A middle school has built a new four story teaching building, with eight rooms on each floor

Solution: (1) Set an average of X students through the main gate and Y students through the side door every minute

Then 2(x+2y)=560 4(x+y)=800, and the solution is x=120 y=80

A: On average, 120 students can pass through one main gate and 80 students through one side gate every minute;

2) Solution 1:

This building has a maximum of 4 8 45 = 1440 (students), and when it is crowded, the four doors can pass through 5 2 (120 + 80) (1-20%) = 1600 (students), 1600 1440

The 4 gates built comply with safety regulations

Solution 2: You can also find out the time taken by students in the whole building to pass through these 4 doors in an emergency: 1 4402 (120+80) (1-20%)=, so it meets the safety regulations

It's not that hard, is it?

1. The main entrance can pass x students per minute, and the side gate can pass through Y students per minute.

According to the title, the system of equations: x+2y=280

x+y=200

Solution: x=120, y=80

2. The maximum number of students in the whole building is 4*10*45=1800.

The total number of students who can pass through the 4 security gates within 5 minutes: 5*(80*2+120*2)=2000>1800

Therefore, it meets the safety regulations. Look".

A front gate has an average of x students per minute, and a side gate has an average of y students per minute.

1）2x+4y=560

2)4x=4y+800

Solve the system of equations to get x=120 y=80

Second question. There is a maximum of 45 * 10 * 4 = 18005 minutes, and the maximum number of students can pass (120 + 80) * 2 * 5 * 80% = 1600, so it is unqualified, and it cannot be passed in 5 minutes.

Math Problem (1) Use the equation to solve a front door can pass through x people per minute, then the side door is [, with 114 minutes can pass 800 students to set the equation, solve x=120 so the main door 120 per minute, side door 80 per minute (2) each classroom has a maximum of 45 students, a total of 1440 students two main doors and two side doors open at the same time can pass through 400 people per minute due to student congestion, the efficiency of going out is reduced by 20%, then 320 people per minute (400x80%) requirements Evacuate safely through 4 gates within 5 minutes, and 1,600 people can pass through, because 1440 is less than 1600, so safety regulations are met.

Solution: (1) If an average of one main door can pass through x students per minute, because when a main door and a side door are opened at the same time, 800 students can pass through within 4 minutes, so it can be known that when a main door and a side door are opened at the same time, 200 students can pass through in 1 minute, and an average of x students can pass through a main door every minute, so a side door can pass through 200 x students per minute on average.

A main door and two side doors can be opened at the same time, and 560 students can pass through in 2 minutes, so the column can be obtained

2*[x+2(200-x)]=560

That is, x+400-2x=280

The solution is x=120

So on average, 120 students can pass through a front gate and a side gate every minute, respectively.

2) Assuming that the building has a maximum of 45 students per classroom, since the building has 4 floors with 8 classrooms on each floor, it can be seen that:

The school building can accommodate up to 4*8*45 1440 students in an emergency, and the efficiency of going out is reduced by 20%, which is to say:

At this time, a main gate can pass through an average of 120 * (1-20%) = 96 students per minute; While a side door can pass through an average of 80 (1-20%) = 80 students per minute.

According to the regulations, the students in the whole building should be safely evacuated through these 4 doors within 5 minutes, so within 5 minutes, a total of students who can be safely evacuated through these 4 gates are: 5 * 2 * (96 + 80) = 1760 people.

Obviously, since 1760>1440, it can be seen that the construction of these 4 gates meets the safety regulations.

(1) Use the equation to solve a main door can pass through x people per minute, then the side door is [, with 114 minutes can pass through 800 students to set the equation, solve x=120, so the main entrance 120 per minute, side door 80 per minute (2) each classroom has a maximum of 45 students, a total of 1440 students, two main doors and two side doors open at the same time, can pass 400 people per minute, due to student crowding, the efficiency of going out is reduced by 20%, then 320 people per minute (400x80%) requirements Evacuate safely through 4 gates within 5 minutes, and 1,600 people can pass through, because 1440 is less than 1600, so safety regulations are met.

Analysis: (1) According to the meaning of the question, there are two unknowns in this question: how many students pass through a main gate and a side gate per minute on average There are two equal relationships:

When a main door and two side doors are opened at the same time, 560 students can be passed within 2 minutes When a main door and a side door are opened at the same time, 800 students can be allowed to solve the equations within 4 minutes according to the above conditions;

2) According to the data of (1), you can find the number of students who can pass through the four doors in 5 minutes when it is crowded, the maximum number of students in the teaching building, and the time that students in the whole building pass through these four gates, and then compare.

Solution: (1) Set an average of X students through the main gate and Y students through the side door every minute

Answer: On average, 120 students can pass through a front gate and 80 students can pass through a side door every minute;

2) Solution 1:

This building has a maximum of 4 8 45 = 1440 (students), and when it is crowded, the four doors can pass through 5 2 (120 + 80) (1-20%) = 1600 (students), 1600 1440

The 4 gates built comply with safety regulations

(1) When opening a main gate and two side doors, students can divide 560 by 2 within 1 minute, which is equal to 280 people; When opening a main gate and a side door, 800 students divided by 4 in 1 minute equals 200 people. 1 minute a side door through 280 200 80 people. 1 minute through the main entrance 200 80 120 people.

2) The number of students in the school is 4x10x45 1800. In case of emergency, the number of people who pass in 5 minutes is (120x2 80x2) x5x (1 20%) 1600 people. 1,600 people are less than 1,800 people, so the 4 gates built do not meet safety regulations.

Set up a front gate with an average of x students per minute, and a side gate with an average of x students per minute 1) 2x+4y=560

2)4x=4y=800

Solve the system of equations to get x=120 y=80

The second question has a total of 45 * 10 * 4 = 18005 minutes, and the maximum number of students can pass (120 + 80) * 2 * 5 * 80% = 16001800>1600

So if you don't qualify, you can't pass it all in 5 minutes.

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Solution: (1) Set an average of X students through the main gate and Y students through the side door every minute

Answer: On average, 120 students can pass through a front gate and 80 students can pass through a side door every minute;

2（x+2y）=560

4（x+y）=800

x=120y=80

2) Solution 1:

This building has a maximum of 4 8 45 = 1440 (students), and when it is crowded, the four doors can pass through 5 2 (120 + 80) (1-20%) = 1600 (students), 1600 1440

The 4 gates built comply with safety regulations

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Good luck with your new semester!

Solution: x=80

A: On average, 120 students can pass through one main gate and 80 students through one side gate every minute.

2) The maximum number of students in this building: 4 8 45 = 1440 people can pass through 4 doors in 5 minutes when crowded 5 2 (120 + 80) (1-20%) = 1600 (students).

Construction of 4 gates to comply with safety regulations.

(1) If the number of students passing through one main door per minute is x, then the number of students passing through one side door per minute is 800 4-x=200-x

According to the problem, there is 2x(x+2x(200-x))=560, and the unary equation is solved to obtain x=120,200-x=80

So 120 students pass through the main entrance every minute, and 80 students pass through the side gate.

2) In case of emergency, when all four doors are open, the number of students who can pass in 5 minutes = 5 x (120 + 120 + 80 + 80) x (1-20%) = 1,600.

The total number of students in the teaching building = 4x10x45 = 1800 Answer: 1600 is less than 1800, so these 4 doors do not meet the safety regulations.

Assuming that the number of people passing through the main gate per minute is x, and the number of people passing through the side gate per minute is y, the first question can list two formulas: 2 (x + 2 y) = 560;x+y=200 gives x=120 and y=80

The second question is very simple. There are 4 floors in total, with 8 classrooms on each floor, and each classroom has a maximum of 45 students, which means that the maximum number of students is 4*8*45=1440 students. The number of people who should be able to pass through the 2 main gates and 2 side gates per minute is 400, and the efficiency is reduced by 20% in case of emergency, that is, the number of people who pass is 400 * 80% = 320 minutes.

Then you can pass 5*320=1600 people in 5 minutes. Comply with safety regulations.