Knowing the sum and difference of two numbers, the application problem of finding these two numbers is called the sum and difference problem. The general relationship is:

(sum and difference) ÷ 2 = smaller number

(sum+difference) ÷ 2 = larger number.

Example: The sum of two numbers, A and B, is 24, and A is smaller than B by 4. What are the numbers a and b?

(24+4)÷2

=28÷2

= 14 → B number

(24-4)÷2

=20÷2

= 10 → Numbers

A: The number A is 10 and the number B is 14.

Difference problem

Given the difference between two numbers and their multiples, the application problem of finding these two numbers is called the difference multiple problem. The basic relationship is:

Difference between two numbers ÷ multiple difference = smaller number

There are two piles of coal, and the second pile is 40 tons more than the first. If you take 5 tons of coal from the second pile to the first pile, the weight of the second pile is exactly three times that of the first pile. How many tons are there in each of the two piles of coal?

Analysis: It turns out that the second pile of coal is 40 tons more than the first pile. After giving the first pile 5 tons, the second pile of coal is only 40-5× 2 tons more than the first pile. The basic relationship is as follows:

(40-5×2)÷(3- 1)-5

=(40- 10)÷2-5

=30÷2-5

= 15-5

= 10 (ton) → weight of the first pile of coal

10+40 = 50 (ton) → the weight of the second coal pile

A: The first coal pile has 10 tons, and the second coal pile has 50 tons.

Reduction problem

The problem of finding the original unknown after knowing the result of some changes in a number is generally called reduction problem.

The reduction problem is the inverse solution of the application problem. Generally speaking, according to the relationship between reciprocal operations of addition, subtraction, multiplication and division. Start with the order of topic description, think in reverse order, start with the last known condition, and get the result in reverse.

Example: There is some rice in the warehouse, and the weight sold on the first day is 12 ton, which is less than half of the total. The weight sold the next day was less than the remaining half 12 tons, leaving 19 tons. How many tons of rice are there in this warehouse?

Analysis: If the remaining half is just sold the next day, it should be 19+ 12 tons. After being sold on the first day, the remaining tonnage is (19+ 12) × 2 tons. The following analogy.

Formula: [(19+12) × 2-12 ]× 2.

=[3 1×2- 12]×2

=[62- 12]×2

=50×2

= 100 (ton)

This warehouse used to have 65,438,000 tons of rice.

Replacement problem

There are two unknowns in the problem, and one of them is often regarded as the other for the time being, and then the hypothetical operation is carried out according to the known conditions. The results are often inconsistent with the conditions, and then appropriate adjustments are made to get the results.

Example: A stamp collector bought stamps of 10 and 20 cents *** 100, with a total value of180 cents in 8 yuan. How many stamps did this stamp collector buy?

Analysis: Assuming that every 100 stamp bought is 20 cents, the total value should be 20× 100 = 2000 (minutes), which is 2000- 1880 = 120 (minutes) more than the original total value. And this extra 120 points means that every point of 10 is regarded as 20 points, and every point is 20- 10 = 10 (points), so how many points can you get from 10 points?

Formula: (2000-1880) ÷ (20-10)

= 120÷ 10

= 12 (sheets) → 10 Number of sheets per sheet.

100- 12 = 88 (sheets) → 20 minutes for each sheet.

Or first find the number of sheets with 20 points, and then find the number of sheets with 10 points. The method is as above. Please note that the total value is less than the original total value.

Profit and loss problem (insufficient profit)

There are often two distribution schemes in the topic, and the result of each distribution scheme will be more (surplus) or less (deficit). This kind of problem is usually called profit and loss problem (also called the problem of insufficient surplus).

To solve this kind of problem, we should first compare the two distribution schemes, find out the change of remainder caused by the change of each share, find out the total number of shares participating in the distribution, and then find out the number of items to be distributed according to the meaning of the question. Its calculation method is:

When one time is left and the other time is insufficient:

Per share = (remainder+deficiency) ÷ twice the difference per share.

When there are two remainders:

Total number of copies = (larger remainder-smaller number) ÷ twice the difference per copy.

When neither time is enough:

Total number of copies = (large shortage-small shortage) ÷ twice the difference per copy.

Example 1. A certain class of a PLA unit participated in afforestation activities. If each person plants 5 seedlings, there are still 14 seedlings left; If everyone plants seven trees, there will be a shortage of four seedlings. How many people are there in this class? A * *, how many seedlings?

Analysis: according to the conditions, this question belongs to the first case.

Formula: (14+4) ÷ (7-5)

= 18÷2

= 9 (person)

5×9+ 14

=45+ 14

= 59 (tree)

Or: 7× 9-4

=63-4

= 59 (tree)

A: There are 9 people in this class, and each class has 59 seedlings.

Age problem

The main feature of the age problem is that the age difference between two people remains the same, but the multiple difference has changed.

The commonly used calculation formula is:

Age times = age difference ÷ (multiple-1)

Age a few years ago = small gift-multiplied by small age

Age in a few years = times its age-its age when it was very young now.

Example 1. Father is 54 years old and son 12 years old. A few years later, the father was four times as old as his son.

(54- 12)÷(4- 1)

=42÷3

= 14 (years old) → the age of the son in a few years.

14- 12 = 2 (year) →2 years later

Two years later, the father was four times older than his son.

Example 2: My father is 54 years old and my son is 12 years old this year. A few years ago, the father was seven times as old as his son.

(54- 12)÷(7- 1)

=42÷6

= 7 (years old) → the age of my son a few years ago

12-7 = 5 (years) →5 years ago

A: Five years ago, my father was seven times older than my son.

Ex. 3: The total age of Wang Gang's parents this year is 148 years old, and the difference between his father's age and his mother's age is more than three times that of his age and four years old. How old are Wang Gang's parents this year?

( 148×2+4)÷(3+ 1)

=300÷4

= 75 (years old) → father's age

148-75 = 73 (years old) → mother's age

A: Wang Gang's father is 75 years old and his mother is 73 years old.

Or: (148+2) ÷ 2

= 150÷2

= 75 years old

75-2 = 73 years old

Chicken and rabbit problem

Knowing the total number of chickens and rabbits and finding the number of chickens and rabbits, there is a kind of application problem called chicken and rabbit problem, which is also called "turtle crane problem" and "replacement problem"

It is generally assumed that they are all chickens (or rabbits), and then rabbits (or chickens) are used instead of chickens (or rabbits). Commonly used basic formulas are:

(Total number of feet-number of chicken feet × total number of chickens) The difference between the number of feet of each chicken and rabbit = number of rabbits.

(The number of rabbits × the total number of rabbits-the total number of rabbits) ÷ The difference between the number of feet of each chicken and rabbit = the number of chickens.

Example: There are 24 chickens and rabbits in the same cage. There are 64 legs. How many chickens and rabbits are there in the cage?

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(64-2×24)÷(4-2)

=(64-48)÷(4-2)

= 16 ÷2

= 8 (only) → Number of rabbits

24-8 = 16 (only) → number of chickens

Answer: There are 8 rabbits and 16 chickens in the cage.

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The problem of cattle eating grass (the problem of ship leaking)

Several cows are grazing on a limited meadow. Cattle eat grass and grass grows on grass. When increasing (or decreasing) the number of cattle, how long will it take before the grass on this grassland has just been eaten?

Example 1. A piece of grassland can feed 15 cows 10 days and feed 25 cows for 5 days. If the grass grows at the same speed every day, how many days can this grass feed 10 cows?

Analysis: Generally, the daily grazing amount of 1 cow is regarded as one share, so 15 cows ate 10 days, including the original grass on the grassland, and the grass grew on this grassland for 10 days, and so on ... Among them, it can be found that the grazing amount of 25 cows for 5 days is/kloc-. The reason is because, first, it takes less time; Second, the corresponding grass grows less. The difference is that grass grows on this grassland for 5 days. The grass that grows every day can feed five Niu Yi days. In this way, when feeding 10 cows, 5 cows are taken out every day to eat the long grass, and the rest cows eat the original grass on the grassland.

( 15× 10-25×5)÷( 10-5)

=( 150- 125)÷( 10-5)

=25÷5

= 5 (head) → You can feed 5 heads in Niu Yi for one day.

150- 10×5

= 150-50

= 100 (head) → The original grass on the grassland can feed 100 cows a day.

100÷( 10-5)

= 100÷5

= 20 days

Answer: If you feed 10 cows, you can eat them for 20 days.

Example 2: A well gushes water upward at a uniform speed, and four pumps can discharge water for 100 minutes; If you use six identical pumps, it can be emptied in 50 minutes. Now, with seven identical pumps, how many minutes can you pump the water out of this well?

( 100×4-50×6)÷( 100-50)

=(400-300)÷( 100-50)

= 100÷50

=2

400- 100×2

=400-200

=200

200÷(7-2)

=200÷5

= 40 minutes

With seven identical pumps, the water in this well can be drained in 40 minutes.

Common divisor and common multiple problem

Solving application problems with the greatest common divisor or the least common multiple is called common divisor and common multiple problem.

Example: 1: a rectangular piece of wood, 2.5m long, 1.75m wide and 0.75m thick. If this piece of wood is sawed into cubes of the same size, there is no surplus, and each piece is as large as possible, then what is the side length of the cube? * * * How many pieces did you saw?

Analysis: 2.5 = 250 cm

1.75 =175cm

0.75 = 75 cm

The greatest common divisor of 250, 175 and 75 is 25, so the side length of the cube is 25 cm.

(250÷25)×( 175÷25)×(75÷25)

= 10×7×3

= 2 10 (block)

Answer: The side length of the cube is 25 cm, and the * * saw is 2 10.

Example 2. Two meshing gears, one with 24 teeth and the other with 40 teeth, how many turns does each gear make from the first contact to the second contact?

Analysis: Because the least common multiple of 24 and 40 is 120, that is, when both gears rotate 120 teeth, a pair of teeth that contact for the first time just contact for the second time.

120 ÷ 24 = 5 (weeks)

120 ÷ 40 = 3 (weeks)

A: Each gear must rotate 5 and 3 times respectively.

Fraction application problem

Refers to the application problem solved by fractional calculation, which is called fractional application problem, also called fractional problem.

Fraction application problems are generally divided into three categories:

1. Find the fraction of one number to another.

2. Find the fraction of a number.

3. Know what the score of a number is, and find out this number.

Among them, each category is divided into two types, one is the general score application problem; Second, the application of scores is more complicated.

Example 1: Yucai Primary School has students 1000, including 250 students with three good qualities. What percentage of the students in the whole school are "Three Good Students"?

A: Miyoshi students account for the whole school.

Example 2: Transporting a pile of 180 tons of coal. How many tons did you walk?

180× = 80 (ton)

A: 80 tons were shipped.

Example 3: An agricultural machinery factory produced 65,438+0,800 agricultural machinery last year, and plans to increase this year compared with last year. How many units are planned to be produced this year?

1800×( 1+)

= 1800×

= 2400 (unit)

A: It is planned to produce 2,400 sets this year.

Example 4: Build a 2400-meter-long expressway. On the first day, complete the whole length, and on the second day, complete the rest. How many meters are left?

2400×( 1-)×( 1-)

=2400××

= 1200 (m)

A: There is 1200m left.

Example 5: There are 168 students in a school, accounting for% of the total number of students in the school. How many students are there in the school?

168 ÷ = 840 (person)

There are 840 students in this school.

Example 6: A's stock of grain 120 tons is less than B's stock. B How many tons of grain are in stock?

120 ÷ =120× =180 (ton)

Answer: B has grain in stock 180 tons.

Exodus 7: A pile of coal was completely shipped the first time, and the second time, which was 8 tons less than the first time. How many tons was this pile of coal?

8÷(-)

= 8÷

= 48 (ton)

A: This pile of coal used to be 48 tons.

Engineering problems

It is a special case of fractional application. It is a problem to find the third quantity from two of the three quantities when the workload, working time and working efficiency are known.

When solving engineering problems, we should generally regard all projects as "1", and then answer them according to the following quantitative relationship:

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Work efficiency × working hours = workload

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Workload ÷ working time = working efficiency

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Workload ÷ work efficiency = working hours

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Example 1: It takes 18 days for Team A to do a project alone, and 24 days for Team B to do it alone. If two teams work together for eight days, how many days will it take for the remaining projects to be completed by team A alone?

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= 4 (days)

A: (omitted).

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Example 2: A pool is equipped with two water inlet pipes A and B and one water outlet pipe. Single-port nail tube can be filled in 2 hours; Single-port B tube can be filled in 3 hours; Only open the outlet pipe for 6 hours. Now that the pool is empty and three pipes are opened together, how many hours can the pool be filled?

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= 1 (hour)

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Percentage application problem

The solution of this kind of application problem is basically the same as that of fractional application problem. When you only ask for "rate", the expression is different and the meaning is different.

Example 1. An agricultural research institute conducted a germination test and planted 250 seeds. 230 seeds germinated. Find the germination rate.

A: The germination rate is 92%.