Three solutions, A, B and C, have concentrations as, b% and cl, respectively. When 100ml of A, 40ml of B and 60ml of C are mixed, we get a solution D of concentration 33.5% When 40ml of A, 100ml of B and 60ml of Care mixed, we get a solution E of concentration 35%. When 60ml of A, 40ml of B and 100ml of C are mixed, we get a solution F of concentration 38.5%. Find the value of (a+bc)

Let's define the concentrations of solutions A, B, and C as follows:

- Solution A has concentration \( a \% \)
- Solution B has concentration \( b \% \)
- Solution C has concentration \( c \% \)

We are given three mixtures:

1. **Mixture D**: 100 ml of A, 40 ml of B, and 60 ml of C, with resulting concentration 33.5%.
2. **Mixture E**: 40 ml of A, 100 ml of B, and 60 ml of C, with resulting concentration 35%.
3. **Mixture F**: 60 ml of A, 40 ml of B, and 100 ml of C, with resulting concentration 38.5%.

The concentration of a mixture is given by the weighted average of the concentrations of its components, where the weights are the volumes of each solution in the mixture.

### Equation for Mixture D:
\[
\frac{100a + 40b + 60c}{100 + 40 + 60} = 33.5
\]
\[
\frac{100a + 40b + 60c}{200} = 33.5
\]
\[
100a + 40b + 60c = 6700 \tag{1}
\]

### Equation for Mixture E:
\[
\frac{40a + 100b + 60c}{40 + 100 + 60} = 35
\]
\[
\frac{40a + 100b + 60c}{200} = 35
\]
\[
40a + 100b + 60c = 7000 \tag{2}
\]

### Equation for Mixture F:
\[
\frac{60a + 40b + 100c}{60 + 40 + 100} = 38.5
\]
\[
\frac{60a + 40b + 100c}{200} = 38.5
\]
\[
60a + 40b + 100c = 7700 \tag{3}
\]

Now, we have a system of three equations:

\[
100a + 40b + 60c = 6700 \tag{1}
\]
\[
40a + 100b + 60c = 7000 \tag{2}
\]
\[
60a + 40b + 100c = 7700 \tag{3}
\]

We will solve this system of equations to find \( a \), \( b \), and \( c \), and then calculate \( a + bc \).

### Step 1: Solve the system of equations

First, subtract equation (2) from equation (1):

\[
(100a + 40b + 60c) - (40a + 100b + 60c) = 6700 - 7000
\]
\[
60a - 60b = -300
\]
\[
a - b = -5 \tag{4}
\]

Next, subtract equation (3) from equation (1):

\[
(100a + 40b + 60c) - (60a + 40b + 100c) = 6700 - 7700
\]
\[
40a - 40c = -1000
\]
\[
a - c = -25 \tag{5}
\]

From equations (4) and (5), we have two relations:
1. \( a - b = -5 \)
2. \( a - c = -25 \)

### Step 2: Express \( b \) and \( c \) in terms of \( a \)

From equation (4): \( b = a + 5 \)

From equation (5): \( c = a + 25 \)

### Step 3: Substitute into one of the original equations

Substitute \( b = a + 5 \) and \( c = a + 25 \) into equation (1):

\[
100a + 40(a + 5) + 60(a + 25) = 6700
\]
\[
100a + 40a + 200 + 60a + 1500 = 6700
\]
\[
200a + 1700 = 6700
\]
\[
200a = 5000
\]
\[
a = 25
\]

### Step 4: Find \( b \) and \( c \)

From \( b = a + 5 \):

\[
b = 25 + 5 = 30
\]

From \( c = a + 25 \):

\[
c = 25 + 25 = 50
\]

### Step 5: Calculate \( a + bc \)

Now, calculate \( a + bc \):

\[
a + bc = 25 + (30 \times 50) = 25 + 1500 = 1525
\]

Thus, the value of \( a + bc \) is \( \boxed{1525} \).

Would you like more details on any step? Here are some follow-up questions to consider:

1. How do weighted averages work in mixtures?
2. What is the method for solving systems of linear equations?
3. Can you explain more about substitution in solving equations?
4. How does the volume ratio affect the concentration in mixtures?
5. What are other real-life examples where weighted averages are used?

**Tip:** When dealing with mixtures, always remember that the total volume of the mixture affects how the individual concentrations contribute to the final concentration.

Three solutions, A, B and C, have concentrations as a%, b% and c%, respectively. When 100ml of A, 40ml of B and 60ml of C are mixed, we get a solution D of concentration 33.5%. When 40ml of A, 100ml of B and 60ml of C are mixed, we get a solution E of concentration 35%. When 60ml of A, 40ml of B and 100ml of C are mixed, we get a solution F of concentration 38.5%. Find the value of (a + bc).

Learn how to solve a system of linear equations to find the value of (a + bc) in a mixture problem. The problem involves mixing three solutions with different concentrations and applying weighted averages to derive the result. Detailed solution with steps for Grades 10-12.

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