# 么么么么么 # When solving Olympic math problems, you should always remind yourself whether the new problems you encounter can be transformed into old problems and whether the new problems can be transformed into old problems. Through the surface, you can grasp the essence of the question and turn it into a familiar question to answer. The types of transformation are conditional transformation, problem transformation, relationship transformation and graphic transformation. The following is the relevant information of "Five Questions and Five Answers to the Olympic Mathematics in the Third Grade of Primary School", hoping to help you.

1. Olympiad math problems and answers in grade three of primary school.

1, the school bought five thermos bottles and 10 teacups, and * * * used 90 yuan money. The price of each thermos is four times that of each teacup. How much is each thermos and teacup? Think about it: according to the calculation that the price of each thermos is four times that of each teacup, the price of five thermos can be converted into the price of 20 teacups. In this way, the 90 yuan money for five thermos bottles and 10 teacups can be regarded as the money for 30 teacups.

Solution: Price of each teacup:

90÷(4×5+ 10)=3 (yuan)

Price of each thermos:

3×4= 12 (yuan)

A: each thermos 12 yuan, each teacup 3 yuan.

2. The sum of two numbers is 572, and one of them is 0 of several digits. After removing 0, it is the same as the second addend. What are these two numbers?

Think about it: it is known that several digits of an addend are 0, and if you remove 0, it will be the same as the second addend. It is known that the first addend is 10 times of the second addend, so the sum of the two addends is 572 times of the second addend.

Solution: The first addend:

572÷( 10+ 1)=52

The second addend:

52× 10=520

A: These two addends are 52 and 520 respectively.

2. Question and answer of Olympiad Mathematics in the third grade of primary school

1, it's easy to think that A and B, C and D should be time-saving, and they only have one flashlight, and they can only pass two people at a time, so one person has to come back to deliver the flashlight after crossing the bridge. In order to save time, people must undertake the task of sending flashlights back and forth as soon as possible. Then you have to let Party A and Party B cross the bridge first, which takes 2 minutes, then Party A returns to send flashlights, which takes 1 minute, and then Party C and Party D cross the bridge together, which takes1minute. Next, B returns, sends a flashlight, which takes 2 minutes, and then crosses the bridge with A, which takes another 2 minutes. So the total time is: 2+1+10+2+2 =17 minutes. Solution: 2+1+10+2+2 =17 minutes.

2. In order to minimize the transit time, we should grasp the following two points:

(1) The time difference between two cows crossing the river at the same time should be as small as possible.

(2) After crossing the river, you should spend the least time riding an ox.

Solution: It takes 2+ 1=3 minutes for Xiaoming to ride A Niu's back to drive Bull B across the river and then ride A Niu back.

Then ride on the back of the third cow to drive the fourth cow across the river, and then ride the second cow back, 6+2=8 minutes.

Finally, I spent 2 minutes riding on the back of A cow and driving B cow across the river.

Total * * * time (2+ 1)+(6+2)+2= 13 minutes.

3. Question and answer of Olympiad Mathematics in the third grade of primary school

1, car A and car B leave from ab respectively. Car A travels 50 kilometers per hour and car B travels 40 kilometers per hour. Ahead of bus B 1 hour. How far is the distance between the two places? Solution: When car A reaches the finish line, car B is 40× 1 = 40km away from the finish line. Car a travels 40 kilometers more than car B.

Then the time for a car to reach the finish line =40/(50-40)=4 hours; Distance between the two places =40×5=200 kilometers.

Two cars leave from both parties at the same time and meet at 4 o'clock. The speed of the local train is three-fifths of that of the express train. When we met, the express train traveled 80 kilometers more than the local train. How far is the distance between the two places?

Solution: When the express train and the local train meet, the speed ratio is = 1: 3/5 = 5: 3. The express train travels 5/8 of the whole journey and the local train travels 3/8 of the whole journey.

Then the whole journey = 80/(5/8-3/8) = 320km.

Party A and Party B set out from A and B at the same time and walked towards each other. Party A walks 65,438+000 meters per minute, and Party B walks 65,438+020 meters per minute. After two hours, the distance between them is150m. What is the shortest distance between a and b? What is the longest distance?

Solution: The shortest distance is that we have met, and the longest distance is the speed at which we haven't met and =100+120 = 220m/min, 2h = 120min. The shortest distance = 220×120-150 = 26400-6540.

4. Question and answer of Olympiad Mathematics in the third grade of primary school

1. Two cars, A and B, start from 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 B It takes 72/ 12=6 hours to catch up with A.

Party A and Party B start from ab, which is 36 kilometers apart, and go 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, and B walked 36× 1/2 = 18km.

Then A walked 20- 18 = 2km more than B.

Then the time taken to meet =2/0.5=4 hours, then the speed of A =20/4=5 km/hour, and the speed of B =5-0.5=4.5 km/hour.

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: The velocity sum =60+40= 100 km/h is divided into two cases, and there is no encounter.

Then the time required for meeting =(400- 100)/ 100=3 hours.

Then the required time =(400+ 100)/ 100=5 hours.

5. Question and answer of Olympiad Mathematics in the third grade of primary school

1, A car and B car leave 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: When the speed ratio of bus and truck is 5: 4, the distance ratio when they meet is 5: 4, which is 4/9 of the whole journey of truck. At this time, the truck has traveled 1/4, and the distance from the meeting point is 4/9- 1/4=7/36, so the whole journey = 28/(7/36) = 16.

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 Party A and Party B = 8: 6 = 4: 3. When meeting, Party B did 3/7 of the whole journey.

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.