Class 9 Olympiad math worksheet featuring multiple-choice questions on circles, functions, family relationships, volume calculations, and number properties.

Let's solve each problem step by step.

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Problem (1):

Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A (or B) intersecting the circles at P and Q. Find the ratio PQ:OO'.

#### Solution:
1. Understanding the Geometry:
- Let the two circles have centers \( O \) and \( O' \).
- The circles intersect at points \( A \) and \( B \).
- A line \( PQ \) is drawn parallel to \( OO' \) through point \( A \) (or \( B \)).
- This line intersects the first circle at \( P \) and the second circle at \( Q \).

2. Key Observations:
- Since \( PQ \) is parallel to \( OO' \), the segment \( PQ \) is a common chord of the two circles.
- The distance between the centers \( O \) and \( O' \) is \( OO' \).
- The length of the chord \( PQ \) can be related to the distance between the centers using properties of circles.

3. Using Circle Properties:
- When a line is drawn parallel to the line joining the centers of two intersecting circles, the length of the chord in each circle is equal to the diameter of the smaller circle if the circles are congruent or proportional otherwise.
- Here, since \( PQ \) is parallel to \( OO' \), the length \( PQ \) is effectively the length of a chord that spans both circles symmetrically.

4. Ratio Calculation:
- The key property here is that the length of the chord \( PQ \) is exactly twice the distance between the centers \( OO' \) when the line is parallel to \( OO' \) and passes through the intersection points.
- Therefore, \( PQ = 2 \cdot OO' \).

5. Final Ratio:
\[
\text{Ratio } \frac{PQ}{OO'} = \frac{2 \cdot OO'}{OO'} = 2
\]

Answer:
\[
\boxed{2}
\]

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Problem (2):

If \( f(x) = 4x + 8 \), find the inverse function \( f^{-1}(x) \).

#### Solution:
1. Start with the given function:
\[
f(x) = 4x + 8
\]

2. Express \( y \) in terms of \( x \):
\[
y = 4x + 8
\]

3. Solve for \( x \) in terms of \( y \):
\[
y = 4x + 8
\]
Subtract 8 from both sides:
\[
y - 8 = 4x
\]
Divide by 4:
\[
x = \frac{y - 8}{4}
\]

4. Replace \( y \) with \( x \) to get the inverse function:
\[
f^{-1}(x) = \frac{x - 8}{4}
\]

Answer:
\[
\boxed{\frac{x - 8}{4}}
\]

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Problem (3):

Uncle is to Aunt as Father is to ?

#### Solution:
1. Analyze the relationship:
- An "uncle" is the brother of one's parent.
- An "aunt" is the sister of one's parent.
- The relationship between "uncle" and "aunt" is that they are siblings.

2. Apply the same logic:
- A "father" is the male parent.
- The female counterpart of a "father" is a "mother."

3. Conclusion:
- The relationship is analogous: Uncle : Aunt :: Father : Mother.

Answer:
\[
\boxed{\text{Mother}}
\]

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Problem (4):

A sphere is just enclosed inside a cube of volume \( 72 \, \text{cm}^3 \). Find the volume of the sphere.

#### Solution:
1. Find the side length of the cube:
- The volume of the cube is given by:
\[
V_{\text{cube}} = s^3
\]
- Given \( V_{\text{cube}} = 72 \, \text{cm}^3 \):
\[
s^3 = 72
\]
\[
s = \sqrt[3]{72}
\]

2. Determine the diameter of the sphere:
- Since the sphere is just enclosed inside the cube, the diameter of the sphere is equal to the side length of the cube:
\[
\text{Diameter of sphere} = s = \sqrt[3]{72}
\]
- The radius \( r \) of the sphere is half of the diameter:
\[
r = \frac{s}{2} = \frac{\sqrt[3]{72}}{2}
\]

3. Calculate the volume of the sphere:
- The volume of a sphere is given by:
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
- Substitute \( r = \frac{\sqrt[3]{72}}{2} \):
\[
V_{\text{sphere}} = \frac{4}{3} \pi \left( \frac{\sqrt[3]{72}}{2} \right)^3
\]
- Simplify the expression:
\[
\left( \frac{\sqrt[3]{72}}{2} \right)^3 = \frac{(\sqrt[3]{72})^3}{2^3} = \frac{72}{8} = 9
\]
\[
V_{\text{sphere}} = \frac{4}{3} \pi \cdot 9 = 12 \pi \, \text{cm}^3
\]

Answer:
\[
\boxed{12 \pi \, \text{cm}^3}
\]

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Problem (5):

The difference between two natural numbers is 37. If the larger number is divided by the smaller one, then the quotient is 5 and the remainder is 5. Find the larger number.

#### Solution:
1. Define the variables:
- Let the smaller number be \( x \).
- Let the larger number be \( y \).

2. Set up the equations based on the problem:
- The difference between the two numbers is 37:
\[
y - x = 37 \quad \text{(1)}
\]
- When the larger number is divided by the smaller one, the quotient is 5 and the remainder is 5:
\[
y = 5x + 5 \quad \text{(2)}
\]

3. Solve the system of equations:
- Substitute equation (2) into equation (1):
\[
(5x + 5) - x = 37
\]
\[
4x + 5 = 37
\]
\[
4x = 32
\]
\[
x = 8
\]

4. Find the larger number \( y \):
- Substitute \( x = 8 \) back into equation (2):
\[
y = 5(8) + 5 = 40 + 5 = 45
\]

Answer:
\[
\boxed{45}
\]

---

Problem (6):

If \( N = 1111^2 \), find the 5th digit in the expansion of \( N \).

#### Solution:
1. Calculate \( 1111^2 \):
- Use the formula for squaring a number:
\[
(a + b)^2 = a^2 + 2ab + b^2
\]
- Here, \( 1111 = 1000 + 111 \):
\[
1111^2 = (1000 + 111)^2 = 1000^2 + 2 \cdot 1000 \cdot 111 + 111^2
\]
- Calculate each term:
\[
1000^2 = 1000000
\]
\[
2 \cdot 1000 \cdot 111 = 222000
\]
\[
111^2 = 12321
\]
- Add these together:
\[
1111^2 = 1000000 + 222000 + 12321 = 1234321
\]

2. Identify the 5th digit:
- The number \( 1111^2 = 1234321 \).
- The digits are: \( 1, 2, 3, 4, 3, 2, 1 \).
- The 5th digit is \( 3 \).

Answer:
\[
\boxed{3}
\]

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Final Answers:

1. \(\boxed{2}\)
2. \(\boxed{\frac{x - 8}{4}}\)
3. \(\boxed{\text{Mother}}\)
4. \(\boxed{12 \pi \, \text{cm}^3}\)
5. \(\boxed{45}\)
6. \(\boxed{3}\)