Match List-I with List-II, List-I: Inequation, List-II: Solution set, (A) 2(2x+3)-10 2+4x/3, (IV) x(4,∞), Choose the correct answer from the options given below: Option 1: (A) - (I), (B) - (III), (C) - (II), (D) - (IV) Option 2: (A) - (III), (B) - (IV), (C) - (II), (D) - (I) Option 3: (A) - (IV), (B) - (III), (C) - (II), (D) - (I) Option 4: (A) - (III), (B) - (I), (C) - (IV), (D) - (II)

Answer: (A) - (IV), (B) - (III), (C) - (II), (D) - (I). Solution: (A) 4x-4 4, so (IV). (B) Multiplying by 12: -2x+3 0, so (III). (C) x/(2x+1)-14≥0 reduces to (2x-1)/(2x+1)≥0, giving x 24+16x gives x<3.9, so (I).

Algebra

Inequalities

Medium

applied

Match List-I with List-II

List-I: Inequation List-II: Solution set
(A) $2(2x+3)-10<6(x-2)$ (I) $x\in(-\infty,3.9)$
(B) $\frac{2x-1}{12}-\frac{x-1}{3}<\frac{3x+1}{4}$ (II) $x\in\left(-\infty,\frac{-1}{2}\right)\cup\left[\frac{1}{2},\infty\right)$
(C) $\frac{x}{2x+1}\ge \frac{1}{4}$ (III) $x\in(0,\infty)$
(D) $\frac{2x-3}{4}+6>2+\frac{4x}{3}$ (IV) $x\in(4,\infty)$

Choose the correct answer from the options given below:

List-I: Inequation	List-II: Solution set
(A) $2(2x+3)-10<6(x-2)$	(I) $x\in(-\infty,3.9)$
(B) $\frac{2x-1}{12}-\frac{x-1}{3}<\frac{3x+1}{4}$	(II) $x\in\left(-\infty,\frac{-1}{2}\right)\cup\left[\frac{1}{2},\infty\right)$
(C) $\frac{x}{2x+1}\ge \frac{1}{4}$	(III) $x\in(0,\infty)$
(D) $\frac{2x-3}{4}+6>2+\frac{4x}{3}$	(IV) $x\in(4,\infty)$

(A) - (I), (B) - (III), (C) - (II), (D) - (IV)

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)

✅ Correct Option: 3

(A) $4x-4<6x-12$ gives $x>4$ , so (IV).

(B) Multiplying by $12$ : $-2x+3<9x+3$ gives $x>0$ , so (III).

(D) Multiplying by $12$ : $6x+63>24+16x$ gives $x<3.9$ , so (I).

2025: 19 May Shift 1

Match List-I with List-II

List-I	List-II
(Inequality)	(Solution Set)
---	---
(A) $2x - 3 < x + 2 \le 3x + 5, x \in \mathbb{R}$	(I) $x \in (-1, \infty)$
(B) $\vert 2x + 3\vert < 7, x \in \mathbb{R}$	(II) $x \in (-\infty, 120]$
(C) $\frac{1}{2}\left(\frac{3}{5}x + 4\right) \ge \frac{1}{3}(x - 6), x \in \mathbb{R}$	(III) $x \in (-5, 2)$
(D) $\frac{\vert x + 1\vert }{x + 1} > 0, x \in \mathbb{R} - \{-1\}$	(IV) $x \in \left[-\frac{3}{2}, 5\right)$

Choose the correct answer from the options given below:

2025: 15 May Shift 1

The inequality $\frac{5x - 2}{3} - \frac{7x - 3}{5} > \frac{x}{4}$ holds when

2026: 29 May Shift 1

Which of the following are correct? (Where $R$ is set of real numbers)

(A) If $x \le 8$ , then $-x \ge -8$

(B) If $x \in R, |x| \le 3$ , then $-3 \le x \le 3$

(D) $x=|x|$ iff $x \ge 0$

Choose the correct answer from the options given below:

2026: 21 May Shift 1

Which one of the following inequalities are satisfied by $x = 10$ ?

2025: 15 May Shift 2

The solution set of $6 \leq -3(2x - 4) < 12$ , $x \in R$ is:

2026: 25 May Shift 1

If $\dfrac{x+3}{2x-1} < 1$ , then the solution set contains :

A. $(-\infty,1/2)$

B. $(-\infty,4)$

C. $(-4,4)$

D. $(4,\infty)$

Choose the correct answer from the options given below:

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