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Inequalities and Modulus questions for IPMAT, with solutions

10 free Inequalities and Modulus practice questions for IPMAT Indore, Rohtak and JIPMAT, 3 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Inequalities and Modulus on Learn.

1. Question 1Multiple choiceWritten by IPMATwayInequalities

   The solution set of $3-2x\ge7$ is

   1. A.$(-\infty,-2)$
   2. B.$(-\infty,2]$
   3. C.$[-2,\infty)$
   4. D.$(-\infty,-2]$
   Show the answer and solution

   Answer: D. $(-\infty,-2]$

   Subtract $3$: $-2x\ge4$. Divide by $-2$ and reverse: $x\le-2$, i.e. $(-\infty,-2]$.

2. Question 2Type the answerWritten by IPMATwayFloor and Fractional Part

   If $[x]$ denotes the greatest integer not exceeding $x$, the number of integers $x$ satisfying $[x]^{2}-5[x]+6=0$ is

   Show the answer and solution

   Answer: 2

   Solve for $t=[x]$: $(t-2)(t-3)=0$. Then $x\in[2,4)$, which contains the integers $2$ and $3$.

3. Question 3Multiple choiceWritten by IPMATwayInequalities

   If $x>y$ and $z<0$, which of the following must be true?

   1. A.$x-z<y-z$
   2. B.$xz>yz$
   3. C.$xz<yz$
   4. D.$\dfrac{x}{z}>\dfrac{y}{z}$
   Show the answer and solution

   Answer: C. $xz<yz$

   Adding or subtracting keeps the direction ($x-z>y-z$); multiplying or dividing by a negative number reverses it, so $xz<yz$ and $\tfrac{x}{z}<\tfrac{y}{z}$.

4. Question 4Multiple choiceAsked in IPMAT Indore 2024Modulus (Absolute Value)

   If $|x+1|+(y+2)^{2}=0$ and $ax-3ay=1$, then the value of $a$ is

   1. A.$\dfrac{1}{5}$
   2. B.$\dfrac{1}{7}$
   3. C.$\dfrac{1}{2}$
   4. D.$2$
   Show the answer and solution

   Answer: A. $\dfrac{1}{5}$

   A modulus and a square are never negative; their sum is $0$ only when each is $0$: $x=-1$, $y=-2$. Substituting: $-a+6a=1$.

5. Question 5Multiple choiceWritten by IPMATwayModulus (Absolute Value)

   The number of real roots of $x^{2}-5|x|+6=0$ is

   1. A.$1$
   2. B.$0$
   3. C.$2$
   4. D.$4$
   Show the answer and solution

   Answer: D. $4$

   With $t=|x|$: $(t-2)(t-3)=0$. Both values are positive, each gives two values of $x$.

6. Question 6Type the answerWritten by IPMATwayFloor and Fractional Part

   If $[t]$ denotes the greatest integer not exceeding $t$, the value of $\left[\dfrac{1}{4}\right]+\left[\dfrac{2}{4}\right]+\cdots+\left[\dfrac{30}{4}\right]$ is

   Show the answer and solution

   Answer: 105

   Group by value: $[k/4]=j$ for $4j\le k\le4j+3$. Complete groups $j=1$ to $6$ give $84$; the partial group $j=7$ gives $3\times7=21$.

7. Question 7Multiple choiceAsked in IPMAT Indore 2023Inequalities

   The set of all real values of $x$ satisfying the inequality $\dfrac{x^{2}(x+1)}{(x-1)(2x+1)^{3}}>0$ is

   1. A.$\left(-1,-\frac{1}{2}\right)\cup(0,+\infty)$
   2. B.$(-\infty,-1)\cup\left(-\frac{1}{2},0\right)\cup(1,+\infty)$
   3. C.$(-1,0)\cup(1,+\infty)$
   4. D.$\left(-1,-\frac{1}{2}\right)\cup(1,+\infty)$
   Show the answer and solution

   Answer: D. $\left(-1,-\frac{1}{2}\right)\cup(1,+\infty)$

   The factor $x^{2}$ is positive except at $0$ (where the expression is $0$, not $>0$). The sign is that of $\dfrac{x+1}{(x-1)(2x+1)}$. Answer: $\left(-1,-\tfrac12\right)\cup(1,\infty)$.

8. Question 8Multiple choiceAsked in IPMAT Indore 2022Modulus (Absolute Value)

   The sum of the squares of all the roots of the equation $x^{2}+|x+4|+|x-4|-35=0$ is

   1. A.$74$
   2. B.$175$
   3. C.$50$
   4. D.$148$
   Show the answer and solution

   Answer: C. $50$

   Between $-4$ and $4$ the two moduli add to $8$; outside they add to $2|x|$. Only the outer case gives roots consistent with its own condition: $(|x|+7)(|x|-5)=0\Rightarrow|x|=5$. Sum of squares $25+25=50$.

9. Question 9Multiple choiceWritten by IPMATwayModulus (Absolute Value)

   The number of real solutions of $x|x|=4x$ is

   1. A.$2$
   2. B.$1$
   3. C.$4$
   4. D.$3$
   Show the answer and solution

   Answer: D. $3$

   Never divide by $x$ - it may be $0$. Factor instead: $x(|x|-4)=0$, so $x=0$ or $|x|=4$, i.e. $x=-4,0,4$.

10. Question 10Multiple choiceWritten by IPMATwayInequalities

    What is the sum of all integers $x$ that satisfy $\dfrac{(x-5)(x+2)^2}{x-1}\le0$?

    1. A.$14$
    2. B.$12$
    3. C.$13$
    4. D.$7$
    Show the answer and solution

    Answer: B. $12$

    The factor $(x+2)^2$ is non-negative, and equals $0$ at $x=-2$, which satisfies $\le0$. For $x\ne-2$, divide by it: $\frac{x-5}{x-1}\le0$, true for $1<x\le5$ (at $x=1$ undefined, at $x=5$ equal to 0). Integer solutions: $2,3,4,5$ and the separate $-2$. Sum $=14-2=12$.