Twenty percent of the number of whites exceeded the number of reds by 10. Also, three fifths of the number of blues was exactly equal to 3 times the number of reds. How many of each were there if ten percent of the sum of the reds and blues was 94 less than the number of whites?

Let's define the variables as follows:
- \( w \) = number of whites,
- \( r \) = number of reds,
- \( b \) = number of blues.

We are given three key pieces of information, which we can convert into equations:

### 1st Condition: "Twenty percent of the number of whites exceeded the number of reds by 10."
This translates to:
\[
0.2w = r + 10 \quad \text{(Equation 1)}
\]

### 2nd Condition: "Three fifths of the number of blues was exactly equal to 3 times the number of reds."
This can be written as:
\[
\frac{3}{5}b = 3r \quad \text{(Equation 2)}
\]

### 3rd Condition: "Ten percent of the sum of the reds and blues was 94 less than the number of whites."
This gives us:
\[
0.1(r + b) = w - 94 \quad \text{(Equation 3)}
\]

Now we have a system of three equations with three unknowns. Let's solve them step by step.

### Step 1: Solve for \( r \) from Equation 1
From Equation 1:
\[
0.2w = r + 10 \quad \Rightarrow \quad r = 0.2w - 10
\]

### Step 2: Substitute \( r \) into Equation 2
Substitute \( r = 0.2w - 10 \) into Equation 2:
\[
\frac{3}{5}b = 3(0.2w - 10) 
\]
\[
\frac{3}{5}b = 0.6w - 30 
\]
Multiply both sides by 5 to eliminate the fraction:
\[
3b = 3(1w - 50) 
\]
\[
b = w - 50 \quad \text{(Equation 4)}
\]

### Step 3: Substitute \( r \) and \( b \) into Equation 3
Substitute \( r = 0.2w - 10 \) and \( b = w - 50 \) into Equation 3:
\[
0.1((0.2w - 10) + (w - 50)) = w - 94
\]
Simplify the expression inside the parentheses:
\[
0.1(0.2w + w - 10 - 50) = w - 94
\]
\[
0.1(1.2w - 60) = w - 94
\]
Distribute the 0.1:
\[
0.12w - 6 = w - 94
\]
Move the terms involving \( w \) to one side and constants to the other side:
\[
0.12w - w = -94 + 6
\]
\[
-0.88w = -88
\]
Solve for \( w \):
\[
w = \frac{-88}{-0.88} = 100
\]

### Step 4: Solve for \( r \) and \( b \)
Now that we know \( w = 100 \), substitute it into the expressions for \( r \) and \( b \):
\[
r = 0.2(100) - 10 = 20 - 10 = 10
\]
\[
b = 100 - 50 = 50
\]

### Final Answer:
- Number of whites (\( w \)) = 100,
- Number of reds (\( r \)) = 10,
- Number of blues (\( b \)) = 50.

Let me know if you'd like further details or have any questions!

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Here are some questions to explore further:
1. What happens if the number of whites were doubled? How would that affect the solution?
2. What if the blues had been 5 fewer than in the original problem—how would that change the solution?
3. Can you generalize the steps taken here to any other scenario where the number of groups follow a linear relationship?
4. If the number of reds increased by 10%, how would that change the final result?
5. How could this system of equations be represented graphically?

**Tip**: When solving word problems, it is helpful to define the unknowns and translate the conditions into equations before solving them systematically!

Learn how to solve a word problem involving whites, reds, and blues using a system of linear equations. The problem includes concepts such as percentages and proportional relationships, making it perfect for students in Grades 9-11. Detailed step-by-step solution included.

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