There is an interesting question Q about the primary school itinerary, and I am eager to explain it

After a little analysis, we know that Xiao Jing completed her homework between 4 and 5 o'clock. In fact, this is a typical catch-up problem, chasing the minute hand, and being chased is the hour hand. Here's the answer:

In any case, the minute hand always travels faster than the hour hand, which travels at 360 degrees per hour and the hour hand at 30 degrees per hour. That is to say, every hour that passes, the minute hand will catch up with the clock hand 330 degrees, now let's take a look at the problem, Xiao Jing from the beginning of the homework to the completion of the homework, the minute hand came a big surpass, not only caught up with the hour hand, but also threw it behind, a total of more than the hour hand (90 + 180) degrees, in this way, Xiao Jing's time to do homework should be: (270 degrees 330 degrees) * 1 hour = (9 11) * 60 minutes = 49 minutes Like the above answer, the difference is that he used the equation solution method, And I used the arithmetic solution.

Since it is between 4 and 5 o'clock, the range of the hour hand is between 4 and 5.

The number of divisions between the minute and hour hands is 12:1.

Let the minute hand be perpendicular to the hour hand after going X divisions at 4:00 (15 divisions apart); At 4:00, the minute hand is 20 squares apart, and the equation x+15=20+x 12 is obtained, and the solution is x60 11 (grid), that is, 4:60 11 minutes;

Let the minute hand start at 4:00 and then move in a straight line with the hour hand (the minute hand is 30 divisions before the hour hand);

The equation y-30=20+y 12 is solved, and x 600 11 (grid) is obtained, that is, 4 points 600 11 points;

4:600 11:40 11:49 (min).

The minute hand travels 6 degrees per minute (360 60) and the hour hand travels 12 60 degrees per minute and travels x minutes at right angles. At 4 o'clock is and the hour hand is 120 degrees.

120+, which gives x as 60 11, which is about 4:05.

When set to a straight line, the minute hand travels for y minutes.

Then 6y-(120+, we get y=600 11, which is about 4.55 points.

The total time is y-x, then it is 540 11 minutes, which is about 49 minutes.

The first case: car A arrives at station B to wait for the train.

Yes: Car A takes 95 to 50 hours.

Car B takes (110+15) 60 25 12 hours, so the waiting time is 25 hours, minutes, and 11 minutes.

The second case: car B arrives at station C to wait for the train.

Car A takes (95+15) 50 hours.

Car B takes 110 60 11 6 hours.

So the waiting time is hours and minutes and 22 minutes.

Therefore, choose to wait at point B, the waiting time is the shortest, and the time for A to wait for B is 11 minutes.

Cars A and B depart from A and B respectively, and continue to travel back and forth between A and B. When it is known that the speed of car A is 15 kilometers and the speed of car B is 25 kilometers, the difference between the third meeting place and the fourth meeting place of car A and B is 100 kilometers. Find the distance between A and B.

Solution: When A and B met for the third time, they had gone through 5 whole journeys, and when they met for the fourth time, they had already gone through 7 full journeys. From the speed ratio of A and B, it can be seen that B walked 25 (15+25) 5 8 full courses for the first time, then walked 25 8 3 1 8 full courses when they met for the third time, and walked 35 8 4 3 8 full courses for the fourth time.

Then the distance between the two places is 1 2 whole journeys, and it can be seen that A and B are 200 kilometers apart.

According to the speed, car B ran a round trip, and car A couldn't finish a one-way trip (speed B was greater than 2 times the speed of A), that is, when they met for the third time, the two cars ran three full journeys. , the two cars ran five full courses when they met each other four times. According to the speed ratio 3:

7 can be known, car B when returning to B, car A is running in the whole process of 6 7, the two cars will meet for the third time in this 1 7 distance, the meeting point is 1 7 * 7 10 = 1 10 in the whole journey from B place, in order to facilitate your analysis of the problem, here we divide the whole journey into 10 parts, four encounters from three encounters, two cars ran two whole journeys, both 20 equal parts, car A 20 * 3 10 = 6 parts, car B 20 * 7 10 = 14 parts, because the three meeting points are 1 10 places away from point B, Therefore, when the four encounters are 14 10-9 10=1 2 or 6 10-1 10=1 2, and the corresponding components are known, the problem is understood: 100 (1 2-1 10)=250

To solve this kind of problem, it is recommended that you draw a diagram for analysis, and the diagram is the kind of simple line segment diagram.

The upstairs idea is correct, but the value is wrong, it should be.

Solution: When A and B met for the third time, they walked a total of 5 whole courses, and when they met for the fourth time, they walked a total of 7 whole courses. From the speed ratio of A and B, it can be seen that B walked 35 (15+35) 7 10 full courses for the first time, then walked 35 10 full courses when they met for the third time, and walked 49 10 full courses for the fourth time.

Then the distance between the two places is a whole journey, and it can be seen that A and B are 250 kilometers apart.

Set: the water volume of the pool is 1, and the seepage water volume per hour is a;

A pipe has a displacement of 1 8 + A per hour

Pipe B has a displacement of 1 10+A per hour

Pipe C has a displacement of 1 12+A per hour

A, B two pipes 4 hours flow 4 (1 8 + A + 1 10 + A) = 1 + 4 A1 2 + 2 5 + 8A = 1 + 4A4A = 1 10, A = 1 40

B, C two pipes per hour flow rate 1 10 + a + 1 12 + a = 1 10 + 1 12 + 1 20 = 7 30

Set: B and C two pipes to empty the pool takes time x;

7/30)x = 1 + x(1/40)(7/30 - 1/40)x = 1

5/24)x = 1

x=24 5=hours.

Open the two pipes B and C, and it takes hours to empty the pool.

If the water volume of the pool is 1 and the scheduled time is 1, then the water filling speed of one water pipe is 1 8 and several water pipes are x

1/8*2/5*x+1/8*3/5*2x=1

Find x=5

It's easy to draw a diagram :

If Xiao Jing walked x meters when they first met, then the distance Xiao Yi walked was (100+x) 2-x=x+200, that is, he walked 200 meters more. Then to the second encounter, Xiao Jing walked 100 + 300 = 400 meters, and Xiao Yi walked 2 x + 400 meters, that is, twice the distance that Xiao Jing walked (go to C and turn back). Since the velocity is constant, the equation can be columned:

x+200) x=(x+400) 400 This equation is very easy to solve, and we get x=200, so the whole process is (200+100) 2=600 meters, and it's over.

Judging from this question, the horse must have walked the whole way, and the distance traveled by the two brothers is also a full journey, and the brother riding the brother walks the stage, and the brother spends 1 5-1 12 = 7 60 hours per kilometer more than the younger brother, which is 7 minutes.

In the brother-brother walking phase, the brother takes 1 4-1 12=10 60 hours per kilometer less than the younger brother, which is 10 minutes.

The distance of these two stages should be distributed in reverse proportion to 7:10, that is, the younger brother rides 51*10 17=30 kilometers, and the brother rides 51-30=21 kilometers, each of which takes 30 12 + 21 4 = 30 5 + 21 12 = 465 60 = 7 hours and 45 minutes.

So they arrived at 13:45.

This question is a bit vague, excluding the possibility that the horse is waiting for the person behind him, and there are two situations: one is to ignore the time of the horse's return and forth, this question is easy, but this is a bit impossible; The second is to count the time when the horse comes back to pick up people, which has a variety of situations, one is to send a person, and then come back to pick up another person, and at the same time to the end point, the second is to pick up multiple round-trips, the number of round-trips is different, and the arrival time is also different. I don't understand the solution of the guy upstairs.

I'm going to do this question in the same way as understanding the round-trip of Erli:

The road taken by Shego is x kilometers.

x/5+(51-x)/12=(51-x)/12+[51-x-(51-x)/12*4]/(12+4)*2+x/12

The trouble is, it wouldn't be a hassle if they were separated, and the solution was x = 95 4 and it would take them about 7 hours to reach the end again. That is, they arrived in town at the same time at 1 p.m.

There are flaws in this question, you can't delve into it, just understand it.

Let's first analyze how much time the car travels less, which is obviously 15 minutes, but in fact, the car usually walks one round trip every day, so one way is one minute less.

Then the distance that the minute car can travel is 30 * 1 8 = miles then Mr. is the station that arrives at 4 o'clock, usually the car arrives at 5 o'clock, and this time it is met on the way, so the time should be earlier than 5 o'clock, Mr. actually walked for 1 hour and less minutes, so the time for Mr. to walk on the road is 1-1 8 = 7 8 hours.

So his average speed is (15 4) (7 8) = 30 7 mph.

The distance traveled by the husband is half the distance traveled by the wife's car in 15 minutes, which is miles minutes * 15 minutes 2 = miles; The time taken is 1 hour and 15 minutes, and the speed is miles per hour = 3 miles per hour.

I think: matthaues - manager of the fifth level he is correct.

The simplest one-dimensional quadratic elementary school solution.

Suppose A and B have walked for x minutes when they meet.

Then A's velocity is: the total distance traveled (the total time taken by A), i.e. 1800 (x+8).

B's velocity is: total distance traveled (total time taken by B), 1800 (x+18).

B's velocity * x + A's velocity * x = 1800

That is, 1800 (x+8)*x+1800 (x+18)*x=18001 (x+8)*x+1 (x+18)*x=1(x+18)*x+(x+8)*x=(x+8)*(x+18)x*x=144

x = 12 A's velocity = 1800 (12 + 8) = 90 m min B's speed = 1800 (12 + 18) = 60 m min.

Let A and B have walked a minute when they meet, A has a speed b meters per minute, and B has a speed c meters per minute a * b 18c

a*c＝8b

8b+18c＝1800

From Eq. 1 we get a 18c b

Substituting 2 formulas.

18c*c＝8b*b

B can be substituted into 3 formulas.

Get 12c+18c 1800

C 60 substitution into 4 formulas.

B 90 A speed 90 m/min.

B speed 60 meters per minute.

The two met in 12 minutes.

The key to solving this kind of problem is to grasp:

1.After the two people walked, the itinerary was the same, both were 1800 meters.

2.After the two people meet, they go back to each other, and the combined journey of the two people is 1800 meters.

That's easy to solve :

Suppose the first speed is a, the second speed is b, and the unit of speed is meters. When they met, both walked for C minutes. List the above two keys as equations.

8a+18b=1800

a*c+8a=1800

b*c+18b=1800

According to A=1800 (8+C).

According to b=1800 (18+c).

Substituting has 8*1800 (8+c) +18*1800 (18+c) =1800

The solution is c = 12, and the other answer is -12, rounded.

Then we get a=90 and b=60

The speed of A is 90 meters, and the speed of B is 60 meters.