If a number is divisible by 5, it either ends in a 0 or a 5. Since this doesn't end in either, we know it isn't divisible by 5. We should find the next lowest number that is divisible by 5. Since 155 ends in a five, we know it is divisible by 5, and because this is the closest number below 157 that is divisible by 155, we know this is the greatest number of students that could be put equally into 5 classes.
Alternatively, we can divide 157 by 5, and we get 31.4. Since we only want something that divides evenly by 5, we round this down to 31, then multiply by 5 to get 155.
18 * 35 = 630 total students 630 / 30 = 21 total classes An additional 3 classes.
A school has 18 classes with 35 students in each class. In order to reduce class size to 30, how many new classes must be formed?
Any number up to & including 14 in the new classes
boomadoomadonga
17>=(s/2)
3
18 classes with 35 students each equals 630 children total. If you put them in classes of 30, you would need to form 21 classes.
18 * 35 = 630 total students 630 / 30 = 21 total classes An additional 3 classes.
A school has 18 classes with 35 students in each class. In order to reduce class size to 30, how many new classes must be formed?
· There are many reasons as to why students are escaping classes: 1. They have a very strict teacher and they have not completed the work assigned by the teacher 2. The teacher teaching them is dull and the students become disinterested in the subjects.
A school has 18 classes with 35 students in each class. In order to reduce class size to 30, how many new classes must be formed?
(18) x (35) = 30 xx = (18) x (35) / (30) = 21
You would first multiply 35 x 18 to find out how many total students there are, which is 630. Then, you would divide 630/30, since you can only have 30 students maximum in a class. This gives you the answer of 21 classes must be formed.
Depends on the school.
6 classes
3
does special ed student's take a few alternative ed classes (alt ed) classes