( 1250-20*50)/ 10=25
2) Xiaoming bought five pencils and four exercise books in the shop and paid 2.50 yuan. It is known that each pencil is 0. 12 yuan. How much is each exercise book?
(2.5-0. 12*5)/4=4.75
3) The fourth-grade students in Dongfeng Primary School 125, and the fifth-grade students are six times less than the fourth grade. How many students are there in Grade 4 and Grade 5?
125+ 125*2-6=369
4) The cement plant will transport 42 tons of cement, 2.5 tons per day. After 6 days, 3 tons will be shipped in the remaining day. How many days does it take to transport the rest?
(42-2.5*6)/3=9
5) A factory wants to manufacture a batch of machine tools, and it is planned to produce 64 machines every day, which can be completed in 15 days. In fact, the task was completed three days ahead of schedule. How many more machine tools are actually produced than planned every day?
64* 15/ 12-64= 16
6) A coal mine originally planned to produce 660,000 tons of coal in the first half of the year, but actually produced 22,000 tons more per month than originally planned. According to this calculation, how many months will it take to complete the production plan for the first half of the year?
66/(66/6+2.2)=5
7) The garment factory used to make a 2.5-meter garment cloth. After adopting the new cutting method, each suit can save 0. 1 meter. It can be used as the original cloth for 120 suits. How many suits can it make now?
120*2.5/2.4= 125
8) Hongshan Primary School15Kg collected tree species180kg, and completed the task five days ahead of schedule. How many kilograms more trees are actually collected every day than originally planned?
180/ 15- 180/20=3
The binding team bound a batch of books. 750 books were bound on the first day and 800 books were bound on the second day. The next day was 20.5 yuan more than the first day. How much is the binding fee for the first day?
20.5/(800-750)*750=307.5
10) The price of kerosene per kilogram is 3.5 yuan, and a barrel of kerosene weighs 8 kilograms. Half of the kerosene was sold, and the remaining barrel of kerosene weighed 4.5 kilograms. How much can this barrel of kerosene sell for?
(8-4.5)*2*3.5=24.5
1 1) The steel output of steel mills in the first half of last year was 360,000 tons, which was108,000 tons more than that in the first half of last year. How many tens of thousands of tons of steel were produced on average every month last year?
(36+36+ 10.8)/ 12=6.9
12) A chemical fertilizer factory produces 97 tons of chemical fertilizer in the first four days of a week and 26 tons of chemical fertilizer every day in the last three days. How many tons of chemical fertilizer does it produce every day this week?
(97+26*3)/7=25
Find out the solution to the problem by determining the corresponding relationship.
The application of scores has an obvious "quantity-rate correspondence" feature. For a unit "1", each fraction corresponds to a specific quantity, and each specific quantity also corresponds to a fraction. Therefore, it is the key to correctly determine the "quantity-price correspondence". It is necessary to guide students to learn and master the problem-solving method of "clear correspondence and accurate corresponding score".
Example: Xiaodong read a story book. On the first day, he read 1/6 of the total number of pages. On the second day, he read 1/3 of the total number of pages, and there were 78 pages left unread. How many pages does this story book have?
Take the total number of pages in this story book as "1", find how many pages there are in this story book, and find the corresponding score of the remaining 78 pages. According to the known conditions, I read (1/6+ 1/3) of the total pages of Grade One and Grade Two, and the corresponding score of the remaining 78 pages is (1- 1/6- 1/3), so I want this story book * *. So the formula is:
78 (1-1/6-1/3) =156 (page)
Second, find out the problem-solving method by unifying the standard quantity
In a fractional application problem, if there are several fractions, and the standard quantities of these fractions are different, the properties of the quantities are also different, then when solving the problem, we must take a certain quantity in the problem as the standard quantity, and unify the scores corresponding to its remainder to this standard quantity, so as to solve the problem in a tabular way.
Example: There are ***420 apple trees and pear trees in the orchard, and 1/3 of the apple trees is equal to 4/9 of the pear trees. How many are these two kinds of fruit trees?
1/3 in the problem is based on apple trees, and 4/9 is based on pear trees, so it must be unified into a standard quantity when solving the problem.
If the unit of apple tree is "1", then 1× 1/3 = pear tree× 4/9, then pear tree is equivalent to 1/3 ÷ 4/9 of the unit of "1",the total number of two kinds of fruit trees.
420÷ (1+1/3÷ 4/9) = 240 (tree) ... apple tree.
240÷ (1/3÷ 4/9) =180 (tree) ... pear tree.
Pear trees can also be regarded as the unit "1", or the total number of two kinds of fruit trees, or the difference between the number of trees can also be regarded as the unit "1".
Third, find out the solution through hypothetical calculation.
Some fractional application problems are difficult to find solutions if you think directly according to the conditions given in the problems. If you assume a subjective condition when solving a problem, and then calculate it according to the quantitative relationship in the problem, the result will be different from the condition of the problem, and then you can find the correct answer by making appropriate adjustments.
Example: Honghua Village built a canal. In the first week, more than 2/5 of the total length of the canal was completed, and in the second week, 1/4 of the total length was completed, less than 5 meters, and 282 meters remained to be treated. How long is this canal?
Assuming that the first week is just 2/5 of the total length, the remaining 282m after the first and second weeks will increase by10m; Assuming that the second week is exactly 1/4 of the total length, the remaining 282 meters after the first and second weeks will be reduced by 5 meters, so the condition becomes "2/5 of the total length was repaired in the first week, 1/4 of the total length was repaired in the second week, and the remaining (282+ 10-5) Taking the total length of this canal as the unit "1", the corresponding score of (282+ 10-5) meters is (1-2/5- 1/4). So the formula is:
(282+10-5) ÷ (1-2/5-1/4) = 820 (m)
Fourth, find out the solution through reverse deduction.
Some fractional application problems, if analyzed from beginning to end, are difficult to achieve the purpose of solving problems, and even desperate. Might as well "think backwards" and push backwards, so it is easy to open your mind and solve the problem smoothly.
Example: There is an oil drum. After the first time, pour out 1/3 and add 20 kg. When pouring out for the second time, 1/6 was more than 5 kg. At this time, there are 95 kilograms of oil in the barrel. How many kilograms of oil were there in the original barrel?
Considering from the last condition: 95+5 = 100 (kg), which is 5/6 of the existing oil, so there is oil in the barrel now, 100 ÷ 5/6 = 120, and then considering from the first condition,120. Comprehensive formula:
[(95+5) ÷ (1-1/6)-20] ÷ (1-1/3) =150 (kg)
Fifth, find out the solution with the help of line segment diagram.
The quantitative relationship of fractional application problems is abstract and hidden. If you draw a line diagram according to the meaning of the problem, the abstraction can become concrete, obscure and clear, so that with the help of the quantitative relationship revealed by the line diagram, you can find the solution intuitively, and even find a simple solution for some problems.
Example: Party A and Party B each deposit several RMB yuan, of which Party A accounts for three fifths. If Party B gives 60 yuan to Party A, the remaining money of Party B accounts for 1/4 of the total. How much RMB do Party A and Party B each deposit?
Draw a line segment according to the meaning of the question: attached drawing (figure)
From the line chart, it is clear at a glance that 60 yuan's corresponding score is (1-3/5- 1/4), so that we can work out how much RMB each party has saved, and then work out how much RMB each party has saved.
60÷ (1-3/5-1/4) = 3,200 yuan ... both parties * * *
3200× 3/5 = 1920 (yuan) ...
3200× (1-3/5) =1280 (yuan) ...
Or 3200- 1920 = 1280 (yuan)
6. Grasp the invariants and find the solution.
If we can find an invariant from the problem of fractional application with different standard quantities, we can quickly find a solution by taking the invariant as a breakthrough.
There are 360 workers in a workshop, three fifths of whom are women. Later, a group of women workers were recruited. At this time, the number of female workers accounted for 5/8 of the total number of workshop workers. How many women workers have been recruited?
From the topic, we can see that the number of female workers has changed, which makes the total number of workers in the whole workshop change, while the number of male workers has never increased but decreased. Therefore, it is necessary to grasp the invariant that the number of male workers has not changed. When there are 360 workers in the whole workshop, women account for 3/5 and men account for 1-3/5 = 2/5, that is, 360 × 2/5 = 144 (person). After a group of female workers are recruited, the number of female workers accounts for 5/8 of the total number of workers in the workshop, and the number of male workers accounts for 1-5/8 = 3/8 of the total number of workers in the workshop. So there is 144 ÷ 3/8 = 384 (people) in the workshop at this time. There used to be 360 workers in the whole workshop, but now it has increased to 384. The reason for the increase is that a group of women workers have been recruited, so 384-360 = 24 (people) have been recruited. Comprehensive formula:
360× (1-3/5) ÷ (1-5/8)-360 = 24 (person)
Seven, find a solution by changing the conditions.
Some fractional application problems can be transformed into another related quantity by changing the perspective of the problem, making it a familiar and simple problem, thus finding a new way to solve the problem.
Example: There are two goldfish bowls. If you take out 15 goldfish from the first tank and put them in the second tank, the goldfish in the second tank is just 5/7 of that in the first tank. It is known that there are 35 goldfish in the second tank. How many goldfish are there in the first fish tank?
This problem can be transformed into the familiar "standardization" problem. According to the meaning of the score, 5/7 in the question means that the mantissa of the goldfish in the first tank is divided into 7 parts on average, and that of the goldfish in the second tank accounts for 5 parts. These 5 parts * * * 35+ 15 = 50 (tail), so each part is 50 ÷ 5 = 10 (tail). Comprehensive formula:
(35+ 15) ÷ 5× 7+ 15 = 85 (mantissa)
Eight, list the corresponding contrast to find solutions.
For some fractional application problems, we can find a solution by comparing the known conditions and studying the changing law between the corresponding quantities through list correspondence.
For example, a workshop held a training course on technological innovation. If 1/3 of the total number of male employees and 1/4 of the total number of female employees are taken out, 90 people will participate; if 1/4 of the total number of male employees and 1/3 of the total number of female employees are taken out, 85 people will participate. How many people are there in this workshop?
Comparative Analysis of List Correspondence: Attached Figure (Figure)
If 1/3 of the number of male employees and female employees is taken out, it is obtained from formula (5): 65438+ 0/3 of the number of male employees+0/3 of the number of female employees = 300× 1/3 = > (number of male employees+number of female employees )×/. This 15 is equivalent to the number of men in the whole workshop (1/3- 1/4), so this workshop has15 ÷ (1/3-1/4) =/.