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Talking about the problem, many people know the problem tactics, and also know that the problem is to further consolidate the knowledge and expand the knowledge. But why talk about "doing a question" again? Before explaining the reason, then ask you a question, how many times do you generally do it?
Certainly someone said it once, someone said it was done many times, and then someone said it was done three times in general. Very good, so why do you say it is usually done three times? What can I do to achieve the best results? These problems are the problems we have to solve here. Do the problem three times. Did you repeat the task three times? no! Doing it three times means doing it in three different stages. Let's analyze it carefully:
That is to say, you can see that this is an engineering problem. The core formula is "total amount of work = working efficiency x working time", and this problem involves alternating and cooperative completion. The problem is not difficult, but the amount of calculation Slightly bigger.
Therefore, when you do the first pass, you basically understand the characteristics and practices of this alternate and cooperatively completed question type, but the core part may not be summarized enough. In the later problems, I will still encounter other problems of the same type, but the data or background may be different.
When you do this question for the second time, you cannot do it retrospectively, but you should think and understand it from a deeper perspective. This set of questions belongs to the "frog jumping well problem" in the completion of alternating cooperation. When this type of problem is done, it must be distinguished from the general alternating cooperation problem.
Not only must we analyze the minimum work cycle and cycle efficiency, but also pay attention to the issue of reservation, that is, to prevent continued work in the event that the work is finally completed in alternate cooperation, so we must reserve the highest efficiency value in a cycle. Secondly, the special value method can basically be used to do this kind of problem. Usually, the total number of works is the least common multiple of the condition number. Finally, pay attention to how many cycles and how long it takes.
Engineering problems include general engineering problems, cooperative completion problems, and alternate cooperative completion problems. Among them, the commonly used methods for solving problems include equation method, special value method, proportional method, and formula method. Some of these question types are somewhat special, including the question of how many labors are completed and the question of frog jumping from the well. What are the practices and precautions of this type of question? This also requires understanding.
Therefore, if the above three levels of problem solving can be achieved, then we will reach the point to point and then face to face, not only returning to the theoretical knowledge points and methods, but also learn to summarize and summarize, so as to achieve the true purpose of doing the problem.
Students often say:
Mathematics, the difficulty of my public examination;
Information, the barrier of my public examination.
Now with these formulas, mathematics is no longer difficult, and the data is not a bump!
Mathematical reasoning formula
If a: b = m: n (m, n coprime), then a is a multiple of m and b is a multiple of n.
If a = m / n × b, then a = m / (m + n) × (a + b), that is, a + b is a multiple of m + n
(1) The mantissas of the options are different, and the algorithm is addition, subtraction, multiplication, and power operation. The mantissa is used for judgment first;
(2) There is a lot of calculation data. When the calculation is complicated, the mantissa is taken into consideration to get the answer quickly. Commonly used in the principle of exclusion.
And = (first term + last term) × number of terms ÷ 2 = average number × number of terms = median × number of terms;
Number of items = (last item-first item) ÷ number of items + 1. Starting from 1, consecutive n odd numbers are added up, and the total is n × n, such as: 1 + 3 + 5 + 7 = 4 × 4 = 16, ...
Related formulas for geometric edge problems bundle
棵树=总长÷间隔+1，总长=（棵树－1）×间隔 (1) One-sided linear tree planting formula (planting trees at both ends): tree = total length ÷ interval + 1, total length = (tree-1) × interval
在一条路的一侧等距离栽种m棵树，然后要调整为种n棵树，则不需要移动的树木棵树为：（m－1）与（n－1）的最大公约数＋1棵； (2) Tree planting without moving formula: m trees are planted at equal distances on one side of a road, and then adjusted to plant n trees, the trees that do not need to be moved are: (m-1) and (n-1) ) The greatest common divisor + 1;
棵树=总长÷间隔，总长=棵树×间隔 (3) One-sided ring tree planting formula (ring tree planting): tree = total length ÷ interval, total length = tree × interval
棵树=总长÷间隔－1，总长=（棵树+1）×间隔 (4) Tree planting formula in single-sided building (no planting at both ends): tree = total length ÷ interval -1, total length = (tree +1) × interval
最外层总人数＝4×（N－1），相邻两层人数相差8人，n阶方阵的总人数为n²。 (5) The square matrix problem: the total number of people in the outermost layer = 4 × (N-1), the number of people in the adjacent two layers is 8 people different, and the total number of people in the n-order square matrix is n².
路程＝桥长＋车长（火车过桥过的不是桥，而是桥长＋车长） (1) The core formula for train crossing the bridge: distance = bridge length + vehicle length (not the bridge, but the bridge length + vehicle length)
相遇距离＝（速度1＋速度2）×相遇时间追及距离＝（速度1－速度2）×追及时间 (2) Encounter and follow-up problem formula: Encounter distance = (speed 1 + speed 2) x encounter time follow-up distance = (speed 1-speed 2) x follow-up time
队首→队尾：队伍长度=（人速＋队伍速度）×时间；队尾→队首：队伍长度=（人速－队伍速度）×时间 (3) The formula of the team's progress problem: team head → team tail: team length = (person speed + team speed) × time; team tail → team head: team length = (person speed-team speed) × time
顺速＝船速＋水速，逆速＝船速－水速 (4) Flowing ship problem formula: forward speed = ship speed + water speed, reverse speed = ship speed-water speed
(5) Round-trip encounter problem formula:
（第一次相遇距离A为S1，第二次相遇距离B为S2） The two sides meet twice: S = 3S1-S2, (the distance A for the first encounter is S1, and the distance B for the second encounter is S2)
Single shore type meets twice: S = (3S1 + S2) / 2, (the first encounter distance A is S1, the second encounter distance A is S2);
Departure from left and right points: Nth head-to-head encounter, distance sum = (2N-1) × full journey; Nth time to catch up with encounter, distance difference = (2N-1) × full journey.
Departure at the same point: the Nth head-on encounter, the distance sum = 2N × the whole journey; the Nth time after the encounter, the distance difference = 2N × the whole journey.
(1) The formula of the relationship between the three sides of a triangle:
The sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side.
(2) Pythagorean theorem:
In a right triangle, the sum of the squares of the two right sides is equal to the square of the hypotenuse. Commonly used stocks: (3, 4, 5); (5, 12, 13); (6, 8, 10).
(3) Internal angle and theorem
Regular polygon internal angle and theorem, the sum of internal angles of n polygons is equal to: (n-2) × 180 ° (n is greater than or equal to 3 and n is an integer).
Given the number of angles within a regular polygon, the number of sides is: 360 ° ÷ (180 ° －number of internal angles).
(1) Common formulas for economic profit
Profit = sale price-purchase price, profit rate = profit ÷ purchase price, total profit = single profit × sales price = purchase price + profit = original price × discount
(2) Basic formula of solution problem
Solution = Solute + Solvent, Concentration = Solute ÷ Solution, Solute = Solution × Concentration Mixed Solution Concentration = (Solution 1 + Solute 2) ÷ (Solution 1 + Solution 2)
Base period volume correlation
Master these formulas,
Your test is no longer afraid!
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