For example, there are 47 apples. There are two fewer apples than oranges. How many oranges are there?
Some students will calculate 47-2=45 (1).
I let the students look at the questions several times, but some students can't do it in another form.
Cause analysis: These students failed to abstract the purpose and significance of the question well, and their thinking was in the stage of thinking in images.
So, how can we make students understand more intuitively? I thought for a while, and thought that the method of drawing line segments can make students understand better.
(1) Learn to find "standards"
Before class, let me give an example, "The teacher is taller than Xiaohong". Who is the standard? Answer: Xiaohong
"Xiaohong is shorter than the teacher", who is the standard? Answer: Teacher.
"Who" is the standard.
(2) Learn to draw line segments.
As mentioned above, there are 47 apples, two fewer apples than oranges, and how many oranges are there?
This question is based on oranges.
Orange:? individual
47 is less than 2
Apple:-
By looking at this line chart, students can easily see that there are fewer apples than oranges, so they need to add oranges.
How to illustrate the application problems that are less than Dobby in grade one, such as: here are five apples, I ate one, and how many are left. `(*∩_∩*)′
What is the sum of the first grade? 10. How much is left in the first grade? 5. How much is the first grade? 20. How many cats have 10 fish and caught 12 * *?
Xiaoxiao had 18 apples and bought 19 apples. How many apples are there in Xiaoxiao?
Is this the kind of question?
There are 30kg pears in the sixth grade, one tenth less than apples. Q: How many kilograms does an apple weigh?
The fifth grade is smaller than Dobby. There are a bunch of 1 yuan coins, five fives and six sixes, all of which are very accurate. At least this pile of coins is missing?
Answer: 5 times 6 equals 30.
What is the difference in the size of the first-grade math application questions? Use subtraction.
Subtraction is used to find a small number.
To find a larger number, add.
How to teach first-year students to do application problems? So, let's work out the answer first ... find the equivalence relationship ... and then put the number in to see if it's right.
Try to simplify the topic when teaching. That's right. Read the topic slowly and give the students a process of thinking.
Also, first teach students what they want, what they know, and what is the relationship between these data (what are the implied conditions, such as unit price × quantity = total price). Let them calculate carefully.
That's about it.
There is an urgent need for fifth-grade mathematics to be less than Dobby's application problem 1, and two cars, A and B, start from AB at the same time. A walked 5/ 1 1 of the whole journey. If A drives at a speed of 4.5 kilometers per hour, B drives for 5 hours. How many kilometers are AB apart?
Solution: AB distance = (4.5× 5)/(5/11) = 49.5 km.
2. A bus and a truck leave from Party A and Party B at the same time. The speed of a truck is four-fifths that of a bus. After a quarter of the journey, the truck and the bus met for 28 kilometers. How many kilometers is it between A and B?
Solution: The speed ratio of passenger cars and trucks is 5: 4.
Then the distance ratio when meeting is 5: 4.
When they met, it was 4/9 of the whole truck journey.
At this time, the truck has traveled all the way 1/4.
4/9- 1/4=7/36 from the meeting point.
Then the whole journey = 28/(7/36) = 144km.
3. Party A and Party B walk around the city, with Party A walking 8 kilometers per hour and Party B walking 6 kilometers per hour. Now both of them start from the same place at the same time. After B meets A, it will take another 4 hours to return to the original starting point. B How long does it take to go around the city?
Solution: The speed ratio of A and B = 8: 6 = 4: 3.
When they met, B walked 3/7 of the way.
Then 4 hours is 4/7 of the whole trip.
Therefore, the time spent on line B in a week =4/(4/7)=7 hours.
4. Party A and Party B walk from place A to place B at the same time. When Party A completes the whole journey of 1\4, Party B is still 640 meters away from B. When Party A completes the remaining 5\6, Party B completes the whole journey of 7\ 10. What's the distance between AB and place?
Solution: After A left 1/4, the remaining 1- 1/4=3/4.
Then the remaining 5/6 is 3/4×5/6=5/8.
At this time, a * * * left 1/4+5/8=7/8.
Then the distance ratio between Party A and Party B is 7/8: 7/ 10 = 5: 4.
So when A goes 1/4, B goes 1/4×4/5= 1/5.
Then AB distance =640/( 1- 1/5)=800 meters.
5. Two cars, A and B, start from A and B at the same time and drive in opposite directions. Car A travels 75 kilometers per hour, and it takes 7 hours for car B to complete the journey. Three hours after the departure of the two cars, the distance is15km. What is the distance between a and b?
Solution: Case A: Party A and Party B have not met yet.
3/7 of the 3-hour journey of the B train.
The three-hour journey is 75×3 = 225 kilometers.
AB distance = (225+15)/(1-3/7) = 240/(4/7) = 420km.
In one case, Party A and Party B have met.
(225- 15)/( 1-3/7)= 2 10/(4/7)= 367.5km。
6. One, two people should go this way. A It takes 30 minutes to walk and 20 minutes to walk. After walking for 3 minutes, A found that she didn't take anything, which delayed for 3 minutes. I walked for a few minutes before I saw him.
Solution: A is 3+3+3=9 minutes later than B.
Think of the whole distance as 1.
Then the speed of a = 1/30.
Speed B = 1/20
When Party A packed up and set out, Party B had already left 1/20×9=9/20.
Then the distance between Party A and Party B is1-9/20 =11/20.
The sum of the speeds of Party A and Party B =1/20+1/30 =112.
Then meet again in (11/20)/(112) = 6.6 minutes.
7. two cars, a and b, start from place a and drive in the same direction. A walks 36 kilometers per hour and B walks 48 kilometers per hour. If car A leaves two hours earlier than car B, how long will it take for car B to catch up with car A?
Solution: distance difference = 36× 2 = 72km.
Speed difference = 48-36 = 12km/h
It takes 72/ 12=6 hours for car b to catch up with car a.
8. Party A and Party B respectively set out from ab, which is 36 kilometers apart, and walked in opposite directions. When Party A departs from A to 1 km, it has been in A until it finds something and returns immediately. After the goods were gone, he immediately went from place A to place B, where Party A and Party B met. He knew that Party A walked 0.5 kilometers more than Party B every hour and asked both of them to walk.
Solution:
A actually walked 36× 1/2+ 1× 2 = 20km when they met.
B walked 36× 1/2 = 18km.
Then A walked 20- 18 = 2km more than B.
Then the meeting time =2/0.5=4 hours.
So A = 20/4 speed = 5 km/h.
Speed B = 5-0.5 = 4.5km/h/h.
9. At the same time, two trains travel in opposite directions from two places 400 kilometers apart. The bus speed is 60 kilometers per hour, and the truck speed is 40 kilometers per hour. A few hours later, did the two trains meet at 100 km?
Solution: velocity sum = 60+40 =100 km/h.
There are two situations,
No encounter
Then the required time =(400- 100)/ 100=3 hours.
Met it.
Then the required time =(400+ 100)/ 100=5 hours.
10, A travels 9 kilometers per hour, and B travels 7 kilometers per hour. They walked back to back at the same time in two places 6 kilometers apart, and a few hours later they were separated by 150 kilometers.
Solution: velocity sum = 9+7 =16 km/h.
Then after (150-6)/16 =144/16 = 9 hours, the distance is150 kilometers.
7. Party A and Party B produce a batch of parts. The efficiency ratio of Party A and Party B is 2: 1. Co-shoot for three days, and Party B will shoot alone for the other two days. At this time, Party A has produced 14 more parts than Party B. How many parts are there in this batch?
Solution: Take the work efficiency of B as the unit 1.
Then a's work efficiency is 2.
B 2 days to complete 1×2=2.
Otsuichi * * * produces 1×(3+2)=5.
A * * * Output 2×3=6
So the work efficiency B = 14/(6-5)= 14/ day.
A's work efficiency = 14×2 = 28/ day.
A * * * has 28×3+ 14×5= 154 parts.
Or let the work efficiency of Party A and Party B be 2a/ day and A/ day respectively.
2a×3-(3+2)a= 14
6a-5a= 14
a= 14
A * * * has 28×3+ 14×5= 154 parts.
8. For a project, the time for Party B to complete the project alone is twice that of Party A's team; It takes 20 days for Team A and Team B to cooperate to complete the project. The daily work cost of Team A is 1 1,000 yuan, and Team B is 550 yuan. From the above information, which company should I choose from the perspective of saving money? How much should be paid to the construction team?
Solution: The sum of the work efficiency of both parties = 1/20.
Working time ratio of Party A and Party B = 1: 2.
Then the work efficiency ratio of Party A and Party B is 2: 1.
So the working efficiency A = 1/20×2/3= 1/30.
Party B's work efficiency =1/20×1/3 =1/60.
A it takes one person 1/( 1/30)=30 days.
B It takes1(1/60) = 60 days to complete it alone.
A need to complete it alone1000× 30 = 30,000 yuan.
B alone needs 550× 60 = 33,000 yuan.
The cooperation needs of Party A and Party B are (1000+550) × 20 = 31000 yuan.
obviously
A needs the least money to finish it alone.
Choose a, you need to pay 30000 yuan for this project.
9. For a batch of parts, if Party A and Party B work together for 5.5 days, it can exceed 0. 1 of the batch of parts. Now Party A works for 2 days, then Party A cooperates for 2 days, and finally Party B works for 4 days to complete the task. If Party B works alone, how many days can this batch of parts be completed?
Solution: treat all parts as a unit 1.
Then the sum of the work efficiency of Party A and Party B = (1+0.1)/5.5 =1/5.
The whole process is that A works 2+2=4 days.
B working 2+4=6 days.
It is equivalent to 4 days of cooperation between Party A and Party B, and 1/5×4=4/5 is completed.
Then B does it alone for 6-4=2 days 1-4/5= 1/5.
So it takes 2/( 1/5)= 10 days for B to complete it alone.
10, there is a project to be completed within the specified date. If Team A does it alone, it will be finished on schedule. If Team B does it alone, it will take more than 5 days to finish. Now Team A and Team B have been working together for three days, and the rest of the projects are completed by Team B alone as planned. How many days is the specified date?
Solution: 3 days of A is equivalent to 5 days of B.
The work efficiency ratio of Party A and Party B is 5: 3.
Then the ratio of completion time between Party A and Party B is 3: 5.
So it takes 3/5 time for A to complete.
So it takes 5/(1-3/5) = 5/(2/5) =12.5 days for B to complete it alone.
Specified time = 12.5-5=7.5 days.
1 1. A project will be completed in 20 days by team A and 30 days by team B. Now team B will finish it in five days, and the rest will be completed by team A and team B. How many days will it take?
Solution: B completed 5× 1/30= 1/6 in 5 days.
The work efficiency of Party A and Party B =1/20+1/30 =1/6.
Then (1-1/6)/(1/6) = (5/6)/(1/6) = 5 days.
14, a project, a team of 20 people do it alone for 25 days. If it takes 20 days to complete, how many people need to be added?
Solution: Take everyone's workload as the unit 1.
It is also necessary to increase/kloc-0 /× 25× 20/(/kloc-0 /× 20)-20 = 25-20 = 5 people.
15. For a project, Party A will do it for 3 days first, and then Party B will join in. Two-thirds of the projects completed after 4 days are 1 and three-quarters of the projects completed after 10. A was transferred because of some things, and B did the rest. How many days did a * * * do?
Solution: according to the meaning of the problem
The cooperation between Party A and Party B began in 4 days13 and ended in 3/4 days 10.
So the cooperation between Party A and Party B is10-4 = 3/4- 1/3=5/ 12 in 6 days.
Therefore, the work efficiency of both parties is =(5/ 12)/6=5/72.
Then the working efficiency a = (1/3-5/72× 4)/3 = (1/3-5/18)/3 =1/54.
Party B's work efficiency = 5/72-1/54 =11/216.
Then B needs to complete the remaining (1-3/4)/(11/216) = 54/11day.
A * * * made 3+10+54/1=17 and10//day.
16, both parties made the same parts. /kloc-After 0/6 days, Party A needs 64 B and 384 B to complete it. The work efficiency of Party B is 40% less than that of Party A, so how can we find the efficiency of Party A?
Solution: Let the working efficiency of A be A/ day, and then B be (1-40%)A = 0.6a/ day.
According to the meaning of the question
16a+64=0.6a× 16+384
16×0.4a=320
0.4a=20
A=50/day
A's work efficiency is 50/ day.
Arithmetic method:
B does 40% less than A every day.
Then 16 days is 384-64= 320 less.
Do 320/ 16 = 20 less every day.
Then the working efficiency of A = 20/40% = 50/ day.
17, Master Zhang has a rest every six days 1 day, and Master Wang has a rest every five days for two days. For an existing project, Master Zhang needs 97 days and 75 days. If two people cooperate, how many days does a project take?
Solution:
97 divided by 7 equals 13, leaving 6 13 * 6 = 78, 78+6 = 84 working days.
75 divided by 7 equals 10,5,10 * 5 = 50,50+5 = 55 working days.
Master Zhang completes 1/84 every working day and 6/84 =114 every week.
Master Wang completes 1/55 every working day and 5/55 =11/every week.
Two people work together to complete 139/4620 every working day and 25/ 154 every week.
Six weeks to complete 150/ 154, leaving 4/ 154.
(4/ 154)/( 139/4620)= 120/ 139
So, six weeks and one day, 43 days.
18, Party A, Party B and Party C jointly complete a project, and complete it all in three days 1/5. Then Party A rested for three days, Party B rested for two days, and Party C didn't rest. If Party A's workload is three times that of Party C and Party B's workload is four times that of Party C, how many days will it take to complete the work from scratch?
Solution: The sum of the working efficiencies of A, B and C = (1/5)/3 =115.
Work efficiency c = (115)/(3+4+1) =1120.
A's work efficiency =1120× 3 =1/40.
Party B's work efficiency =1120× 4 =1/30.
Here, the working efficiency of C is regarded as a multiple of 1.
A rest for 3 days, B rest for 2 days, and a * * * is finished.
1/30+ 1/ 120×3=7/ 120
Then the remaining needs (1-1/5-7/120)/(115) = 89/8 days.
A * * * takes 3+3+89/8= 17 and 1/8 days.
19. Party A works alone for 30 days and Party B works alone for 20 days. After Party A works alone for a few days, Party B takes over, Party A works alone for 22 days, and Party A works alone for several days.
Solution: work efficiency B = 1/20.
B Completed in 22 days1/20× 22 =1110.
Complete1110-1=1/kloc-0.
The work efficiency difference between Party B and Party A =1/20-1/30 =1/60.
So A did (110)/(1/60) = 6 days.
B did it for 22-6= 12 days.
Considering that chickens and rabbits are in a cage.
How to teach the first-grade children to learn Olympic Mathematics? Just go to Xinhua Bookstore and buy an Olympic Mathematics textbook and exercises. It's not bad to teach children by the book, and it's also good to have Olympic math books synchronized with the school.
Compared with the number of application problem teaching plans, the goal of the first-year mathematics teaching plan is correct, that is, the teaching goal should meet the requirements of curriculum standards and the actual situation of students. Teaching goal is the basis of designing teaching process, the general guiding ideology of classroom teaching, and the starting point and final destination of classroom teaching. How to make a concrete and feasible teaching goal? First of all, we should carefully study the teaching materials, combine the objectives and teaching contents of the mathematics course, and work out the teaching plan of this lesson: what knowledge students should master, what skills they should form, what proficiency they should achieve, and what methods to solve problems. This is the double-base target. Secondly, we should consider what kind of thinking ability students can cultivate through the teaching of this knowledge, which is the goal of thinking ability. Thirdly, through the imparting of these knowledge, we should think about what kind of ideological education and good moral quality students should have, which is the requirement of infiltrating ideological education. Finally, we should consider where students can receive innovative education and how to cultivate their innovative consciousness and creativity. This is the requirement of innovative education and the most important goal of classroom teaching.