Basic Math Application Questions for Grade 7 and Methods, thank you

I'm good at math, and doing the math problems is very easy for me.

How to solve equation application problems:

1.Review the question and find the equivalent relationship (2).

2.Set the unknown according to an equiquantitative relationship, of course, to find a simple one.

3.Column the equation according to another quantitative relation.

4.Solve equations.

5.Testing, which brings the answer into the equation first and then into the question.

6.Write an answer.

Note: Units, some of the question conditions are kilometers, and the requirements may be meters.

Some problems, such as counting the number of people, cannot be calculated as a decimal or negative number.

Graphical problems: sets of formulas.

Drawing questions: Follow the topic, don't forget the conclusion and the right angle symbol or something.

My grades are still okay, but my method may not be suitable for you, I hope it will help you somewhat.

1.Graphical application problems: set of formulas. It depends on whether you usually memorize it well. Also watch out for pitfalls, such as unit disunity.

One, divide by two, and so on.

2.Equation application questions: follow the topic.

I'll give you six steps: province (find the equivalent relationship, very important), set (generally ask what to set what), column (according to the set of equal relations), solution (careful), test (1.).Check if the equation is correct.

2.Check that the equation sum is in line with the question. Very important.

Answer. 3.Drawing Questions: Be careful, do what you are asked to do, and pay attention to details.

4.Reasoning application questions: Do not let go of any condition, and adjust whether to think backwards or follow the topic according to the question.

5.These methods are all based on the basics, and they are used in normal times and before the exam.

I'm also a seventh-grade student, and my grades are okay, but I don't get off work and go to the top three anyway, hehe.

A good study method is that I just listen carefully in class, complete my homework alone, and review exams carefully. I hope your results are on the top!

It's the same as me, it's good to overcome it yourself, and I wish you good luck in the exam.

1. The analysis method is actually a simulation method, which has a strong intuitiveness and pertinence, and is very commonly used in mathematics teaching. For example, engineering problems, speed problems, deployment problems, etc., are mostly analyzed by drawing pictures, and through **, help students understand the meaning of the problem, so as to set up unknowns and list the solutions of equations according to the content of the problem.

2. Hands-on experience of the Fa is like talking about the problem of sailing against the current and sailing along the current. There are many students who have never been on a boat before, and it is difficult for students to understand the speed of sailing with the current, sailing against the current, and the current. In order for students to understand, for example, riding a bicycle (because most students can ride a bicycle), students have first-hand experience that riding with the wind is easy and against the wind is difficult, which is the effect of wind speed.

At the same time, it is clear that boating and cycling are the same thing, and the different factors that affect one are the water flow speed and the other is the wind speed. In this way, it is easy for students to understand.

3. For intuitive analysis methods such as concentration problems, we must first clarify the meaning of percent concentration, and at the same time explain the calculation method of percentage concentration. Secondly, it is important to prepare a few cups, weigh a certain weight of water, and a few small packets of salt into the classroom before class, so that you can use them for examples.

Profit = Sales Volume ** - Sales Volume Unit ** - Expense Set Sales Volume to a

Suppose the profit of the first type is greater than the profit of the second type.

Equation: Profit (1st type) - Profit (2nd type) = (64a-30a-6000) - (54a-30a)>0

34a-6000)-24a>0

34a-6000-24a>0

10a-6000>0

a>600

Answer, when the sales volume is greater than 600, choose the first way, when the sales volume is 600, the profit of the two sales methods is the same, when the sales volume.

If the number is less than 600, the second method is selected.

The first way is sold by the sales department of our factory.

The second way is to sell wholesale to shopping malls.

Let the profit be y x pieces sold.

Type 1: y=(64-30)x-6000 Type 2: y=(54-30)x

Simplification Type 1: y=30x-6000 Type 2: Y=20x

You can just bring in a few pieces that you have sold, and then compare which one is bigger.

Let x be the number of pieces produced and y be the profit.

y=64x-30x-6000=34x-6000y=54x-30x=24x

When x>600, the store department of the factory makes more money.

When x=600, there are as many of both.

When x<600 Wholesale to others makes more money.

This question is mainly to test your ideas for categorical discussion.

You may want to start by setting the number of pieces to x and the profit to y. Then:

Scheme 1: y=(64-30)x-6000=34x-6000Scheme 2: y=(54-30)x

Subtraction of the two = 10x-6000

It is concluded that when x is greater than 600 pieces, scheme 1 has a large profit; When x is equal to 600 pieces, the profit is equal; When x is less than 600 pieces, option 2 has a large profit.

You can do this in the future, and classification discussion is an important idea in mathematics learning.

Let the profit be equal when x pieces are sold.

64-30）x-6000=（54-30）x x=600

That is, when the number of sales is 600 pieces, the two ways are equal, when the second way is less than 600 pieces, the second way is more profitable, and when the number of sales is greater than 600 pieces, the first way is more profitable.

Set up class A x people, class B y people, then.

8x+10y=920

x+y=515

The solution is x=55 y=48

In order to stimulate domestic demand, Chongqing launched the "household appliances to the countryside" mutual verification car leasing campaign, a household appliance company sold a total of 960 types of refrigerators to farmers in the month before the launch of the activity, and the sales of type 1 and type 2 refrigerators sold to farmers in the first month after the launch of the activity increased by % respectively compared with the month before the launch of the activity, and a total of 1,228 units of these two types of refrigerators were sold.

1. How many Type 1 and Type 2 refrigerators were sold to farmers one month before the launch campaign?

2. If the ** of the 1 type refrigerator is 2200 yuan, and the ** of the 2 type refrigerator is 2000 yuan, according to the relevant policy of "household appliances to the countryside", 13% of each refrigerator will be subsidized to the farmers who buy the refrigerator, and the 1228 1 and 2 refrigerators sold to the farmers in the first month after the start of the activity will be subsidized (the result is that Baofan is leaving 2 significant digits)?

Solution: Solution: Before the launch of the activity, Type 1 sales x units, Type 2 sales Y units.

x+y=960

x(1+30%)+y(1+25%)=1228 to get x=560, y=400

2. After the launch of the activity, Type 1 sales are 560*, and Type 2 sales are 400*** subsidy amount.

728 2200 + 500 2000) 13% = 338208 (yuan) thank you!! Please!!

1. The speed of the two: between 10 o'clock and 12 o'clock, they each walked 36 kilometers, and the speed was 18 kilometers per hour.

Then at 11 o'clock, when the two met, they each walked 18 3 = 54 kilometers.

then AB distance = 54 2 = 108 km.

2. The original price of each pencil is RMB.

6 divided by 50 = yuan divided by (yuan.

3. Set a total length of x kilometers.

1/3x+(1-1/3)x*1/3+20=x

x=45 4, if you have a card, and buy clothes at 20% off:

A spent a total of 200+(1200*<1200 and earned 40 yuan.

B: A total of 200+ (500*>500 lost 100 yuan.

When equal, ** is 1000 yuan. That is, more than 1,000 yuan, it is cost-effective to apply for a card; 1,000 yuan is not cost-effective to apply for a card.

Of course, this refers to spending once with a concession card, and if you can use it consistently, that's another story.

1.Untie; Set AB two places to gather x kilometers.

x
*4-36=x

Solved down x is equal to 108 meters.

Answer. AB The two places meet 108 meters.

1. 108km

2.Yuan. 3. 45km

4.Xiaofang is cost-effective, Xiaomin is not cost-effective, and it is cost-effective to buy goods greater than 1000 yuan.

1. Let the distance between A and B be X kilometers.

x-36）/（10-8）=（36+36）/（12-10）x-36/2=72/2

x-36=36*2

x=108A: The distance between A and B is 108 km.

2. Set: The original price is X yuan.

Each pencil element.

3. Set a total length of x kilometers.

1 3x+(1-1 3)x*1 3+20=xx=454, if you buy x yuan, you can save 200 yuan, and it is cost-effective to buy a card if you buy more than x yuan.

x>=1000

If you buy goods of more than 1,000 yuan, it is cost-effective to buy a card.