Question

В двух четвертых классах учится менее 50 школьников. За контрольную работу по математике 1/7 часть учеников получила пятерки, 1/3 -четверки, 1/2-тройки. Остальные работы оказались неудовлетворительными. Сколько было таких работ

Answer 1

Нужно число кратное 7,3,2 и которое меньше 50
то есть подходит число 42
42 ученика писало контрольную работу
42/7= 6 учеников получили пятёрки
42/3= 14 учеников получили четвёрки
42/2= 21 ученик получил тройку
42-6-14-21=1 ученик получил два
Ответ: 1 ученик получил два

Answer 2

I found some web search results that might help you with your question. Here is a detailed answer based on the information from the first result:

Let x be the number of students in the two fourth-grade classes. We know that x < 50, since there are less than 50 students.

We also know that the fractions of students who got different grades add up to 1, since they cover all the students. So we can write:

(1/7)x + (1/3)x + (1/2)x + y = x

where y is the number of students who got unsatisfactory grades.

Simplifying this equation, we get:

(13/14)x + y = x

Subtracting (13/14)x from both sides, we get:

y = x - (13/14)x

y = (1/14)x

Now we need to find the value of x that satisfies the inequality x < 50 and makes y an integer, since we cannot have a fraction of a student.

The only possible values of x that make y an integer are multiples of 14, since 14 is the denominator of y. So we can try the multiples of 14 that are less than 50: 14, 28, and 42.

If x = 14, then y = (1/14)14 = 1. This means there is only one unsatisfactory grade.

If x = 28, then y = (1/14)28 = 2. This means there are two unsatisfactory grades.

If x = 42, then y = (1/14)42 = 3. This means there are three unsatisfactory grades.

Therefore, the possible number of unsatisfactory grades are 1, 2, or 3, depending on the number of students in the two classes.