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Mathematical problems.
Solution: Set X students in this middle school on duty.

There are 70 people left.

* * * There are (X-70) people on duty at the intersection.

Arrange four people at each intersection.

* * There is a general intersection (X-70)÷4.

Also, there are 8 people at each intersection.

And there is an intersection with less than 8 people.

∴ There are (total intersections-1) intersections executed by 8 people, (negative 1 means subtracting the last intersection with less than 8 people).

∴ The last intersection with less than 8 people is: total number of students -8× (total number of students-1).

∴ Yes: X-[(X-70)÷4- 1]×8 people.

=X-(2X- 140-8)

= 148-X people are on duty at the last intersection.

∵ There are less than 8 people at the last intersection, but not less than 4 people.

∴ inequality: 4 ≤ 148-x < 8.

The solution is 140 < x ≤ 144.

∵X is a positive integer (people can't be decimals or fractions)

∴X= 14 1 or 142 or 143 or 144

① When X= 14 1, there are total intersections =(X-70)÷4= 17.75 intersections for duty.

② When X= 142, the total number of intersections =(X-70)÷4= 18 is on duty.

③ When X= 143, the total intersection of duty =(X-70)÷4= 18.25 intersection.

④ When X= 144, there are total intersections =(X-70)÷4= 18.5 intersections for duty.

The intersection is a positive integer.

∴ Excluding ① ③ ④.

To sum up, this middle school has selected 142 students to be on duty, and there are 18 intersections to be on duty.

A: There are 142 students on duty in this middle school, and there are 18 intersections on duty.

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