Q.a) Solve the inequality, $\dfrac{x}{3} > \dfrac{x}{2} + 1$; $x \in \mathbb{R}$, the real numbers.

Showing the 12 most recent of 35 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Solution of $4x + 3 < 6x + 7$, $x \in R$ is

   (a) $(-\infty, -2)$

   (b) $(-2, \infty)$

   (c) $(2, \infty)$

   (d) $(-\infty, 2)$

   ›Reveal solutionSolution

   Collect $x$-terms on one side and constants on the other, keeping the inequality direction (no division by a negative here).

   $4x + 3 < 6x + 7$

   Subtract $4x$ from both sides:

   $3 < 2x + 7$

   Subtract 7:

   $-4 < 2x$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Solution of $\dfrac{x}{3} > \dfrac{x}{2} + 1$, $x \in R$ is

   (a) $(-6, \infty)$

   (b) $(-\infty, -6)$

   (c) $(-\infty, 6)$

   (d) None of these

   ›Reveal solutionSolution

   Clear the fractions by multiplying by the LCM (6), then isolate $x$.

   $\dfrac{x}{3} > \dfrac{x}{2} + 1$

   Multiply every term by 6 (positive, direction unchanged):

   $2x > 3x + 6$

   Subtract $3x$:

   $-x > 6$ …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Solution of $-8 \le 5x - 3 < 7$, $x \in R$ is

   (a) $[-1, 2)$

   (b) $(-1, 2)$

   (c) $(-1, 2]$

   (d) $[-1, 2]$

   ›Reveal solutionSolution

   Add 3 to all three parts of the compound inequality, then divide by 5, preserving direction and endpoint inclusivity throughout.

   $-8 \le 5x - 3 < 7$

   Add 3 to all parts:

   $-5 \le 5x < 10$

   Divide by 5 (positive):

   $-1 \le x < 2$ …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $x > -4$ then the solution of the inequality will be —

   (a) $(-4, \infty)$

   (b) $(2, 2)$

   (c) $(1, 4)$

   (d) $(\infty, -4)$

   ›Reveal solutionSolution

   The solution set of $x > -4$ is the open interval $(-4, \infty)$, option (a).

   The inequality $x > -4$ describes every real number strictly greater than $-4$. In interval notation, a round bracket is used at $-4$ because $-4$ itself is NOT included (strict inequality), and the interval extends w …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.From the following, which value of $x$ satisfies the inequality $x + 4 > 7$?

   (a) $-3$

   (b) $1$

   (c) $4$

   (d) $3$

   ›Reveal solutionSolution

   Only $x=4$ satisfies $x+4>7$, option (c).

   Solve the inequality: $x + 4 > 7 \Rightarrow x > 3$.

   Now test each option:

   • $x=-3$: $-3 < 3$ — fails
   • $x=1$: $1 < 3$ — fails
   • $x=4$: $4 > 3$ — satisfies ($4+4=8>7$ ✓) …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The solution set of the inequation $24x < 100$; $x$ is a natural number, is

   (a) $\{0, 1, 2, 3, 4\}$

   (b) $\{1, 2, 3, 4\}$

   (c) $\{0, 1, 2, 3\}$

   (d) $\{4\}$

   ›Reveal solutionSolution

   The solution set is $\{1, 2, 3, 4\}$.

   Solve $24x < 100$: dividing both sides by 24 (positive, so the inequality direction is unchanged), $x < \dfrac{100}{24} = 4.1\overline{6}$.

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Let $a$ is a positive integer then $|x| > a$ is

   (a) $x < -a$ or $x > a$

   (b) $x > -a$ or $x < a$

   (c) $x < -a$ or $x > -a$

   (d) None of these

   ›Reveal solutionSolution

   $|x| > a$ (with $a>0$) means $x$ lies outside the interval $[-a,a]$, i.e. $x<-a$ or $x>a$.

   By definition, $|x|$ is the distance of $x$ from $0$. $|x| > a$ means this distance exceeds $a$, so $x$ must lie strictly beyond $a$ on either side of $0$:

   $x > a \quad \text{or} \quad x < -a$

   …

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8. CBSE 2025Set sz1 markMCQ
   Q.The solution of the inequality $7x - 8 \geq 6$ is :

   (a) [2, \infty)

   (b) (2, 8)

   (c) [4, \infty)

   (d) (14, \infty)

   ›Reveal solutionSolution

   $7x-8\ge 6 \implies 7x\ge14 \implies x\ge2$, so the solution set is $[2,\infty)$.

   Start with $7x - 8 \ge 6$.

   Add 8 to both sides:

   $7x \ge 14.$

   Divide both sides by 7 (a positive number, so the inequality direction is unchanged): …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $-3x + 17 < -13$, then

   (a) $x \in (10, \infty)$

   (b) $x \in [10, \infty)$

   (c) $x \in (-\infty, 10)$

   (d) $x \in [-10, 10]$

   ›Reveal solutionSolution

   Isolate $x$, remembering to flip the inequality when dividing by a negative number.

   $-3x+17<-13$

   $-3x<-13-17=-30$ …

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10. CBSE 2025Set ANNUAL1 mark
    Q.For a real number $x$, write the solution of the inequality $3x - 7 > 5x - 1$ in interval form.

    ›Reveal solutionSolution

    Solving $3x - 7 > 5x - 1$ gives $x < -3$, i.e. the interval $(-\infty, -3)$.

    $3x - 7 > 5x - 1$

    $3x - 5x > -1 + 7$

    $-2x > 6$

    Dividing both sides by $-2$ (a negative number reverses the inequality):

    $x < -3$.

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Solution of $3(2-x) \ge 4x-9$ is

    (a) $\left(-\infty, \dfrac{15}{7}\right]$

    (b) $\left[\dfrac{15}{7}, \infty\right)$

    (c) $[-3, \infty)$

    (d) None of these

    ›Reveal solutionSolution

    Expand the bracket, collect the $x$ terms on one side and constants on the other, keeping the inequality direction (no sign flip since we never multiply/divide by a negative).

    Start with $3(2-x) \ge 4x - 9$.

    Expand: $6 - 3x \ge 4x - 9$

    Add $3x$ to both sides: $6 \ge 7x - 9$

    Add $9$ to both sides: $15 \ge 7x$

    …

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12. CBSE 2024Set sz1 markMCQ
    Q.The value of $-12x > 30$ when $x$ is a natural number is:

    (a) 3

    (b) $< 3$

    (c) No solution

    (d) 0

    ›Reveal solutionSolution

    $-12x > 30$ requires $x < -2.5$, which no natural number can satisfy, so there is no solution.

    We are given $-12x > 30$ with $x$ restricted to be a natural number.

    Divide both sides by $-12$. Since we are dividing by a negative number, the inequality sign flips:

    $x < \dfrac{30}{-12} = -2.5$

    …

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