Question

per 13, 2003 rd problem To Students: 3.5 play Hooky a game many play both th is many play only Hokey there are so employeer our and is drink tea IADS per a hou y drink both A 52 both the games​

Answer 1

It appears to be a problem related to a group of students playing two games, Hooky and Tea. The number of students who play Hooky is 3.5 times the number of students who play Tea. Additionally, there are some students who play both games, and the total number of students who play both games is 52.

Let's denote:

- H as the number of students who play Hooky,

- T as the number of students who play Tea.

Given that
`\(H = 3.5T\)` and 52 students play both games, we can set up an equation:

`\[ H + T - 52 = \text{Total number of students} \]`

Now, substitute
`\(H = 3.5T\)`:

`\[ 3.5T + T - 52 = \text{Total number of students} \]`

Given that
`\( H = 3.5T \)`, substitute this into the equation:

`\[ 3.5T + T - 52 = \text{Total} \]`

Combine like terms:

`\[ 4.5T - 52 = \text{Total} \]`

Now, you mentioned that there are 52 students who play both games. If these students are counted twice (once for Hooky and once for Tea), we need to subtract them once:

`\[ 4.5T - 52 - 52 = \text{Total} \]`

Simplify:

`\[ 4.5T - 104 = \text{Total} \]`

This equation can be used to find the total number of students given the values for T.