If $\mathbb{Z}$ and $\mathbb{R}$ denote set of integers and set of real numbers respectively, then match List I with List II.

List-I	List-II
(A) $5x - 3 \leq 3x + 1$, $x \in \mathbb{Z}$	(I) $x \in (-\infty, -3]$
(B) $3x + 17 \leq 2(1 - x)$, $x \in \mathbb{R}$	(II) $x \in (-\infty, -1)$
(C) $13x + 17 \leq 2(1 - x)$, $x \in \mathbb{R}$	(III) $\{......, -4, -3, .......,0,1\}$
(D) $\frac{2x + 3}{5} - 2 > \frac{3(x - 2)}{5}$, $x \in \mathbb{Z}$	(IV) $\{......, -4, -3, -2\}$

Choose the correct answer from the options given below:

(A) $5x - 3 \leq 3x + 1$, where $x \in \mathbb{Z}$

$5x - 3x \leq 1 + 3$

$2x \leq 4$

$x \leq 2$

Since $x \in \mathbb{Z}$, the solution set is $\{\ldots, -4, -3, -2, -1, 0, 1, 2\}$

This matches (III)

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(B) $3x + 17 \leq 2(1 - x)$, where $x \in \mathbb{R}$

$3x + 17 \leq 2 - 2x$

$3x + 2x \leq 2 - 17$

$5x \leq -15$

$x \leq -3$

This matches (I): $x \in (-\infty, -3]$

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(C) $13x + 17 \leq 2(1 - x)$, where $x \in \mathbb{R}$

$13x + 17 \leq 2 - 2x$

$13x + 2x \leq 2 - 17$

$15x \leq -15$

$x \leq -1$

This matches (II): $x \in (-\infty, -1]$

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(D) $\dfrac{2x + 3}{5} - 2 > \dfrac{3(x - 2)}{5}$, where $x \in \mathbb{Z}$

$(2x + 3) - 10 > 3(x - 2)$

$2x - 7 > 3x - 6$

$-7 + 6 > 3x - 2x$

$-1 > x$

$x < -1$

Since $x \in \mathbb{Z}$ and the inequality is strict, $x = -1$ is not included.

The solution set is $\{\ldots, -4, -3, -2\}$

This matches (IV)

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List-I	List-II
(A)	(III) $\{\ldots, -4, -3, \ldots, 0, 1, 2\}$
(B)	(I) $x \in (-\infty,\ -3]$
(C)	(II) $x \in (-\infty,\ -1]$
(D)	(IV) $\{\ldots, -4, -3, -2\}$

Correct Answer: Option 3