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Each of them wants to get in a car with fewer people" is key. Because "each of them wants to ride in a car with fewer people", there is only one difference between the number of people in the car. That is, there will be no empty seats for car A 3 and car B for 1 empty seat.
They still have 35 people on the bus, and if it is known that all buses have 1 less empty seats than the initial waiting number, then 34 people is just right. There are 17 cars, which are exactly divisible by 34, so each car has an equal number of empty seats. So 34 empty seats per car divided by 17 = 2
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Here's why: The title says, "Everyone wants to ride in a car with fewer people.""This means that everyone will run to the last car at the beginning, if we have 3 cars.
When the number of people in the last car is the same as the number of people in the penultimate car, the number of people in the penultimate car will be minimized as long as the previous person is present, and then someone will board the car until the last car and the penultimate car have the same number of people as the first car, and the number of seats in the three cars is the same. Then the remaining number of people waiting for the bus is the number of empty seats on the car + the number of overloaded. The same is true for 4 cars.
So we can generalize this logic; When there are n buses and the overloaded number of people is m, as long as it is satisfied that each car is manned, then when the number of people waiting for the bus is t*n + m people, t is the number of vacant seats of each bus.
So according to the title t=2
Good luck with your studies! (Anyway, didn't you ask...)
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(82 + 90 + 118 + 67) 3 = 119 books (the book that each class should get) is divided into 1st grade = 119-82 = 37 books.
Assigned to the second grade = 119-90 = 29 books.
Assigned to the third grade = 119-118 = 1 book.
I hope it can help you, @数学辅导团 I wish you progress in your learning, please ask if you don't understand, please adopt in time if you understand! (*
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(82 + 90 + 118 + 67) 3 = 357 3 = 119 books for the first grade: 119-82 = 37 books.
Then to the second grade: 119-90 = 29 books.
Give the third grade again: 119-118 = 1 book.
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Set up x borrowed in the third grade.
82+90+67-x=2*(118+x)
Solve it. x=1 set the first grade to borrow y.
82+y=66-y+90
y=37, 2nd grade, 29
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82+90+118+67=119
The first grade is divided into 119-82 = 37 books.
The second grade is divided into 119-90 = 29 books.
The third grade is divided into 119-118 = 1 book.
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Solution: Suppose each grade borrows x books, according to the problem, x-82) + (x-90) + (x-118) = 67 to get x = 119
Then it is divided into the first grade: 119-82 = 37 (book) the second grade: 119-90 = 29 (book) and the third grade: 119-118 = 1 (book).
A: ......
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First of all, the sum of the original books of the three grades is 290
Buy another 67 copies for 357 stupid and divide by 3 to get 119
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(82+90+118+67) 3=119 (Root) First Grade: 119-82=37 (Root);
2nd grade: 119-90=29 (Ben);
Year 3: 119-118=1 (Ben);
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