A: No.3 is Europe and No.2 is America;
B: No.4 Asia, No.2 Oceania;
C: 1 number is Asia, and number 5 is Africa;
D: No.4 is Africa and No.3 is Oceania;
E: Europe on the 2nd and America on the 5th.
The teacher said that each of them was only half right. No.65438+0 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Answer and analysis: 1 is Asia; No.2 is Oceania; The third place is Europe; The fourth place is Africa; The fifth is the United States.
Assuming that the first half of A is right, then No.3 is Europe, then Ding's conclusion that No.3 is Oceania is wrong. Since everyone is only half right, it can be seen that Ding is right to say that No.4 is Africa, and thus B is wrong to say that No.4 is Asia and No.2 is Oceania. It is also known that E is wrong to say that No.2 is Europe and No.5 is America, which leads C to say that No.5 is Africa and 1 is Asia. Finally, the correct conclusion is: No.65438 +0 is Asia; No.2 is Oceania; The third place is Europe; The fourth place is Africa; The fifth is the United States.
Q&A 2 1 on the fourth grade Olympic Mathematics. Tourism problem.
Party A and Party B practice running. If Party A lets Party B run 10 meter first, Party A can catch up with Party B after running for 5 seconds. If A lets B run for 2 seconds first, A can catch up with B in 4 seconds. Q: What are the speeds of A and B respectively?
Answer: The analysis shows that if A lets B run10m first, then10m is the distance difference between A and B, and 5 seconds is the catching-up time. Based on this, we can find that their speed difference is 10÷5=2 (m/s); If A lets B run for 2 seconds first, A can catch up with B in 4 seconds. In this process, the catching-up time is 4 seconds, so the distance difference is 2×4=8 (meters), that is, B ran 8 meters in 2 seconds, so the speed of B can be calculated, and the speed of A can also be calculated. The comprehensive formula is as follows:
Solution: The speed of B is: 10÷5×4÷2=4 (m/s).
The speed of A: 10÷5+4=6 (m/s).
A: A's speed is 6 meters per second, and B's speed is 4 meters per second.
2. Travel issues
At 8: 08 in the morning, Xiaoming set off from home by bike. Eight minutes later, his father came after him on a motorcycle and caught up with him 4 kilometers away from home. Then his father went home at once. When he got home, he went back to chase Xiao Ming and caught up with him. Just 8 kilometers from home. What time is it?
Answer: During the period from the first time my father caught up with Xiaoming to the second time, Xiaoming walked 8-4=4 (km), while my father walked 4+8= 12 (km), so the speed ratio of motorcycle and bicycle was 12: 4 = 3: 1. Xiaoming walked the whole journey. Also because of the late departure, there are 8 minutes less. From the speed ratio calculated above, it can be seen that Xiao Ming rides 8 kilometers, and Dad should ride 24 kilometers if he starts at the same time. Now it takes 8 minutes less to ride and 24- 16=8 (km) less, so the speed of the motorcycle is calculated to be 1 km per minute. Dad always rides 18.
Q&A on the fourth grade Olympiad 3 Q&A on the fourth grade Olympiad: simple operation. The study of Olympic Mathematics should consolidate the learned knowledge and open up new ideas through continuous practice. Here, the column of Mathematical Network Olympic Mathematics Question Bank collects and sorts out the questions about elementary arithmetic in the fourth grade Olympic Mathematics for students, and attaches the answers to the questions for students to practice.
Simple operation:
Test center: the operation is regular and simple.
Analysis:
(1) First decompose 32 into 4×8, and then simplify it by using the multiplicative associative law.
(2) Calculate division first, and then simplify according to the nature of subtraction.
Comments: This question is about elementary arithmetic. We should carefully observe the characteristics of the formula and use some laws flexibly to make simple calculations.
The fourth grade Olympic math problem and answer 4 let both A and B represent numbers, and stipulate that A △ B = 3× A-2× B
① Find 3 △ 2,2 △ 3;
② Does this operation "△" have a commutative law?
③ Find (17△6)△2,17 △ (6 △ 2);
④ Does this operation "△" have an associative law?
⑤ If 4△b=2 is known, find b. ..
answer
Analysis:
The key to analyze and solve the problem of defining new operations is to grasp the essence of definition. The essence of the operation specified in this question is: subtract the 2 times after the symbol from the 3 times before the operation symbol.
Solution: ①3△2=3×3-2×2=9-4=5.
2△3=3×2-2×3=6-6=0。
② From the example of ①, it can be seen that "△" has no commutative law.
③ To calculate (17△6)△2, first calculate the number in brackets, that is,17 △ 6 = 3×17-2× 6 = 39; Recalculate the second step
39△2=3×39-2×2= 1 13,
So (17 △ 6) △ 2 =113.
For 17△(6△2), the number in brackets is also calculated first, 6△2=3×6-2×2= 14, and then
17△ 14=3× 17-2× 14=23,
So 17△(6△2)=23.
④ From the example of ③, it can be seen that△ has no associative law.
⑤ Because 4△b=3×4-2×b= 12-2b, then 12-2b=2 and b=5.
Fill in the blanks about the fourth grade Olympic math problem and answer 5 operation symbols: (medium difficulty)
Fill in the four operation symbols+,-,×, ÷ in the following equation respectively to make the equation hold (each operation symbol can only be used once): (5 ○13 007) ○ (17 009) =12.
Fill in the blanks with operation symbols Answer:
Because the result of the operation is an integer, only the division operation in the four operations may have a score, so we must first determine the position of "∫".
When "÷" is in the first χ, because the divisor is 13, if you want to get an integer, only the second bracket is a multiple of 13. At this time, there is only one filling method, which is irrelevant.
(5÷ 13-7)×( 17+9)。
When "∫" is within the second or fourth χ, the operation result cannot be an integer.
When "∫" is in the third ○, the following filling method can be obtained: (5+13× 7) (17-9) =12.
Questions and answers about the fourth grade Olympiad 6 A and B running around the track with a circumference of 400 meters. If they leave the same place with their backs to each other, they will meet in 2 minutes. If two people start from the same place and walk in the same direction, after 20 minutes, it is known that A is faster than B. What are the running speeds of A and B respectively?
Answer:
From the same place, two people walk with their backs to each other. After a two-minute meeting, they knew that they walked 400÷2=200 meters per minute. After meeting for 20 minutes, they knew that A walked 400÷20=20 (meters) more than B. According to the solution of sum-difference problem, A's speed was 20 (meters) per minute.
Q&A on Grade Four Olympiad 7. The breeder Xiao Wang keeps 40 chickens and rabbits in his yard. Their feet are 108. How many chickens and rabbits does Xiao Wang keep?
Answer and analysis:
Suppose Xiao Wang has 40 rabbits, and one * * has 4×40= 160 (only) feet, which is more than the actual 160- 108 = 52 (only). The extra 52 feet are caused by understanding the chickens as rabbits, that is, each chicken is 4-2=2 feet, so how many chickens will there be in 2 feet of 52 chickens, that is, 52÷2=26 chickens. Number of rabbits: 40-26= 14 (only)
Solution:
Number of chickens: (4×40- 108)÷(4-2)=26 (only)
Number of rabbits: 40-26= 14 (only)
A: Xiao Wang keeps 26 chickens and 14 rabbits.
About the fourth grade Olympiad 8, the spider has 8 legs, the dragonfly has 6 legs and 2 pairs of wings, and the cicada has 6 legs 1 pair of wings. At present, these three insects have 17, 120 legs and1/wings. How many insects do you want for each kind?
Hugging: There are three kinds of insects, which have legs and wings, and it is more complicated than the problem of chickens and rabbits in the same cage. After careful analysis, we will find that if the number of legs of insects is classified, they can be divided into two categories: 8 legs and 6 legs. But only hexapod insects have wings, so we know the total number of legs and the total number of legs of hexapod and hexapod insects. According to the basic formula of chicken and rabbit in the same cage, the sum of the number of spiders with 8 legs and dragonflies and cicadas with 6 legs can be obtained. In this way, by using the basic formula of the problem of chickens and rabbits in the same cage, we can know the total number of wings and the respective number of wings of dragonflies and cicadas, and then we can get the respective number of wings of dragonflies and cicadas.
Solution: Spider number: (120-17× 6) ÷ (8-6) = 9 (only)
Number of insects with six legs: 17-9=8 (only)
Number of cicadas: (8× 2-11) ÷ (2-1) = 5 (only)
Number of Dragonflies: 8-5=3 (only)
A: There are 9 spiders, 5 cicadas and 3 dragonflies.
Q&A on the fourth grade Olympic Mathematics 9 female squirrels can pick pine nuts every day 16 in sunny days and 1 1 in rainy days. For several days, there were sunny days and rainy days, among which rainy days were 3 days more than sunny days, but rainy days were 27 days less than sunny days. Ask a * * * how many days?
Note: As mentioned above, rainy days are 3 days more than sunny days, but the number of picking is 27 less than sunny days. If rainy days are the same as sunny days, rainy days need to be reduced by 1 1×3=33 (pieces); There are more rainy days than sunny days, but there are still 27 days less. This time, it is not an increase in the number of days, but a decrease of three days. At this time, the number of picks in rainy days is 33+27=60 (pieces) less than that in sunny days. It can be seen that these 60 flowers are caused by picking five more flowers in sunny days than in rainy days, so picking five more flowers in sunny days and picking them for several days.
Finally, it is 60 more than rainy days, that is, the number of sunny days: 60÷5= 12 (days). Therefore, we can work out the number of rainy days: 12+3= 15 (days), so that we can know that the number of days in a * * * mine is 15+ 12=27 (days).
Solution: There are as many rainy days as sunny days, and the number of sunny days is 27+ 1 1×3=60 (one).
Days of sunny days: 60 ÷ (16-11) =12 (days)
A * * * Mining days: 12+3+ 12=27 (days).
A: I took it for 27 days.
The question and answer about the fourth-grade Olympic Mathematics began on June 5438+00 20x65438+1 0/.Employee A has a rest every three days 1 day, and Employee B has two days off every five days. A and B work in the same position. If both A and B have a day off and it is agreed that employee C will take their place, it will be 20xx years.
analyse
In the 28 days numbered 1, 2, 3 ... 28,
The rest days of employee A are numbered 4, 8, 12, 16, 20, 24 and 28.
The rest days of employee B are numbered 6, 7, 13, 14, 20, 2 1, 27 and 28.
Therefore, the * * * numbered 20 and 28 as A and B are the same as rest days.
365 ÷ 28 = 13...7.
So in 20xx, C has to work 13×2=26 days.
The fourth grade Olympic math test questions and answers 1 1. If the sum of a four-digit number and a three-digit number is 1999, and the four-digit number and the three-digit number are composed of seven different numbers, how many such four-digit numbers can there be at most?
Answer and analysis:
Four thousand digits are 1, and the hundred digits (set as a) can be selected from 0, 2, 3, 4, 5, 6 and 7, and then the three hundred digits are 9-a; The ten digits of the four digits are set to B, and the remaining six digits can be selected. The tenth digit of a three-digit number is 9-B, and the unit number c of four digits can be selected from the remaining four digits. The unit number of three digits is 9-C, so there are four digits of 7×6×4= 168.