Choose the correct answer. If $x$ is a real number and $|x| < 3,$ then:
- A$\text{x}\geq3$
- ✓$-3<\text{x}<3$
- C$\text{x}\leq-3$
- D$-3\leq\text{x}\leq3$
Answer: B.
View full solution →Question types
Linear Inequalities question types
235 questions across 9 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
235
Questions
9
Question groups
5
Question types
01
M.C.Q (1 Marks)
145 Q→02Assertion (A) & Reason (B) MCQ
4 Q→03True False[1 Marks ]
11 Q→041 Marks Question
13 Q→052 Marks Questions
31 Q→063 Marks Question
19 Q→07Case study (4 Marks)
2 Q→08Fill In The Blanks[1 Marks ]
7 Q→095 Marks Questions
3 Q→Sample Questions
Linear Inequalities questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
If $−5\leq\frac{5 – 3\text{x}}{2}\leq8, $ then $\text{x}\in$
- ✓$\big[\frac{11}{3},5\big]$
- B$\big[-5,5\big]$
- C$\big[\frac{-11}{3},\infty\big]$
- D$\big(-\infty,\infty\big)$
Answer: A.
View full solution →The solution of $\Big|\frac{2}{(\text{x} – 4)}\Big|>1$ where $\text{x}\neq4$ is:
- A$(2, 6)$
- ✓$(2, 4)\cup(4, 6)$
- C$(2, 4)\cup(4,\infty)$
- D$(-\infty, 4)\cup(4, 6)$
Answer: B.
View full solution →Write the solution of inequality $\frac{1}{5}\bigg(\frac{3\text{x}}{5}+4\bigg)\geq\frac{1}{3}(\text{x}-6).$
- ✓$\text{x}\leq\frac{105}{8}$
- B$\text{x}\geq\frac{105}{8}$
- C$\text{x}\geq120$
- D$\text{x}\leq120$
Answer: A.
View full solution →If the roots of the equation $x^2- bx + c = 0$ be two consecutive integers, then $b^2 - 4c$ equals:
- ✓$1$
- B$2$
- C$3$
- D$-2$
Answer: A.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: If $11\text{x}-9\leq68,$ then $\text{x}\in(-\infty,7).$
Reason: If an inequality consist of signs $\leq$ or $\geq,$ then the point on the line are also included in the solution region.
Assertion: If $11\text{x}-9\leq68,$ then $\text{x}\in(-\infty,7).$
Reason: If an inequality consist of signs $\leq$ or $\geq,$ then the point on the line are also included in the solution region.
- AAssertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- BAssertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- CAssertion is correct statement but Reason is wrong statement.
- ✓Assertion is wrong statement but Reason is correct statement.
Answer: D.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion:$|38x - 5| > 9\Rightarrow\text{x}\in\Big(-\infty,\frac{-4}{3}\Big)\cup\Big(\frac{14}{3},\infty\Big).$
Reason: The region containing all the solutions of an inequality is called the solution region.
Assertion:$|38x - 5| > 9\Rightarrow\text{x}\in\Big(-\infty,\frac{-4}{3}\Big)\cup\Big(\frac{14}{3},\infty\Big).$
Reason: The region containing all the solutions of an inequality is called the solution region.
- AAssertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- ✓Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- CAssertion is correct statement but Reason is wrong statement.
- DAssertion is wrong statement but Reason is correct statement.
Answer: B.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: If f $a < b, c < 0,$ then $\frac{\text{a}}{\text{c}}<\frac{\text{b}}{\text{c}}.$
Reason: If both sides are divided by the same negative quantity, then the inequality is reversed.
Assertion: If f $a < b, c < 0,$ then $\frac{\text{a}}{\text{c}}<\frac{\text{b}}{\text{c}}.$
Reason: If both sides are divided by the same negative quantity, then the inequality is reversed.
- AAssertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- BAssertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- CAssertion is correct statement but Reason is wrong statement.
- ✓Assertion is wrong statement but Reason is correct statement.
Answer: D.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: If $\text{x}\geq-3,$ then $\text{x}+5\geq2.$
Reason: Same number can be added to both sides of the inequality without changing the sign of inequality.
Assertion: If $\text{x}\geq-3,$ then $\text{x}+5\geq2.$
Reason: Same number can be added to both sides of the inequality without changing the sign of inequality.
- ✓Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- BAssertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- CAssertion is correct statement but Reason is wrong statement.
- DAssertion is wrong statement but Reason is correct statement.
Answer: A.
View full solution →State which of the following statements is True or False.
If $|\text{x}|\leq4$ then $\text{x}\in[-4, 4].$
View full solution →If $|\text{x}|\leq4$ then $\text{x}\in[-4, 4].$
State which of the following statements is True or False.
If x < y and b < 0, then $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}.$
View full solution →If x < y and b < 0, then $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}.$
State which of the following statements is True or False.
If |x| > 5, then $\text{x}\in(-\infty, -5)\cup(5, \infty).$
View full solution →If |x| > 5, then $\text{x}\in(-\infty, -5)\cup(5, \infty).$
State which of the following statements is True or False.
Solution set $\text{x}\geq0$ and $\text{y}\leq1$ is:
View full solution →Solution set $\text{x}\geq0$ and $\text{y}\leq1$ is:
State which of the following statements is True or False.
Solution set of $\text{x}\geq0$ and $\text{y}\leq0.$
View full solution →Solution set of $\text{x}\geq0$ and $\text{y}\leq0.$
Solve: 3x + 8 >2, when x is a real number.
Solve: 3x + 8 >2, when x is an integer.
Solve: 5x – 3 < 7, when x is a real number.
Solve: 5x – 3 < 7, when x is an integer.
Solve: –12x > 30, when x is an integer.
Solve the inequality: $7 \leq \frac{(3 x+11)}{2} \leq 11$
View full solution →Solve the inequality: $-12<4-\frac{3 x}{-5} \leq 2$
View full solution →Solve the inequality: $-15<\frac{3(x-2)}{5} \leq 0$
View full solution →Solve the inequality: $-3 \leq 4-\frac{7 x}{2} \leq 18$
View full solution →Solve the inequality: $6 \leq- 3(2x - 4) < 12$
View full solution →Solve the inequality and represent the solution graphically on number line: 3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x
Solve the inequality and represent the solution graphically on number line: 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
Solve the inequality and represent the solution graphically on number line: 5x + 1 > – 24, 5x – 1 < 24
A solution is to be kept between 68°F and 77°F. What is the range of temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) convension formula is given by $F = \frac{9}{5}C + 32$
View full solution →Solve the inequality and represent the solution graphically on number line: 5 (2x – 7) – 3 (2x + 3) $\le$ 0 , 2x + 19 $\le$ 6x + 47.
View full solution →A manufacturing company produces certain goods. The company manager used to make a data record on daily basis about the cost and revenue of these goods separately. The cost and revenue function of a product are given by $C(x)=20 x+4000$ and $R(x)=60 x+2000$, respectively, where $x$ is the number of goods produced and sold.
Based on above information, answer the following questions.
(i) How many goods must be sold to realise some profit?
(a) $\mathrm{x}<\mathbf{5 0}$
(b) $x>50$
(c) $x \geq 50$
(d) $\mathbf{x} \leq \mathbf{5 0}$
(ii) If the cost and revenue functions of a product are given by $C(x)=3 x+400$ and $R(x)=$ $5 x+20$ respectively, where $x$ is the number of items produced by the manufacturer, then how many items must be sold to realise some profit?
(a) $x \leq 190$
(b) $x \geq 190$
(c) $x<190$
(d) $x>190$
(iii) Let $\mathbf{x}$ and $\mathbf{b}$ are real numbers. If $\mathbf{b} > \mathbf{0}$ and $\mathbf{x}< \mathbf{b}$, then
(a) $x$ is always positive
(b) $\mathbf{X}$ is always negative
(c) $\mathrm{x}$ is real number
(d) None of these
(iv) The solution set of $\mathbf{3}-\mathbf{5}<\mathrm{x}+\mathbf{7}$, when $\mathrm{x}$ is a whole number is given by
(a) $\{0,1,2,3,4,5\}$
(b) $(-\infty, 6)$
(c) $[0,5]$
(d) None of these
(v) Graph of inequality $x>2$ on the number line is represented by
(a)
(b)
(c)
(d) None of the above
View full solution →Based on above information, answer the following questions.
(i) How many goods must be sold to realise some profit?
(a) $\mathrm{x}<\mathbf{5 0}$
(b) $x>50$
(c) $x \geq 50$
(d) $\mathbf{x} \leq \mathbf{5 0}$
(ii) If the cost and revenue functions of a product are given by $C(x)=3 x+400$ and $R(x)=$ $5 x+20$ respectively, where $x$ is the number of items produced by the manufacturer, then how many items must be sold to realise some profit?
(a) $x \leq 190$
(b) $x \geq 190$
(c) $x<190$
(d) $x>190$
(iii) Let $\mathbf{x}$ and $\mathbf{b}$ are real numbers. If $\mathbf{b} > \mathbf{0}$ and $\mathbf{x}< \mathbf{b}$, then
(a) $x$ is always positive
(b) $\mathbf{X}$ is always negative
(c) $\mathrm{x}$ is real number
(d) None of these
(iv) The solution set of $\mathbf{3}-\mathbf{5}<\mathrm{x}+\mathbf{7}$, when $\mathrm{x}$ is a whole number is given by
(a) $\{0,1,2,3,4,5\}$
(b) $(-\infty, 6)$
(c) $[0,5]$
(d) None of these
(v) Graph of inequality $x>2$ on the number line is represented by
(a)
(b)
(c)
(d) None of the above
Shweta was teaching "method to solve a linear inequality in one variable" to her daughter.
Step I Collect all terms involving the variable (x) on one side and constant terms on other side with the help of above rules and then reduce it in the form $\mathbf{a x}<\mathbf{b}$ or $\mathbf{a x} \leq \mathbf{b}$ or $\mathbf{a x}>\mathbf{b}$ or $\mathbf{a x} \geq \mathbf{b}$.
Step II Divide this inequality by the coefficient of variable (x). This gives the solution set of given inequality.
Step III Write the solution set.
Based on above information, answer the following questions.
(i) The solution set of $\mathbf{2 4 x}<\mathbf{1 0 0}$, when $\mathrm{x}$ is a natural number is
(a) $\{1,2,3,4\}$ (b) $(1,4)$ (c) $[1,4]$ (d) None of these
(ii) The solution set of $24100 \mathrm{x}<$, when $\mathrm{x}$ is an integer is
(a) $\{\ldots \ldots-4,-3,-2,-1,0,1,2,3,4\}$ (b) $(-\infty, 4]$ (c) $[4, \infty]$ (d) None of the above
(iii) The solution set of $-\mathbf{5 x}+\mathbf{2 5}>0$ is
(a) $[5, \infty)$ (b) $(-\infty, 5]$ (c) $(5, \infty)$ (d) $(-\infty, 5)$
(iv) The solution set of $\mathbf{3 x}-\mathbf{5}<\mathbf{x + 7}$ is
(a) $(6, \infty)$ (b) $[6, \infty)$ (c) $(-\infty, 6)$ (d) $(-\infty, 6]$
(v) The solution set of $x+\frac{x}{2}+\frac{x}{3}<11$ is
(a) $(-\infty, 6]$ (b) $(-\infty, 6)$ (c) $[6, \infty)$ (d) None of these
View full solution →Step I Collect all terms involving the variable (x) on one side and constant terms on other side with the help of above rules and then reduce it in the form $\mathbf{a x}<\mathbf{b}$ or $\mathbf{a x} \leq \mathbf{b}$ or $\mathbf{a x}>\mathbf{b}$ or $\mathbf{a x} \geq \mathbf{b}$.
Step II Divide this inequality by the coefficient of variable (x). This gives the solution set of given inequality.
Step III Write the solution set.
Based on above information, answer the following questions.
(i) The solution set of $\mathbf{2 4 x}<\mathbf{1 0 0}$, when $\mathrm{x}$ is a natural number is
(a) $\{1,2,3,4\}$ (b) $(1,4)$ (c) $[1,4]$ (d) None of these
(ii) The solution set of $24100 \mathrm{x}<$, when $\mathrm{x}$ is an integer is
(a) $\{\ldots \ldots-4,-3,-2,-1,0,1,2,3,4\}$ (b) $(-\infty, 4]$ (c) $[4, \infty]$ (d) None of the above
(iii) The solution set of $-\mathbf{5 x}+\mathbf{2 5}>0$ is
(a) $[5, \infty)$ (b) $(-\infty, 5]$ (c) $(5, \infty)$ (d) $(-\infty, 5)$
(iv) The solution set of $\mathbf{3 x}-\mathbf{5}<\mathbf{x + 7}$ is
(a) $(6, \infty)$ (b) $[6, \infty)$ (c) $(-\infty, 6)$ (d) $(-\infty, 6]$
(v) The solution set of $x+\frac{x}{2}+\frac{x}{3}<11$ is
(a) $(-\infty, 6]$ (b) $(-\infty, 6)$ (c) $[6, \infty)$ (d) None of these
Fill in the blanks.
If $-\frac{3}{4}\text{x}\leq-3,$ then $\text{x}.......4.$
View full solution →If $-\frac{3}{4}\text{x}\leq-3,$ then $\text{x}.......4.$
Fill in the blanks.
If x > -5 then 4x .....-20.
If x > -5 then 4x .....-20.
Fill in the blanks.
If $-4\text{x}\geq12,$ then $\text{x}.......-3.$
View full solution →If $-4\text{x}\geq12,$ then $\text{x}.......-3.$
Fill in the blanks.
If $\frac{2}{\text{x}+2}>0,$ then $\text{x}.......2.$
View full solution →If $\frac{2}{\text{x}+2}>0,$ then $\text{x}.......2.$
Fill in the blanks.
If |x + 2| > 5, then x ...... -7 or x ......-3.
If |x + 2| > 5, then x ...... -7 or x ......-3.
IQ of a person is given by the formula
$\mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}} \times 100$
where MA is mental age and CA is chronological age. If 80 $\le$ IQ $\le$ 140 for a group of 12 years old children, find the range of their mental age.
View full solution →$\mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}} \times 100$
where MA is mental age and CA is chronological age. If 80 $\le$ IQ $\le$ 140 for a group of 12 years old children, find the range of their mental age.
How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
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