Usually called the typical application problems.
(1) average number of questions: average number of division is the attainment of development.
solving key: is to determine the total number and the corresponding total number of copies.
arithmetic mean: a few are not the same kind of known volume and corresponding number of copies, find the average of each is. Number of relationship: the number of quantities and the number of ÷ = arithmetic mean.
weighted average: two or more known to the average number of copies, find the overall average is.
number of relations (part of the average × weight) of the total ÷ (weight and) = weighted average.
mean difference: It is all more or less than the standard number of parts and the total number of copies to be shared equally, find the standard number and the number of and the average difference.
number of relations: (large numbers - decimal) ÷ 2 = maximum number of decimal places to be thanks to a number of differences with the number and the maximum number ÷ total number of copies = number should be given the maximum number and the number of poor and the total number of copies = minimum number of ÷ to be counts.
Example: a car speed of 100 km per hour from point A to point B bound on, he speed of 60 kilometers per hour from A to B to bound. Find the average speed of the car.
analysis: find the average speed of vehicles can use the same formula. This title can be A to B to the distance set to ground speed of 60 km, the time spent is a total trip time of cars + =, the average speed of vehicle 2 ÷ = 75 (km)
(2) normalization issues: Known each other associated with two quantities, one of which the amount of change, another volume will be changed,cheap UGG boots, the change in the law is the same, this problem is called normalization problem.
according to strive for
crazy single volume under the ball, the problem-solving using multiplication or division, the problem can be divided into the normalized normalization problem is, anti-normalization problem.
a normalization problem,UGGs, one step operation can be calculated using a Also known as the Also known as
anti-normalization issues: Equal division obtained with a
solving the key: from a known amount of a corresponding division obtained using a number of equal portions (one volume), and then use it as the standard, according to the requirements of Title calculated results.
number of relationship: a single volume × number of copies = total number of (positive normalized)
single volume = ÷ total number of shares (anti-normalization)
cases a weavers, weaving 4774 m in July, according to this calculation, weaving 6930 meters, how many days?
analysis: you must first calculate how many meters of cloth per day, that is, a single volume. 693 0 ÷ (477 4 ÷ 31) = 45 (days)
(3) and belongs question: is a known quantity the number of units and the number of units of measurement, and the different number of units (or the number of units number), obtained by calculating the total number of number of number of units (or number of units).
features: two related to the amount of change in one volume, along with other changes in volume, but the variation of contrast, and inverse algorithms belong.
number of relationship: the number of units to another unit × the number of units of the number of ÷ the number of units = the number of units to another unit number × ÷ number of units = one unit to another number.
cases repair a ditch, originally planned to repair 800 meters per day, 6 days completing. 4 days completing the actual day, the number of meters repaired?
analysis: because the required length of the daily repair, you must first find the length of canals. Applied to such problems so it is called The difference is 80 0 × 6 ÷ 4 = 1200 (m)
(4) and poor problem: given the size of two numbers, and their differences, find the two numbers is the number of application questions is called and differential problems.
key problem solving: is the size of the two numbers into two large numbers and (or two decimal places and), and then find another number.
problem solving rules: (and + poor) ÷ 2 = Large Number of Large Numbers - SD = decimal
(and - worse) ÷ 2 = decimal and - decimal = large numbers
cases a processing plant Jiaban and Yiban total of 94 workers, because of work temporarily transferred from 46 to Jiaban Yiban work, then Yiban 12 Bijia the small number of classes, find the original Jiaban and B How many people have their own classes?
analysis: 46 from Yi Ban transferred to Jiaban, no change in the total, and now the number of B into two Yiban that 94--12, which is for now and Yi Ban (9 4 - 12) ÷ 2 = 41 (people), Yi Ban before the recall should be 46 to 41 +46 = 87 (people), Jiaban to 94--87 = 7 (person)
( 5) and fold problem: given two numbers and their multiple relationships between, find the two numbers is the number of application questions, call and fold problem.
critical problem solving: identify the standard number (ie 1 times) Generally speaking, the title says is Find multiples and after, and then find the standard number is. According to another number (the number may be several) and the standard relationship between the number of times, go find another number (or some number) number.
problem solving rule: and ÷ and = standard number of times the standard number × multiplier = number of
another example: the size of motor transport field has 115 trucks, big trucks small truck than 5 times more than seven, transport farm trucks and cars have a large number of vehicles?
Analysis: big trucks small truck than 5 times more than 7, which seven are within the total number of 115, in order to bring the total of (5 +1) times the corresponding total number of vehicles should be (115 - 7) vehicles.
formulation for the (115-7) ÷ (5 +1) = 18 (units), 18 × 5 +7 = 97 (units)
(6) times the problem of difference: is know the difference between two numbers, and the relationship between the two multiple of the number, find two numbers is the number of the application questions.
problem solving rule: the difference between two numbers ÷ (factor -1) = standard number of standard digital × multiplier = another number.
patients with B two ropes, a rope 63 meters long, 29 meters long rope B,UGG boots, two rope cut the same length, the result is the length of a remaining length of the rope 3 times B, and B the two ropes The remaining length of the number of meters? The less the number of meters?
Analysis: The same two cut a rope, the length difference has not changed, a length of rope remaining rope B 3 times more real than the B string (3-1) times, the B string The length of the standard number. Formulation (63-29) ÷ (3-1) = 17 (m) ... the rest of the length of rope B, 17 × 3 = 51 (m) ... the rest of the length of a rope, 29-17 = 12 (m) ... cut to length.
(7) travel question: on the walk, traffic and other issues, are generally calculate distance, time, speed, called the stroke problem. Answer these questions first to find out the speed, time, distance, direction, speed, and Du, the concept of speed and poor, to understand the relationship between them, then answer according to the laws of such issues. The key problem solving and rule
:
the same time and contrary to the line: distance = speed × time.
also meet each other halfway: meeting time = speed × time
the same time and to the line (the first slow, fast in the post): tracking and time = distance speed difference.
the same time and in the same direction and the line (in the slow, the fast first): distance = velocity × time difference.
cases the back of a 28 km in B, both at the same time in the same direction and the line, a line of 16 kilometers per hour, B line 9 kilometers per hour, a few hours to catch up with B?
Analysis: A more per hour than the B line (16-9) km, which is a nearly an hour to catch up with B (16-9) km, which is the speed difference. B
known in the back of a 28 km (chasing away), 28 contains a number of one thousand meters (16-9) km, that is, the time required to pursue. Formulation 2 8 ÷ (16-9) = 4 (h)
(8) water issues: research vessel is generally in the It is the journey of the more specific issue of a type, it is also a problem and the poor. It is characterized mainly on account of the water velocity in the retrograde and anterograde different role.
speed of the ship: sail boat speed in still water.
water velocity: the speed of water flow.
downstream speed: the speed of the ship sailing downstream.
against the current speed: the speed of the ship sailing upstream.
cis-speed = speed of the ship speed
+ water = boat speed against the speed - the key to solving water speed
: Because speed is the speed of the ship and the water downstream speed and , counter-current speed boat speed and water speed difference, so the water problem and the poor as questions. Problem solving as a clue when to water.
problem solving rules: ship speed = (speed + upstream downstream speed) ÷ 2
flow rate = (downstream speed counter-current speed) ÷ 2
downstream distance = the time required for downstream navigation speed × distance =
reflux reflux velocity × time navigation of a ship
cases of B from A bound for the line down the river every hour for 28 thousand m to B after he sailed against the current back to a ground. Downstream more than the line against the current 2 hours, 4 km per hour water velocity is known. Find how many kilometers away from the two B?
Analysis: This question must first know the downstream speed and the time required for downstream, or against the current speed, and against the current time. Known downstream speed and flow speed, so not difficult to calculate the speed against the current, but the time spent at little cost, the time spent against the current do not know, only know little cost less than 2 hours against the current, to seize on this to be able to calculate the downstream from A to B in the time spent so that we can calculate the distance between B. Formulation for the 284 × 2 = 20 (one thousand meters) 2 0 × 2 = 40 (one thousand meters) 40 ÷ (4 × 2) = 5 (h) 28 × 5 = 140 (km).
(9) Restore problem: given an unknown, after a certain amount of arithmetic the result after, find the unknown word problems, we called the reduction of the problem.
solving the key: to understand every step of the relationship between change and unknown.
problem solving rule: starting from the final results, use contrary to the original question in the operation (inverse) method, and gradually the original number is derived.
computation based on the original title listed in order of number of relationships, and then calculated using inverse operations derive the original number.
answer the question to observe the operation to restore order. If you need to count addition and subtraction, multiplication and division after the operator do not forget to write when the brackets.
cases of the four third grade classes in a total of 168 students who transferred if the four classes to three classes of 3, 6 transferred to the second class three shifts, two classes were transferred to a group of 6, a group of transfer 2 to four classes, then an equal number of four classes, four classes of students how many of the original?
analysis: the time when an equal number of four classes, should be 168 ÷ 4, the four classes, for example, adjusting to the three classes of 3, 2 transferred to and from class, so the original four classes minus the number of plus 2 equals 3, average. Four classes of the original formula for the number of columns 168 ÷ 4-2 +3 = 43 (people)
number of formulations for the original group of 168 ÷ 4-6 +2 = 38 (people); second class the original number of columns formula 168 ÷ 4-6 +6 = 42 (people) the number of three shifts of the original formulation for the 168 ÷ 4-3 +6 = 45 (people).
(10) planting questions: This type of application questions is Students of the total distance, spacing, sections, the number of trees between the application of the four questions, called tree-planting problem.
solving key: answer planting questions we must first determine the terrain, to distinguish whether the closed figure to determine a section of tree planting along the perimeter or along the tree-planting, then the basic formula.
problem solving rule:
tree planting along the section of the number of segments = tree = total distance ÷ +1 +1
spacing spacing = total distance ÷ (tree -1 ) total distance = spacing × (tree -1)
tree planting along the perimeter of the total distance ÷ =
spacing spacing = distance ÷ total tree
total distance = spacing × tree
cases buried along the road side poles 301, each of two adjacent spacing is 50 meters. Later all the modifications, only buried 201. Request modified the spacing of each adjacent two.
analysis: the problem is buried along the section of wire rod, take the root of the number of poles minus one. Column-type is 50 × (301-1) ÷ (201-1) = 75 (m)
(11) profit and loss issues: the division in equal portions on the basis of developed. His characteristic is to a certain number of items, the average allocation to a certain number of people in the two distribution, one more than, less than once (or twice have I), or twice is enough), and deficiencies in the remainder of the known The number, find the number of items assigned amount and to participate in the issue, called the profit and loss issues.
key problem-solving: Problem profit and loss point is to find two copies of the distribution of income distribution are not the difference between the number of items, and then find the distribution of the sub-divided into two items the difference (also known as the difference between the total ), a difference with the former after the removal of a difference to the number of those who receive a distribution, which can then obtain the number of items.
problem solving rule: the total difference = difference between the number per ÷
method for finding the difference between the total can be divided into the following four conditions:
first redundant, less than a second The total difference = extra +
for the first time just less than a second redundant or insufficient, excess or shortage of the total difference =
first redundant, the second is also redundant, with a total difference = big excess - excess
little less than the first, the second is also inadequate, with a total lack of difference = large - less than
cases of small groups of students to participate in art, everyone points the same number of support color pen, if the group of 10 people, many of 25, if the group of 12 people, extra five colored pens. Find a few pieces per share? The total number of branch color pencils?
analysis: color pen for each student assigned to the same. This activity group of 12 people, 2 more than 10 people, while more of the color pen (25-5) = 20, 2 more than the 20 individuals, one share of 10. Formulation for the (25-5) ÷ (12-10) = 10 (support) 10 × 12 +5 = 125 (branch).
(12) age: the difference is a constant theme in the two numbers as a condition of this application title is called
solving key: age and and the bad, and times,UGG shoes, bad times the problem is similar to the main feature is the change over time, increasing age, but the size difference between the two different age groups will not change and, therefore, age is a
patients 48 years old father, son, 21 years old. Q. A few years ago the son of his father's age is 4 times?
analysis: the age difference between father and son is 48-21 = 27 (years old). Since a few years ago my father was the son of 4 times the age, we can see a multiple of the age difference between father and son is (4-1) times. This can work out a few years ago his son's age, which can be calculated a few years ago the son of his father's age is 4 times.
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