Match the LIST-I with LIST-II, LIST-I (Inequality), LIST-II (Solution set), A., x-3 < 1, I., x (-2/5, 8/5), B., 3-5x < 5, II., x (-∞, -1/2), C., 2x+1 < 2x+1, III., x (-1/3, 5), D., 7-3x < 8, IV., x (2,4), Choose the correct answer from the options given below: Option 1: A-I, B-II, C-III, D-IV Option 2: A-II, B-IV, C-I, D-III Option 3: A-I, B-IV, C-III, D-II Option 4: A-IV, B-I, C-II, D-III

Answer: A-IV, B-I, C-II, D-III. Solution: A: -1<x-3<1 gives (2,4) (IV). B: -5<3-5x<5 gives -2/5<x<8/5 (I). C: a number is less than its modulus only when it is negative, so 2x+1<0, giving x<-1/2 (II). D: -8<7-3x<8 gives -1/3<x<5 (III).

Algebra

Inequalities

Medium

applied

Match the LIST-I with LIST-II

LIST-I (Inequality) LIST-II (Solution set)
A. $\vert x-3 \vert < 1$ I. $x \in \left(-\dfrac{2}{5}, \dfrac{8}{5}\right)$
B. $\vert 3-5x \vert < 5$ II. $x \in \left(-\infty, -\dfrac{1}{2}\right)$
C. $2x+1 < \vert 2x+1 \vert$ III. $x \in \left(-\dfrac{1}{3}, 5\right)$
D. $\vert 7-3x \vert < 8$ IV. $x \in (2,4)$

Choose the correct answer from the options given below:

	LIST-I (Inequality)		LIST-II (Solution set)
A.	$\vert x-3 \vert < 1$	I.	$x \in \left(-\dfrac{2}{5}, \dfrac{8}{5}\right)$
B.	$\vert 3-5x \vert < 5$	II.	$x \in \left(-\infty, -\dfrac{1}{2}\right)$
C.	$2x+1 < \vert 2x+1 \vert$	III.	$x \in \left(-\dfrac{1}{3}, 5\right)$
D.	$\vert 7-3x \vert < 8$	IV.	$x \in (2,4)$

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

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

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

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

✅ Correct Option: 4

A: $-1<x-3<1$ gives $(2,4)$ (IV).

B: $-5<3-5x<5$ gives $-\dfrac{2}{5}<x<\dfrac{8}{5}$ (I).

C: a number is less than its modulus only when it is negative, so $2x+1<0$ , giving $x<-\dfrac{1}{2}$ (II).

D: $-8<7-3x<8$ gives $-\dfrac{1}{3}<x<5$ (III).

2026: 19 May Shift 1

The number that satisfies the inequality $\frac{29}{4} < \frac{x+20}{4} < 8$ is

2026: 22 May Shift 1

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:

2025: 26 May Shift 2

The solution of $\frac{7x+12}{x-9} < 4$ ; $\neq 9$ is:

2025: 27 May Shift 1

If $x, y \in \mathbb{R}$ then match List-I with List-II

List-I	List-II
(A) $\vert x\vert < \vert y\vert$	(I) iff $x^2 > y^2$
(B) $\vert x\vert > \vert y\vert$	(II) iff $x^2 \le y^2$
(C) $\vert x\vert \le \vert y\vert$	(III) iff $x^2 < y^2$
(D) $\vert x\vert \ge \vert y\vert$	(IV) iff $x^2 \ge y^2$

Choose the correct answer from the options given below:

2025: 15 May Shift 2

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

2025: 21 May Shift 2

If a, b, c are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:

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