MCQ 1011 Mark
If Ram has $x$ rupees and he pay $40$ rupees to shopkeeper then find range of $x$ if amount of money left with Ram is at least $10$ rupees is given by inequation, $.........?$
- A
$x ≥ 10$
- B
$x ≤ 10$
- C
$x ≤ 50$
- ✓
$x ≥ 50$
AnswerCorrect option: D. $x ≥ 50$
Amount left is at least $10$ rupees i.e.amount left $≥ 10.$
$x - 40 \geq 10$
$\Rightarrow x \geq 50.$
View full question & answer→MCQ 1021 Mark
Consider the linear inequations and solve them graphically: $3\text{x−y}-2>0,\text{x+y}\leq4:\text{x}>0\text{y}\geq0$ Which of the following points belong to the feasible solution region?
- A
$\Big(\frac{1}{2},0\Big)$
- B
$\Big(\frac{1}{2},\frac{1}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{5}{2}\Big)$
- ✓
View full question & answer→MCQ 1031 Mark
Choose the correct answer. If $|x - 1| > 5,$ then:
- A
$\text{x}\in(-4, 6)$
- B
$\text{x}\in[-4,6]$
- ✓
$\text{x}\in[-\infty,-4)\cup(6,\infty) $
- D
$\text{x}\in[-\infty,-4)\cup[6,\infty) $
AnswerCorrect option: C. $\text{x}\in[-\infty,-4)\cup(6,\infty) $
Given that $|x - 1| > 5$
$\Rightarrow (x - 1) < -5$ or $(x - 1) > 5$
$\Rightarrow x < -5 + 1$ or $x > 5 + 1$
$\Rightarrow x < -4$ or $x > 6$
$\Rightarrow\text{x}\in[-\infty,-4)\cup(6,\infty) $
View full question & answer→MCQ 1041 Mark
The inequality $\frac{2}{\text{x}}<3$ is true, when $x$ belongs to:
- A
$\big[\frac{2}{3},\infty\big) $
- B
$\big(-\infty,\frac{2}{3}\big)$
- ✓
$\big(-\infty ,0\big)\cup\big(\frac{2}{3},\infty\big) $
- D
$\text{None of these}$
AnswerCorrect option: C. $\big(-\infty ,0\big)\cup\big(\frac{2}{3},\infty\big) $
View full question & answer→MCQ 1051 Mark
If $7\text{x}+3<5\text{x}+9$ then $\text{x}\in$
AnswerCorrect option: C. $\big(-\infty, 3\big)$
View full question & answer→MCQ 1061 Mark
If $\frac{\text{x} – 2}{\text{x} + 5}> 2 ,$ then $\text{x}\in$
- A
$(–12, 5)$
- ✓
$(–12, –5)$
- C
$(–5, 12)$
- D
$(5, 12)$
AnswerCorrect option: B. $(–12, –5)$
View full question & answer→MCQ 1071 Mark
Solution of $11 +\frac{3}{\text{xl}}> 2 $ is:
AnswerCorrect option: C. $\big(-1, 0\big)\cup\big(0, 3\big)$
View full question & answer→MCQ 1081 Mark
Solving $ – 8 \leq 5\text{x} – 3<7, $ we get:
AnswerCorrect option: C. $ –1\leq\text{x}<2$
Given,
$-8\leq5\text{x}-3$ and $5\text{x}-3<7$
Let us solve these two inequalities simultaneously.
$-8\leq5\text{x}-3$ and $5\text{x}-3<7$ can be written as:
$– 8\leq5\text{x} –3 < 7$
Adding $3,$ we get
$– 8 + 3\leq5\text{x}-3+3<7+3$
$–5\leq5\text{x}<10$
Dividing by $5,$ we get
$–1\leq\text{x}<2$
View full question & answer→MCQ 1091 Mark
The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is $160\ cm,$ then:
- A
breadth $> 20\ cm$
- B
length $< 20\ cm$
- ✓
breadth $\geq 20\ cm$
- D
length $\leq 20\ cm$
AnswerCorrect option: C. breadth $\geq 20\ cm$
View full question & answer→MCQ 1101 Mark
Number of integral solutions satisfy inequality $|\text{x-3}|-|2\text{x}+5|\geq|\text{x}+8|$
View full question & answer→MCQ 1111 Mark
If $x > 7$ then $-x > -7$ is, .$............?$
AnswerIf we multiply by negative number on both sides of inequality then,
sign of inequality will change
i.e. if $x > 7$ then $(-1) x < (-1)^7$
$\Rightarrow -x < -7.$
View full question & answer→MCQ 1121 Mark
If the equations $\text{x}^2+2\text{x}+3=0 $ and $\text{ax}^2+\text{bx}+\text{c}=0,\text{abc}\in\text{R},$ I have a common root, then $a : b : c$ is.
- ✓
$1 : 2 : 3$
- B
$3 : 2 : 1$
- C
$1 : 3 : 2$
- D
$3 : 1 : 2$
AnswerCorrect option: A. $1 : 2 : 3$
View full question & answer→MCQ 1131 Mark
If $\frac{\text{lx}-2\text{l}}{\text{x}-2}\geq0$ then, $\text{x}\in$
- A
$\big[-2,\infty\big)$
- ✓
$\big(2,\infty\big)$
- C
$\big(\infty, 2\big)$
- D
$\big(-\infty, 2\big]$
AnswerCorrect option: B. $\big(2,\infty\big)$
View full question & answer→MCQ 1141 Mark
If $x$ is a positive integer and $20 x < 100$ then find solution set of $x.$
- A
$(0, 1, 2, 3, 4, 5)$
- B
$(1, 2, 3, 4, 5)$
- ✓
$(1, 2, 3, 4)$
- D
$(0, 1, 2, 3, 4)$
AnswerCorrect option: C. $(1, 2, 3, 4)$
$20\text{x}<100$
Dividing by $20$ on both sides, $\text{x}<\Big(\frac{100}{20}\Big)$
$\Rightarrow\text{x}<5$
Since $x$ is a positive integer so $x = 1, 2, 3, 4.$
View full question & answer→MCQ 1151 Mark
Choose the correct answer: $x$ and $b$ are real numbers. If $b > 0$ and $|x| > b,$ then:
- A
$\text{x}\in(-\text{b},\infty)$
- B
$\text{x}\in(\infty,-\text{b})$
- C
$\text{x}\in(-\text{b},\text{b})$
- ✓
$\text{x}\in(-\infty,-\text{b})\cup(\text{b},\infty)$
AnswerCorrect option: D. $\text{x}\in(-\infty,-\text{b})\cup(\text{b},\infty)$
Given that $|x| > b, b > 0$
$\Rightarrow x < -b$ or $x > b$
$\Rightarrow\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$
View full question & answer→MCQ 1161 Mark
Solve : $2\text{x}+1>3$
- A
$\big[-1,\infty\big]$
- ✓
$(1,\infty)$
- C
$(\infty,\infty)$
- D
$(\infty,1)$
AnswerCorrect option: B. $(1,\infty)$
Given, $2\text{x}+1>3$
$\Rightarrow2\text{x}>3-1$
$\Rightarrow2\text{x}>2$
$\Rightarrow\text{x}>1$
$\Rightarrow\text{x}\in(1,\infty)$
View full question & answer→MCQ 1171 Mark
If $2x + 1 > 5$ then which is true?
- A
$x > 4$
- B
$x < 4$
- ✓
$x > 2$
- D
$x < 2$
AnswerCorrect option: C. $x > 2$
$2x + 1 > 5$
$\Rightarrow 2x > 5 - 1$
$\Rightarrow 2x > 4$
$\Rightarrow x > 2.$
View full question & answer→MCQ 1181 Mark
If $1\leq|\text{x} -2|\leq3,$ then $\text{x}\in$
- A
$\big[–1, 5\big]$
- ✓
$\big[–1, 1\big]\cup\big[3, 5\big]$
- C
$\big(–1, 0\big)\cup\big(2, 5\big)$
- D
$\big(–1, 5\big)$
AnswerCorrect option: B. $\big[–1, 1\big]\cup\big[3, 5\big]$
View full question & answer→MCQ 1191 Mark
If $x^2 < 4$ then the value of $x$ is:
- A
$(0, 2)$
- ✓
$(-2, 2)$
- C
$(-2, 0)$
- D
AnswerCorrect option: B. $(-2, 2)$
Given, $x^2 < 4$
$\Rightarrow x^2– 4 < 0$
$\Rightarrow (x – 2) \times (x + 2) < 0$
$\Rightarrow -2 < x < 2$
$\Rightarrow x \in (-2, 2)$
View full question & answer→MCQ 1201 Mark
Solve: $\frac{-1}{(|\text{x}| – 2)}\geq1$ where $\text{x}\in\text{R}, \text{x}\neq\pm2$
- A
$(-2, -1)$
- B
$(-2, 2)$
- ✓
$(-2, -1)\cup(1, 2)$
- D
$\text{None of these}$
AnswerCorrect option: C. $(-2, -1)\cup(1, 2)$
Given, $\frac{-1}{(|\text{x}| – 2)}\geq1$
$\Rightarrow\frac{-1}{(|\text{x}|-2) -1}\geq0$
$\Rightarrow\frac{-1 – (|\text{x}| – 2)}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{1 – |\text{x}|}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{-(|\text{x}| – 1)}{(|\text{x}| – 2)}\geq0$
Using number line rule:
$1\leq |\text{x}|<2$
$\Rightarrow\text{x}\in(-2, -1)\cup(1, 2)$ View full question & answer→MCQ 1211 Mark
Which of the following statements is correct:
- ✓
If $x > y$ and $b < 0,$ then $bx < by$
- B
If $x > y,$ then $x > 0$ and $y < 0$
- C
If $xy < 0,$ then $x > 0$ and $y > 0$
- D
AnswerCorrect option: A. If $x > y$ and $b < 0,$ then $bx < by$
View full question & answer→MCQ 1221 Mark
Rahul obtained $20$ and $25$ marks in first two tests. Find the minimum marks he should get in the third test to have an average of at least $30$ marks.
AnswerAverage is at least $30$ marks.
Let $x$ be the marks in $3^\text{rd}$ test.
Average $=\frac{(20+25+\text{x})}{3}\geq30$
$\Rightarrow45+\text{x}\geq90$
$\Rightarrow\text{x}\geq90-45$
$\Rightarrow\text{x}\geq45.$
Minimum marks in $3^\text{rd}$ test should be $45.$
View full question & answer→MCQ 1231 Mark
If $|\text{x}| = -5$ then the value of $x$ lies in the interval:
- A
$(-5,\infty)$
- B
$(5,\infty)$
- C
$(\infty,-5)$
- ✓
$\text{No Solution}$
AnswerCorrect option: D. $\text{No Solution}$
Given, $|x| = -5$
Since $|x|$ is always positive or zero.
So, it can not be negative.
Hence, given inequality has no solution.
View full question & answer→MCQ 1241 Mark
If $-2<2\text{x}-1<2$ then the value of $x$ lies in the interval:
- A
$\Big(\frac{1}{2},\frac{3}{2}\Big)$
- ✓
$\Big(\frac{-1}{2},\frac{3}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{1}{2}\Big)$
- D
$\Big(\frac{3}{2},\frac{-1}{2}\Big)$
AnswerCorrect option: B. $\Big(\frac{-1}{2},\frac{3}{2}\Big)$
Given, $-2<2\text{x}-1<2$
$\Rightarrow2+1< 2\text{x}<2+1$
$\Rightarrow-1<2\text{x}<3$
$\Rightarrow\frac{-1}{2}<\text{x}<\frac{3}{2}$
$\Rightarrow\text{x}\in\Big(\frac{-1}{2},\frac{3}{2}\Big)$
View full question & answer→MCQ 1251 Mark
If the roots of the quadratic equation $x^2+ px + q = 0$ are $\tan 30^\circ$ and $\tan 15^\circ$ then the value of $2 + q - p$ is:
View full question & answer→MCQ 1261 Mark
Consider the linear inequations and solve them graphically: $3\text{x−y}-2>0,\text{x+y}\leq4:\text{x}>0\text{y}\geq0$ Which of the following points belong to the feasible solution region?
- A
$\Big(\frac{1}{2},0\Big)$
- B
$\Big(\frac{1}{2},\frac{1}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{5}{2}\Big)$
- ✓
View full question & answer→MCQ 1271 Mark
If $\frac{\text{lx}-2\text{l}}{\text{x}-2}\geq0$ then, $\text{x}\in$
- A
$\big[2,\infty\big)$
- ✓
$\big(2,\infty\big)$
- C
$\big(\infty, 2\big)$
- D
$\big(-\infty, 2\big]$
AnswerCorrect option: B. $\big(2,\infty\big)$
View full question & answer→MCQ 1281 Mark
The marks obtained by a student of Class $XI$ in first and second terminal examination are $62$ and $48,$ respectively.Find the minimum marks he should get in the annual examination to have an average of at least $60$ marks.
View full question & answer→MCQ 1291 Mark
The inequality representing the following graph is:
- A
$\text{|x|}<3$
- ✓
$\text{|x|}\leq3$
- C
$\text{|x|}>3$
- D
$\text{|x|}\geq3$
AnswerCorrect option: B. $\text{|x|}\leq3$
As according to the graph,
$x$ lies between $−3$ and $3$
$\Rightarrow-3\leq\text{x}\leq3$
$\Rightarrow|\text{x}|\leq3$
View full question & answer→MCQ 1301 Mark
Solution of $0<|3\text{x}+1\big|<\frac{1}{3} $ is:
- A
$(\frac{-4}{9},\frac{2}{9})$
- B
$[\frac{-4}{9},\frac{-2}{9}]$
- ✓
$(\frac{-4}{9},\frac{2}{9})-(\frac{-1}{3})$
- D
$[\frac{-4}{9},\frac{-2}{9}]-(\frac{-1}{3})$
AnswerCorrect option: C. $(\frac{-4}{9},\frac{2}{9})-(\frac{-1}{3})$
View full question & answer→MCQ 1311 Mark
Choose the correct answer. Given that $x, y$ and $b$ are real numbers and $x < y, b < 0,$ then:
- A
$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
- B
$\frac{\text{x}}{\text{b}}\leq\frac{\text{y}}{\text{b}}$
- ✓
$\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
- D
$\frac{\text{x}}{\text{b}}\geq\frac{\text{y}}{\text{b}}$
AnswerCorrect option: C. $\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
Given that $x < y, b < 0$
$\Rightarrow\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}},\text{b}<0$
View full question & answer→MCQ 1321 Mark
The sum of four numbers in $AP$ is $20.$The numbers are such that the ratio of the product of first and fourth is to the product of second and third as $2 : 3.$The greatest number is:
AnswerLet the four terms be $a − 3d, a − d, a + d,$ and $a + 3d$ with common difference $2d.$
sum $= 4a = 20a = 5a = 5$
$\frac{\text{(a-3d)}\text{(a+3d)}}{\text{(a-d)}\text{(a+d)}}=\frac{2}{3}$
$\frac{25-9\text{d}^2}{25-\text{d}^2}=\frac{2}{3}$
$\text{d}^2=1$
$\Rightarrow\text{d}=1$
Largest term $= a + 3d = 5 + 3(1) = 8$
View full question & answer→MCQ 1331 Mark
The set of values of $x$ which satisfy the inequations $5\text{x}+2<3\text{x}+8$ and $\text{x}+\frac{\text{x}+2}{\text{x}-1}<4$ is:
- A
$(-\infty, 1)$
- B
$(2, 3)$
- C
$\big(-\infty, 3)$
- ✓
$(-\infty, 1)\cup(2, 3)$
AnswerCorrect option: D. $(-\infty, 1)\cup(2, 3)$
View full question & answer→MCQ 1341 Mark
If $\text{x}^2<-4$ then the value of $x$ is:
- A
$(-2,2)$
- B
$(2,\infty)$
- C
$(-2,\infty)$
- ✓
$\text{No solution}$
AnswerCorrect option: D. $\text{No solution}$
Given, $\text{x}^2<-4$
$\Rightarrow\text{x}^2+4<0$
Which is not possible.
So, there is no solution.
View full question & answer→MCQ 1351 Mark
If $\frac{\big[\text{x} – 7\big]}{(\text{x} – 7)\geq 0}$ then:
- A
$\text{x}\in\big[7,\infty)$
- ✓
$\text{x}\in(7,\infty)$
- C
$\text{x}\in(\infty, 7)$
- D
$\text{x}\in(-\infty, 7)$
AnswerCorrect option: B. $\text{x}\in(7,\infty)$
Given,
$\frac{|\text{x}-7|}{(\text{x}-7)}\geq0$
This is possible when $\text{x}-7\geq0,$ and $\text{x}-7\neq0.$
Here, $\text{x}\geq7$ but $\text{x}\neq7$
Therefore, $\text{x}> 7,$
i.e $\text{x}\in(7,\infty).$
View full question & answer→MCQ 1361 Mark
If $3\text{x}-7<5+\text{x},11-5\text{x}\leq1,$, then $\text{x}\in$
- A
$\big[2, 6\big]$
- B
$\big[–2, 6\big]$
- ✓
$\big[2,6\big)$
- D
$\big(-2,6\big)$
AnswerCorrect option: C. $\big[2,6\big)$
View full question & answer→MCQ 1371 Mark
If $|\text{x} + 3|\geq10,$ then:
- A
$\text{x}\in(-13,7\big)$
- B
$\text{x}\in(-13,7\big]$
- C
$\text{x}\in(-\infty -13\big]\cup \big[7,\infty)$
- ✓
$\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
AnswerCorrect option: D. $\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
$|\text{x} + 3|\geq10$
$\Rightarrow\text{x} + 3\leq-10$ or $\text{x}+3\geq10$
$\Rightarrow\text{x}\leq -13 $ or $\text{x}\geq7$
$\Rightarrow\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
View full question & answer→MCQ 1381 Mark
The cost and revenue functions of a product are given by $C(x) = 20x + 4000$ and $R(x) = 60x + 2000,$ respectively, where $x$ is the number of items produced and sold. How many items must be sold to realise some profit?
- A
Less than $40$
- ✓
More than $50$
- C
Less than $50$
- D
Exactly $50$
AnswerCorrect option: B. More than $50$
View full question & answer→MCQ 1391 Mark
Choose the correct answer. If $– 3x + 17 < – 13,$ then:
- ✓
$\text{x}\in(10, \infty)$
- B
$\text{x}\in[10, \infty)$
- C
$\text{x}\in(-\infty\text{j},10]$
- D
$\text{x}\in[-10, 10)$
AnswerCorrect option: A. $\text{x}\in(10, \infty)$
Given that $- 3x + 17 < - 13$
$\Rightarrow - 3x < - 17 - 13$
$\Rightarrow -3x < - 30$
$\Rightarrow 3x > 30$
$\Rightarrow x > 10$
$\Rightarrow\text{x}\in(10, \infty)$
View full question & answer→MCQ 1401 Mark
Solve: $2\text{x}+1>3$
AnswerCorrect option: B. $\big(1,\infty\big)$
Given, $2\text{x}+1>3$
$\Rightarrow2\text{x}>3-1$
$\Rightarrow2\text{x}>2$
$\Rightarrow\text{x}>1$
$\Rightarrow\text{x}\in\big(1,\infty\big)$
View full question & answer→MCQ 1411 Mark
$y < -2$ involves region are, $...........?$
- A
above dotted line $y = -2$
- ✓
below dotted line $y = -2$
- C
above complete line $y = -2$
- D
below complete line $y = -2$
AnswerCorrect option: B. below dotted line $y = -2$
$y < -2$ does not satisfy $(0, 0)$
so, region is below $y = -2.$
Since only inequality sign given,
so dotted line $y = -2.$
View full question & answer→MCQ 1421 Mark
Value of $\text{i(iota)}$ is, $...........?$
- A
$-1$
- B
$1$
- ✓
$(-1)^\frac{1}{2}$
- D
$(-1)^\frac{1}{4}$
AnswerCorrect option: C. $(-1)^\frac{1}{2}$
Explanation: Iota is used to denote complex number.
The value of $\text{i (iota)}$ is $\sqrt{-1}$
$\text{ i.e.}(-1)^\frac{1}{2}$
View full question & answer→MCQ 1431 Mark
Find the number of real numbers in the solution set of following $\frac{2\text{x}}{5}+1<-3.$
Answer$\frac{2\text{x}}{5}+1<-3.$
$\frac{2\text{x}}{5}<-3-1$
$2\text{x}<-4\times5$
$\text{x}<-\frac{20}{2}$
$\text{x}<-10$
As we know the number of rational numbers on either side of a number on the number line is infinite.
$\therefore$ set $x = ($infinite number of integers which are $< - 10)$
View full question & answer→MCQ 1441 Mark
Formulate the equations for the above problem.$(x$ and $y$ are the number of units of $A$ and $B$ manufactured in a day respectively$)$
- A
$15\text{x}+5\text{y}\leq10:24\text{x}+14\text{y}\geq1000$
- ✓
$15\text{x}+5\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
- C
$15\text{x}+15\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
- D
$15\text{x}+15\text{y}\leq10:24\text{x}+14\text{y}\geq1000$
AnswerCorrect option: B. $15\text{x}+5\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
View full question & answer→MCQ 1451 Mark
$x > 5$ is, $............?$
AnswerSince $a$ variable $'x\ '$ is compared with number $'5\ '$ with inequality sign so it is called literal inequality.
View full question & answer→