will 3 pieces cost? Answer: 20 * 3 = 60 rubles.
2) 3 chocolates were bought for 60 rubles. How much does one cost? Answer: 60 / 3 = 20 rubles.
3) The price of one chocolate bar is 20 rubles. How many chocolates can you buy for 60 rubles? 60 / 20 = 3 pcs.
Mathematics grade 3, part 1, Dorofeev, page 82
6. Equal figures: No. 1 = No. 6 (12 cells), No. 2 (3 * 4 = 12 cells), No. 3 (5 * 2 + 2 = 12 cells), No. 4 (5 + 6 = 11 cells), No. 5 (6 + 4 * 2 = 14 cells), No. 7 (2 * 8 + 6 = 16 + 6 = 24 cells)
7. Add the cubes in all columns: 1 + 2 * 2 + 3 * 3 + 4 * 2 + 5 = 1 + 4 + 9 + 8 + 5 = 27 cubes.
8. Find out how much is in one package: 6 / 3 = 2 kg. In 5 bags: 2 * 5 = 10 kg. In 8 bags: 2 * 8 = 16 kg.
9. Fill in: 5 * 4 = 20; 24 / 3 = 8; 9 * 3 = 27; 60 / 6 = 10.
10. 1st pack \u003d 2nd pack + 18 notebooks, in order for the 1st to become more by 10, you need to remove the difference 18 - 10 = 8 notebooks. Divide it in half into 2 packs: 8 / 2 = 4 pcs. Answer: 4 notebooks must be shifted.
Mathematics grade 3, part 1, Dorofeev, page 83
Multiplying the number 4. Dividing by 4.
1. Count: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40.
2. 3 times: 4 + 4 + 4 = 4 * 3 = 12; 4 times: 4 + 4 + 4 + 4 = 4 * 4 = 16; 5 times: 4 + 4 + 4 + 4 + 4 = 4 * 5 = 20.
3. Do the calculations.
4 * 5 + 4 = 20 + 4 = 24, 4 * 6 = 24; 4 * 6 + 4 = 24 + 4 = 28, 4 * 7 = 28;
4 * 7 + 4 = 28 + 4 = 32, 4 * 8 = 32; 4 * 8 + 4 = 32 + 4 = 36, 4 * 9 = 36.
The sum of the terms of one number is equal to the product of this number by the number of terms.
4. Make a table of multiplication and division by 4 in your notebook.
Multiplication: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40. Division: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
5. Find out how much 1 candy costs: 28/4 = 7 rubles. 2 candies: 7 * 2 = 14 rubles. 5 sweets: 7 * 5 = 35 rubles.
Mathematics grade 3, part 1, Dorofeev, page 84
6. Cookies cost: 20 * 3 = 60 rubles. Buns cost: 2 * 8 = 16 rubles. Whole purchase: 60 + 16 = 76 rubles.
7. Take action.
(10 + 2) * 5 = 10 * 5 + 2 * 5 = 50 + 10 = 60; (3 + 20) * 4 = 3 * 4 + 20 * 4 = 12 + 80 = 92;
(7 + 10) * 2 = 7 * 2 + 10 * 2 = 14 + 20 = 34; (4 + 10) * 3 = 4 * 3 + 10 * 3 = 12 + 30 = 42;
4 * (10 + 9) = 4 * 10 + 4 * 9 = 40 + 36 = 76; 2 * (6 + 30) = 2 * 6 + 2 * 30 = 12 + 60 = 72.
20 + 3 = 23; 40 + 6 = 46; 10 + 8 = 18; 30 + 9 = 39; 80 + 4 = 84; 50 + 2 = 52; 10 + 1 = 11.
9. Viti should get a rectangle with a width of 2 cm, a length of 12 cm.
10. Answer: how many feeders did Vanya hang?
When the tits sat in groups of 2, 1 feeder was not enough, if they sat in groups of 4, then 1 feeder was extra.
Using the selection method, we find out the number of feeders, if the birds would sit 4 on 2 feeders: 4 * 2 = 8, one feeder was left extra. If the birds sat in 2: 3 * 2 = 6, one feeder was not enough.
Answer: Vanya hung 3 feeders, 8 birds arrived.
Mathematics grade 3, part 1, Dorofeev, page 85
1. Decrease each of the numbers by 50, reduce the results by 4 times: (62 - 50) / 4 = 12 / 4 = 3;
(74 – 50) / 4 = 24 / 4 = 6; (82 – 50) / 4 = 32 / 4 = 8; (58 – 50) / 4 = 8 / 4 = 2; (90 – 50) / 4 = 40 / 4 = 10.
2. Multiply the quantity by the price of one bun. Sasha needs to take: 4 * 8 = 32 rubles.
3. Find out how many kilograms are in one box: 27/3 = 9 kg. In 2x: 9 * 2 = 18 kg., In 4x: 9 * 4 = 36 kg.
4. Calculate the values of the expressions.
32 / 4 = 8; 28 / 4 = 7; 24 / 4 = 6; 40 / 4 = 10;
51 + 4 * 9 = 51 + 36 = 87; 53 – 3 * 7 = 53 – 21 = 32; 20 + 4 * 6 = 20 + 24 = 44; 87 – 9 * 2 = 87 – 18 = 69;
(10 + 4) * 4 = 40 + 16 = 56; (3 + 20) * 2 = 6 + 40 = 46; (30 + 30) / 3 = 60 / 3 = 20; (16 + 4) / 4 = 20 / 4 = 5.
5. Make three tasks according to the table.
1) The price of one notebook is 9 rubles. Bought 4 notebooks. How much did all the notebooks cost? 9 * 4 = 36 rubles.
2) We bought 4 notebooks for 36 rubles. How much did one notebook cost? 36 / 4 \u003d 9 rubles.
3) The price of one notebook is 9 rubles. How many notebooks can be bought for 36 rubles? 36 / 9 = 4 pcs.
6. Find out how many notebooks are in the second pile: 28 / 4 = 7 pcs. Find out how many notebooks are in the third pile: (28 / 4) * 3 = 7 * 3 = 21 pcs. Find out how many notebooks are in the first and second piles together: 28 + 28 / 4 = 28 + 7 = 35 pcs. Find out how many more notebooks are in the first pile than in the second: 28 - 28 / 4 = 28 - 7 = 21 pcs. Find out how many more notebooks are in the first pile than in the third: 28 - (28 / 4) * 3 = 28 - 21 = 7 pcs. Find out how many notebooks there are: 28 + 28 / 4 + (28 / 4) * 3 = 28 + 7 + 21 = 56 pcs.
7. Calculate in a convenient way.
(5 + 7) * 2 = 12 * 2 = 24; (5 + 5) * 3 = 10 * 3 = 30;
(14 + 6) * 2 = 20 * 2 = 40; (3 + 17) * 5 = 20 * 5 = 100;
(8 + 12) * 3 = 20 * 3 = 60; (4 + 26) * 2 = 30 * 2 = 60.
Mathematics grade 3, part 1, Dorofeev, page 86
8. Replace each number with the sum of the bit terms.
50 + 6 = 56; 60 + 5 = 65; 30 + 3 = 33; 90 + 8 = 98; 70 + 1 = 71; 10 + 7 = 17.
9. All the fruits can be in the ladle together or separately. 123, 12, 23, 13, 1, 2, 3.
Multiplication check.
1. Do the multiplication and check in 2 ways. 2 * 8 = 16, check 16 / 2 = 8, 16 / 8 = 2;
5 * 4 = 20, check 20 / 5 = 4, 20 / 4 = 5; 3 * 7 = 21, check 21 / 3 = 7, 21 / 7 = 3;
4 * 6 = 24, check 24/4 = 6, 24/6 = 4; 4 * 9 = 36, check 36 / 4 = 9, 36 / 9 = 4.
2. Solve the problem and check.
1) For 9 bedside tables you will need: 3 * 9 = 27 m of boards. Check: 27 / 3 = 9, 27 / 9 = 3.
2) In 8 jugs: 4 * 8 = 32 liters. milk. Check: 32 / 4 = 8, 32 / 8 = 4.
Mathematics grade 3, part 1, Dorofeev, page 87
3. Do the calculations.
20 * 2 = 40; 3 * 2 = 6; 23 * 2 = 46; 30 * 3 = 90; 3 * 3 = 9; 33 * 3 = 99; 40 * 2 = 80; 4 * 2 = 8; 44 * 2 = 88;
20 * 4 = 80; 1 * 4 = 4, 21 * 4 = 84; 20 * 3 = 60, 3 * 3 = 9, 23 * 3 = 69.
4. Find out how many cucumbers are in 1 jar: 12/4 = 3 kg. To decompose 27 kg. needed: 27 / 3 = 9 cans.
5. Expression: 2 * 10 = 20 kg. wheat flour was brought to the canteen. 2 * 6 = 12 kg. rye flour was brought to the dining room. 2 * 10 + 2 * 6 = 20 + 12 = 32 kg. all the flour was brought to the dining room. 10 + 6 = 16 bags of flour were brought in total to the canteen. 2 * (10 + 6) = 20 + 12 = 32 kg. everything was brought to the dining room.
6. Compare.
5 dm. 3cm 30 cm; 5 dm. 3 cm. 2 m. 4 dm. 2 m
7. There will be 18 cells in the “P” figure.
8. From dishes 3 l. and 5l. you need to pour 2 liters. To do this, pour a vessel of 5 liters, fill it with 3 liters, and the rest 5 - 3 \u003d 2 liters. pour into a saucepan. To pour 4 liters, repeat the filling of 2 liters twice. (5 - 3) + (5 - 3) \u003d 2 + 2 \u003d 4 l. To pour 1 liter, pour a vessel of 3 liters. pour into 5 liters, pour 3 liters again. from it we fill 5 l. to the end, and in 3 l. remains 1 liter. (3 + 3) - 5 = 6 - 5 = 1 liter.
Mathematics grade 3, part 1, Dorofeev, page 88
Multiply a two-digit number by a one-digit number.
1. Replace each number with the sum of the bit terms:
13 = 10 + 3; 56 = 50 + 6; 28 = 20 + 8; 67 = 60 + 7; 92 = 90 + 2; 55 = 50 + 5; 36 = 30 +6.
2. Calculate the values of the expressions in the first line and write the results in the second line:
(30 + 5) * 2 = 30 * 2 + 5 * 2 = 60 + 10 = 70; (6 + 10) * 4 = 6 * 4 + 10 * 4 = 24 + 40 = 64;
(20 + 7) * 3 = 20 * 3 + 7 * 3 = 60 + 21 = 81. This can be done using the technique of multiplying a two-digit number by a one-digit number.
3. Compare: 93 min. = 1 h, 33 min. > 1 hour; 93 cm. 1 dm. = 10 cm.
Mathematics grade 3, part 1, Dorofeev, page 89
4. Solve problems and check.
1) In 7 boxes there will be: 4 * 7 = 28 balls, check: 28 / 7 = 4, 28 / 4 = 7;
2) In 8 days, the carpenter makes: 3 * 8 = 24 frames, check: 24 / 8 = 3, 28 / 3 = 8.
17 * 2 = (10 + 7) * 2 = 10 * 2 + 7 * 2 = 20 + 14 = 34;
24 * 4 = (20 + 4) * 4 = 20 * 4 + 4 * 4 = 80 + 16 = 96;
4 * 16 = 16 * 4 = (10 + 6) * 4 = 10 * 4 + 6 * 4 = 40 + 24 = 64;
7 * 12 = 12 * 7 = (10 + 2) * 7 = 10 * 7 + 2 * 7 = 70 + 14 = 84;
25 * 3 – 40 = (20 + 5) * 3 – 40 = 20 * 3 + 5 * 3 – 40 = 60 + 15 – 40 = 75 – 40 = 35;
11 * 8 + 2 = (10 + 1) * 8 + 2 = 10 * 8 + 1 * 8 + 2 = 80 + 8 + 2 = 88 + 2 = 90;
32 * 2 + 9 = (30 + 2) * 2 + 9 = 30 * 2 + 2 * 2 + 9 = 60 + 4 + 9 = 64 + 9 = 73;
6 * 14 – 70 = 14 * 6 – 70 = (10 + 4) * 6 – 70 = 10 * 6 + 4 * 6 – 70 = 60 + 24 – 70 = 84 – 70 = 14.
6. Explain what the expressions mean:
3 * 6 = 18 kg. papa bought all the potatoes; 2 * 4 = 8 kg. papa bought all the cabbages; 3 * 6 + 2 * 4 = 18 + 8 = 26 kg. papa bought vegetables; 3 * 6 - 2 * 4 \u003d 18 - 8 \u003d 10 kg. Papa bought so many more potatoes than cabbages.
7. Compile two problems according to the table and solve them:
1) What is the weight of one box of juice if 3 boxes weigh 6 kg? Answer: 6 / 3 = 2 kg.
2) 5 boxes of ice cream with a total weight of 10 kg were brought to the warehouse. How much does one box weigh? Answer: 10 / 5 = 2 kg. + Task. In the dining room there are 4 boxes with vegetables with a total weight of 8 kg. How much does one box weigh? Answer: 8 / 4 = 2 kg.
8. How many cubes are used to build the figure? 3 * 5 + 6 + 3 + 3 = 15 + 12 = 27 dice.
Mathematics grade 3, part 1, Dorofeev, page 90
9. There were 5 people in the first row, in each next row there were 2 more people. There were 6 rows in total. We find out how many were in the 6th row, each following the 1st row is 2 more, so the 6th row: 5 * 2 + 5 = 10 + 5 = 15 people. How many athletes participated in total?
1 row = 5;
2nd row = 5 + 2 = 7;
3rd row = 5 + 2 * 2 = 5 + 4 = 9;
4th row = 5 + 2 * 3 = 5 + 6 = 11;
5 row = 5 + 2 * 4 = 5 + 8 = 13;
6 row = 5 + 2 * 5 = 5 + 10 = 15.
Answer: 5 + 7 + 9 + 11 + 13 + 15 = 12 + 20 + 28 = 40 + 20 = 60 people.
1. Calculate.
10 + 7 = 17; 3 + 40 = 43; 8 + 50 = 58; 70 + 2 = 72; 1 + 60 = 61.
2. Replace each of the numbers with the sum of the bit terms:
16 = 10 + 6; 18 = 10 + 8; 23 = 20 + 3; 47 = 40 + 7; 29 = 20 + 9; 51 = 50 + 1; 96 = 90 + 6.
3. Solve the pattern according to which the products of each column are composed.
10 * 2 = 20; 3 * 2 = 6; 13 * 2 = 26;
20 * 2 = 40; 5 * 2 = 10; 25 * 2 = 50;
10 * 3 = 30; 3 * 3 = 9; 13 * 3 = 39;
20 * 3 = 60; 5 * 3 = 15; 25 * 3 = 75;
10 * 4 = 40; 3 * 4 = 12; 13 * 4 = 52;
20 * 4 = 80; 5 * 4 = 20; 25 * 4 = 100.
4. Fill in the gaps in the table by doing the calculations.
11 * 2 = 22; 12 * 4 = (10 + 2) * 4 = 10 * 4 + 2 * 4 = 40 + 8 = 48; 13 * 3 = (10 + 3) * 3 = 10 * 3 + 3 * 3 = 30 + 9 = 39; 14 * 2 = 10 * 2 + 4 * 2 = 20 + 8 = 28; 15 * 3 = 45; 16 * 4 = 10 * 4 + 6 * 4 = 40 + 24 = 64.
5. There were 23 buttons in the first box. Then the second one had: 23 * 2 = 46 buttons. Let's find out how many were in the third: 46 - 16 \u003d 30 buttons. There were: 23 + 46 + 30 = 53 + 46 = 99 buttons.
Mathematics grade 3, part 1, Dorofeev, page 91
6. Colored pencils: 5*6=30pcs; 3*12=36pcs Total: 30 + 36 = 66 pencils.
7. Write the expressions shorter, using the rule of multiplying a sum by a number.
7 * 4 + 9 * 4 = (7 + 9) * 4 = 16 * 4 = 64; 2 * 3 + 5 * 3 = (2 + 5) * 3 = 7 * 3 = 21;
4 * 2 + 8 * 2 = (4 + 8) * 2 = 12 * 2 = 24; 6 * 4 + 4 * 4 = (6 + 4) * 4 = 10 * 4 = 40;
5 * 3 + 4 * 3 = (5 + 4) * 3 = 9 * 3 = 27; 2 * 4 + 5 * 4 = (2 + 5) * 4 = 7 * 4 = 28.
8. To get a rectangle, you can connect segments AB, ER. Also FP, KL. 2 pcs.
ABRE perimeter: 3 * 2 + 6 * 2 = (3 + 6) * 2 = 9 * 2 = 18 cm; FKLP: 3 * 2 + 2 * 2 = 5 * 2 = 10 cm.
9. Ten matches were laid out in 4 boxes and the number of matches was written on each. Can the product of these numbers be odd (i.e., not divisible by two) Using the selection method, we find out:
Sum: 3 + 3 + 3 + 1 = 10 Product: 3 * 3 * 3 * 1 = 9 * 3 * 1 = 27 * 1 = 27 (odd)
Mathematics grade 3, part 1, Dorofeev, page 92
Tasks for bringing to unity
1. 18 cakes were divided equally into 6 plates. How many of them are on 4 plates?
The first step is to find out how many cakes are on one plate: 18 / 6 = 2 pcs.
Now let's calculate how many of them are on 4 plates: 2 * 4 = 8 pcs.
Mathematics grade 3, part 1, Dorofeev, page 93
2. In five identical jars there were 10 kg of jam, equally in all. How many kilograms of jam are in 3 such jars? The solution to the problem by the expression looks like this: (10 / 5) * 3 = 2 * 3 = 6 kg.
3. Compose and solve the problem according to the drawing.
Five balloons cost 15 rubles. How much will two of these balls cost? The solution to the problem by the expression looks like this: (15 / 5) * 2 = 3 * 2 = 6 rubles.
4. In six bottles of 12 liters. milk, all equally. Used up 4 bottles of milk. Let's find out how much we spent: (12 / 6) * 4 = 2 * 4 = 8 liters.
5. Calculate the values of the expressions.
4 * 7 = 28; 3 * 9 = 27; 4 * 8 = 32; 3 * 7 = 21;
28 / 4 = 7; 27 / 3 = 9; 32 / 4 = 8; 21 / 3 = 7;
90 / 3 = 30; 40 / 2 = 20; 60 / 1 = 60; 100 / 5 = 20;
4 * 6 / 3 = 24 / 3 = 8; 3 * 8 / 4 = 24 / 4 = 6; 4 * 4 / 8 = 16 / 8 = 2; 4 * 3 / 6 = 12 / 6 = 2.
6. Draw a segment AB, the length of which is 1 dm. 5 cm. \u003d 15 cm. Divide it with dots into 5 equal parts. One piece length: 15 / 5 = 3 cm Two pieces: 3 * 2 = 6 cm Three pieces: 3 * 3 = 9 cm
7. There were 5 liters in the first bucket. water, in the second - 3 times more than in the first, and in the third - by 6 liters. less than in the second. In the second: 5 * 3 \u003d 15 liters. In the third: 15 - 6 \u003d 9 liters. Total: 5 + 15 + 9 = 29 liters.
Mathematics grade 3, part 1, Dorofeev, page 94
8. How many triangles are shown in the drawing?
Write down: ABD, FBC, FCA, ABC, ACE, ECD, ACD - 7 triangles.
9. How many two-digit numbers are there in which all the digits are odd and do not repeat? Brute force, odd numbers 1, 3, 5, 7, 9 write: 13, 15, 17, 19, 31, 35, 37, 39, 51, 53, 57, 59, 71, 73, 75, 79, 91, 93, 95, 97. There are 20 two-digit numbers in total.
1. Fill in the gaps in the tables by doing the calculations.
Table 1) 3 * 6 = 18; 4 * 8 = 32; 7 * 3 = 21; 4 * 9 = 36; 10 * 3 = 30; 7 * 4 = 28;
Table 2) 16 / 4 = 4; 36 / 9 = 4; 24 / 8 = 3; 40 / 10 = 4; 80 / 8 = 10; 24 / 4 = 6.
2. There are 50 flags in 5 garlands, equally in all. How many flags are in 7 such garlands?
Let's find out how many flags are in one garland: 50 / 5 = 10 flags. Then in 7 * 10 = 70 pcs.
Mathematics grade 3, part 1, Dorofeev, page 95
3. Make a task for each schematic entry. Write down:
1) Three jars hold 9 liters of juice. How much juice will five of these jars hold?
The solution to the problem with an expression looks like this: (9 / 3) * 5 = 3 * 5 = 15 liters.
2) Three jars hold 9 liters of juice. How many cans will fit 15 liters of juice?
The solution to the problem with an expression looks like this: 15 / (9 / 3) \u003d 15 / 3 \u003d 5 cans.
3) How many liters of juice are contained in three jars if there are 15 liters of juice in five jars?
The solution to the problem by expression looks like this: (15 / 5) * 3 = 3 * 3 = 9 liters.
Problems are similar to data, but with different unknowns. Such tasks are called “reduction to unity”. Such tasks have a similar solution. A schematic notation can be proposed with an unknown number of cans for 9 liters of juice: 9 / (15 / 5) = 9 / 3 = 3 cans.
4. The price of a notebook is 27 rubles. Find out how much change from 100 rubles if you buy 3 notebooks. Solution: 27 * 3 - 100 = (20 + 7) * 3 - 100 = (20 * 3 + 7 * 3) - 100 = 60 + 21 - 100 = 81 - 100 = 19 rubles.
Tasks inverse to this one:
1) How much money was there if, after buying three notebooks for 27 rubles. each, 19 rubles left?
(3 * 27) + 19 = 81 + 19 = 100 rubles.
2) How much does one notebook cost if from 100 rubles, after buying three notebooks, 19 rubles remain?
(100 - 19) / 3 \u003d 81 / 3 \u003d 27 rubles.
5. Fill in the gaps with the following numbers:
4 * 6
90 / 3 > 20;
28 / 4 > 6;
4 * 0 = 0;
16 / 4 * 8 = 32;
4 / 4 * 39 6. There are 80 apartments in the first building. In the second: 80 / 4 = 20 apartments. In the third 80 + 20 = 100 apartments.
7. To fold the square, you need to cut the figure so that the three cells on the far right fit up, next to one individual cell. The second way is to cut off the leftmost four cells and place them up in the fourth row.
Mathematics grade 3, part 1, Dorofeev, page 96
1. Is it true that:
1) Product of numbers: 3 * 6 = 18, even number;
2) The sum of numbers 3 + 9 + 7 = 19, an odd number;
3) The number 6 / (10 - 7) \u003d 6 / 3 \u003d 2, is divisible;
4) Private 27 / 3 2. Write down each of the numbers:
(64 – 40) / 4 = 24 / 4 = 6; (56 – 40) / 4 = 16 / 4 = 4; (72 – 40) / 4 = 32 / 4 = 8; (80 – 40) / 4 = 40 / 4 = 10.
3. First, find out how many liters of water were poured into the bucket: 27/3 = 9 liters. Now we can find out how many liters of water were poured into the trough: 4 * 9 = 36 liters. Total: 27 + 9 + 36 = 36 + 36 = 72 liters.
4. Do the calculations.
32 / 4 = 8; 20 / 5 = 4; 18 / 9 = 2; 7 / 7 = 1;
29 – 3 * 7 = 29 – 21 = 8; 40 – 4 * 9 = 40 – 36 = 4; 26 – 3 * 8 = 26 – 24 = 2; 25 – 4 * 6 = 25 – 24 = 1;
(8 + 16) / 3 = 24 / 3 = 8; (7 + 9) / 4 = 16 / 4 = 4; (23 – 17) / 3 = 6 / 3 = 2; (30 – 26) / 4 = 4 / 4 = 1;
27/3 - 5/5 = 9 - 1 = 8; 20 * 2 / (70 / 7) = 40 / 10 = 4; 60/6 - (81 - 73) = 10 - 8 = 2; 9 / (33 - 6 * 4) \u003d 9 / (33 - 24) \u003d 9 / 9 \u003d 1. You can see that as a result of the calculations, a series is obtained: 8, 4, 2, 1.
5. The expression 4 * 8 = 32 meters of fabric was required to sew all the children's coats.
The expression 6 * 3 = 18 meters of fabric was required to sew all the adult coats.
The expression 4 * 8 + 6 * 3 = 32 + 18 = 50 meters of fabric was required for the entire tailoring.
The expression 4 * 8 - 6 * 3 \u003d 32 - 18 \u003d 14 meters, so much more fabric was required for sewing children's coats.
Mathematics grade 3, part 1, Dorofeev, page 97
6. Compare.
1 dm. 6 cm = 16 cm 1 dm. 6 cm = 16 cm > 10 cm
1 dm. 6 cm. = 16 cm. 3 m. 7 dm. = 37 dm. > 3 m. = 30 dm.
3 m. 7 dm. 37 dm. > 30 dm.
3 dm. 7 cm = 37 cm. 7. The boy had 50 rubles. He bought 6 stamps, 4 rubles each: 4 * 6 = 24 rubles.
1) The boy has left: 50 - 24 = 26 rubles;
2) If we subtract the cost of the purchased stamps from the remaining money, we will find out if he can buy the same number of stamps: 26 - 24 = 2 rubles will remain after buying 12 stamps.
8. Draw in a notebook a figure in the shape of the letter “O”, consisting of 16 cells.
9. The name of the ancient Greek scientist - mathematician: 17 - P; 10 - I, 22 - F, 1 - A, 4 - D, 16 - O, 18 - R. "PYTHAGORAS".
Mathematics grade 3, part 1, Dorofeev, page 98
Multiplying the number 5. Dividing by 5.
1. Count and write: “5, 10, 15, 20, 25, 30, 35, 40, 45, 50”.
2. If the number 5 is taken as a summand 3 times: 5 + 5 + 5 = 5 * 3 = 15, 4 times: 5 + 5 + 5 + 5 = 5 * 4 = 20.
3. Write down: 4 * 6 = 24; 3 * 8 = 24; 4 * 7 = 28; 4 * 9 = 36;
6 * 4 = 24; 8 * 3 = 24; 7 * 4 = 28; 9 * 4 = 36. By changing the multiplier, the product will not change.
4. Calculate according to the sample: 5 * 5 \u003d 5 * 4 + 5 \u003d 20 + 5 \u003d 25;
5 * 6 = 5 * 5 + 5 = 25 + 5 = 30; 5 * 7 = 5 * 6 + 5 = 30 + 5 = 35; 5 * 8 = 5 * 7 + 5 = 35 + 5 = 40; 5*9=5*8+5=40+5=45
5. Make a multiplication table for the number 5 in your notebook.
Multiplication: 5 * 1 = 5; 5 * 2 = 10; 5 * 3 = 15; 5 * 4 = 20; 5 * 5 = 25; 5 * 6 = 30; 5 * 7 = 35; 5 * 8 = 40;
5 * 9 = 45; 5 * 10 = 50;
Division: 5 / 5 = 1; 10 / 5 = 2; 15 / 5 = 3; 20 / 5 = 4; 25 / 5 = 5; 30 / 5 = 6; 35 / 5 = 7; 40 / 5 = 8; 45 / 5 = 9; 50 / 5 = 10.
Mathematics grade 3, part 1, Dorofeev, page 99
6. Find out how many guys will fit on one bench: 20/4 = 5 people. To plant 45 guys, this is 45 - 20 = 25, 25 people more. Let's put more: 25 / 5 = 5 benches.
7. Find out how many boxes can be loaded onto the second car: 12 + 8 = 20 pcs. Now we find out how many boxes were transported by each car, 1st: 12 * 3 = 10 * 3 + 2 * 3 = 30 + 6 = 36 pieces; 2nd car: 20*2=40pcs In total they transported: 36 + 40 = 76 boxes.
8. One set contains 3 baskets and 2 eclairs. Total 3 + 2 = 5 cakes in one set.
In 6 such sets there were: 5 * 6 = 30 cakes.
1. Name the numbers from 20 to 40 that are divisible by 4: 20, 24, 28, 32, 36, 40.
2. Name the numbers from 40 to 50 that are divisible by 5: 40, 45, 50.
3. Each of the numbers, increase by 5 times, then reduce the result by 19.
6 * 5 = 30, 30 – 19 = 11;
8 * 5 = 40, 40 – 19 = 21;
5 * 5 = 25, 25 – 19 = 6;
14 * 5 = (10 + 4) * 5 = 10 * 5 + 4 * 5 = 50 + 20 = 70; 70 – 19 = 51;
7 * 5 = 35, 35 – 19 = 16.
4. Find out how old Vasya is by dividing dad's age by 5: 30 / 5 = 6 years.
5. Calculate and write down:
2 * 7 = 14; 4 * 9 = 36; 3 * 8 = 24; 5 * 7 = 35;
14 + (10 + 4) = 14 + 14 = 28; 36 + (30 + 6) = 36 + 36 = 72; 24 + (20 + 4) = 24 + 24 = 48; 35 + (30 + 5) = 35 + 35 = 70;
28 – (20 + 8) = 28 – 28 = 0; 72 – (70 + 2) = 72 – 72 = 0; 48 – (40 + 8) = 48 – 48 = 0; 70 – (60 + 10) = 70 – 70 = 0;
6. Draw a square ABCD and calculate the perimeter: 4 * 4 \u003d 16 cm, this is equal to 1 dm. and 6 cm.
Mathematics grade 3, part 1, Dorofeev, page 100
7. Find out how many kilograms of fish give the bear in one day: 24 / 4 = 6 kg.
To find out how many days 60 kg will last, we divide everything into fish in one day: 60 / 6 \u003d 10 days.
8. First, find out how many buttons were sewn onto one raincoat: 24/3 = 8 buttons. Now we will find out how many buttons are needed for 5 raincoats: 8 * 5 = 40 buttons.
9. First, find out how much gasoline was in total: 15 + 20 = 35 liters. Now let's divide all the gasoline by the amount of gasoline in one can: 35/5 = 7 cans.
10. On flipped cards: 99 + 1 = 100.
1. Fill in the gaps in the tables:
Table 1) 5 * 5 = 25; 5 * 6 = 30; 5 * 7 = 35; 5 * 8 = 40; 5 * 9 = 45; 5 * 10 = 50.
Table 2) 40 / 4 = 10; 36 / 4 = 9; 32 / 4 = 8; 28 / 4 = 7; 24 / 4 = 6; 20 / 4 = 5.
1) The product has increased by 5 because the second factor has increased by 1.
2) The dividend has decreased by 4 because the quotient has decreased by 1.
Mathematics grade 3, part 1, Dorofeev, page 101
2. Find out how many gingerbreads are in one box: 50 / 5 = 10 gingerbread.
a) For 60 gingerbread you will need: 60/10 = 6 boxes;
b) For 40 gingerbread you will need: 40 / 10 = 4 boxes.
3. Make up a problem for each schematic entry and solve:
1) 2 pens cost 14 rubles, how much do 5 pens cost. Solution: 14 / 2 = 7 rubles. one pen costs. 5 * 7 = 35 rubles cost 5 pens;
2) 2 pens cost 14 rubles, how many pens can you buy for 35 rubles? Solution: 14 / 2 = 7 rubles. one pen costs. 35 / 7 = 5 pens can be bought.
3) How many pens can you buy for 14 rubles if 5 pens cost 35 rubles? Solution: 35 / 5 = 7 rubles. one pen costs. 14 / 7 = 2 pens can be bought.
Tasks are similar in condition, one price and number of pens. Such tasks are called bringing to one, first we find out how much one unit is. A schematic entry can be offered with an unknown price for 2 pens, (35 / 5) * 2 = 7 * 2 = 14 rubles.
4. Calculate the values of expressions:
4 * 7 = 28; 3 * 9 = 27; 5 * 8 = 40;
(10 + 7) * 5 = 10 * 5 + 7 * 5 = 50 + 35 = 85; (10 + 2) * 4 = 10 * 4 + 2 * 4 = 40 + 8 = 48; (20 + 6) * 3 = 20 * 3 + 6 * 3 = 60 + 18 = 78;
15 * 3 = (10 + 5) * 3 = 10 * 3 + 5 * 3 = 30 + 15 = 45; 14 * 2 = (10 + 4) * 2 = 10 * 2 + 4 * 2 = 20 + 8 = 28; 23 * 4 = (20 + 3) * 4 = 20 * 4 + 3 * 4 = 80 + 12 = 92;
(52 – 20) / 4 = 32 / 4 = 8; (70 – 40) / 5 = 30 / 5 = 6; (60 – 36) / 3 = 24 / 3 = 8.
5. The expression 4 * 8 = 32 meters of fabric was spent on all duvet covers.
Expression 52 - 4 * 8 \u003d 52 - 32 \u003d 20 meters of fabric left after sewing duvet covers.
The expression (52 - 4 * 8) / 10 = (52 - 32) / 10 = 20 / 10 = 2 meters of fabric was spent on one sheet.
6. Find out how much 3 toothbrushes cost: 18 * 3 = (10 + 8) * 3 = 10 * 3 + 8 * 3 = 30 + 24 = 54 rubles. Now let's find out how much we spent on all the toothpaste: 94 - 54 = 40 rubles. If 2 pastes cost 40 rubles, then one toothpaste: 40 / 2 = 20 rubles. Answer: 20 rubles.
Mathematics grade 3, part 1, Dorofeev, page 102
7. Measure the length of the broken line from the textbook with a ruler. Divide this length by 5, draw a segment of the resulting length.
8. Since all the boys sit between the girls, and the girls between the boys, their number at the table is equal. From the conditions of the problem, the total number of boys and girls can be from 4 or more by 2 (boy and girl). Let's get the number of children at the table: 4, 4 + 2 = 6, 6 + 2 = 8, 8 + 2 = 10 and so on. Since the number always increases by 2, the total number of boys and girls at the table is even.
Multiplying the number 6. Dividing by 6.
1. Count in sixes to 60, write down: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
2. If the number 6 is taken as a summand 3 times: 6 + 6 + 6 = 6 * 3 = 18, 4 times: 6 + 6 + 6 + 6 = 6 * 4 = 24.
3. Calculate the values of expressions:
4 * 6 = 24; 6 * 4 = 24; 3 * 9 = 27; 9 * 3 = 27; 5 * 6 = 35; 6 * 5 = 35; 5 * 7 = 35; 7 * 5 = 35.
By changing the multipliers, the product does not change.
4. Perform calculations according to the sample.
6 * 6 = 6 * 5 + 6 = 30 + 6 = 36;
6 * 7 = 6 * 6 + 6 = 36 + 6 = 42;
6 * 8 = 6 * 7 + 6 = 42 + 6 = 48;
6 * 9 = 6 * 8 + 6 = 48 + 6 = 54.
The product could be calculated by addition, the first method is more convenient.
Mathematics grade 3, part 1, Dorofeev, page 103
5. Make a table in your notebook for multiplying the number 6 and dividing by 6.
Multiplication: 6 * 1 = 6; 6 * 2 = 12; 6 * 3 = 18; 6 * 4 = 24; 6 * 5 = 30; 6 * 6 = 36; 6 * 7 = 42; 6 * 8 = 48; 6 * 9 = 54; 6 * 10 = 60.
Division: 6 / 6 = 1; 12 / 6 = 2; 18 / 6 = 3; 24 / 6 = 4; 30 / 6 = 5; 36 / 6 = 6; 42 / 6 = 7; 48 / 6 = 8; 54 / 6 = 9; 60 / 6 = 10.
6. The expression 28 / 4 \u003d 7 toys, means how many toys the boys put in each box.
The expression 12 / 4 = 3 toys means how many toys the girls put in each box. The expression 28 + 12 = 40 toys were laid out together by boys and girls.
The expression (28 + 12) / 4 = 40 / 4 = 10 toys were placed in each of the 4 boxes.
7. Green rectangle: 5 + 5 + 7 + 7 = 10 + 14 = 24 meters; 5 * 2 + 7 * 2 = 10 + 14 = 24 m; (5 + 7) * 2 \u003d 12 * 2 \u003d 24 m., the perimeter of the rectangle.
Blue square: 6 + 6 + 6 + 6 = 24 meters; 6 * 4 \u003d 24 m., the perimeter of the square.
Pink rectangle: 8 + 8 + 10 + 10 = 16 + 20 = 36 meters; 8 * 2 + 10 * 2 = 16 + 20 = 36 meters, (8 + 10) * 2 = 8 * 2 + 10 * 2 = 16 + 20 = 36 meters, the perimeter of the rectangle.
Mathematics grade 3, part 1, Dorofeev, page 104
8. Let's solve the problem by the selection method. From the conditions of the problem, the first head ate the most candies, let's take this amount by 3 more than half (the third one ate 3 less than the second) of all candies: 48 / 2 + 3 = 24 + 3 = 27 kg. Then the third head ate: 27/3 = 9 kg. And the second: 9 + 3 = 12 kg.
Let's check: 27 + 12 + 9 = 27 + 21 = 48 kg. candy was given for a birthday.
1. Fill in the gaps in the tables by doing the calculations:
Table 1) 6 * 10 = 60; 6 * 9 = 54; 6 * 7 = 42; 6 * 6 = 36; 6 * 5 = 30; 6 * 4 = 24.
Table 2) 20 / 5 = 4; 25 / 5 = 5; 30 / 5 = 6; 35 / 5 = 7; 40 / 5 = 8; 45 / 5 = 9.
1) The product has decreased by 6 because the second factor has decreased.
2) The quotient has increased by 1 because the dividend has increased by 5.
2. The expression 36 / 6 \u003d 6 pieces, means the number of packages in which Santa Claus packaged 6 mint gingerbread cookies.
The expression 24 / 6 \u003d 4 pieces, means the number of packages in which Santa Claus packaged 6 chocolate gingerbread cookies.
The expression 36 + 24 \u003d 60 pieces, means the total number of gingerbread.
The expression 36 - 24 \u003d 12 pieces, means how much more mint gingerbread was than chocolate.
The expression 36 / 6 + 24 / 6 \u003d 6 + 4 \u003d 10 pieces, means the total number of packages with gingerbread.
The expression 36 / 6 - 24 / 6 \u003d 6 - 4 \u003d 2 pieces, means how many more packages there were with mint gingerbread.
The expression (36 + 24) / 6 = 60 / 6 = 10 pieces, means the total number of packages with gingerbread.
3. Calculate the values of the expressions.
5 * 6 = 30; 4 * 9 = 36; 6 * 7 = 42;
(10 + 4) * 6 = 10 * 6 + 4 * 6 = 60 + 24 = 84; (10 + 2) * 5 = 10 * 5 + 2 * 5 = 50 + 10 = 60; (10 + 3) * 6 = 10 * 6 + 3 * 6 = 60 + 18 = 78;
17 * 4 = (10 + 7) * 4 = 10 * 4 + 7 * 4 = 40 + 28 = 68; 11 * 6 = 66; 21 * 3 = 63;
(68 – 41) / 3 = 27 / 3 = 9; (23 + 17) / 5 = 40 / 5 = 8; (40 + 14) / 6 = 54 / 6 = 9.
Mathematics grade 3, part 1, Dorofeev, page 105
4. Question, a) How many mushrooms did Kostya collect? Solution: first we find out how many mushrooms Yura found: 20 / 4 = 5 mushrooms. Now let's find out how much Kostya collected: 20 + 5 = 25 mushrooms.
Question b) How many mushrooms did the boys pick together? Solution: first we find out how many mushrooms Yura found: 20 / 4 = 5 mushrooms. Let's find out how much Kostya collected: 20 + 5 = 25 mushrooms. Then together they collected: 20 + 5 + 25 = 50 mushrooms.
5. Make up a problem for each schematic entry and solve:
1) It takes 25 meters of fabric to sew 5 coats. How much fabric is needed for 8 coats? Solution: find out how much is required for one coat: 25 / 5 = 5 meters of fabric. Now let's find out how much it takes for 8 coats: 5 * 8 = 40 meters.
2) It takes 25 meters of fabric to sew 5 coats. How many coats can be made from 40 meters of fabric? Solution: find out how much is required for one coat: 25 / 5 = 5 meters of fabric. Now let's find out how much can be sewn from 40 meters: 40 / 5 = 8 coats.
3) It takes 40 meters of fabric to sew 8 coats. How much fabric is needed for 5 coats? Solution: find out how much is required for one coat: 40 / 8 = 5 meters of fabric. Now let's find out how much it takes for 5 coats: 5 * 5 = 25 meters.
Problems are similar to data, but with different unknowns. Such tasks are called “reduction to unity”. You can suggest an entry with an unknown number of coats that are required for sewing from 25 meters of fabric: 25 / (40 / 8) = 25 / 5 = 5 coats.
6. Find out the length of the first broken line: 5 * 6 = 30 cm. We also find out the length of the second: 6 * 8 = 48 cm.
Now let's find out how much the second is larger than the first: 48 - 30 \u003d 18 cm.
7. Let's calculate how much Kolya spent on notebooks: 4 * 9 = 36 rubles.
1) Kolya has left: 50 - 36 \u003d 14 rubles;
2) With the remaining money, Kolya will be able to buy: 14 / 7 = 2 portions of ice cream.
8. 100 is the sum of all numbers from the subtraction example. Take the minuend equal to half of this sum 50. Then the subtrahend and the difference in the sum are equal to 50, divide them in half 50 / 2 = 25. We get an example: 50 - 25 = 25, check: 50 + 25 + 25 = 100.
Mathematics grade 3, part 1, Dorofeev, page 106
1. Name all the figures in the drawing: 1. Cube; 2. Tetrahedron; 3. Triangle; 4. Square; 5. Rectangle; 6. Pentagon.
2. Fill in the gaps in the tables by doing the calculations:
Table 1) 7 * 6 = 42; 6 * 5 = 30; 6 * 6 = 36; 4 * 6 = 24; 9 * 3 = 27; 4 * 7 = 28.
Table 2) 48 / 8 = 6; 25 / 5 = 5; 35 / 7 = 5; 54 / 6 = 9; 42 / 7 = 6; 50 / 10 = 5.
3. Express.
a) in minutes: 1 hour 7 minutes = 67 min.; 1 hour 28 minutes = 88 min.; 1 hour 10 minutes = 70 min.;
b) in hours and minutes: 70 min. = 1 hour 10 minutes; 99 min. = 1 hour 39 minutes; 62 min. = 1 hour 02 minutes;
c) in decimeters and centimeters: 65 cm = 6 dm. 5 cm; 86 cm = 8 dm. 6 cm; 94 cm. = 9 dm. 4 cm; 77 cm. = 7 dm. 7 cm;
d) in meters and decimeters: 21 dm. \u003d 2 m. 1 dm.; 36 dm. \u003d 3 m. 6 dm.; 55 dm. \u003d 5 m. 5 dm.; 89 dm. = 8 m. 9 dm.
Mathematics grade 3, part 1, Dorofeev, page 107
4. Calculate the values of the expressions. Underline the results that are even numbers:
45 / 5 = 9; 35 / 5 = 7; 27 / 3 = 9; 48 / 6 = 8;
30 – 4 * 6 = 30 – 24 = 6; 60 – 5 * 9 = 60 – 45 = 15; 80 – 6 * 10 = 80 – 60 = 20; 50 – 4 * 9 = 50 – 36 = 14;
3 * 6 / 2 = 18 / 2 = 9; 4 * 6 / 3 = 24 / 3 = 8; 5 * 4 / 2 = 20 / 2 = 10; 6 * 5 / 3 = 30 / 3 = 10;
40 / 5 / 8 = 8 / 8 = 1; 32 / 4 * 2 = 8 * 2 = 16; 25 / 5 * 4 = 5 * 4 = 20; 36 / 6 / 3 = 6 / 3 = 2.
5. Find out how much flour was poured: 2 * 9 = 18 kg. Then:
1) 40 - 18 = 22 kg. flour left in the bag;
2) 22 / 2 = 11 bags will be needed to pour the remaining flour.
6. Which of the statements are true?
1) True, 28 / 4 = 7;
2) Not true, 6 * 8 = 48;
3) True, 2 * 4 = 8, 24 / 8 = 3;
4) True, 18/6 = 3, 27/3 = 9;
5) Not true, An even number can be divisible by 2.
7. One division of the diagram is 15 flowers.
1) Lilies: 3 * 15 = 3 * (10 + 5) = 3 * 10 + 3 * 5 = 30 + 15 = 45. Chrysanthemums: 8 * 15 = 8 * (10 + 5) = 8 * 10 + 8 * 5 = 80 + 40 = 120 Carnations: 10 * 15 = 150 Roses: 6 * 15 = 6 * (10 + 5) = 6 * 10 + 6 * 5 = 60 + 30 = 90 Total: 90 + 150 + 120 + 45 = 240 + 165 = 405 flowers of each type were sold;
2) 150 - 90 = 60, 60 rose flowers sold less than carnations;
3) How many flowers of roses and lilies were sold together? 90 + 45 = 135 colors;
4) How many fewer lilies were sold than roses? 90 - 45 = 45 colors.
Mathematics grade 3, part 1, Dorofeev, page 108
8. Make a task for each schematic entry. Decide:
1) If 6 meters of fabric costs 48 rubles, how much will 4 meters of fabric cost?
Solution: 48 / 6 = 8 rubles. costs one metre. 4 * 8 = 32 rubles. costs 4 meters of fabric;
2) How many meters of fabric can be bought for 48 rubles, if 4 meters can be bought for 32 rubles?
Solution: 32 / 4 = 8 rubles. costs one metre. 48 / 8 \u003d 6 m. of fabric can be bought for 48 rubles;
3) How much does 6 meters of fabric cost if 4 meters costs 32 rubles?
Solution: 32 / 4 = 8 rubles. costs one metre. 6 * 8 = 48 rubles. costs 6 meters of fabric.
Schematic notation: an unknown amount of fabric for 32 rubles. 32 / (48 / 6) = 32 / 8 = 4 meters.
9. Write an expression using the rule for multiplying a sum by a number.
6 * 3 + 6 * 4 = 6 * (3 + 4) = 42; 5 * 6 + 5 * 3 = 5 * (6 + 3) = 45;
8 * 7 + 8 * 3 = 8 * (7 + 3) = 80; 4 * 4 + 4 * 16 = 4 * (4 + 16) = 80;
12 * 2 + 12 * 4 = 24 + 48 = 72; 17 * 2 + 17 * 3 = 34 + 34 + 17 = 68 + 17 = 85.
10. First, let's find out how many kilograms of Yubileinoye cookies were brought to the buffet: 5 * 8 = 40 kg. Now let's find out how many cookies "Maria" 6 * 8 = 48 kg. Total: 40 + 48 = 88 kg. cookies.
11. In total, there are 24 rectangles in a rectangular tile. Imagine that the tiles are in 4 rows of 6 pieces. To free one row of tiles, you need to make 6 breaks. First, along the entire tile in one row. The rest, between adjacent pieces across. To get the last row, you don't have to break the tile. Between 6 pieces, 5 breaks. Row: 1 + 5 = 6.
For a tile 4 * 6 = 24, we get 6 + 6 + 6 + 5 = 23 times Dima will have to break the chocolate.
1. Name two numbers from the following: a) 9, 12, 15; b) 8, 12, 16; c) 12, 18, 24.
2. Numbers that are divisible: a) 5, 10, 15, 20, 25, 30, 35, 40, 45, 50; b) 6, 12, 18, 24, 30, 36, 42, 48.
3. Increase the number by 10, and reduce the result by 6 times:
(26 + 10) / 6 = 36 / 6 = 6; (32 + 10) / 6 = 42 / 6 = 7; (50 + 10) / 6 = 60 / 6 = 10; (38 + 10) / 6 = 48 / 6 = 8; (44 + 10) / 6 = 54 / 6 = 9.
4. Find out how many meters of fabric it took for one overall: 30 / 6 = 5 m. Now we find out how much it took for 5: 5 * 5 = 25 m.
Mathematics grade 3, part 1, Dorofeev, page 109
5. Calculate the values of expressions:
(10 + 6) * 3 = 3 * 10 + 3 * 6 = 30 + 18 = 48; (2 + 10) * 6 = 6 * 2 + 6 * 10 = 12 + 60 = 72; (8 + 10) * 2 = 2 * 8 + 2 * 10 = 16 + 20 = 36; (10 + 5) * 4 = 4 * 10 + 4 * 5 = 40 + 20 = 60;
16 * 3 = 48; 12 * 6 = 72; 18 * 2 = 36; 15 * 4 = 60;
8 * 2 * 3 = 16 * 3 = 48; 4 * 3 * 6 = 12 * 6 = 72; 9 * 2 * 2 = 18 * 2 = 36; 5 * 3 * 4 = 15 * 4 = 60.
It can be seen that numbers multiplied individually or by the sum of one factor remain with the same result.
6. Find out how long the two sides of one side are: 17 + 17 \u003d 34 cm. Now we find out the length of the other two sides: 74 - 34 \u003d 40 cm. Accordingly, one side is 40 / 2 \u003d 20 cm. Rectangle: 20 X 17
Check: 17 + 17 + 20 + 20 \u003d 34 + 40 \u003d 74 cm. The perimeter of the rectangle.
7. Write down: 7 + 8 = 15; 15 + 16 + 17 = 48; 7 + 8 +9 + 10 = 34.
8. Draw figures in a notebook and draw lines: the first figure (pink), horizontally 4 cells, between the vertical sides of the figure; the second figure (blue), horizontally 2 cells, so that a rectangle of 8 cells is obtained in the lower part and the same on top; the third figure (yellow), vertically 3 cells, from the highest point to the middle of the bottom side of the figure.
Mathematics grade 3, part 1, Dorofeev, page 110
9. All the way from the house to the stream 40 m. Half the way 40 / 2 = 20 m. When Gosha returned for the straw, he twice walked half the way 20 * 2 = 40 m. Answer: extra 20 m.
1. Is it true that:
1) True, 45 / 5 = 9;
2) That's right, 15 and 18 is divisible by 3;
3) That's right, 24/6 = 4, 24/8 = 3.
2. Choose the numbers: a) 6, 12; b) 6, 10, 14, 15, 27; c) 10, 15, 20; d) 4, 8, 10, 12, 16, 20.
3. Name two-digit numbers: a) 12, 18; b) 18; at 10 o'clock.
4. Find out how many liters of juice are in one jar: 25/5 = 5 liters. Then in two banks: 2 * 5 = 10 l.
Task 1. How many cans are needed for 10 liters. vegetable juice, if 25 liters. fit in 5 jars? Solution: 25 / 5 = 5 liters. placed in one jar. 10 / 5 = 2 cans required.
Problem 2. How many liters of vegetable juice will fit in 5 jars if two jars contain 10 liters?
Solution: 10 / 2 = 5 liters. placed in one bank. 5 * 5 = 25 l. will fit in 5 cans.
Mathematics grade 3, part 1, Dorofeev, page 111
5. Calculate the value of the expressions.
(10 + 3) * 5 = 10 * 5 + 3 * 5 = 50 + 15 = 65; (4 + 10) * 6 = 4 * 6 + 10 * 6 = 24 + 60 = 84; (7 + 10) * 3 = 7 * 3 + 10 * 3 = 21 + 30 = 51; (10 + 6) * 4 = 10 * 4 + 6 * 4 = 40 + 24 = 64;
18 * 3 = (10 + 8) * 3 = 10 * 3 + 8 * 3 = 30 + 24 = 54; 12 * 5 = (10 + 2) * 5 = 10 * 5 + 2 * 5 = 50 + 10 = 60; 15 * 4 = (10 + 5) * 4 = 10 * 4 + 5 * 4 = 40 + 20 = 60; 19 * 2 = (10 + 9) * 2 = 10 * 2 + 9 * 2 = 20 + 18 = 38.
6. Mark the following numbers on the beam, every three cells: 0, 3, 6, 9, 12, 15, 18, 21.
7. Let's calculate if the hen could carry every second simple, and the third - golden: 2 - simple, 3 - golden, 4 - simple, 5 - golden, 6 - simple (should be golden) Answer: no, it cannot.
Division check.
The correctness of the division can be checked in two ways.
Mathematics grade 3, part 1, Dorofeev, page 112
1. Perform division and check in two ways:
27 / 3 = 9, check: 1) 3 * 9 = 27, 2) 27 / 9 = 3;
30 / 5 = 6, check: 1) 5 * 6 = 30, 2) 30 / 6 = 5;
18 / 6 = 3, check: 1) 6 * 3 = 18, 2) 18 / 3 = 6;
32 / 4 = 8, check: 1) 4 * 8 = 32, 2) 32 / 8 = 4.
2. Solve the problem and check:
1) 21/7 = 3 kg, check: 1) 7 * 3 = 21, 2) 21/3 = 7;
2) 16 / 2 = 8 blouses, check: 1) 2 * 8 = 16, 2) 16 / 8 = 2.
3. Name the numbers that divide 20: 2, 4, 5, 10.
4. Three numbers into which numbers are divisible: a) 36: 6, 12, 18; b) 45: 5, 9, 15; c) 100: 10, 20, 50.
5. Write down expressions using the property of multiplying a sum by a number and calculate:
6 * 5 + 6 * 7 = 6 * (5 + 7) = 6 * 12 = 72; 5 * 6 + 5 * 3 = 5 * (6 + 3) = 5 * 9 = 45;
14 * 2 + 14 * 3 = 14 * (2 + 3) = 14 * 5 = 70; 9 * 6 + 1 * 6 = 6 * (9 + 1) = 6 * 10 = 60;
8 * 5 + 8 * 1 = 8 * (5 + 1) = 8 * 6 = 48; 3 * 4 + 3 * 5 = 3 * (4 + 5) = 3 * 9 = 27.
6. Find out how many boys are involved in the ensemble: 18 + 7 = 25 boys. Now let's find out how many guys are involved in the ensemble: 18 + 25 = (18 + 2) + (25 - 2) = 20 + 23 = 43.
7. Write down the number that must be increased by 6 times to get: a) 3; b) 6; in 2; d) 9; e) 1.
8. First, find out how much the mass of one pack of paper is: 12 / 6 = 2 kg. Then three packs: 3 * 2 = 6 kg. Difference between 6 and 3 reams of paper: 6 / 3 = 2. Answer: 2 times.
9. Calculate the values of expressions:
36 / 6 = 6; 42 / 6 = 7; 24 / 6 = 4; 60 / 6 = 10;
28 + 5 * 7 = 28 + 35 = 63; 73 – 6 * 3 = 73 – 18 = 55; 30 + 4 * 6 = 30 + 24 = 54; 62 – 8 * 2 = 62 – 16 = 46;
(10 + 7) * 4 = 10 * 4 + 7 * 4 = 40 + 28 = 68; (2 + 30) * 2 = 32 * 2 = 64; (23 + 7) * 3 = 30 * 3 = 90; (60 – 40) * 5 = 20 * 5 = 100.
Mathematics grade 3, part 1, Dorofeev, page 113
Tasks for multiple comparison.
To find out how many times one number is greater or less than another, you need to divide the larger number by the smaller one.
Mathematics grade 3, part 1, Dorofeev, page 114
1. The length of the green strip is 6 times longer. The length of the red stripe is 6 times less (6 / 1 = 6)
2. There are 3 times more circles than squares. Squares are 3 times less than circles (9 / 3 = 3)
3. There are 12 / 3 = 4 times more brown bears. Polar bears are less than 12 / 3 = 4 times.
4. Lyosha made 15/5 = 3 times more snowballs than Katya.
5. The number 24 is greater: a) 24/4 = 6; b) 24 / 3 = 8.
6. Compare:
(10 + 4) * 3 = 10 * 3 + 4 * 3 = 30 + 12 = 42 (2 + 10) * 5 = 2 * 5 + 10 * 5 = 10 + 50 = 60 (10 + 10) * 2 = 20 * 2 = 40 > than 19 * 2 = 38;
(6 + 10) * 2 = 16 * 2 = 16 * 2 = 32;
(3 + 10) * 5 = 3 * 5 + 10 * 5 = 15 + 50 = 65 (4 + 20) * 3 = 4 * 3 + 20 * 3 = 12 + 60 = 72 7. Draw shapes in your notebook and count number of cells: 1 = 14, 2 = 12, 3 = 14, 4 = 17, 5 = 17.
Mathematics grade 3, part 1, Dorofeev, page 115
8. First, find out how many students are involved in karate and volleyball: 18 + 20 = 38.
Of the total number of karate students, 6 volleyball players: 38 - 6 = 32 students are engaged in karate and volleyball. Now we can find out how many students in the class are not studying: 40 - 32 = 8. Answer: 2 students.
1. Number 27: 1) 27 / 3 = 9, 9 times the number 3; 2) 27 - 3 \u003d 24, 24 more than the number 3.
2. Vera and grandmother peeled potatoes: 1) 12 / 4 = 3, 3 times more potatoes were peeled by grandmother; 2) 12 - 4 = 8, 8 fewer potatoes were peeled by Vera.
3. Baby strollers were brought to the store: 1) 30 / 5 = 6, 6 times fewer strollers were sold than brought; 2) 30 - 5 = 25, 25 more wheelchairs were brought in than sold.
4. Draw two segments FD 3 cm long and KL 1 dm long. 5 cm = 15 cm. 1) 15 / 3 = 5, 5 times the length of segment FD is less than the length of segment KL; 2) 15 - 3 \u003d 12 cm, 12 cm. The length of the segment FD is less than the length of the segment KL.
5. Measure the lengths of the sides of the triangle and quadrilateral with a ruler and add to find the perimeter.
Mathematics grade 3, part 1, Dorofeev, page 116
6. Write the numbers in the empty boxes: 45 / 5 = 9; 6 * 6 = 36; 28 / 4 = 7; 5 * 8 = 40.
7. The girl bought 2 simple pencils, 3 rubles each. and 10 colored ones also for 3 rubles. We find out the cost of simple pencils: 2 * 3 = 6 rubles, colored: 10 * 3 = 30 rubles.
1) 2 pencils cost: 2 * 3 = 6 rubles;
2) 10 pencils cost: 10 * 3 = 30 rubles;
3) 10 / 2 = 5, the girl bought 5 times more colored pencils than ordinary ones;
4) 30 / 6 = 5, 5 times more she paid for colored pencils than for ordinary ones;
5) How much more expensive are colored pencils: 30 - 6 \u003d 24 rubles.
6) How much do all the pencils cost together: 30 + 6 = 36 rubles.
8. Find out how much a meter of nylon tape costs: 48/6 = 8 rubles. Then 5 m. 8 * 5 = 40 rubles.
9. The sum of three numbers is an even number, 1 + 2 + 3 = 6, or 10 + 30 + 40 = 80. Then their product: 1 * 2 * 3 = 6 or 10 * 30 * 40 = 120, will also be even number.
1. Compare without calculating.
15*3 18/9; 0 * 8 (4 + 10) * 3 > 14 * 2; (10 - 2) * 6 2. 1) 45 / 5 \u003d 9, the lesson lasts 9 times longer than the break; 2) 45 - 5 = 40, for 40 minutes. recess is shorter than the lesson.
Mathematics grade 3, part 1, Dorofeev, page 117
3. Find out how many pears grow in the garden: 28 - 7 = 21 trees. Then: 21 / 7 \u003d 3, 3 times more pears than apple trees grow in the garden.
4. Draw a segment CD 2 cm long.
Below it are segments: a) AB, 2 * 2 = 4 cm; b) MN, 4/4 = 1 cm; c) OP, 1 + 6 = 7 cm.
5. Find out how much they paid for one sheet of whatman paper: 40 / 5 = 8 rubles. Now let's find out how much one ballpoint pen costs: 8 / 2 \u003d 4 rubles. Then for 40 rubles you can buy: 40/4 = 10 ballpoint pens.
6. Calculate and compare:
45 / 9 = 5 54 / 9 = 6 64 - 44 = 20 76 + 4 = 80 > 40 * 2 / 10 = 80 / 10 = 8, 80 / 8 = 10 times.
7. Count how many triangles the figure holds: 1) 11 pcs. = 3) 11 pieces; 2) 11 pcs., 4) 11 pcs.
8. Make up all possible two-digit numbers using the numbers: 2, 4, 6, 8 and 0.
Write down: 24, 26, 28, 20, 42, 46, 48, 40, 62, 64, 68, 60, 82, 84, 86, 80.
Mathematics grade 3, part 1, Dorofeev, page 118
1. Write the numbers in the boxes to get the correct entries:
35/5 > 35/7; 3 * 8 + 3 6 * 7 2. Answer the following questions:
1) The number 7 * 4 = 28 is conceived;
2) 12/4 = 3 times more;
3) 30 - 10 = 20 more;
4) 30/3 = 10 times less;
5) It is necessary to reduce by 7, 34 - 7 \u003d 27, 3 * 9 \u003d 27.
3. Write down the numbers that divide 24: 8, 6, 4, 2, 1.
4. Answer the questions according to the table:
1) 15 / 5 = 3, bought notebooks in a cage;
2) 4 * 6 = 24 rubles, paid for 6 lined notebooks;
3) 15 + 24 = 39 rubles, paid for the entire purchase;
4) 3 + 6 = 9, they bought a total of notebooks.
5) 4 + 5 = 9 rubles, there is a notebook in a ruler and a notebook in a cage together;
6) 24 - 15 \u003d 9 rubles, all notebooks in a line cost so much more;
7) How many checkered notebooks can you buy for 100 rubles? 100 / 5 = 20 pieces;
8) How much will 9 lined notebooks cost? 4 * 9 = 36 rubles.
Mathematics grade 3, part 1, Dorofeev, page 119
5. Draw in a notebook a segment AB 1 dm long. 2 cm. \u003d 12 cm. Draw under it:
1) CD, 12/4 = 3 cm; 2) EC, 12 - 5 = 7 cm.
6. Make a task according to the schematic notation:
1) 5 cans of paint, weighing 10 kg, were brought to the warehouse. How much will 7 cans of paint weigh?
Answer: (10 / 5) * 7 = 2 * 7 = 14 kg;
2) 7 cans of paint were brought to the warehouse, with a total weight of 14 kg. How many jars will weigh 10 kg.?
Answer: 10 / (14 / 7) = 10 / 2 = 5 cans.
3) 7 cans of paint were brought to the warehouse, with a total weight of 14 kg. How much will 5 cans of paint weigh?
Answer: (14/7) * 5 = 2 * 5 = 10 kg.
4) How many cans weigh 14 kg if 5 cans have a mass of 10 kg?
Answer: 14 / (10 / 5) = 14 / 2 = 7 cans.
7. Name the figure that has a right angle: 1, 2, 4. All angles are right at the square number 2.
8. Insert the desired word instead of the gap:
1) 5 + 14 = 19 - odd;
2) 30 - 18 = 12 - even;
3) 6 * 10 = 60 - even;
4) 54 / 6 = 9 - odd.
9. Make all possible two-digit numbers from the numbers: 2, 3, 4, 5 and 6, which are divisible by 6.
Write down: 24, 36, 42, 54, 66.
Mathematics grade 3, part 1, Dorofeev, page 120
Material for repetition and self-control
1. 1) Name the even numbers: 6, 8, 10, 12, 14;
2) Name all odd numbers: 13, 15, 17, 19.
2. Calculate the values of expressions:
3 * 7 + 9 = 21 + 9 = 30; 9 / 3 + 38 = 3 + 38 = 41;
83 – (7 + 23) = 83 – 30 = 53; (63 + 9) + 11 = 72 + 11 = 83;
(38 + 9) – 8 = 47 – 8 = 39; 59 – (7 + 29) = 59 – 36 = 23.
3. There are 3 rubber balls and 2 plastic balls in the box. How many balls are in 7 such boxes?
1) (3 + 2) * 7 = 5 * 7 = 35 balls;
2) 3 * 7 + 2 * 7 = 21 + 14 = 35 balls.
4. Compare.
(10 + 4) * 5 = 10 * 5 + 4 * 5 = 50 + 20 = 70 > 60;
(3 + 20) * 3 = 3 * 3 + 20 * 3 = 9 + 60 = 69
(7 + 10) * 4 = 7 * 4 + 10 * 4 = 28 + 40 = 68
(5 + 20) * 4 = 5 * 4 + 20 * 4 = 20 + 80 = 100.
5. Calculate the values of expressions:
32 / 4 = 8, 40 / 5 = 8, 27 / 3 = 9;
50 – 4 * 7 = 50 – 28 = 22, 60 – 3 * 8 = 60 – 24 = 36, 70 – 5 * 9 = 70 – 45 = 25;
(9 + 26) / 5 = 35 / 5 = 7, (18 + 18) / 4 = 36 / 4 = 9; (40 – 19) / 3 = 21 / 3 = 7;
5 * 8 / 4 = 40 / 4 = 10, 6 * 4 / 3 = 24 / 3 = 8, 3 * 8 / 4 = 24 / 4 = 6;
54 / 6 * 4 = 9 * 4 = 36, 45 / 5 / 3 = 9 / 3 = 3, 32 / 4 * 7 = 8 * 7 = 56.
6. From 36 kg. peas poured 8 packages of 3 kg.
1) Find out how much was poured 3 * 8 = 24 kg. Then 36 - 24 = 12 kg remained;
2) It will take 12 / 3 = 4 packets to pour the remaining peas.
Mathematics grade 3, part 1, Dorofeev, page 121
7. 1) Write down the numbers that divide the number 36: 2, 3, 4, 6, 9, 12, 36.
8. Decrease each of the numbers by 40, and reduce the result by 5 times.
65 – 40 = 25, 25 / 5 = 5; 55 – 40 = 15, 15 / 5 = 3; 80 – 40 = 40, 40 / 5 = 8; 75 – 40 = 35, 35 / 5 = 7.
9. Perform calculations, compare the values of expressions.
25 / 5 = 5; 18 / 3 = 6; 42 / 6 = 7; 32 / 4 = 8;
41 – 6 * 6 = 41 – 36 = 5, 30 – 3 * 8 = 30 – 24 = 6, 37 – 5 * 6 = 37 – 30 = 7, 36 – 4 * 7 = 36 – 28 = 8;
(62 – 47) / 3 = 15 / 3 = 5, (12 + 18) / 5 = 30 / 5 = 6, (35 – 7) / 4 = 28 / 4 = 7, (16 + 24) / 5 = 40 / 5 = 8.
You can notice that the result of all calculations in columns: 5, 6, 7, 8.
10. There are 24 felt-tip pens in 4 boxes. Find out how many felt-tip pens are in one box: 24 / 4 = 6 pcs.
a) In 6 boxes: 6 * 6 = 36 markers; b) In 3 boxes: 3 * 6 = 18 markers.
11. 1) In the buffet, 3 waffles cost 18 rubles, how much will 6 waffles cost? First, find out how much 1 waffle costs: 18 / 3 = 6 rubles. Then 6 waffles: 6 * 6 = 36 rubles.
2) In the buffet, 3 waffles cost 18 rubles, how many waffles can you buy for 36 rubles? One waffle costs: 18 / 3 = 6 rubles. Then for 36 rubles you can buy 36 / 6 = 6 waffles.
3) In the buffet, 6 waffles cost 36 rubles, how much will 3 waffles cost? One waffle costs: 36 / 6 = 6 rubles. Then 3 waffles cost: 3 * 6 = 18 rubles.
Problems are similar in conditions, differ in different unknowns. You can make a schematic notation: If 6 wafers cost 36 rubles, how many wafers can you buy for 18 rubles? Let's find out how much one waffle costs: 36/6 = 6 rubles, then 18/6 = 3 waffles.
12. One broken line consists of 5 links of 6 cm each: 6 * 5 \u003d 30 cm. The second of 8 links of 6 cm each: 6 * 8 \u003d 48 cm. Let's find out the difference: 48 - 30 \u003d 18 cm. Answer: the length of the second is 18 cm longer than the first.
13. From 10 kg. fresh apples are obtained 2 kg. dried, in 30 kg. 3 x 10 kg. 30 / 10 \u003d 3, multiply by the number of dried apples out of 10: 2 * 3 \u003d 6 kg. Answer: from 30 kg. apples can be obtained 6 kg. dried.
14. Find out how many kilograms of apples were sold: 6 * 10 = 60 kg. Cherries sold: 6 * 4 = 24 kg. Then apples and cherries were sold: 60 + 24 = 84 kg.
Mathematics grade 3, part 1, Dorofeev, page 122
Practical work.
Draw a rectangle of 16 squares. You can also build a rectangle with a length of 16 and a width of 1 cell or a length of 2 cells, a width of 8, and also a length of 1, a width of 16 cells.
Perimeter: (2 * 8) + (2 * 2) / 2 = 16 + 4 / 2 = 20 / 2 = 10 cm, (2 * 16) + 2 / 2 = 32 + 2 / 2 = 34 / 2 = 17 cm
In December 2012, Russian legislation adopted the Federal Law. It is considered the main regulatory legal act in the field of education.
General education in Russia
Education in our country is aimed at personal development. And also in the learning process, the child must learn the basic knowledge, skills and abilities that will be useful to him in the future for adaptation among people and the right choice of profession.
Stages of general education:
- preschool;
- general primary (grades 1-4);
- basic general (grades 5-9);
- general secondary (grades 10-11).
Thus, it becomes clear that education in Russia is divided into 2 types:
- preschool - children receive it in kindergartens and schools;
- school - from 1st to 11th grade, children study in educational institutions, schools, lyceums, gymnasiums.
Many children, coming to grade 1, begin to study according to the educational program "Perspective Primary School". Reviews about it are different, teachers and parents discuss the program in various forums.
The main provisions of the program include all the requirements of state standards for primary general education. The system-active approach to the development of the child's personality became the basis.
Promising Elementary School Program in Grade 1
Reviews of parents and teachers in elementary school about the program "Perspective" are diverse, but in order to understand its whole essence, you need to get to know it in more detail.
What the program is learning:
- philology;
- mathematics;
- informatics;
- social science;
- art;
- music.
The child, studying the program, as a whole can form his own opinion about the environment and get a complete scientific picture of the world.
The program "Perspective" has a number of textbooks. Among them:
- Russian language - alphabet;
- literary reading;
- mathematics;
- informatics and ICT;
- the world;
- foundations of religious cultures and secular ethics;
- art;
- music;
- technology;
- English language.
All textbooks included in the "Perspective Primary School" curriculum have been certified for compliance with the Federal State Educational Standard. And they were recommended by the Ministry of Education and Science for use in teaching children in educational institutions.
The main objective of the entire program "Perspective Primary School" is the full development of the child based on the support of teachers of his individual characteristics. At the same time, the program is designed to ensure that each student will be able to visit different roles. Thus, at one time he will be a student, at another - a teacher, and at certain moments - an organizer of the educational process.
Like any program, the "Promising Elementary School" has its own principles in teaching children. The main ones are:
- the development of each individual child must be continuous;
- in any situation, the child must formulate a holistic picture of the world;
- the teacher must take into account the characteristics of each student;
- the teacher protects and strengthens the physical and mental condition of the child;
- a student for education should receive a good example.
Main properties of the program "Perspective"
- Completeness - at the time of learning, the child learns to find data from different sources. Such as a textbook, a reference book, the simplest equipment. Children develop business communication skills, as the program has developed joint tasks, working in pairs, solving problems in small and large teams. The teacher, while explaining new material, uses several points of view on one task, this helps the child to consider the situation from different angles. The textbooks have the main characters that help children learn to perceive information while playing.
- Instrumentation - specially developed mechanisms for children that help to put into practice the acquired knowledge. It was made so that the child could, without outside help, look for the necessary information not only in the textbook and dictionaries, but also beyond them, in various teaching aids.
- Interactivity - each textbook has its own Internet address, thanks to which the student can exchange letters with the heroes of the textbooks. This program is mainly used in schools where computers are widely used.
- Integration - the program is designed so that the student can get a general picture of the world. For example, in the classroom of the world around the child will be able to obtain the necessary knowledge from different areas. Such as natural science, social science, geography, astronomy, life safety. And also the child receives an integrated course at the lessons of literary reading, since there the teaching of language, literature and art is included in the basis of education.
The main features of the program "Perspective"
For teachers, the developed methodological manuals have become great helpers, as lesson plans are detailed in them. Most parents and teachers are satisfied with the program.
Peculiarities:
- in addition to textbooks for each subject, an anthology, a workbook, an additional teaching aid for the teacher are attached;
- The curriculum for students consists of two parts. In the first part, the teacher is offered theoretical classes, while the second part helps the teacher to build a lesson plan separately for each lesson. And also in the methodological manual there are answers to all the questions that are asked in the textbook.
It should be understood that education in primary school is a very important process in which the child builds the foundation for all subsequent education. The curriculum "Perspective Primary School", reviews confirm this, has many positive aspects. The child is quite interested in getting new knowledge.
How do the authors see the future of their program?
When developing the program, the authors tried to include in it all the key points that will help the child in later life. After all, just in elementary school, children must learn to comprehend the correctness of their actions, to get a more complete picture of the world around them.
In our time, virtually all school programs are aimed at personal development. "Perspective" was no exception. Therefore, as teachers who have encountered work on this program say, there is nothing complicated about it. The main thing is that the child is engaged not only at school, but also at home.
Is it worth it to study this system?
Whether or not to go to school with the Promising Primary School program is up to each parent to decide for themselves. In any case, the child must receive primary education.
Teachers try not to leave negative feedback about the Promising Elementary School program, as they will continue to work with it. But the opinions of parents are ambiguous, some like it, some do not.
What you need to know about the Perspective program:
- the program is designed very close to traditional;
- should help the child to become independent;
- parents will not be able to relax, throughout the entire education the child will need their help.
A little about "Promising Primary School"
If a student is enrolled in an elementary school under the Perspective program, feedback for parents very often becomes a powerful argument to think about whether he will be able to understand all aspects of education.
The entire program is one large system of interconnected subroutines. At the same time, each discipline is a separate link and is responsible for a certain direction in activity. For many parents, reviews of the "Primary School Perspective" curriculum help to correctly assess their abilities and the abilities of their child.
- the child must be ready to develop independently;
- the child must comprehend and understand the basic values in life;
- It is necessary to motivate the child to learn and learn.
For many parents, these goals seem out of place and quite difficult for first graders. That is why reviews of the training program "Perspective" (elementary school) are far from unambiguous. Some people like textbooks and the material presented in them, some do not. But this is true for all tutorials. Each of them has its pros and cons, and the task of parents is to understand what is more.
If we consider the program1 "Perspective Primary School", Grade 1, the feedback from the authors will help to understand the principles on which the entire educational process is built. What are the creators hoping for?
- Personal development in this program is given the most attention. The child must understand which of the human values should be above all.
- Education of patriotism. From childhood, a child must be hardworking, respecting human rights and freedoms, showing love for others, nature, family, and the Motherland.
- Unification of cultural and educational process. Protection of national culture and understanding of the significance of all cultures, different nations for the entire state as a whole.
- Self-realization of personality. The child should be able to develop independently and participate in various creative tasks.
- Formation of the correct point of view and a general picture of the world.
- One of the main goals is to help the child learn to live in society with other people.
From the feedback on the "Primary School Perspective" program, you can understand how completely different children learn information and how adaptation takes place at school. It should be noted that this largely depends on the teacher (sometimes much more than on the program).
Achievements of schoolchildren
Primary school under the program "Perspective", reviews of employees of the Ministry of Education confirm this, contributes to the harmonious development of students.
Achievements:
- In meta-subject results - students quite easily cope with mastering
- In the subject results - children learn new knowledge and try to apply it based on the general picture of the world.
- Personal results - students easily study and find the necessary material on their own.
These are the main achievements aimed at by the elementary school with the "Perspective" program. Feedback on the project is often positive, as parents notice changes in children for the better. Many become much more independent.
School program "Perspective elementary school": feedback from teachers
Despite the fact that the Perspektiva program appeared relatively recently, many teachers are already working on it.
It is very important for parents to have feedback on the program "Perspective Primary School" (Grade 1) from teachers. Since they work with her and know all the pitfalls that they will have to face.
With the advent of a large number of school programs for elementary school in the learning process, it is impossible to say unequivocally which one will be better. So in "Perspective" there are both minuses and pluses.
The pluses of the teacher include methodological aids for conducting lessons. They are divided into two parts, one of which contains theoretical material, the other contains a detailed lesson plan for the school program "Perspective Primary School".
