Is it used for application problems in elementary school? How to do primary school math application

It is a matter of course to do practical problems in elementary school.

From the fourth grade, students will learn to do practical problems.

The premise of doing practical problems is to lay a good foundation, and the foundation is calculation, equations, etc.

If lz is asking if it is useful to do application problems.

Of course, there are many types of application problems, and if you do more application problems, you can get more exposure to some aspects of application problems.

If you encounter one kind of question, you can ask teachers, classmates, parents or ask for help online.

In short, I hope it can help LZ, and if it's not perfect, you can continue to ask.

There are many benefits to doing word problems:

1. Consolidate what you have learned.

2. Connect with the reality of life, cultivate the good quality of using your brain and thinking more, and apply what you have learned.

3. Let abstract knowledge be concretized and visualized, and cultivate the interest of learning.

4. Exercise rigorous thinking habits, quantify, precise, and simplify the daily fuzzy things, and discover mathematical laws from production practice, so as to lay a good foundation for the study, research, and further study in many science and engineering fields such as machinery, chemical engineering, electronics, computer, finance, and even astronomy, geography, archaeology, exploration, etc., and lay the foundation for further development.

So instead of hesitating, you must learn it unswervingly!

Hand hit, hehe...

Of course, it is useful to exercise children's thinking skills.

Speaking of mathematics, especially mathematics in junior high school and primary school has nothing to do with whether you are smart or not, if you pay attention to it, you will find that your final exam, all the types of questions are what you have done, therefore, it is very simple to improve your grades, I don't have to say much about this in class, mainly to do the questions, targeted questions, the practice papers sent by the teacher and the usual homework are very good, you just need to understand, do the type again, basically no problem. You said that grades 1-5 are not bad, which means that you have a good foundation in calculation, and it is very simple to learn mathematics. And I ask the teacher almost half-understood, sometimes it feels like listening to a book from heaven, and when the teacher talks about the difficulty, I don't understand.

For a manuscript, it takes 6 hours for A to type alone, and 10 hours for B to type alone. Now, after A has been playing alone for a few hours, B has continued to play for some reason, and it has taken 7 hours together. How many hours did A take to type?

Of course, you don't need to write answers at the beginning of the grade, and you can use them from the third grade, but you don't need to write the explanations.

There are a total of 28 apples in two baskets A and B, and after taking a few from basket A and putting them into basket B, basket B has 10 more apples than basket AThere are 9 baskets.

The calculation process is as follows:

9 (pcs) <>

Classification of primary school application problems.

Normalization Problem, Summing Problem, Difference Problem, Sum Factor Problem, Difference Ratio Problem, Encounter Problem, Chase Problem, Tree Planting Problem, Age Problem, Boat Problem, Train Problem, Clock Problem, Profit and Loss Problem, Engineering Problem, Positive and Negative Proportion Problem, Proportional Distribution, Percentage Problem, Cattle Grazing Problem.

The problem of chickens and rabbits in the same cage, the problem of square matrix, the problem of commodity profits, the problem of deposit interest rate, the problem of solution concentration, the problem of composition of the number of cloths, the problem of the magic square, the problem of the drawer principle, the problem of convention common times, and the problem of maximum value.

The above content refers to: Encyclopedia - Application Questions.

Analysis: 1. Because the total amount of grass can be divided into two parts: the original grass and the new grass. Although the newly grown grass is changing, it should be noted that it grows at a uniform rate. Therefore, the amount of new grass in this meadow every day is also constant.

2. Suppose that the amount of grass eaten by 1 cow in a week is 1 serving, then 17 cows need to eat 17x6=102 (parts of grass) for 6 weeks, and the new grass and the original grass are also eaten; 13 cows need to eat 13x9=117 (parts of grass) for 9 weeks, and the new grass and the original grass are also eaten. The 102 servings of grass are the sum of the amount of original grass and the number of new grass that grows in 6 weeks.

Servings is the sum of the amount of original grass and the number of new grass growing in 9 weeks, so the number of new parts of grass growing each week is: (117-102) 9-6) = 5 (parts).

4. The amount of original grass is 102-5x6=72 (parts)5. This grass can be eaten by 11 cows: 72 (21-15) =12 (weeks).

Take the amount of 1 unit of grass that a cow eats in a week.

17 cows ate for 6 weeks, 17 6 = 102 units of grass, 13 cows ate for 9 weeks, 13 9 = 117 units of grass, indicating that in 9-6 = 3 weeks, this grass grew 117-102 = 15 units of grass. That is, 15 3 = 5 units of grass per week.

This grassland originally had 102-6 5=72 or 117-9 5=72 units of grass for 11 cows to eat, and 11 units of grass per week, and 5 units of grass per week, then 11-5=6 units of grass needed to be consumed every week.

Then you can eat 72 6 = 12 weeks.

When 17 cows eat for 6 weeks, divide the grass into two pieces, the large pieces for 13 cows to eat for 6 weeks, and the small pieces for 4 cows to eat for 6 weeks.

When the 13 cows eat for 9 weeks, the grass is also divided into two identical pieces, and the large pieces are eaten for the 13 cows for 6 weeks, and they are finished. Suppose that all the grass that grows after that is moved to small pieces. Then 13 cows can be eaten on small pieces for 3 weeks.

It stands to reason that there is enough grass for 4 cows to eat for 6 weeks, and only enough for 8 cows to eat for 3 weeks, but now it is enough for 13 cows to eat for 3 weeks, and the grass eaten by 5 cows is grown in 3 weeks. That is, the grass grows enough for 5 cows to eat for a week.

So, the original grass was enough for 12 cows for 6 weeks and 8 cows for 9 weeks.

Of the 11 cows, five have eaten the grass that grows, and the remaining six can eat for 12 weeks, so they can eat for a total of 12 weeks.

This is a sixth-grade application problem, first of all, you need to know how much grass there is in the pasture;

Then calculate how much grass the pasture can produce per week;

In the end, 11 cows could be fed for a few weeks.

First, 13 cows ate for 9 weeks, 17 cows ate for 6 weeks, 9-6 = 3 weeks, and the grass produced in 3 weeks is equal to 13 9-17 6 = 15

Grass 15 = 5 produced in a week

Second, calculating how much grass there was in the pasture, 13 cows ate for 9 weeks, a total of 13 9 = 117 produced 5 per week, and 5 9 = 45 in 9 weeks

The grass originally had in the pasture was 117-45=72

Thirdly, calculate how many weeks 11 cows can eat.

Suppose 11 cows can eat x weeks.

11x=72+5x

6x = 72x = 12 weeks.

So I can eat it for 12 weeks.

Solution: Assuming that the amount of grass eaten by a cow in a week is in units of 1, then 17 cows eat a total of 6 weeks: 17*6=102

13 cows eat a total of 9 weeks: 13*9=117

During the period from 6 to 9 weeks, the total score of new grass growth was: 117-102 = 15 Weekly grass growth: 15 (9-6) = 5

Then 17 cows eat for 6 weeks, and the new grass grows in total: 5 * 6 = 30 then 17 cows eat for 6 weeks, and the total amount of original grass is: 102-30 = 72 for 11 cows to eat

Assuming that 11 cows eat for 10 weeks, they need to eat 11 * 1 * 10 = 110 parts of grass, and the total amount of grass for 10 weeks is: 72 + 10 * 5 = 122110≠122, assuming that eating for 10 weeks is not valid.

Assuming that 11 cows eat for 11 weeks, they need to eat 11 * 1 * 11 = 121 parts of grass, and the total amount of grass for 11 weeks is: 72 + 11 * 5 = 124121≠124, assuming that eating 11 weeks is not valid.

Assuming that 11 cows eat for 12 weeks, they need to eat 11 * 1 * 12 = 142 parts of grass, and the total amount of grass in 12 weeks is: 72 + 12 * 5 = 132132 = 132, assuming that eating for 12 weeks holds

This is a problem of elementary school cattle eating grass.

The key is to ask for the amount of growth per day.

The same is after eating, the difference between the two before and after: 17 * 6-13 * 9 = 15 servings. It shows that 15 parts of grass grew in three days.

Amount of grass per day: 15 (9-6)=15 3=5 servings of original grass: 17*6-5*6=12*6=72 servings per cow per day:

11 cows can eat:

72 6 = 12 (days).

This is the Newtonian problem (the problem of cows grazing), which is similar to the catch-up problem.

Assuming that the amount of grass eaten per cow per week is 1, then.

1 17 6 = 102, 1 13 9 = 117, 17 cows eat 102 grass for 6 weeks, and 13 cows eat 117 grass for 9 weeks.

From these two data, it can be calculated that in 3 weeks, the grass grew 15, with an average of 5 (117 102) (9 6) per week

5. And then you can calculate the original amount of grass:

or 117 5 9 = 72

The 11 cows eat 11 grass per week, which is 6 faster than grass grows. When the amount of grass eaten per week is 6 and the amount of grass is 72, it is just in time for the newly grown grass to be eaten.

12 (weeks) can feed 11 cows for 12 weeks.

Let's say the cow eats 1 serving of grass per week.

At 6 weeks, the cow ate 17*6 servings of grass (102).

At 9 weeks, the cow ate 13*9 servings of grass (117).

The speed of grass growth: (117-102) (9-6) = 5 (parts week) Therefore, the grassland has 117-5 * 9 = 72 (parts).

11 cows per week is equivalent to eating 11-5=6 (portions).

72 6 = 12 (weeks).

A: Eat for 12 weeks.

The grass grows at a uniform rate, with 17 cows eating for 6 weeks, 13 cows for 9 weeks, and 11 cows for a few weeks, typical of cows grazing. The crux of this kind of problem is that the grass eaten by the cows grows at a uniform rate, so think of it this way, so that the grass that grows new each week is allocated to the cow so that the remaining cows can eat the grass on the grass.

Set up an ox to eat the grass that grows newly. The total amount of grass in this meadow is a(17-a)*6=a to get 17*6-6a=a(13-a)*9=a to get 13*9-9a=a Comparing the two equations, we can see that a=(13*9-17*6) (9-6)=5 cows.

17-5)*6 (11-5)=12 weeks A: 11 cows can be eaten for 12 weeks.

17 cows eat for 6 weeks, indicating that the total edible grass is, 17x6=10213 cows eat 9 weeks, indicating that the total edible grass is, 13x9=117 The difference between the number of weeks they eat is 9-6=3 weeks, indicating that the grass grows 117-102=15 in 3 weeks, and grows 15 3=5 per week

So when 17 cows eat for 6 weeks, the grass that grows is 5x6=30, and the original grass is 102-30=72

Suppose 11 cows can eat x weeks.

72 + 5x = 11x (original grass 72 + grown grass = grass that can be eaten by cows), solution x = 12 weeks.

Available for the week! "17 Ox 6 weeks" to "13 Ox 9 weeks" shows that 4 Ox less has increased by 3 weeks, and "11 Ox" has 2 Ox less than "13 Ox", and the natural number of weeks should naturally be increased by half of 3 weeks on the basis of "13 Ox 9 Weeks", that is: increase the week, so the number of weeks corresponding to 11 Ox is:

9+ weeks! Therefore, 11 cows can eat weekly!

From the title, it can be found that 17 cows eat 6 weeks and 13 cows eat 9 weeks, the total amount is different, the latter has 15 more, because after another 3 weeks, the average growth rate of 15, so the grassland grows 5 per week.

Returning to the first case, 17 cows eat for 6 weeks, the grass grown is 5*6=30, and the grass consumed is 17*6=102, and the original grass in the grassland is 72

At this time, 11 cows eat n weeks, and there is an equation.

11n=5n + 72

n =12 weeks.

Solution: If the grass eaten by 1 cow for 1 week is regarded as 1 serving, then 17 cows eat 17x6=102 (portions) for 6 weeks, and 13 cows eat for 9 weeks.

13x9 = 117 (parts), so we know that the pasture grows grass 117-102 = 15 (parts) in 3 weeks, then the grass grows every week.

15 3 = 5 (parts).

The pasture originally had grass 102-5x6=72 (parts).

5 out of 11 cows eat the newly grown grass and the rest.

11-5=6 (head) cattle eat the original grass and need to eat it.

72 6 = 12 (weeks).

A: 11 cows eat for 12 weeks.

Let this grass grow "1" per week, and 1 cow will eat x per week

Then 6(17-x) = 9(13-x) = total amount of grass (well understood.) The actual reduction in a week is what you eat minus the long one) to solve the equation to get 102-6x=117-9x

3x=15x=5, so a cow eats 5 per week

Bring in the original equation.

6(17-x)=6*12=72

There are a total of 72 grasses.

The substitution of 11 cows can be calculated.