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The problem states that \( Y \) is inversely proportional to the square of \( x \), and it provides the condition \( Y = 6 \) when \( x = 4 \). It asks for the value of \( Y \) when \( x = 8 \).

### Step-by-step Solution:

1. Since \( Y \) is inversely proportional to \( x^2 \), we can write the relationship as:
   \[
   Y \propto \frac{1}{x^2}
   \]
   This means:
   \[
   Y = \frac{k}{x^2}
   \]
   where \( k \) is a constant.

2. To find \( k \), we use the condition that \( Y = 6 \) when \( x = 4 \):
   \[
   6 = \frac{k}{4^2}
   \]
   \[
   6 = \frac{k}{16}
   \]
   Solving for \( k \):
   \[
   k = 6 \times 16 = 96
   \]

3. Now that we know \( k = 96 \), we can find \( Y \) when \( x = 8 \). Using the equation:
   \[
   Y = \frac{96}{8^2} = \frac{96}{64}
   \]
   Simplifying:
   \[
   Y = \frac{3}{2}
   \]

Thus, the value of \( Y \) when \( x = 8 \) is \( \frac{3}{2} \), which is \( 1 \frac{1}{2} \).

### Answer:
The correct option is:
\[
\boxed{1 \frac{1}{2}}
\]

Would you like further details or have any questions?

Here are some related questions to explore:
1. How do you solve proportionality problems in general?
2. What happens when \( Y \) is directly proportional to \( x^2 \) instead?
3. How do you find the constant \( k \) in an inverse proportionality problem?
4. What is the difference between inverse and direct proportionality?
5. How does the behavior of \( Y \) change as \( x \) increases in inverse square proportionality?

**Tip:** When dealing with proportionality, always identify the relationship and find the constant before applying it to new values.

Y is inversely proportional to the square of x. If Y = 6 when x = 4, then what is y when x is 8?

Solve the problem where Y is inversely proportional to the square of X. Learn step-by-step how to find Y when X changes from 4 to 8, using the inverse proportionality formula Y = k/x^2. This is a core algebraic concept for students in Grades 8-10.

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