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Sagot :
Answer:
Mr. Pinter's class has 42 students, Mrs. Rupert's class has 21 students, and Mrs. Althouse's class has 43 students.
Step-by-step explanation:
This question is solved using a system of equations.
I am going to say that:
Mr. Pinter's class has x students.
Mrs. Rupert's class has y students.
Mrs. Althouse's class has z students.
Mr. Pinter's class has twice as many students as Mrs. Rupert's class.
This means that:
[tex]x = 2y[/tex]
Mrs. Althouse's class has 20 less than three times Mrs. Rupert's class.
This means that:
[tex]z = 3y - 20[/tex]
Together they have 106 students.
This means that:
[tex]x + y + z = 106[/tex]
We have x and z has a function of y, so:
[tex]2y + y + 3y - 20 = 106[/tex]
[tex]6y = 126[/tex]
[tex]y = \frac{126}{6}[/tex]
[tex]y = 21[/tex]
And:
[tex]x = 2y = 2(21) = 42[/tex]
[tex]z = 3y - 20 = 3(21) - 20 = 63 - 20 = 43[/tex]
Mr. Pinter's class has 42 students, Mrs. Rupert's class has 21 students, and Mrs. Althouse's class has 43 students.
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