Two boxes, A and B, contain some red and blue beads. In Box A, the ratio of the number of red beads to the number of blue beads is 2: 1. In Box B, the ratio of the number of red beads to the number of the blue beads is 1: 3. Box A has thrice as many beads as Box B.
(a) Find the ratio of the number of red beads in Box A to the number of blue beads in B operatorname Box B.
(b) If 105 red beads are taken out from Box A and placed in Box B, the ratio of the number of red beads to the number of blue beads in Box B becomes 3: 2. What is the total number of beads in the 2 boxes?
To solve this problem, let's break it down into steps:
Step 1: Define the variables.
Let's assume that the number of red beads in Box A is represented by 'x', and the number of blue beads in Box A is represented by 'y'. Similarly, let's represent the number of red beads in Box B as 'a' and the number of blue beads in Box B as 'b'.
Step 2: Set up the given conditions as equations.
From the problem statement, we have the following information:
- In Box A, the ratio of the number of red beads to the number of blue beads is 2:1. This can be expressed as x:y = 2:1, or x = 2y.
- In Box B, the ratio of the number of red beads to the number of blue beads is 1:3. This can be expressed as a:b = 1:3, or a = (1/3)b.
- Box A has thrice as many beads as Box B. This can be expressed as x + y = 3(a + b).
Step 3: Solve the system of equations.
We now have three equations with three variables:
Equation 1: x = 2y
Equation 2: a = (1/3)b
Equation 3: x + y = 3(a + b)
We can substitute Equation 1 and Equation 2 into Equation 3 to solve for the variables:
2y + y = 3((1/3)b + b)
3y = 3((4/3)b)
3y = 4b
Now, we have a relationship between y and b.
Step 4: Solve for y in terms of b.
Dividing both sides of the equation 3y = 4b by 3, we get:
y = (4/3)b
Step 5: Solve for x in terms of b.
Using Equation 1: x = 2y
Substituting y = (4/3)b, we get:
x = 2(4/3)b
x = (8/3)b
Step 6: Solve for a in terms of b.
Using Equation 2: a = (1/3)b
Step 7: Set up the second part of the problem.
In the second part of the problem, 105 red beads are moved from Box A to Box B. This means that the new number of red beads in Box A is (x - 105), and the new number of red beads in Box B is (a + 105). The ratio of red beads to blue beads in Box B becomes 3:2. Therefore, we can set up the following equation:
(a + 105)/(b + 105) = 3/2
Step 8: Solve for b.
Using the equation from Step 7, we can cross-multiply and solve for b:
2(a + 105) = 3(b + 105)
2a + 210 = 3b + 315
2a - 3b = 105
Step 9: Solve the system of equations to find the values of a and b.
We have two equations:
Equation 1: 3y = 4b
Equation 2: 2a - 3b = 105
We can substitute y = (4/3)b from Equation 1 into Equation 2:
2a - 3(4/3)b = 105
2a - 4b = 105
Multiplying Equation 2 by 2 gives us:
4a - 8b = 210
Now, we have a system of equations:
2a - 4b = 105
4a - 8b = 210
By subtracting the first equation from the second equation, we get:
2a = 105
Solving for a, we find:
a = 105/2 = 52.5
Since the number of beads must be a whole number, we can conclude that the values for a and b are not integers. This means that the problem has no valid solution with the given parameters.
Therefore, we cannot determine the total number of beads in the two boxes.