Q.Find all pairs of consecutive odd natural numbers both of which are larger than 10, such that their sum is less than 40.

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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