Engineering problem+trip problem
First of all, I'll tell you something about fractional engineering. Generally, the total amount will not be given. The solution to this problem focuses on:
1 Take the total workload as the unit "1"
2 work efficiency * working hours = workload
3 Variable relationship: workload ÷ work efficiency = working hours; Workload ÷ working time = working efficiency
4 For example, it takes 6 days for a project to be completed alone, and 65,438+00 days for B to be completed alone, so the work efficiency of A is 65,438+0/6, and that of B is 65,438+0/65,438+00 (that is, all projects working for 65,438+0 days are 65,438).
Let's learn from examples.
Example 1
Team A and Team B can finish a project in 20 days. * * * We worked together for 8 days. A left and B continued to work. 18 days later. If this project is completed by Team A or Team B alone, how many days will it take?
Train of Thought Navigation: Take this project as "1",
When Party A leaves his post, the workload of Party B is: 1- 1/20*8=3/5.
B the time to do this project alone is
18 divided by 3/5 18÷3/5=30 days.
Time for Party A to do it alone:1÷ (1/20-1/30) = 60 days.
Example 2
It takes 15 days for a master and an apprentice to do a job together. If the master is asked to do it for 10 days first, it will take 17 days for the apprentice to do the rest of the work alone. How many days does it take for the apprentice to do this work alone?
Train of Thought Navigation: Since the given condition is "cooperation will be completed in 15 days", then the individual work will be transformed into cooperation in 10 * * How much to do:115 *10; How much is left:1-115 *10 =1/3. How many days does the apprentice work alone: (17-10)/1/3 = 21day.
Write down the analysis:1-115 *10 =1/3.
17- 10=7
7÷ 1/3=2 1
Of course, the equation can be solved, but it is more troublesome:
1/X+ 1/Y = 1/ 15
10/X+ 17/Y= 1
Example 3
A batch of manuscripts, finished in 20 minutes at a time; B 30 minutes alone. Now the two of them are typing these manuscripts together. During the collaboration, Party A walked for 5 minutes because of something, and Party B took a few minutes off, so it took 16 minutes to finish typing. B how many minutes did you rest?
Train of Thought Navigation: Since we don't know how much of 16 minutes is cooperation, and how many minutes Party A and Party B have worked alone, assuming that they have neither left nor rested, then the next question will be easy to do.
Party A and Party B can play without a break 16 minutes: (1/20+1/30) *16 = 4/3.
4/3- 1= 1/3 -
How much can I play in five minutes? 5* 1/20= 1/4
B How long can I play during the break? 1/3- 1/4= 1/ 12
B how long did you rest? 1/ 12÷ 1/30=5/2
That is, B rested for 5/2 minutes.
Example 4
An assignment can be completed in 7 days, followed by 14 days. If A does 10 days first, B can do it for 2 days. Now, in five days, A will finish all the remaining work. How many days will it take to complete?
Mental navigation: General solution: Let A do 1/X and B do 1/Y every day.
Then we can get the equation: 7/X+ 14/Y= 1.
10/X+2/Y= 1
Solution 2: Equivalent substitution method
A (10-7) day workload = B (14-2) day workload.
Namely: A 1 day workload = B 4 days workload.
A (7-5) days' workload = B's 8 days' workload
So B needs 8+ 14=22 days.
The solution will give the answer soon.
Example 5
9 hours for Party A, 0/2 hours for Party B/kloc, and 0/8 hours for Party B to transport the goods to the warehouse. There are the same warehouses A and B. A is in warehouse A and B is in warehouse B, and the goods are moved at the same time. C began to help A lift, and turned to help B on the way. The last three people moved at the same time. Q: How long did C help A and B respectively?
Train of Thought Navigation: Set the total cargo volume of a warehouse as "1". Although it is difficult to determine the processing time of two warehouses C in AB, we still have to "find the variant and keep the same". What remains the same? Because the three of them moved at the same time, that's when the three of them moved.
2 ÷ (1/9+112+118) = 8 hours.
The time for Party C to help Party A move is (1-1/9 * 8) ÷118 = 2.
So what helps B is 8-2=6 hours.
The second part: Travel problem.
Example 1
Two cars, A and B, leave from A and B relatively simultaneously. They first met 50 kilometers away. After meeting, they went on to their destination and returned immediately. The second time they met was 26 kilometers away from B. How many kilometers are there between A and B?
Train of thought navigation: According to the condition of "the first meeting distance is 50 kilometers from A", A traveled 50 kilometers during the first meeting time. So, A and B began to leave each other at the same time. When they met for the second time, A and B went through three whole journeys. That is to say, it took three meeting times, that is, it took three meeting times for A to reach the second meeting place.
So A-B is separate.
50*3-26= 124
Formula s= 3a-b
A is the distance a walks, that is,
B is the remaining distance.
Example 2
Xiao Li went up the mountain from A, crossed the top of B, and went down to C, * * * trip 18km, which took 5 hours. I also know that he goes up the mountain at 3 kilometers per hour and down the mountain at 5 kilometers per hour. How long does it take for Xiao Li to go up the mountain from C and cross the top of B to return to A?
This problem can be solved by "chicken and rabbit in the same cage"
Let all downhill: 5*5=25.
Gap with reality: 25- 18=7
Then the uphill time is 7÷(5-3)=3.5 hours.
Downhill time: 5-3.5= 1.5 hours.
So the distance between AB and BC can be calculated.
The remaining problems will be solved easily.
Example 3
A and B started to climb the mountain from the foot of the mountain at the same time, and went down immediately after reaching the top of the mountain. Both of them went down the mountain twice as fast as themselves. When A reached the top of the mountain, B was 500 meters away from the top of the mountain. When A returned to the bottom of the mountain, B just came down halfway up the mountain. Find the distance from the foot of the mountain to the top of the mountain
Train of Thought Navigation: Assuming A continues to climb the mountain after reaching the top, he can still climb 1/2 and B can also climb 1/4.
At this point, the distance difference is; 500*( 1+ 1/2)=750
750÷( 1/2- 1/4)=3000
Write down the general solution below:
S/V A =(S-500)/V B
S/2V A = 1/2S/2V B +500/V B。
Example 4 (Old question, but classic)
Students in Class A and Class B leave school for the park at the same time. A-class speed is 4 kilometers per hour and B-class speed is 3 kilometers per hour. There is a car in the school with a speed of 48 kilometers per hour. This car can just take a class of students. In order to let the students of two classes arrive in the shortest time, what is the proportion of the distance that students of Class A and Class B need to walk?
It is estimated that many people remember the answer 15: 1 1.
Let's see ...
To get there in the shortest time, only A needs to drive to Park A and B needs to walk at the same time.
Suppose B takes the bus first and A walks. The bus will take Class B to Point C, the students of Class B will get off and walk, and the bus will return to Class A students at Point B. According to a certain time, the distance ratio is equal to the speed ratio:
Simplify the following picture
a……B……C……..D
In fact, this is a proportional solution:
AB(AC+BC)= 4； 48= 1: 12
AB:2BC = 1: 1 1-①
At point C, Class B gets off and walks, and the bus returns to pick up the bus. Then the bus and Class B arrive at a park at the same time.
(BC+BD):CD = 48:3 = 16: 1
2BC:CD= 15: 1 - ②
Compare ① and ②.
AB:CD= 15: 1 1
Trajectory tracking method for solving travel problems (original)
The so-called trajectory tracking method is a method to grasp the relationship between trajectory and S by drawing and solve the answer.
An example is given to illustrate this problem.
Example 1: Party A and Party B set out from A and B at the same time and walked towards each other. When they met for the first time, A was 0/04m away from B/KLOC-,then they went on and returned immediately after reaching their destination. When they met for the second time, B and B were 40 meters apart, and AB was how many meters apart?
a . 176 b . 144 c . 168d . 186m。
Kakashi analysis:
This topic is entitled "The most basic problem of multiple encounters: grasping the meeting time is the key to solving the problem".
This must be: the first meeting took a meeting time t and the second meeting took three meeting times 3t.
Trajectory tracking method:
A - C - D - B
Let C be the first meeting place and D the second meeting place.
Judging from the topic "When we first met, the distance between A and B was104m", it means that B walked104m in a meeting time t.
Track b: BC-CA-AD
We found that when we met for the second time, B took less BD segment than two full S segments, and the BD segment was exactly 40 meters. According to the third meeting time of the second meeting, we can know that B left 104*3.
So 104*3+40=2S S= 176.
It is estimated that some new Q friends will ask: "Why did it take three times to meet for the second time? Why not meet twice? " . Let me deduce this question.
A - C - D - B
Let C be the first meeting place and D the second meeting place.
A leaves for the first time: AC B leaves BC A and B to meet for the first time 1 time, * * * leaves 1 time.
When we met for the second time, A left AC+CB+BD-①.
B is gone. BC+CA+AD-②.
①+②=3S (Party A * * * leaves for 3S)
Party A and Party B left 1S and 1t when they met for the first time.
When Party A and Party B met for the second time, they walked for 3S. Because the speed remained unchanged, the walking time was 3t.
Summarize the following formula: At the nth meeting, Party A and Party B * * * walked (2N- 1) s and spent (2N- 1) meeting time t.
Example 2: Two ships, A and B, set out from the north and south banks respectively, met for the first time at a distance of 260 kilometers from the north bank, continued to drive, and met again at a distance of 200 kilometers from the south bank when returning, seeking the width of the river.
Kakashi analysis:
Drawing: nan-c-d-d.
Similarly, C means the first meeting and D means the second meeting.
According to: "Meet for the first time 260 kilometers away from the north shore", so the track of tracking B is
North C+C south+south d, it is observed that the south d segment is more than1s.
So: 3*260-200=S
Exercise: A and B are driving in opposite directions at the same time, meet at a distance of B 54 kilometers, return immediately after arriving at the other station, and meet at a distance of 42 kilometers. How many kilometers is it between A and B?
Two important ideas of pursuing problems
1, let the interval distance be regarded as the unit 1.
2. Distance difference = speed difference × time
Explain a few examples:
1、
Someone is walking along the tram line. Every 12 minutes, a tram catches up from behind, and every 4 minutes, a tram comes on. The departure interval between the two departure stations is the same, so what is the interval?
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1, let the interval distance be regarded as the unit 1.
2. Distance difference = speed difference × time
Draw a simple picture to help you understand.
Catch up later:->-b-> -(speed difference)
Face to face:->-<; -b-(speed and)
So according to the chart, we can get the following equation.
(1) followed: (v-v) =112.
(2) oncoming: (V +V) = 1/4
( 1)+(2)= = & gt； 2V electricity =112+1/4 =1/3 (the problem is to calculate the departure interval, so the speed should be calculated).
V = 1/6
Display time = distance/speed
Interval =1present1/6
T=6
PS:1/[(112+1/4)/2] = 6.
2、
On a street, a cyclist and a pedestrian walk in the same direction. The speed of cyclists is three times that of pedestrians, and a bus passes by a pedestrian every 10 minute. There is a bus carrying more than one cyclist every 20 minutes. If the bus leaves the departure station at the same time, how many minutes does it leave?
A 10 B 8 C 6 D 4
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1, let the interval distance be regarded as the unit 1.
2. Distance difference = speed difference × time
So there is the following equation:
(1) (V steam -V step) =110
(2) (V steam -3V step) = 1/20
Calculate v steam = 1/8.
T= 1/( 1/8)=8
Solution to the problem of hour hand.
Clockwise problem has two key points.
1 Minute hand moves 6 degrees per minute; The hour hand moves 0.5 per minute (or the minute hand moves 1 grid and the hour hand moves112 grid).
2 The minute hand moves 5.5 squares more than the hour hand (or1112 squares); Think of the pursuit of the hour hand as the pursuit of the degree.
Example 1
At the instant of 14 and 16, the included angle between the hour hand and the minute hand on the clock face is () degrees.
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Analysis: This problem can be regarded as a pursuit problem: at 14, there is a distance between the minute hand and the hour hand, and then find the distance between the minute hand and the hour hand after 16 minutes.
14, the minute hand and the hour hand form 60.
After another 16 minutes, the minute hand moved more than the hour hand in 16 minutes: 16*5.5=88.
88-60=28
Example 2
After 4 o'clock, when the minute hand and the hour hand coincide, it should be 4 o'clock ()?
a 2 1 * 9/ 1 1 B 2 1 * 8/ 1 1 C 2 1 * 7/ 1 1D 2 1 * 6/ 1 1
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Analysis: 4 o'clock, the minute makes an angle with the hour hand 120 degrees, and the minute hand is 6-0.5=5.5 degrees.
When the total distance is 120, the speed difference is 5.5.
So the time is 120÷5.5=2 1 and 9/1.
Example 3
It's 2: 00 15, and in () minutes, the hour hand and the minute hand will reunite for the first time.
a 60/ 1 1 b . 14/ 1 1 c . 264/ 1 1d . 675/ 1 1
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Reference answer: 2 o'clock, 15 minutes, minutes and hour hands coincide between 1 and 2 o'clock, so the next coincidence should be after 3 o'clock, after 3 o'clock, 90/5.5 =180/1minute, kloc. It can also be seen as the time when the hour hand and the minute hand overlap for the second time from 2 o'clock, and then subtract 15 minutes. At 2 o'clock, the angle between the hour hand and the minute hand is 60 degrees. By the second coincidence, the pursuit distance is 360+60=420 degrees, the angular velocity difference is 5.5 degrees/minute, and 420/5.5-15 = 840/1-165/65438+. You can also directly calculate: (2 * 30+360)/5.5-15 = 675/11minute.
Personal answer: 2 o'clock 15 minutes, the degree between the hour hand and the minute hand is 90-(60+ 15*0.5)=22.5 degrees.
But the distance of clockwise pursuit is 360-22.5=337.5 degrees (because it is clockwise pursuit).
337.5/5.5=675/ 1 1
Take the stairs.
1. The escalator in the shopping center runs at a constant speed from bottom to top. The two children thought the escalator was going too slowly, so on the escalator, the boy went up two levels every second and the girl went up three levels every two seconds. As a result, boys arrived in 40 seconds and girls in 50 seconds. Then when the escalator is stationary, the visible escalator steps are ()
Proportional method is really everywhere, and this kind of problem can also be done by proportional method. Set the speed ratio of the three, boys: girls: elevator = 2:1.5: x.
People go up from the bottom, the elevator goes up, and a * * * is the number of steps that the elevator is exposed.
It takes 40 seconds for boys and 50 seconds for girls.
Indeed
40*2+40*x=50* 1.5+50*x gives x=0.5, then all steps 40 * 2+40 * x = 80+40 * 0.5 = 80+20 =100.
2. The escalator runs upward at a constant speed. A boy and a girl walked up the escalator at the same time. Boys are twice as fast as girls. It is known that the boy took 27 steps to reach the top of the escalator, while the girl took 18 steps to reach the top. How many steps are there in the exposed part of the escalator?
Similarly, set the speed to 2:1:x.
The solution of 27/2 * x+27 =18/* x+18 is x=2, so a * * * has 54 levels.
The key to many encounters is the relationship between speed ratio and distance multiple.
Left at the first meeting 1.
When we met for the second time, we walked for 3 s.
When we met for the third time, we walked for 5 seconds.
..............
When they met for the nth time, they walked 2*N- 1 time, and it took 2*N- 1 time to meet.
"Why did the second meeting take three meetings? Why not meet twice? " . Let me deduce this question.
A - C - D - B
Let C be the first meeting place and D the second meeting place.
A leaves for the first time: AC B leaves BC A and B to meet for the first time 1 time, * * * leaves 1 time.
When we met for the second time, A left AC+CB+BD-①.
B is gone. BC+CA+AD-②.
①+②=3S (Party A * * * leaves for 3S)
Party A and Party B left 1S and 1t when they met for the first time.
When Party A and Party B met for the second time, they walked for 3S. Because the speed remained unchanged, the walking time was 3t.
Summarize the following formula: At the nth meeting, Party A and Party B * * * walked (2N- 1) s and spent (2N- 1) meeting time t.
Two cars, A and B, depart from A and B respectively, and keep going back and forth between A and B. It is known that the speed of car A is 15km/h, and the speed of car B is 35km/h, and the meeting place of the third car and the fourth car of car A and B is different by 100km. Find the distance between a and b.
A, 200 km b, 250 km c, 300 km d, 350 km.
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Draw a sketch
A - C - D - B
C stands for the place where we met for the third time and D stands for the place where we met for the fourth time.
The speed ratio is 15: 35 = 3: 7.
The whole process is divided into 10 parts.
The distance of the third trip is: 3*(2*2+ 1)= 15 copies (equivalent to 1.5S).
The distance of the fourth trip is: 3*(2*3+ 1)=2 1 copy.
The distance between two times is 5- 1=4 copies, corresponding to 100KM.
So 10 corresponds to 250KM.
Tell you 2 1 and 15.
Oh, oh, oh, oh, oh, oh, oh, oh
← C
D→
D and C stand for the third and fourth meetings respectively.
The arrow indicates the direction.
1 Simple exercises to consolidate for everyone:
Two cars, A and B, travel in opposite directions from A and B at the same time, meet at 54 kilometers away from B, return immediately after arriving at the other station, and meet at 42 kilometers away from A ... How many kilometers is there between A and B?
Summary of the discussion itinerary of real questions all over the country over the years (1)
1. Party A and Party B walk from place A to place B at the same time, with Party A's 60 m/min and Party B's 90 m/min, and Party B will turn back after arriving at place B..
When meeting A, A needs to walk for another 3 minutes to reach B. How to find the distance between AB and B?
1350
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Kakashi analysis:
Draw a sketch (m points indicate where they met)
A - M - B
According to the proportional method, the time is fixed and the distance ratio is equal to the speed ratio.
So AM:AM+2MB=60:90=2:3.
AM:MB=4: 1
MB=3*60= 180
So the whole journey is 180*(4+ 1)=900.
2. In the morning, Party A starts from a certain place and advances at a constant speed. After a while, Party B started from the same place and moved in the same direction at the same speed. On 10 in the morning, Party B walked 6 kilometers, and they went on. When Party B arrived at the location where Party A arrived 10 in the morning, Party A left16.8km.. Q: At this time, Party B left.
a . 1 1.4 b . 1.4 c . 1.8d . 5.4
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Kakashi analysis:
This problem seems complicated, but it is easy to work out as long as you draw carefully.
Draw their location map at 10.
A - B - C - D -
It can be known that ab = 6bc = xcd = x.
So 6+2X= 16.8.
X=5.4
5.4+6= 1 1.4
3. Party A, Party B and Party C, Party A walks 50 meters per minute, Party B walks 40 meters per minute, and Party C walks 35 meters per minute. Party A and Party B start from place A, and Party C walks in the opposite direction at the same time. After 2 minutes, Party C meets Party B.. So, how many meters is there between A and B?
A.250m B.500 C.750m D. 1275m
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Kakashi analysis:
When A meets C, the distance from B to C is 2*(40+35)= 150.
A walks 10 meter more than B every minute, so when we meet, A walks 150/ 10= 15 minutes.
Total distance S=(V A +V C) * 15.
So the total length is (50+35)* 15= 1275.
This problem can also kill multiples of (50+35=85)85.
4. There is a railway between Station A and bilibili. Two trains, A and B, stop at Station A and bilibili respectively. The 4-minute journey of Train A is equal to the 5-minute journey of Train B. Train B left Station A at 8: 00 a.m. on time. After a while, Train A left Station A for Station B, and the two trains met at 9: 00 a.m. on time. The distance from the venue to Station A and bilibili is 650.
a . 8: 12； b . 8: 15； c . 8:24； D. 8: 30.
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Kakashi analysis:
The speed ratio of Party A and Party B is 5: 4 and the distance ratio is 15: 16, so the time ratio is 3: 4.
3:4=X: 1
X=0.75*60=45
Start at 8: 15.
Simply put, the relationship between distance, speed and time also applies to the proportional algorithm.
For example, distance/speed = time.
Distance ratio/speed ratio = time ratio
5.AB is connected by a highway. Car A starts from place A, and car B starts from place B, and travels at a constant speed along the expressway. When two cars meet, they turn around and drive at each other's speed. After returning to place A, car A turned around again and set off for place B along the expressway at the same speed. Finally, a car and b car arrive at b at the same time. If the speed of car A is x m/s at first, then the speed of car B is () at first.
A.4m/s B. 2X m/s C. 0.5X m/s D.
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Kakashi analysis:
This problem seems complicated, but as long as you analyze it carefully, you can draw the following conclusions:
At the same time, I walked the distance of AB at the speed of A and the distance of 2 AB at the speed of B,
When the time is fixed, the distance ratio is equal to the speed ratio.
So V A: V B = 1: 2.
V b =2X
6. When a person travels by bus, the bus goes13 and falls asleep. When he wakes up, the bus needs to keep moving.
He drove a distance of 1/3 when he was asleep, and he drove a fraction of the whole distance when he was asleep. ()
A.3/8 B.3/7 C. 1/2 D.3/5
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Kakashi analysis:
There is really nothing to say about this problem.
1/3+X+ 1/3 * X = 1
X= 1/2
7. A 280-meter-long train with a speed of 20 meters per second passes through a 2800-meter-long bridge. How long does it take for the train to cross the bridge completely? ( )
A.48b.2 minutes 20 seconds C.2 minutes 28 seconds D.2 minutes 34 seconds.
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Kakashi analysis:
This question is very simple and belongs to the scope of seconds.
Time from starting to getting off the bridge completely = (bridge length+vehicle length)/vehicle speed;
(2800+280)/20= 154s=2 minutes and 34 seconds
8. On the same circular runway, Xiao Chen runs slower than Xiao Wang. When they both run in the same direction, every once in a while.
Meet every 12 minutes; If the speed of two people is constant and one of them runs in the opposite direction, then they will meet every 4 minutes. How many minutes did it take Xiao Chen to finish a lap than Xiao Wang? ()
A.5 B.6 C.7 D.8
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Kakashi analysis:
There is nothing to say about this topic.
Let the total distance be 1, the speed of Xiao Chen be y, and the speed of Xiao Wang be x, then:
4X+4Y= 1
12X- 12Y= 1, X= 1/6, y =112, so 12-6= 6 minutes more.
9. The speed of a ship sailing along the river is 30 km/h, and it is known that sailing along the river for 3 hours at the same speed is equal to sailing against the current for 5 hours, so the voyage of the ship drifting along the river for half an hour is;
A, 1 km b, 2km c, 3km d, 6km.
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Kakashi analysis:
Water speed = (forward speed-backward speed) /2,
(30- 18)/2=6,
So drifting for half an hour is 6 * 1/2 = 3km.
10.a Start from a certain place and move at a constant speed. After a period of time, B moves in the same direction from the same place at the same speed. At K, B is 3 0 meters away from the starting point. They moved on. When B reaches the position of A in K, A is away from the starting point108m. Q: How many meters is B from the starting point at this time?
A.39 B.69 C.78 D. 138
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Kakashi analysis:
This question is similar to the second one.
2X+30= 108
X=39
39+30=69
"Class A and Class B students go to XX with only one car. Party A takes the bus first. . . "Today, I specially summarized five similar topics and dedicated them to you. I hope you can study hard! They are all classic topics!
First of all, let me talk about my solution "three-segment diagram method"
I usually calculate the ratio of three distances by proportional method according to the speed ratio.
a……B……C……..D
That is, the person who takes the bus first gets off at point C and then walks to terminal D.
The car will pick up the first person at B.
As long as the proportion of the three paragraphs is calculated, this problem will be solved easily.