A small school has 130 students who occupy three classrooms: A, B, and C. After the first period of the school day, half the students in room A move to room B, one-fifth of the students in room B move to room C, and one-third of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room?
To solve this problem, let's break it down into steps:
1. Start by assigning variables to represent the number of students in each room before the movement. Let's call the number of students in room A, B, and C as A, B, and C respectively. We know that A + B + C = 130 since there are 130 students in total.
2. According to the given conditions, after the movement, half the students in room A move to room B. This means that B = A + (1/2)A. Simplifying this expression, we get B = (3/2)A.
3. Additionally, one-fifth of the students in room B move to room C. Therefore, C = B + (1/5)B. Simplifying this expression, we get C = (6/5)B.
4. Lastly, one-third of the students in room C move to room A. Therefore, A = C + (1/3)C. Simplifying this expression, we get A = (4/3)C.
5. We can now solve these equations simultaneously. Since we have expressions for B in terms of A and C in terms of B, we can substitute these values to obtain an equation that only involves one variable.
Substituting B = (3/2)A into C = (6/5)B, we get C = (6/5)(3/2)A => C = (9/5)A.
Substituting C = (9/5)A into A = (4/3)C, we get A = (4/3)(9/5)A => A = (12/5)A.
6. Now, we can solve for A:
A = (12/5)A
1 = 12/5
5 = 12
A = (5/12) * 130 (since A represents a fraction of the total number of students)
A ≈ 54
7. Using the value of A, we can find the values of B and C:
B = (3/2)A ≈ 81
C = (9/5)A ≈ 108
Therefore, there are approximately 54 students in room A, 81 students in room B, and 108 students in room C.
Let's assume the number of students in room A is x, in room B is y, and in room C is z.
According to the given information:
After the first period, half the students in room A move to room B, so the new number of students in room A is x/2, and the new number of students in room B is y + x/2.
One-fifth of the students in room B move to room C, so the new number of students in room B is (y + x/2) - (1/5)*(y + x/2), which simplifies to (4/5)*(y + x/2), and the new number of students in room C is z + (1/5)*(y + x/2).
One-third of the students in room C move to room A, so the new number of students in room A is x/2 + (1/3)*(z + (1/5)*(y + x/2)), which simplifies to x/2 + (1/3)*(z + (1/5)*(y + x/2)), and the new number of students in room C is z + (1/5)*(y + x/2) - (1/3)*(z + (1/5)*(y + x/2)), which simplifies to z + (1/5)*(y + x/2) - (1/3)*(z + (1/5)*(y + x/2)).
Equating the number of students before and after the movements, we can set up the following equations:
x = x/2 + (1/3)*(z + (1/5)*(y + x/2))
y = (4/5)*(y + x/2)
z = z + (1/5)*(y + x/2) - (1/3)*(z + (1/5)*(y + x/2))
To solve these equations, we can simplify and rearrange them:
2x = x + (2/3)*(z + (1/5)*(y + x/2))
5y = 4y + 4/5x
3z = 5/3(z + 1/5(y + x/2))
Simplifying further:
x = (2/3)*(z + (1/5)*(y + x/2))
y = 4/5y + 4/5x
z = (5/3)*(z + 1/5(y + x/2))
Multiplying through by the denominators:
6x = 2z + 2/5*(y + x/2)
5y = 4y + 4x/5
15z = 5z + 1/3*(y + x/2)
Simplifying further:
6x - 2z = 2/5*y + 4/5*x
y = 4y + 4x/5
15z - 5z = 1/3*y + 1/6*x
Combining like terms:
12x - 2z = 2/5*y
4y = 4x/5
10z = 1/3*y + 1/6*x
From the second equation, we get:
y = x/5
Substituting this value of y in the first and third equation, we get:
12x - 2z = 2/5 * (x/5)
10z = 1/3 * (x/5) + 1/6 * x
Simplifying further:
12x - 2z = 2/25 * x
10z = 1/15 * x + 1/6 * x
Multiplying through by 25, 15, and 30 respectively to eliminate fractions, we get:
300x - 50z = 2x
300z = 2x + 5x
Consolidating terms:
298x - 50z = 0
300z = 7x
Now we can find the values of x and z by solving this system of equations.
Dividing the two equations, we get:
x/z = 50/7
Since x and z are integers, the only possible values are x = 50 and z = 7.
Substituting these values back into the second equation, we get:
300z = 7*50
300z = 350
z = 350/300
z = 1.17
Since z is not an integer, this is not a valid solution.
Therefore, there is no solution to this problem, or there may be missing or contradictory information in the given problem statement.
to start,
a+b+c = 100
After the room changes, we have
a/2 + c/3 = a
4b/5 + a/2 = b
2c/3 + b/5 = c
or,
a/2 = c/3
a/2 = b/5
b/5 = c/3 = a/2
so, substituting in,
a + 5a/2 + 3a/2 = 100
2a + 5a + 3a = 200
a = 20
b = 50
c = 30
during the move, 10 students move to each room.