The ratio of students Ms.Finn has to Mr.D is 7:4.Ms.finn has 15 more students than Mr.D.After sending some students to Mr.D,Ms.finn will have 5/6 as many students as Mr.Finn.
a)How many students did Ms.Finn have in the beginning?
b)How many students did Ms.Finn send to Mr.D?
To solve this problem, we can use algebraic equations to represent the given information. Let's assign variables to the unknown quantities.
Let's say Ms. Finn has x students and Mr. D has y students.
According to the given information, the ratio of the number of students Ms. Finn has to Mr. D is 7:4. This can be written as:
x/y = 7/4
4x = 7y (Equation 1)
It's also given that Ms. Finn has 15 more students than Mr. D, which can be written as:
x = y + 15 (Equation 2)
After sending some students to Mr. D, Ms. Finn will have 5/6 as many students as Mr. D. Let's say she sends z students to Mr. D. Therefore, the number of students Ms. Finn will have after sending z students is (x - z), and the number of students Mr. D will have after receiving z students is (y + z). According to the given information, this can be written as:
(x - z) = (5/6) * (y + z) (Equation 3)
Now we can solve these equations to determine the values of x, y, and z.
To solve the system of equations (Equations 1, 2, and 3), we can use the substitution method:
From Equation 2, we know that x = y + 15. We can substitute this value of x into Equation 1:
4(y + 15) = 7y
4y + 60 = 7y
60 = 3y
y = 20
Substituting the value of y back into Equation 2:
x = 20 + 15
x = 35
Therefore, Ms. Finn had 35 students in the beginning (a).
To find out how many students Ms. Finn sent to Mr. D (b), we substitute the values of x and y into Equation 3:
(35 - z) = (5/6)(20 + z)
210 - 6z = 100 + 5z
11z = 110
z = 10
Therefore, Ms. Finn sent 10 students to Mr. D.