How much does a child need to know and learn in a short time:
Moreover, all children have different abilities.
Some people grasp everything on the fly, while others need a little more time.
To consolidate and improve children's initial counting skills, the website has created online - Generator, which creates examples and equations in mathematics for preschool and primary school children.
Using this online generator, you can create, download and print ready-made examples for addition and subtraction, multiplication and division absolutely free of charge.
Ready-made examples in mathematics are generated on a checkered page, which allows the child to train not only mental calculation, but also the correct writing of numbers.
The generator of examples and equations has internal settings, by changing which you can create examples for children of different ages and levels of training (from 5 years to 2-3 grades).
To obtain and print examples in mathematics, you need:
1. Set (select) parameters for tasks
- by number of examples: 10, 20, 30, 60 (2 sheets), 90 (3 sheets)
- by type of task: example or equation
- by functions of mathematical operations: addition, subtraction, multiplication and division.
- by number range: from 1 to 100 (for example - from 5 to 10, from 10 to 50, etc.)
2. Print the resulting file. You can first save the file with the tasks to your computer or flash drive.
GENERATOR OF EXAMPLES AND EQUATIONS
* If you generate examples in the Firefox browser, pdf files may not be displayed correctly as a result of generation (a blank checkered page is generated, or there are no symbols for mathematical operations)
In this case you need:
1. Save the resulting (incorrect) document to your computer, and then open and print the file with examples from your computer.
2. Open this page in another browser (Chrome, Yandex) by copying the page address and pasting it into the address bar.
Use the online math example generator if:
Your child has just started learning to count. Select the very initial parameters for generation. To get the simplest examples in mathematics.
Your child needs additional training in mathematics.
You are going on a long trip. Solving examples and equations will be a useful activity that will help pass the time on the road.
The math example generator will be very convenient for both parents and teachers. Thanks to the selection parameters, you can create as many tasks of different levels of complexity for preparation.
Advantages of the mathematical example generator.
There is no need to buy problem books and math manuals with examples and equations in advance.
To get examples for solutions, you do not need to first download the program to your computer. All examples are generated online.
You can download the example file to your computer and print it at any time.
Examples are generated on a page in a box, which is very convenient for a child to correctly write numbers.
You can select tasks individually for your child depending on his level of preparation.
If you have any difficulties or questions about using the example generator, do not hesitate to ask questions in the comments.
When solving complex problems, pay attention to the solution plan and composition of the problem, the arithmetic expression that resolves it, and the calculation of the required value itself. The following problem should be classified as problems involving complex arithmetic operations.
Problem 20. Someone, having a capital of 8998 rubles, bought 15 acres of arable land for 125 rubles, 37 acres of meadow for 112 rubles, 5 horses for 147 rubles. With all the rest of the money he bought timber for 132 rubles. for a tithe. How many acres of forest were purchased?
Problem solution plan. To determine how many acres of forest a person bought, you need to find how much money he had left from previous purchases.
To do this, you need to find out how much he spent on these purchases.
Composition of the task. It is easy to determine the composition of this complex problem. Our complex task breaks down into the following 6 simple tasks, of which:
First task determines how much he paid for the meadow and decides multiplication.
Second task determines how much he paid for the horses and decides multiplication.
Third task determines how much he paid for the horses, and also decides multiplication.
Fourth task determines how much money he spent on all these purchases, and decides addition.
Fifth task determines how much money he has left after these purchases and decides by subtraction.
Sixth task determines how many acres of forest he bought with the rest of the money, and decides division.
Arithmetic expression of the problem. It is very easy to find an arithmetic expression that solves our problem if arithmetic expressions that solve all simple problems are found.
The 1st problem is solved by the arithmetic expression: 125 × 15.
2nd problem: 112 × 37.
3rd problem: 147 × 5.
4th problem: 125 × 15 + 112 × 37 + 146 × 5 (a).
The arithmetic expression solving the 5th problem will be obtained if we subtract the arithmetic expression (a) from 8998. To indicate this, we enclose it in parentheses. Having done this, we get the expression:
8998 - (125 × 15 + 112 × 37 + 147 × 5).
The 6th problem can be solved if we divide the last arithmetic expression by 132.
The arithmetic expression that solves our problem will be
÷ 132
Computing the problem. We can find a numerical solution to a given problem, either by determining the numerical value of the arithmetic expression that solves the problem, or by separately finding solutions to all simple problems into which our complex problem breaks down.
At the beginning of the calculation, the data quantities of the problem are usually arranged in a known order.
So, in our example, these tasks can be arranged as follows:
Data: capital 8998 rub.
What you are looking for: the number of acres of forest.
The progress of the calculation is written down:
Answer: 17 acres of forest were purchased.
Here we put a number for each separate calculation. It indicates the order of calculation and denotes the simple problem that is solved by each individual action. When solving problems, it is common to keep in mind those preliminary considerations that we have outlined, and proceed directly to the calculation itself.
Order in calculations. When solving problems, you should always maintain order in the arrangement of calculations. This order allows you to clearly see the connection between the data and the required problems, makes it possible to easily review the entire problem, find errors in calculations, and speeds up the process of calculations.
Section 1 NATURAL NUMBERS AND ACTIONS WITH THEM. GEOMETRIC FIGURES AND QUANTITIES
§ 15. Examples and problems for all operations with natural numbers
When calculating the values of numerical expressions, you should not forget about the order of actions.
The order of actions is determined by the following rules:
1. In expressions with parentheses, the values of the expressions in parentheses are evaluated first.
2. In expressions without parentheses, exponentiation is performed first, then multiplication and division, in order from left to right, and then addition and subtraction.
Example 1. Calculate: 8 ∙ (27 + 13) - 144: 2.
Solutions.
1) 27 + 13 = 40;
2) 8 ∙ 40 = 320;
3) 144: 2 = 72;
4) 320 - 72 = 248.
Example 2. Find the value of the expression (x2 - y: 13) ∙ 145, if x = 12, y = 91.
Solutions. If x = 12, y = 91, then (x2 - y: 13) ∙ 145 = (122 - 91: 13) ∙ 145 = (144 - 7) ∙ 145 = 137 ∙ 145 = 19,865.
Action properties can be used where appropriate. For example, the value of the expression 438 ∙ 39 - 338 ∙ 39 can be calculated as follows:
438 ∙ 39 - 338 ∙ 39 = (438 - 338) ∙ 39 = 100 ∙ 39 = 3900.
What rules are used to determine the order of actions when calculating numerical expressions?
Entry level
522. Count (orally):
1) 42 + 38 - 7; 2) 24 ∙ 10: 2;
3) 27 - 30: 5; 4) 42: 6 + 35: 7;
5) 8 (23 - 19); 6) (12 + 18) : (12 - 7).
Intermediate level
523. Calculate:
1) 426 ∙ 205 - 57 816: 72;
2) (362 195 + 86 309) : 56;
3) 2001: 69 + 58 884: 84;
4) 42 275: (7005 - 6910).
524. Calculate:
1) 535 ∙ 207 - 32 832: 76;
2) 1088: 68 + 57 442: 77;
3) (158 992 + 38 894) : 39;
4) 249 747: (4905 - 1896).
525. In 5 hours, the ship traveled 175 km, and the train covered 315 km in 3 hours. How many times is the speed of the train greater than the speed of the ship?
526. In 5 hours, a freight train traveled 280 km, and a fast train traveled 255 km in 3 hours. How much faster is the speed of a fast train than a freight train?
527. Find the meaning of the expression:
1) 78 ∙ x + 3217, if x = 52;
2) a: 36 + a: 39, if a = 468;
3) x ∙ 37 - c: 25, if x = 15, y = 2525.
528. Find the meaning of the expression:
1) 17 392 + 15 300: and, if a = 25, 36;
2) m ∙ 155 - t ∙ 113, if m = 17, t = 22.
529. Paid for 5 pens and 3 common notebooks
16 UAH 70 kopecks How much does a notebook cost if a pen costs 2 UAH? 50 kopecks?
530. Three boxes of apples and two boxes of bananas together weigh 144 kg. How much does a box of apples weigh if a box of bananas weighs 24 kg?
531. The elder brother collected 12 baskets of cherries, and the younger brother collected 9 baskets. In total they collected 105 kg of cherries. How many kilograms of cherries did each brother pick if the weight of all baskets was the same?
532. 27 packs of squared notebooks and 25 packs of lined notebooks were delivered to the store - 2600 pieces in total. How many notebooks were brought in a cage and how many in a line, if there are the same number of notebooks in all packs?
533. One computer-controlled machine produces 12 parts per minute, and the second produces 3 more parts. In how many minutes will both machines, when turned on simultaneously, produce 945 parts?
Sufficient level
534. Collected 830 kg of apples. Of these a kilograms were given to a kindergarten, and those that remained were divided equally into 30 baskets. How many kilograms were in each basket? Write down the letter expression and calculate its value if a = 110.
535. Calculate in a convenient way:
1) 742 + 39 + 58; 2) 973 + 115 - 273;
3) 832 - 15 - 32; 4) 2 ∙ 115 ∙ 50;
5) 29 ∙ 19 + 71 ∙ 19; 6) 192 ∙ 37 – 92 ∙ 37.
536. The television repair shop planned to repair 180 televisions in 12 days, but every day they repaired 3 more televisions than planned. In how many days was the task completed?
538. Find the meaning of the expression:
1) (21 000 - 308 ∙ 29) : 4 + 14 147: 47;
2) 548 ∙ 307 - 8904: (33 ∙ 507 - 16 647);
3) (562 + 1833: 47) ∙ 56 - 46 ∙ 305;
4) 1789 ∙ (1677: 43 - 888: 24)∙500.
539. Find the meaning of the expression:
1) (42 + 9095: 85) ∙ (7344: 36 - 154);
2) 637 ∙ 408 - 54 036: (44 ∙ 209 - 9117);
3) (830 - 17 466: 82) ∙ 65 + 57 ∙ 804;
4) 197 ∙ (588: 49 + 728: 56) ∙ 40.
540. 1506 kg of butter was delivered to three stores. After the first store sold 152 kg, the second - 183 kg, and the third - 211 kg, all stores had the same amount of butter left. How many kilograms of butter were brought to each store?
541. From cities A and B , the distance between them is 110 km, two cyclists rode towards each other at the same time. The speed of one of them is 15 km/h, and the other is 3 km/h less. Will the cyclists meet in 4 hours?
542. High school students Ivan and Vasily worked on a farm in the summer. Ivan worked 4 hours every day for 16 days, and Vasily worked 3 hours every day for 18 days. Together the guys earned 944 UAH. Ask intelligent questions and answer them.
543. Two workers, one of whom worked 12 days, 8 hours daily, and the other 8 days, 7 hours daily, together produced 1368 parts. Find the labor productivity of the workers if they have the same. How many parts did each worker make?
544. Compose and solve a problem involving all four operations with natural numbers.
High level
545. Find roots for the equations:
1) x - x = x ∙ x; 2) m: m = m ∙ m.
546. Find roots for the equations:
1) x: 8 = x ∙ 4; 2) y: 9 = in: 11.
547. What number must be multiplied by 259 259 to obtain a product that is written only in digits 7?
548. What number must be multiplied by 37,037 to obtain a product that is written only in digits 3?
Exercises to repeat
549. Solve the equations:
1) 4x - 2x + 7 = 19; 2) 8x + 3x - 5 = 39.
550. To get to the city, a peasant traveled 3 hours by bus, whose speed is a km/h, and 2 hours by truck, whose speed b km/h He covered the return journey in 4 hours on a motorcycle. Find the speed of the motorcycle. Write down the literal expression and calculate its value if a = 40, b = 32.
Let us consider in detail each of the simple arithmetic operations and give several simple problems that clarify the use of each action.
Addition problems
You need to add the number every time:
when one number is needed increase some number, or when one number needs add other;
when several numbers need to be combined into one.
Problem 1. A person has property consisting of a house, furniture, paintings and horses. The house costs 47,215 rubles, furniture 2,215 rubles, paintings 5,207 rubles, horses 1,925 rubles. How much is all the property worth?
Answer: 56562 rubles.
Problem 2. One library has 1015 books, the other has 117 books more. How many books are in the second library?
Answer: 1132.
Subtraction problems
Subtract every time:
when you need to determine the difference between numbers;
when you need to reduce one number by another.
Problem 3. There are 927 thousand inhabitants in St. Petersburg, 750 thousand in Moscow. How many thousand fewer inhabitants are there in Moscow?
Answer: 177 thousand.
Problem 4. The first crusade was in 1096, and the last in 1270. How many years did the Crusades last?
Answer: 174 years.
Multiplication problems
Multiply numbers whenever required:
increase one number several times;
repeat one number as many times as the other number contains units.
In any multiplication, the product is homogeneous with the factor, and the factor is an abstract number.
Problem 5. In the workshop, each of the 28 workers receives a monthly salary of 15 rubles. How much do all workers earn?
Answer: 420 rubles.
Problem 6. The book has 175 pages. Each page has 22 lines. How many lines are there in the book?
Answer: 3850 lines.
Division problems
Division of integers is needed whenever required:
divide the number into several equal parts;
determine how many times the smaller number is contained in the larger one;
reduce one number several times.
Problem 7. Someone earned 3,648 rubles a year. How much does he earn per month?
Answer: 304 rubles.
Problem 8. A piece of cloth measuring 26 arshins costs 468 rubles. How much does an arshin cost?
Answer: 18 rubles.
Problem 9. Find a number less than 175 25 times.
Arithmetic problems with named numbers
Splitting named numbers.
Problem 10. One person dies every second on the globe. How many will die in 17 days 5 hours. 1 sec?
Answer: 1,486,801 people.
Converting named numbers.
Problem 11. Having pound, pound and spool weights, determine the smallest number of weights required to weigh 5000 spools.
The answer is 5000 gold. = 1 p. 12 f. 8 gold You need 1 + 12 + 8 = 21 weights.
Compound addition.
Problem 12. How much gold is in three bars if the first weighs 3 p. 12 lb. 17 l. 1 gold, second 2 p. 35 f. 11 l. 1 gold and the third 17 f. 2 gold
Answer: 6 p. 24 f. 29 l. 1 gold
Compound subtraction.
Problem 13. From a piece of matter in 5 s. 3 f. 2 limes. a piece of 2 s is cut off. 5 f. 7 d. 1 l. Determine how much matter remains?
Answer: 2 s. 4 f. 5 d. 1 l.
Timed arithmetic problems
Named number addition and subtraction problems involving time have some special features.
Ways to Express Time. Time is usually expressed as a compound named number. This number means how many years, months, days have passed since the Nativity of Christ, the beginning of the Christian era. Thus, 1860 May 17, 7 am is designated by a composite named number:
1859 l. 4 m. 16 d. 7 h.,
and, conversely, a composite named number 1839 l. 11:00 15:00 18:00 denotes the year 1840 December 16th 6 pm, because the day is counted from midnight. 12 hours passed from midnight to noon, and 6 hours passed from noon to 6 pm.
When solving problems involving the addition of named numbers expressing time, you usually have to determine from one event and the time interval between this and the subsequent event the time of the second.
Problem 14. Someone was born on April 14, 1827. Determine when he was 32 years 5 months 25 days.
Adding two composite named numbers, we have:
The required time is 1859 October 9th.
When calculating over time, you need to pay attention to the fact that the months of the year do not have the same number of days. The number of days in a month varies; Therefore, when you have to add up days and turn them into months, you need to take into account the size of one or several recent months.
In the proposed problem, if we add 1826 l to the compound named number. 3 m. 13 d. only 32 g. 5 m., we will have 1858 l. 8 m. 13 days, that is, 1859, September 14th.
After this, you need to add another 25 days. September has 30 days, therefore October 9, 1859 will arrive in 25 days.
If we have one event on August 26, 1812, and another occurs a year later, 6 months and 23 days, the calculation will take a different form.
Applying 1811 l to the compound named number. 7 m. 25 days only 1 year 6 months, we get the composite named number 1813 years 1 month 25 days, meaning February 26, 1814. If another 23 days pass after this time, the event time is calculated as follows. February 1814 has 28 days, therefore, when adding the named numbers we have:
that is, the time of another event will be March 21, 1814.
If, when adding and subtracting named numbers containing time, it is necessary to pay attention to the value of the last month, it is necessary to add only years and months, and then, having determined which month the calculation of the day refers to, add or subtract days and hours.
Subtracting named numbers expressing time. When subtracting named numbers containing time, you have to:
determine the time interval between two given events, or
according to the time interval between the data and the previous event - the time of the last one.
The first kind belongs to
Problem 15. A person set off on a trip around the world on June 14, 1839 and returned on April 15, 1844. How long did the journey last?
In this case, time is usually expressed as a composite named number containing only years and days. This is done because the months of the year do not contain the same number of days. We express the beginning of the journey on June 14, 1839 as follows: adding up all the days contained in the months that have passed since January, we have:
in January 31, in February 28 days (1839 - simple), in March 31, in April 30, in May 31 days, a total of 151 days.
Adding the 13 days of June, we have 164 days, therefore, the beginning of the journey is determined by the composite named number 1838 l. 164 days.
Similarly, for the end of the journey we have January 31, February 29 (1844 is a leap year), March 31 and April 14, for a total of 105 days. The end of the journey is expressed by a composite named number: 1843 105 days.
Subtracting these named numbers, we get:
The journey lasted 4 years 306 days.
The second kind refers to
The time of July 27, 1872, is expressed in days and mountains by the composite named number 1871, 208 days. Subtracting 27 l. 165 days, we have 43 days left in 1844. This number is expressed as February 13, 1845.
Multiplying named numbers.
Problem 17. Purchased 7 pieces of copper, each weighing 4 pounds. 15 l. 1 z. 15 d. Find the weight of these 7 pieces.
Answer: 31 f. 12 l. 1 gold 9 d.
Division of named numbers.
a) Divide a named number by a named number.
Problem 18. How many spoons will be produced from a piece of silver weighing 2 pounds? 30 l. 48 d., if each spoon weighs 4 lot. 2 gold 12 dollars?
Answer: 20 spoons.
b) Dividing a named number by an abstract one.
Problem 19. The train runs in 8 hours 185 ver. 423 pp. 6 f. 4 d. How much does he run in an hour?
Answer: 23 ver. 115 soot 3 f. 5 days
