Q.Solve $\dfrac{5 - 2x}{3} \le \dfrac{x}{6} - 5$.

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1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The solution set of the inequality $6(2x+3)+x>53-2x$ is (A) $\left(\dfrac{7}{3},\infty\right)$ (B) $\left(-\infty,\dfrac{7}{3}\right)$ (C) $\left(\dfrac{7}{3},2\right)$ (D) $\left(-2,\dfrac{7}{3}\right)$ (E) $\left(-\dfrac{7}{3},\infty\right)$

   ›Reveal solutionSolution

   Expand and collect terms: $15x>35$, so $x>\tfrac73$.

   Expand the left side:

   $6(2x+3)+x=12x+18+x=13x+18.$

   The inequality becomes …

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2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.In a right-angled trapezium $ABCD$, $\angle A=90^\circ$, $\angle D=90^\circ$, $AB=7a+1$, $CD=3a+1$ and $AD=3a$. If the perimeter of the trapezium is greater than 56 but less than 92, then the range of possible values of $a$, is (A) $4<a<\dfrac{11}{2}$ (B) $\dfrac{7}{2}<a<5$ (C) $3<a<\dfrac{9}{2}$ (D) $3<a<7$ (E) $3<a<5$

   ›Reveal solutionSolution

   The perimeter is $18a+2$; solving $56<18a+2<92$ gives $3<a<5$.

   In the right-angled trapezium, $AB\parallel CD$ with the perpendicular leg $AD=3a$. The horizontal offset between the parallel sides is $AB-CD=(7a+1)-(3a+1)=4a$, so the slant side is

   $BC=\sqrt{(4a)^2+(3a)^2}=\sqrt{25a^2}=5a.$

   The perimeter is …

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3. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The set of all $x$ satisfying the inequality $8+3x>4(x-3)+2$ is (A) $(18,\infty)$ (B) $(20,\infty)$ (C) $(-\infty,18)$ (D) $(-\infty,20)$ (E) $(-20,18)$

   ›Reveal solutionSolution

   Solving $8+3x>4(x-3)+2$ yields $x<18$.

   $8+3x>4(x-3)+2=4x-12+2=4x-10.$

   Bring terms together:

   $8+10>4x-3x\;\Rightarrow\;18>x.$ …

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4. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $x$ be a real number such that $5 < |x - 1| < 15$. Then (A) $-18 < x < -3$ or $3 < x < 19$ (B) $-14 < x < -3$ or $6 < x < 17$ (C) $-16 < x < -2$ or $6 < x < 20$ (D) $-14 < x < -4$ or $6 < x < 16$ (E) $-10 < x < -1$ or $3 < x < 18$

   ›Reveal solutionSolution

   Solve the two-sided inequality: $-14<x<-4$ or $6<x<16$.

   Upper bound $|x-1|<15$: $-15<x-1<15\Rightarrow -14<x<16$.

   Lower bound $|x-1|>5$: $x-1>5$ or $x-1<-5\Rightarrow x>6$ or $x<-4$. …

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5. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $x$ be a real number such that $\frac{x - 3}{x - 2} \geq 1$. Then the solution set of the inequality is (A) $(-\infty, 3)$ (B) $(-\infty, 2)$ (C) $[0, \infty)$ (D) $(-9, \infty)$ (E) $(0, 8)$

   ›Reveal solutionSolution

   Subtract 1 and analyse the sign: solution is $(-\infty,2)$.

   Rearrange.

   $\frac{x-3}{x-2}-1\ge0\Rightarrow \frac{(x-3)-(x-2)}{x-2}\ge0\Rightarrow \frac{-1}{x-2}\ge0.$ …

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6. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.If $\dfrac{2x - 10}{x + 2} \geq x - 5$ then $x$ lies in (A) $(-\infty, -1) \cup [0, 6]$ (B) $(-\infty, -2) \cup [-1, 5]$ (C) $(-\infty, -2) \cup [0, 10]$ (D) $(-\infty, 0) \cup [0, 5]$ (E) $(-\infty, -2) \cup [0, 5]$

   ›Reveal solutionSolution

   Factoring $2x-10=2(x-5)$ and moving everything to one side yields $-\frac{x(x-5)}{x+2}\ge0$, i.e. $\frac{x(x-5)}{x+2}\le0$. A sign chart gives $x<-2$ or $0\le x\le5$.

   Start from $\dfrac{2x-10}{x+2}\ge x-5$. Since $2x-10=2(x-5)$:

   $\frac{2(x-5)}{x+2}-(x-5)\ge0\;\Rightarrow\;(x-5)\left[\frac{2}{x+2}-1\right]\ge0.$

   Simplify the bracket:

   $\frac{2-(x+2)}{x+2}=\frac{-x}{x+2},$

   so

   $(x-5)\cdot\frac{-x}{x+2}\ge0\;\Longleftrightarrow\;\frac{x(x-5)}{x+2}\le0.$

   Critical points: $x=-2$ (excluded), $x=0$, $x=5$. Sign of $\frac{x(x-5)}{x+2}$: …

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7. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.If $10 < |x + 10| \leq 25$, then $x$ lies in (A) $[-45, -20) \cup (0, 15]$ (B) $[-35, -25) \cup (0, 15]$ (C) $[-35, -20) \cup (0, 15]$ (D) $[-35, -20) \cup (0, 25]$ (E) $[-35, -10) \cup (0, 15]$

   ›Reveal solutionSolution

   Substitute $y=x+10$. The double inequality $10<|y|\le25$ means $y\in[-25,-10)\cup(10,25]$. Converting back ($x=y-10$) gives $x\in[-35,-20)\cup(0,15]$.

   Let $y=x+10$, so the condition is $10<|y|\le25$.

   This breaks into:

   $|y|>10 \Rightarrow y<-10 \text{ or } y>10,$

   $|y|\le25 \Rightarrow -25\le y\le25.$ …

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8. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.A number $x$ is randomly chosen from the set of natural numbers less than or equal to $100$. Then the probability of the event that the chosen number satisfies the inequality $\frac{(x-15)(x-70)}{x-30}\ge 0$, is (A) $0.36$ (B) $0.47$ (C) $0.48$ (D) $0.49$ (E) $0.46$

   ›Reveal solutionSolution

   Sign-analyse $\frac{(x-15)(x-70)}{x-30}\ge 0$ over integers $1$–$100$ ($x=30$ excluded).

   Critical points: $15,30,70$ (with $x=30$ making the denominator zero).

   • $x<15$: expression $<0$.
   • $15\le x<30$: $\ge 0$ (equals $0$ at $x=15$). Integers $15,\dots,29$ → $15$ values.
   • $30<x<70$: $<0$. …

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9. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If $(x - 1)(x^2 - 5x + 7) < (x - 1)$, then $x$ belongs to (A) $(-\infty, -1) \cup (2, 3)$ (B) $(-\infty, -1] \cup [2, 3]$ (C) $(-\infty, 1) \cup (2, 3)$ (D) $(-\infty, 1) \cup [2, 3)$ (E) $(-\infty, 1) \cup (2, 3]$

   ›Reveal solutionSolution

   Bringing everything to one side factors as $(x-1)(x-2)(x-3)<0$, giving $(-\infty,1)\cup(2,3)$.

   $(x-1)(x^2-5x+7)-(x-1)<0\;\Rightarrow\;(x-1)(x^2-5x+6)<0\;\Rightarrow\;(x-1)(x-2)(x-3)<0.$ …

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10. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.The solution set of $\left| x + \dfrac{1}{x} \right| > 2$ is (A) $\mathbb{R}$ (B) $\mathbb{R} - \{0\}$ (C) $\mathbb{R} - \{1, -1\}$ (D) $\mathbb{R} - \{-1\}$ (E) $\mathbb{R} - \{-1, 0, 1\}$

    ›Reveal solutionSolution

    By AM–GM $|x+1/x|\ge2$ with equality only at $x=\pm1$; excluding those and $x=0$ gives $\mathbb{R}-\{-1,0,1\}$.

    For real $x\neq0$, $\left|x+\dfrac1x\right|\ge2$, with equality exactly when $x=\pm1$. The strict inequality $>2$ therefore holds for all $x$ except $x=\pm1$; al …

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11. KEAM 2026Set eng-2026-04224 marksMCQ
    Q.If $\frac{21x - 6}{4} - 9 \leq 0$ and $\frac{x - 1}{3} + 1 \geq 0, x \in \mathbb{R}$, then $x$ lies in the interval (A) $[-2, 1]$ (B) $[-2, 2]$ (C) $[-1, 2]$ (D) $[2, 4]$ (E) $[-2, 4]$

    ›Reveal solutionSolution

    Solve each inequality separately and intersect.

    First: $\frac{21x-6}{4}\le9\Rightarrow 21x-6\le36\Rightarrow 21x\le42\Rightarrow x\le2$. …

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12. KEAM 2026Set eng-2026-04224 marksMCQ
    Q.If $|2x - 3| < 5, x \in \mathbb{R}$, then $x$ lies in the interval (A) $[-1, 1]$ (B) $[-1, 4]$ (C) $(-1, 4)$ (D) $(-2, 4)$ (E) $[-2, 4]$

    ›Reveal solutionSolution

    Remove the modulus with a two-sided inequality; strict $<$ gives an open interval.

    $|2x-3|<5$ means $-5<2x-3<5$.

    Add $3$: $-2<2x<8$. Divide by $2$: $-1<x<4$. …

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