Question

A school principal plans to form teams form 390 third graders, 234 fourth graders, and 1365 fifth-graders so that thereis the same number of students form each grade level on each team.If all students participate, what isthe largestpossible number of teams and now manystudents will there be on each team?

Answer 1

- There are 39 children in each group

- The number of groups that can be formed is 51 groups

SOLUTION

Problem Statement

The question tells us to form teams from 3 groups of children with 390, 234, and 1365 children respectively. We are to arrange them such that they have the same number of children a group. We are asked to find:

1. The number of children in each group.

2. The largest number of groups

Solution

Question 1:

- The number of children in each group must be the same number. That is, if there are x number of children in group 1, there must be x children in all the other groups as well.

- This demands that each group must contain the same number of children from each grade. That is, if we take x children from the third graders to fill up group 1, then, we must also take x children from the third graders for all the other groups.

This is true for all the other grades.

- This means that we are looking for a number common to all the 3 grades. If we can find the largest number common to the number all the grades, then, that number represents the total number of children in each group. Thus, if we find that y is the number common to 390, 234, and 1365, then, the number of children in a group must be y.

- This value of y is, by definition, the Highest common factor of the 3 numbers, 390, 234, and 1365. This is because the Highest Common Factor is simply the largest number that is common to a set of numbers.

- Thus, to find the number of children in each group, we simply find the Highest Common Factor of the numbers 390, 234, and 1365.

`\begin{gathered} 390=2*5*3*13 \\ 234=2*3*3*13 \\ 1365=5*7*3*13 \\ \\ \text{Notice that the number common to all 3 numbers is:} \\ 3*13=39 \\ \\ \text{Thus, the HCF is 39} \end{gathered}`

- This means that there are 39 children in each group

Question 2:

- If each group has 39 children from the 3 grades and each group takes an equal number of children from the grades, we can easily calculate the number groups can be formed.

`\begin{gathered} \text{There are 39 children in each group.} \\ \text{But we are taking equal number of kids from each grade.} \\ \\ \text{Thus, the number of groups formed from the third graders are:} \\ (390)/(39)=10groups \\ \\ \text{The number of groups formed from the fourth graders are:} \\ (234)/(39)=6\text{groups} \\ \\ \text{The number of groups formed from the fifth graders are:} \\ (1365)/(39)=35\text{groups} \\ \\ \text{Thus, the total number of groups formed is:} \\ (10+6+35)=51groups \end{gathered}`

- The number of groups that can be formed is 51 groups

Final Answer

- There are 39 children in each group

- The number of groups that can be formed is 51 groups