Solution: If Party A and Party B go in opposite directions, it will take 600 ÷ 1000 ÷ (4+5) × 60 = 4 minutes to meet. When 1-3+5-7+9 = 5 minutes, we will meet in 1 minute. So 1+3+5+7+9- 1 = 24 minutes. So we met at 8: 24.
Solution: Correct understanding of "Turn around and walk for odd minutes in turn according to 1, 3, 5 and 7" should be 1 minute, 3 minutes backward, 5 minutes forward, 7 minutes backward and 9 minutes forward. ...
Speed of A train: 4000/60 = 200/3 (m/min) Speed of B train: 5000/60=250/3 (m/min) The normal meeting of two trains is 600/(200/3+250/3)=4 minutes1-3. If it is less than 9, 1 point, 600+150 * (3+7-1-5) =1200m, 1200/ 150=8 points.
2. There are two engineering teams to complete a project. Team A has a rest 1 day after working for 6 days, and it takes 76 days to finish it alone. Team B has two days off every five days of work, and it takes 89 days for one person to finish the work. According to this calculation, the two teams cooperated, and the construction started from1998165438+129 October, and was not completed until 1999.
Solution: The two teams do it alone: 6+ 1 = 7, 5+2 = 7, which means that both Team A and Team B have a 7-day cycle.
Group A: 76 ÷ 7 = 10 weeks ... 6 days. Explain that Team A has worked for 76- 10 = 66 days in 76 days.
Group B: 89 ÷ 7 = 12 weeks ... 5 days. It shows that Team B worked 89- 12× 2 = 65 days in 89 days.
Cooperation between the two teams: 1 ÷ (6/66+5/65) = 5+23/24, which means * * 5 cycles.
There are also1-6/66× 5-5/65× 5 = 23/143.
It takes 23/1 43 ÷ (1/66+1/65) = 5+35/131,that is, after five days of cooperation, the remaining A works at1.
* * * It takes 7× 5+5+ 1 = 4 1 day to complete. So it is 4 1-2-3 1 = 8, which means 1999 1.8.
3. In a math contest, Xiao Wang did 2/3 of the total questions, Xiao Li made 5 mistakes, and both of them did 1/4 of the total questions. How many questions did Xiao Wang get right?
Solution: Xiao Wang's correct answer accounts for 2/3 of the total number of questions, indicating that the total number of questions is a multiple of 3. Xiao Li made five mistakes, which means that neither of them will make more than five. That is, the total number of topics will not exceed 5 ÷ 1/4 = 20.
And because all the wrong questions are 1/4 of the total number of questions, it means that the total number of questions is a multiple of 4.
Only 3× 4 = 12 channels meet the requirements, which is both a multiple of 3 and a multiple of 4, not exceeding 20.
So Xiao Wang got the question 12× 2/3 = 8 right.
Solution: Xiao Li made five mistakes, and their mistakes accounted for 1/4 of the total number of questions, so there were at most 20 questions.
Because they are both natural numbers, the number of questions that both of them did wrong may be {1, 2, 3, 4, 5}.
The total number of corresponding questions is {4, 8, 12, 16, 20} respectively.
Among them, only 12 is satisfied: 2/3 of the questions that Xiao Wang answered correctly are natural numbers. So Xiao Wang answered eight questions correctly.
Solution: Let two people have the same number of wrong questions as A,
Then a \( 1/4)×(2/3)= a×8/3 equals the number of questions that Xiao Wang did correctly.
It can be concluded that a is a multiple of 3 (a
4. There are 100 coins (1 minute, 2 points, 5 points). If you change all 2 cents into the equivalent 5 cents, the total number of coins will become 79, and then change all 1 cents into the equivalent 5 cents, and the total number of coins will become 63. What is the value of the original 2 cents and 5 cents?
Solution: According to the meaning of the question, if 2: 5 is replaced by 2: 00, and one group is missing 3 points, then the total * * * is 100-79=2 1, that is, 2 1/3=7 groups, so there are seven coins in five points.
According to the meaning of the question, the five points of 1 are replaced by five points of 1, and one group is missing four points, so the total * * * is missing 79-63= 16 = 4 groups, so if there are five points of/kloc,
5. Two objects A and B, starting from two points that are 150m apart (the length of the small arc on the circular runway), move relatively along the circular runway. If they move along a small arc, A and B meet at 10s; If they move along a big arc, they meet at14 s. It is known that when A runs the circular runway, B only runs 90m.
Solution: The sum of the speeds of Party A and Party B is150 ÷10 =15 m/s. The circumference of the circular runway is15× (/kloc-0+14) = 360 m.
Line A runs 360 meters a week and B runs 90 meters, which means that the speed of A is 360 ÷ 90 = 4 times that of B.
So the speed of B is 15 ÷ (4+ 1) = 3m/s, and the speed of A is15-3 =12m/s. ..
6. In the ranking of competition results, the average score of the top seven is less than that of the top four 1 point, and the average score of the top seven 10 is 2 points less than that of the top seven. How many points are the scores of the fifth, sixth and seventh players more than those of the eighth, ninth and tenth players?
Solution 1: Because the average score of the top seven is less than that of the top four 1 point, the total scores of the fifth, sixth and seventh places are less than that of the top four 1×7=7 points; Because the average score of the former 10 is 2 points lower than that of the former 7, the total score of the 8th, 9th and 10 is 2× 10=20 points lower than that of the former 7, so it is 20+ 1×3=23 points lower than that of the former 4. Therefore, the total score of the 5th, 6th and 7th places is 23-7 = 16 points more than that of the 8th, 9th and 10 places.
Scheme 2: Based on the average score of 10 people, the 8th, 9th and 10 people should be given to the top 7×2 = 14. Then the three of them will be less than the standard total score 14 points. The 5th, 6th and 7th places were originally 3 × 2 = 6 points more than the standard total score, but 1 × 4 = 4 should be given to the top 4. Then the three of them are 6-4 = 2 points more than the standard total score. Therefore, the sum of the scores of the 5th, 6th and 7th players is 2+ 14 = 16 points more than that of the 8th, 9th and 10 players.
Solution: Because the average score of the top seven is less than that of the top four 1 point, the average score of the top seven 10 is 2 points less than that of the top seven.
Therefore, the total score of the fifth, sixth and seventh place is three times less than the average score of the first four places, 1*7=7 points; The total score of the eighth, ninth and tenth places is 2* 10=20 points less than the average score of the first seven places, and 20+ 1*3=23 points less than the average score of the first four places.
So: the total score of the fifth, sixth and seventh places minus the total score of the eighth, ninth and tenth places = 23-7 = 16 points.
Respondent: Uinaf-Juren Grade 5 1-24 23: 17.
Solution: suppose the average score of the top four is a, according to the meaning of the question:
The total score of the top four is 4A, and the total score of the top seven is (A- 1)*7.
The scores of the fifth, sixth and seventh places are 7A-7-4A = 3A-7;
The total score of the top ten is (A-3)* 10,
The scores of 8, 9 and 10 are10a-30-(7a-7) = 3a-23;
Then the sum of the scores is 3A-7-(3A-23)= 16.
7. To finish a job alone, Party A can finish it three days in advance and Party B can finish it five days later. If Party A and Party B continue to do the remaining work independently after three days of cooperation, it will be completed within the specified time. How many days does it take for Party A and Party B to cooperate?
Solution: Three days in A is equivalent to five days in B, so the time ratio for completing the whole project is 3: 5. The time difference between A and B is 3+5 = 8 days. So,
A It takes 8 ÷ (5-3) × 3 = 12 days to complete the whole project alone.
B It takes 12+8 = 20 days to complete the whole project alone.
So it takes two people to cooperate1÷ (112+1/20) = 7.5 days.
8. Both Party A and Party B start from place A at the same time and advance to place B at the same speed. Party A takes a 2-minute break every 5 minutes, and Party B takes a 3-minute break at 2 10 meters. Party A arrived at B 50 minutes after departure, and Party B arrived at B 65,438+00 minutes later than Party A. It is known that their last resting place is 70 meters apart. What is their speed?
Solution: A 50 ÷ (5+2) = 7 times ... 1 minute, indicating that A rested for 7 times * * * 2× 7 = 14 minutes.
B rested 14+ 10 = 24 minutes, and rested for 24 ÷ 3 = 8 times.
When Line B reached the last stop of Party A, it covered 2 10×8+70 = 1.750 meters, and actually covered 5× 7 = 35 points.
So the actual speed is1750 ÷ 35 = 50m/s.
The whole journey is 50× (50-14) =1800m.
Average speed: A 1800 ÷ 50 = 36m/s, b1800 ÷ (50+10) = 30m/s.
Solution: A took 50 minutes, then walked for 7 5 minutes, rested for 7 2 minutes, and finally walked 1 minute. The effective travel time is 36 minutes.
Since both parties have the same speed, the effective time for Party B to walk is 36 minutes, and the effective travel time for walking to the last rest point of Party A is 36- 1=35 minutes;
Because Otsuichi * * * took 60 minutes, he took a rest for 24 minutes, and * * * took eight breaks, during which he walked 2 10*8= 1680 meters, plus 70 meters between their last resting places, which is 1750 meters.
Therefore, Party B can travel 1 750m within 35 minutes, and the speed of both parties is 1, 750/35 = 50m/min. It can be calculated that the AB distance is 50*36= 1800 meters.
So:
A The average speed of completing this distance is 1800/50 = 36m/min.
B The average speed of completing this distance is 1800/60 = 30m/min.
9. There are two bags of rice, A and B. There is 20 kilograms more rice in bag A than in bag B. If 1/3 of the rice in bag A is put into bag B, the rice in bag B is more than that in bag A 10 kg. How many kilograms was there in a bag of rice?
Solution: To make the B bag more than the A bag 10 kg, you have to take it out of the A bag (10+20) ÷ 2 = 15 kg.
It shows that this 15 kg is equivalent to 1/3 of package A, so package A has 15 ÷ 1/3 = 45 kg.
10. There are two piles of coal * * * weighing 8. 1 ton, the first pile uses 2/3, the second pile uses 3/5, and the remaining two piles are combined, which is less than the original first pile16. How many tons was the first pile of coal?
Solution: After being used up, the first pile of coal remains 1/3, and the second pile of coal remains 2/5.
The remaining two piles together account for 1- 1/6=5/6 of the original pile.
Where 1/3 is left over from the original first pile, and the remaining 5/6- 1/3= 1/2 is left over from the original second pile.
That is, 2/5 of the second pile is equal to 1/2 of the first pile.
So the total number of the original second pile is 1/2÷2/5=5/4 times that of the original first pile.
So the original first coal pile is: 8. 1÷( 1+5/4)=3.6 tons.
Solution: If the first pile is 2/3- 1/6 = 1/2,
This used 1/2 is equal to the remaining 1-3/5 = 2/5 in the second pile.
So the second heap is 1/2 ÷ 2/5 = 5/4 of the first heap.
So the first coal pile is 8. 1 ÷ (1+5/4) = 3.6 tons.