Super difficult mathematical problems (solved by equation of degree 1 with one variable) 1. The train was blocked in the middle of the way and delayed for six minutes. Then it increased its speed from 40 kilometers per hour to 50 kilometers per hour. How many kilometers can we make up for the delay? 2. A and B set out from a to B at the speed of 4 kilometers per hour at the same time. After 2.5 kilometers, a had to go back to get a document. He went back at the speed of 6 kilometers per hour and chased B at the same speed. As a result, they arrived at B at the same time. It was known that a had been delayed for 15 minutes in the office to get the document, so they could find the distance between a and B 3. Party A and Party B process 284 parts. Party A processes 48 parts per hour, and Party B processes 70 parts per hour. Party a first processes one hour, and then Party B cooperates with Party A. how many hours has Party B finished the task? Please write the process! Thank you~~~~~~

Super difficult mathematical problems (solved by equation of degree 1 with one variable) 1. The train was blocked in the middle of the way and delayed for six minutes. Then it increased its speed from 40 kilometers per hour to 50 kilometers per hour. How many kilometers can we make up for the delay? 2. A and B set out from a to B at the speed of 4 kilometers per hour at the same time. After 2.5 kilometers, a had to go back to get a document. He went back at the speed of 6 kilometers per hour and chased B at the same speed. As a result, they arrived at B at the same time. It was known that a had been delayed for 15 minutes in the office to get the document, so they could find the distance between a and B 3. Party A and Party B process 284 parts. Party A processes 48 parts per hour, and Party B processes 70 parts per hour. Party a first processes one hour, and then Party B cooperates with Party A. how many hours has Party B finished the task? Please write the process! Thank you~~~~~~

Super difficult?. 1. Set the distance of s km to make up for the delay: that is to say, within s km, 50 km / h is 6 minutes less than 40 km / h: S / 40 = (s / 50) + (6 / 60) = > s = 20 km 2. Set the distance of a and B as s, and the total driving time of a and B is equal: S / 4 = (s / 6) + (2.5 / 6) + (2.5 / 4) +

A mathematical problem ` 1-variable quadratic equation application problem
When someone starts from a, he walks 4 kilometers, then takes a car 10 kilometers to B. then he goes back to B by bike. As a result, the round trip time is equal, and the walking speed of that person can be calculated? (action x)
14 / (x + 8) = 4 / x + 10 / (x + 16)

First of all, the total distance is 14 kilometers
14/ (X+8) = 4/X + 10/(X+16)
After that, it's easy to ask. It's asked according to the same time

A mathematics problem in grade one of junior high school (equation solution)
Salesperson: come and buy the special eggs. The original price is 14 yuan per box, and the current price is 12 yuan per box. There are 30 eggs in each box. Customer B: I bought two boxes of special eggs. After 18 days, the remaining 20 eggs are all broken
Q: is it worthwhile for customer B to buy 2 boxes of eggs? Why?
Fang Chengjie

It's not worth it
After 18 days, the remaining 20 eggs were all broken, that is, only 30 * 2-20 = 40 eggs were used
Set the original price of each egg X Yuan, B bought two boxes of eggs, the actual price is y yuan
If 30x = 14, then x = 0.46666 yuan
40y = 12 * 2 is y = 0.6 yuan
X＜Y
So it's not cost-effective

Xiao Ming and his classmates went to the forest park during the May Day holiday. They rented a small boat at pier a by the stream and rowed upstream at a speed of about 4 km / h. after arriving at destination B, they returned along the same road. The speed increased by 50%. It took 20 minutes less to return to pier a than to go there

Suppose the distance between a and B is x km, then: X4 − X4 × (1 + 50%) = 2060, the solution is: x = 4, answer: the distance between a and B is 4 km

Solving equations. One variable quadratic~
(x-3)(x+5)=4x-7

x²+2x-15=4x-7
x²-2x-8=0
（x-4）（x+2）=0
∴x1=4,x2=-2

Use arithmetic solution and equation solution to solve the problem. 15 points will be awarded to those who answer according to the requirement!
If the car goes from city a to city B and the bus goes from city B to city a 10 hours later, the two cars will meet in city C. if the bus goes from city B to city a and the car goes from city a to city B 10 hours later, the two cars will meet in city D. if ac-cb = 350 km and bd-da = 130 km, the speed of the bus is 2 / 3 of the speed of the car, What's the speed of the car and the bus? Use the arithmetic solution and the equation solution to solve the problem. Those who don't answer according to the requirements won't get any points, and those who answer according to the requirements will get another 15 points!
There is a question like this: for a batch of parts, if Party A processes them 10 days later and Party B completes the joint production, Party A makes 350 more parts than Party B; if Party B processes them 10 days later and Party A completes the joint production, Party B makes 130 more parts than Party A. if this batch of parts is completed by Party A and Party B, how many more parts does Party A make than Party B, The second plan is to make the second batch. At the end of the project, Party A made 220 more parts than Party B. that is to say, when these two batches of parts were completed by two people, Party A made 220 more parts than Party B. then half of them were completed by two people, and Party a made 110 more parts than Party B. the two-step arithmetic synthesis formula solves the problem! The work problem can be transformed into the travel problem, Then, if the car and bus run from AB to ab at the same time, the car runs 110 kilometers more than the bus
Whole journey: 110 / [3 / (3 + 2) - 2 / (3 + 2)] = 550 (km) when meeting according to the first condition (550-350) / 2 = 100 (km) car speed: [550-100 * (1 + 3 / 2)] / 10 = 30 (km) bus speed: 30 * (2 / 3) = 20 (km)

(1) Let the speed of the car and the bus be x and Y respectively, and the total mileage be s. The speed of the bus is 2 / 3 of the speed of the car, y = 2 / 3xac CB = 350 bd-da = 130 (s-10x) / (x + y) * (X-Y) + 10x = 35010y - (s-10y) / (x + y) * (X-Y) = 130y = 2 / 3x. The solution is x = 30y = 20 (2). The arithmetic solution is no! I hope the owner will tell me the answer

Use arithmetic solution and equation solution to solve the problem. 10 points will be awarded to those who answer according to the requirements!
The speed of B is 6 / 7 of that of A. when B meets, the speed of B increases by 1 / 5km. When B arrives at a, a arrives at B at the same time. How many kilometers are the original speed of a and B? Use arithmetic solution and equation solution to solve the problem. Those who don't answer according to the requirements will get 10 points!

Although some complex, but creative! There is a simpler way! I'll add another 10 points when I come up with it! Would you like to have a try?

Arithmetic or equation solution,
Car a and car B leave from a and B at the same time. Car B travels 10% of the whole journey every hour. Car a reaches the midpoint of a and B hours earlier than car B. when car B reaches the midpoint, car a continues to drive 25 kilometers to C. how many kilometers are there between a and B?
Car a and car B leave from a and B at the same time. Car B travels 10% of the whole journey every hour. Car a reaches the midpoint of a and B one third of an hour earlier than car B. when car B reaches the midpoint, car a continues to drive 25 kilometers to C. how many kilometers are there between a and B?

Suppose the total length is 10 parts, then B drives 5 parts to the middle point, divided by the time (each hour is 10% of the whole journey, that is, 10 hours in total, half is 5 hours)
It takes 5 hours for B to reach the destination and 5-1 / 3 = 14 / 3 hours for a
B is 1 / 10 of the whole journey per hour, and a is 3 / 28
It took five hours for B to reach the midpoint, and a drove 15 / 28 of the whole journey
So the AB distance is 25 (1 - 15 / 28) = 25 × 28 / 13 = 700 / 13 km

It's urgent! Use arithmetic,
Dongdong and Honghong start to go upstairs at the same time. It takes 240 seconds for Honghong to reach the 4th floor and 300 seconds for Dongdong to reach the 7th floor. When Honghong reaches the 6th floor, which floor is Dongdong on?

It took 240 seconds for Honghong to reach the 4th floor and 300 seconds for Dongdong to reach the 7th floor. When Honghong reached the 6th floor, which floor was Dongdong on?
Because you don't have to climb the first floor, do you understand!
So Xiao Hong's speed: 240 / (4-1) = 80 seconds
So Xiaodong's speed: 300 / (7-1) = 50 seconds
80 * (6-1) = 400 seconds
400 / 50 = 8
8 + 1 = 9 floors

Equation and arithmetic solution! How to find the equation's equivalent relation,
For a pile of pears, three more than two, four more than three, five less than one, how many pears does this pile need?
I know the answer is 59, but how to calculate? One variable equation and arithmetic solution!

The least common multiple of 3, 4 and 5 is 60
60-1 = 59
Equation——
Let there be at least X
60=x+1
x =59