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

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:

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!