Match List-I with List-II, List-I, List-II, Inequalities, Solution set, (A) If 3 - 5x 5x - 1, then x, (II) (-∞, -3), (C) If x/3 > x/2 + 1, then x, (III) (-4/5, 2), (D) If 5x + 4 ≤ 7x + 9, then x, (IV) (-∞, -6), Choose the correct answer from the options given below: Option 1: (A) - (I), (B) - (IV), (C) - (II), (D) - (III) Option 2: (A) - (III), (B) - (II), (C) - (IV), (D) - (I) Option 3: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 4: (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: (A) - (III), (B) - (II), (C) - (IV), (D) - (I). Solution: (A): -7 5x-1 gives x 3x+6 gives x<-6, so (IV). (D): 5x+4≤7x+9 gives x≥-5/2, so (I).

Algebra

Inequalities

Medium

applied

Match List-I with List-II

List-I List-II
Inequalities Solution set
(A) If $\left\vert 3 - 5x \right\vert < 7$ , then $x \in$ (I) $\left[-\dfrac{5}{2}, \infty\right)$
(B) If $3x - 7 > 5x - 1$ , then $x \in$ (II) $(-\infty, -3)$
(C) If $\dfrac{x}{3} > \dfrac{x}{2} + 1$ , then $x \in$ (III) $\left(-\dfrac{4}{5}, 2\right)$
(D) If $5x + 4 \leq 7x + 9$ , then $x \in$ (IV) $(-\infty, -6)$

Choose the correct answer from the options given below:

List-I	List-II
Inequalities	Solution set
(A) If $\left\vert 3 - 5x \right\vert < 7$ , then $x \in$	(I) $\left[-\dfrac{5}{2}, \infty\right)$
(B) If $3x - 7 > 5x - 1$ , then $x \in$	(II) $(-\infty, -3)$
(C) If $\dfrac{x}{3} > \dfrac{x}{2} + 1$ , then $x \in$	(III) $\left(-\dfrac{4}{5}, 2\right)$
(D) If $5x + 4 \leq 7x + 9$ , then $x \in$	(IV) $(-\infty, -6)$

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

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

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

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

✅ Correct Option: 2

(A): $-7<3-5x<7$ gives $-\frac{4}{5}<x<2$ , so (III).

(B): $3x-7>5x-1$ gives $x<-3$ , so (II).

(C): multiplying by $6$ , $2x>3x+6$ gives $x<-6$ , so (IV).

(D): $5x+4\le7x+9$ gives $x\ge-\frac{5}{2}$ , so (I).

2026: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) The solution set of the inequation $3x-6\le 0$ is	(I) $(3,\infty)$
(B) The solution set of the inequation $-5x+20<0$ is	(II) $[3,\infty)$
(C) The solution set of the inequation $3x-9\ge 0$ is	(III) $(-\infty,2]$
(D) The solution set of the inequation $8x+11>35$ is	(IV) $(4,\infty)$

Choose the correct answer from the options given below:

2025: 14 May Shift 1

Which of the following inequalities holds true?

(A) $\sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2}$ .

(B) If $a > b$ and $c < 0$ , then $\frac{a}{c} < \frac{b}{c}$ .

(D) If $a$ and $b$ are positive integers and $\frac{a - b}{6.25} = \frac{4}{2.5}$ , then $b > a$ .

Choose the correct answer from the options given below:

2026: 31 May Shift 2

Which of the following facts are not true ?( Where $R$ is set of real numbers)

A. $|x+y| \le |x|+|y|, x,y \in R$

B. $|x-y| \ge |x|+|y|, x,y \in R$

C. $|x+y| \ge |x|+|y|, x,y \in R$

D. $|x-y| \le |x|+|y|, x,y \in R$

Choose the correct answer from the options given below:

2024: 16 May Shift 1

The solution set of the inequality $|3 x| \geq|6-3 x|$ is :

2026: 18 May Shift 1

For the inequality $3-2x \geq x-32$ , which of the following are true?

(A) When $x \in N$ (set of natural numbers), the solution set is $\{1,2,3,\dots,11\}$ .

(B) When $x \in I$ (set of integers), the solution set is $\{-\infty,\dots,-3,-2,-1,0,1,2,3,\dots 10,11\}$ .

(D) When $x \in R$ (set of real numbers), the solution set is $(-\infty,\infty)$ .

Choose the correct answer from the options given below:

2026: 29 May Shift 2

Which of the following shaded regions represents the solution to the inequation $|x|+|y| \leq 1$ ?