Engineering problem: working efficiency × working time = workload
Product matching problem: the total amount of processing is proportional
Speed problem: This kind of problem can be divided into two categories: underwater speed and wind speed.
Downstream (wind): speed = speed in still water (no wind)+speed of water (wind).
Countercurrent (wind): speed = speed in still water (no wind)-speed of water (wind)
Profit problem: profit = selling price-buying price, profit rate = (selling price-buying price) ÷ buying price × 100%.
There is no need to take the senior high school entrance examination. It's not that hard.
1. Engineering problems
2. Support issues
3. Economic problems
4. Meeting issues
5. Water velocity problem
6. Travel issues
1. current price divided by original price = how much discount?
2. The principal deposited in the bank is multiplied by the annual interest rate, and then multiplied by time = interest due.
3. Earned interest-Earned interest multiplied by tax rate = Earned interest
4. Speed times time = distance
5. Work efficiency times time = total workload.
6. the problem of chickens and rabbits in the same cage: the total difference divided by the number of copies = one of them, the total-one of them = the other.
7. Tree planting problem: head-to-tail planting: the total length divided by the interval length between each tree, plus 1, = a * * * number of trees.
Not planting one head: the total length divided by the interval length between each tree = the number of trees.
Do not plant the head and tail: divide the total length by the interval length between each tree, and then subtract 1, =1* * the number of plants.
8. Wood sawing problem: the number of saws-1 (the number of saws) times the time taken to saw a section = a * * * time.
9. The number of surviving plants divided by the total number of plants = survival rate
10. Qualified quantity divided by total quantity = qualified rate
1 1. Work efficiency of engineering problems × time = total workload.
Textbooks explain binary linear equations as follows:
An equation group consisting of two linear equations with two unknowns is called a binary linear equation group.
The key here (1) is: two unknowns.
The key (2) is a set of equations composed of two linear equations.
Seeing this, you may think you understand, please continue to analyze it carefully.
The key in key (2) is two linear equations.
Let's see what he looks like in the title. As follows:
When goods with a unit price of 40 yuan are sold in 50 yuan, 500 units can be sold. It is understood that every time the price of this commodity increases by 65,438+0 yuan, its sales volume will decrease by 65,438+00 units. In order to earn a profit of 8000 yuan, what should the price be?
Analysis: the key to the topic is to determine the goal of earning 8,000 yuan in profits, so what is the relationship between profits?
Profit = gross profit of each commodity × sales volume
Gross profit of each commodity = selling price-unit purchase price
The price of each commodity here is X (the topic is important) and the sales volume is Y (two unknowns).
Answer: Let the selling price be X and the buying price be Y..
Single product profit: X-40
The change of sales volume is (quantity of existing commodities): y = 500-(x-50) ÷1×10 (the first linear equation).
(X-50) Rising prices
1 is 1 yuan here, which can be 2 yuan and 3 yuan according to the change of the problem.
10 is the sales volume that decreases with the price increase of 1 yuan.
Then the profit is: (X-40)×Y=8000 (the second linear equation).
This is the binary linear equation in the exam! !
Then, in your textbook, you said that the general solution of binary linear equations is:
Substitution elimination method, (commonly used)
Addition, subtraction and elimination, (commonly used)
Sequential exclusion method, (this method is not commonly used)
I think you can do this problem.
After learning these, look at the following questions.
A shopping mall sells a certain commodity, and several pieces are sold in April, with a gross profit of 30,000 yuan (gross profit of each commodity = sales price of each commodity-cost price of each commodity). In May, with the same cost and price, the mall reduced the sales price of this commodity by 4 yuan, but the sales volume increased by 500 pieces compared with April, so the gross profit increased by 2,000 yuan compared with April. What is the gross profit of each commodity sold before the price adjustment?
Maybe you won't do this topic again, please see the following analysis:
The key to this kind of problem is:
The first is the change of single items.
Assuming that A is the purchase price and B is the selling price, the profit of a commodity is B-A..
(Gross profit of each commodity = sales price of each commodity-cost price of each commodity)
The second is the change in sales volume.
The influence of sales volume on price Suppose that the sales volume is C before the price increase and D after the price increase,
Then the difference in sales volume is C-D.
Third total profit
Total profit = gross profit of each commodity × sales volume
Grasp these three items, and then look at what you have already told, what you don't know and what you don't know in the topic. Set it as X and Y.
Let the gross profit in April be X and the sales volume in April be Y.
The profit of single product in May is: X-4.
The sales volume in May is: Y+500.
The total profit in April is: X Y=30000 (linear equation).
The total profit in May is: (x-4) (y+500) = 32,000 (second linear equation).
The above is the study and understanding of binary linear equation, grasp the key sentences, followed by the solution of a kind of problem, from which you will find and understand that you will do it no matter how the problem changes! ! !
Are you paralyzed? Is this your MA problem in senior one?
1. Workload = working efficiency * working time = total workload of all parts = 1.
2.
3. List price * discount/10- cost = profit rate * cost
4. the distance of a+the distance of b = the total distance.
5. Downwind speed = still water speed+current speed = downstream speed-current speed = countercurrent speed+current speed.
Countercurrent velocity = still water velocity-water velocity-water velocity = counter current velocity+water velocity.
6. Distance = speed * time