https://file.aiquickdraw.com/mathbot/file/1729438035965zenna5l2.jpg

Let's solve each question from the image step-by-step.

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### **Q11**  
**If:**  
\[
(X - Y) \times Y = \{(1, 2), (1, 3)\} \quad \text{and} \quad n(X \times Y) = 6
\]
then \(X =\) ?

1. We know that \((X - Y) \times Y\) contains the pairs \((1, 2)\) and \((1, 3)\).
   - This suggests that element 1 is in \(X - Y\) and both 2 and 3 are in \(Y\).

2. From \(n(X \times Y) = 6\), the product set \(X \times Y\) has 6 elements.
   - Possible pairs could be from \(X = \{1, 3, 6\}\) and \(Y = \{2, 3\}\).

Thus, the answer is:
\[
\boxed{(c) \{1, 3, 6\}}
\]

---

### **Q12**  
**The point \((-5, 7)\) lies in which quadrant?**

- Since \(x < 0\) and \(y > 0\), this point lies in the **second quadrant**.

Answer:
\[
\boxed{(b) \text{second}}
\]

---

### **Q13**  
**The point \((-3, -4)\) lies in which quadrant?**

- Since \(x < 0\) and \(y < 0\), this point lies in the **third quadrant**.

Answer:
\[
\boxed{(c) \text{third}}
\]

---

### **Q14**  
**The Cartesian product \(\{2\} \times \mathbb{R}\)** represents graphically...  
- This product set means all points where the \(x\)-coordinate is always 2, but the \(y\)-coordinate can be any real number.
- This corresponds to a **vertical line passing through the point (2, 0)**.

Answer:
\[
\boxed{(a) \text{(2, 0)}}
\]

---

### **Q15**  
**The point \(A(5, -3)\) lies in which quadrant?**

- Since \(x > 0\) and \(y < 0\), the point lies in the **fourth quadrant**.

Answer:
\[
\boxed{(d) \text{fourth}}
\]

---

### **Q16**  
**If the point \((5, b - 5)\) is on the X-axis, what is \(b\)?**

- For a point to be on the \(x\)-axis, its \(y\)-coordinate must be 0.
- Thus:
  \[
  b - 5 = 0 \quad \Rightarrow \quad b = 5
  \]

Answer:
\[
\boxed{(c) 5}
\]

---

### **Q17**  
**If the point \((5, b - 7)\) is on the X-axis, what is \(b\)?**

- Similarly, for a point on the \(x\)-axis, the \(y\)-coordinate must be 0:
  \[
  b - 7 = 0 \quad \Rightarrow \quad b = 7
  \]

Answer:
\[
\boxed{(c) 7}
\]

---

### **Q18**  
**If \(\left(|x|, 4\right) = (3, y^2)\) and the point \((x, y)\) lies in the second quadrant, then \(x + y =\) ?**

1. From \(\left(|x|, 4\right) = (3, y^2)\), we get:
   - \(|x| = 3\) \(\Rightarrow x = \pm 3\)
   - \(4 = y^2\) \(\Rightarrow y = \pm 2\)

2. Since the point \((x, y)\) lies in the second quadrant, \(x < 0\) and \(y > 0\).
   - Thus, \(x = -3\) and \(y = 2\).

3. Now:
   \[
   x + y = -3 + 2 = -1
   \]

Answer:
\[
\boxed{(c) -1}
\]

---

### **Q19**  
**If the point \((X - 5, 3 - X)\) is in the third quadrant and \(X \in \mathbb{Z}\), what is \(X\)?**

1. In the third quadrant, both \(x\)- and \(y\)-coordinates must be negative:
   - \(X - 5 < 0 \quad \Rightarrow \quad X < 5\)
   - \(3 - X < 0 \quad \Rightarrow \quad X > 3\)

2. Since \(X \in \mathbb{Z}\) (integers), the only possible value is:
   \[
   X = 4
   \]

Answer:
\[
\boxed{(c) 4}
\]

---

Let me know if you need further clarifications on any question! Here are some related questions you might want to explore:

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1. What are the properties of Cartesian products and how are they used?
2. How do you determine the quadrant of a point given coordinates?
3. What are the general rules for absolute values in equations?
4. How can you graph vertical and horizontal lines efficiently?
5. What strategies help solve integer-based logic problems quickly?

**Tip:** When working with absolute values, always consider both positive and negative solutions!

If (X − Y) × Y = {(1, 2), (1, 3)} and n(X × Y) = 6, then X = ?

Learn how to solve a set theory problem involving Cartesian products and cardinality. Explore how to determine the set X given (X − Y) × Y and the total number of elements in X × Y. Suitable for high school students in Grades 10-12.

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