Box B contains 1/2 as many bows as Box A. Box C contains 2/3 as many bows as Box B. The average number of bows per box is 143. How many bows are there in Box A.

Let's solve the problem step by step.

### Step 1: Define Variables
Let:
- \( x \) be the number of bows in Box A.
- \( y \) be the number of bows in Box B.
- \( z \) be the number of bows in Box C.

### Step 2: Establish Relationships
From the problem, we know:
1. \( y = \frac{1}{2}x \) (Box B has half as many bows as Box A)
2. \( z = \frac{2}{3}y \) (Box C has \(\frac{2}{3}\) as many bows as Box B)

### Step 3: Express All Variables in Terms of \( x \)
Using the first relationship:
\[
y = \frac{1}{2}x
\]
Now, substitute this into the second relationship:
\[
z = \frac{2}{3} \times \frac{1}{2}x = \frac{1}{3}x
\]

### Step 4: Calculate the Average Number of Bows
The average number of bows per box is given as 143. Therefore:
\[
\frac{x + y + z}{3} = 143
\]
Substitute the expressions for \( y \) and \( z \) in terms of \( x \):
\[
\frac{x + \frac{1}{2}x + \frac{1}{3}x}{3} = 143
\]

### Step 5: Solve the Equation
First, combine the terms inside the numerator:
\[
\frac{x \left(1 + \frac{1}{2} + \frac{1}{3}\right)}{3} = 143
\]
To simplify, find a common denominator for the fractions:
\[
1 + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6}
\]
Substitute back into the equation:
\[
\frac{x \times \frac{11}{6}}{3} = 143
\]
Multiply both sides by 3 to eliminate the denominator:
\[
\frac{11x}{6} = 429
\]
Now, solve for \( x \):
\[
11x = 429 \times 6 = 2574
\]
\[
x = \frac{2574}{11} = 234
\]

### Conclusion:
The number of bows in Box A is **234**.

Do you want more details or have any questions?

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Here are 8 related questions for further exploration:

1. If Box B contains 1/4 as many bows as Box A, what would be the new average number of bows per box?
2. How many bows are in Box C if the number of bows in Box A is doubled?
3. What happens to the average number of bows if Box C had twice as many bows as Box B?
4. If the average number of bows increased to 150, how many bows would be in Box A?
5. What is the total number of bows in all three boxes?
6. How would the problem change if Box C contained the same number of bows as Box B?
7. What if Box B contained 10 more bows than half the number of bows in Box A?
8. How would the solution change if Box B contained 2/3 as many bows as Box A?

**Tip:** When solving problems involving averages, expressing all quantities in terms of one variable can simplify calculations significantly.

Learn how to solve a problem involving the average number of bows per box. This problem covers key algebraic concepts, including equations and averages. Suitable for students in Grades 7-9.

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