Algebra

Inequalities

Medium

applied

If, in a pair of consecutive positive integers, both numbers are greater than 5 and their sum is less than 23, then the number of such pairs are:

Two

Three

Four

Five

✅ Correct Option: 4

Let the consecutive positive integers be $n$ and $n+1$ , where $n$ is the smaller number.

Both numbers must be greater than 5:

$n > 5$

$n + 1 > 5$

The second condition is automatically satisfied if $n > 5$ .

The sum must be less than 23:

$n + (n + 1) < 23$

$2n + 1 < 23$

$2n < 22$

$n < 11$

Combining the conditions:

$5 < n < 11$

Since $n$ must be a positive integer, the possible values are:

$n = 6, 7, 8, 9, 10$

The valid pairs are:

$(6, 7)$ with sum $13$

$(7, 8)$ with sum $15$

$(8, 9)$ with sum $17$

$(9, 10)$ with sum $19$

$(10, 11)$ with sum $21$

Therefore, there are 5 such pairs.

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:

2025: 30 May Shift 2

The solution set of the inequality $\frac{2x+3}{x-1} < 0$ 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:

2026: 22 May Shift 2

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:

2026: 12 May Shift 2

The solution set of the inequation $|8x-7|\le\dfrac{2}{3}$ is

2026: 19 May Shift 2

Match the LIST-I with LIST-II

LIST-I Inequality		LIST-II Solution set
A.	$4x + 3 \le 6x + 7$	I.	$x \in (-\infty, -2)$
B.	$5x - 5 > 3x - 1$	II.	$x \in (-\infty, 2]$
C.	$4x - 9 > 6x - 5$	III.	$x \in [-2, \infty)$
D.	$3x - 1 \le 5$	IV.	$x \in (2, \infty)$

Choose the correct answer from the options given below:

If, in a pair of consecutive positive integers, both numbers are greater than 5 and their sum is less than 23, then the number of such pairs are:

Solution

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