MCQ Questions for Class 11 Engineering Maths Linear Inequalities Quiz 2

Which of the following inequations represents the shaded region?

0%
$\displaystyle 2x+y\leq 4$
0%
$\displaystyle 2x+y\geq 4$
0%
$\displaystyle x+2y\leq 4$
0%
$\displaystyle x+2y\geq 4$

Explanation

The line passes through

$(0,4)$ and

$(2,0)$
Hence equation of the line

$\dfrac{y}{4}+\dfrac{x}{2}=1$

$2x+y=4$
Now in the above inequality graph the shading is done towards the origin.
Hence by substituting

$x=0,y=0$ we get an inequality which is true.
Hence the possible inequality is

$2x+y<4$
Now the straight line is dark and not dotted. This indicates that the points on the line are a part of the inequality.
Hence the inequality can be written as

$2x+y=4$ or

$2x+y<4$

$2x+y\leq 4$ .

The locus represented by equation |z i| + |z + i|

$=$ 1 on Argand plane

0%
ellipse
0%
hyperbola
0%
straight line
0%
no locus

The shaded region is represented by the inequation:

0%
$\displaystyle y\geq x$
0%
$\displaystyle y\geq- x$
0%
$\displaystyle y\geq \left | x \right |$
0%
$\displaystyle y\leq \left | x \right |$

Explanation

The equations of both the lines in the above graph are

$y=-x$ and

$y=x$
Hence if we put them together we get

$y=|x|$
Now
Let us take a point inside the shaded region.
Let it be

$(0,2)$
Now

$2>0$

$y>|x|$
Hence the required inequality is

$y\geq |x|$ .

The table shown here contains pricing information for the chopper (a kitchen tool) that is advertised on television The price depends on how soon a customer calls after the end of the advertisement Approximately, how much more would a customer who orders 3 choppers, 23 minutes after the end of the television advertisement pay than a customer who orders 2 chopper, 17 minutes after the advertisement ?

Minutes (m) after television advertisement ends	Price per chopper (In Rs.)
$\displaystyle 0\leq m\leq10$	34.95
$\displaystyle 10< m\leq20$	39.95
$\displaystyle 20< m\leq30$	44.95
$\displaystyle 30< m\leq40$	49.95
$\displaystyle m> 40$	59.95

0%
Rs 15
0%
Rs 35
0%
Rs 40
0%
Rs 55

The inequalities in which terms compared are never equal to each other is classified as

0%
strict equality
0%
strict inequality
0%
strict quadres
0%
differential quadrates

Explanation

The inequalities in which the terms are never equal to each other are termed as

$strict$

$inequality$

The inequality symbol in the strict inequality is either

$\gt$ or

$\lt$ i.e., it doesn't have any equality conditions.

The point which does not belong to the feasible region of the LPP:
Minimize:

$Z=60x+10y$
subject to

$3x+y \ge 18$

$2x+2y \ge 12$

$x+2y\ge 10$

$x,y \ge 0$ is

0%
$(0,8)$
0%
$(4,2)$
0%
$(6,2)$
0%
$(10,0)$

Explanation

We test whether the inequalitiies are satisfied or not

$(0,8)$ ,

$3(0)+8 \ge 8$

$8\ge 8$ is true.

$2(0)+2(8)=16\ge 12$ is true.

$0+2(8)=16 \ge 10$ is true.

$\therefore$

$(0,8)$ is in the feasible region.

$(4,2)$ ,

$3(4)+2=14 \ge 8$

$2(4)+2(2)=16\ge 12$

$4+2(2)=8\ge 10$ is not true

$\therefore$

$(4,2)$ is not a point in the feasible region

$\therefore$ (2) is correct

Which of the following number line represents the solution of the inequality

$-6x + 12 > -7x + 17$ ?

Explanation

Given,

$-6x+12>-7x+17$

$\Rightarrow -6x+7x>17-12$

$\Rightarrow x > 5$

Hence, option A is correct.

Which region is described by the shade in the graph given?

0%
$2x+3y=3$
0%
$2x+3y>3$
0%
$2x+3y<3$
0%
None of these

Explanation

The graph of the function

$2x+3y=3$ is shown in the figure.

It is clear from the graph,

$(0,0)$ does not lie on the line.

Now, put

$(0,0)$ in the equation of line.

$2(0)+3(0)=3$

$0\neq 3$ hence,

$0<3$ is true

Hence, region which lies below the graph is denoted by

$2x+3y<3$

It is clear from the graph,

$(0,2)$ does not lie on the line.

Now, put

$(0,2)$ in the equation of line,

$2(0)+3(2)=3$

$0\neq 6$ hence,

$6>3$ is true.

Hence, the region which lies above the graph is denoted by

$2x+3y>3$ .

Sketch the solution to system of ineqalities

$x\le -3$

$y < \frac{5}{3}x+2$

Explanation

Given,

$x\leq -3$ and

$y<\dfrac{5}{3}x+2$

first, draw the graph for equations

$x= -3$ and

$y=\dfrac{5}{3}x+2$

$x=-3$ is the line which is parallel to the y-axis and passes through (-3,0).

Hence,

$x\leq -3$ includes the left side region of the line.

similarly, for

$y=\dfrac{5}{3}x+2$

substitute y=0 we get,

$\dfrac{5}{3}x=-2 \implies x=-\dfrac{6}{5}=-1.2$

substitute x=0 we get,

$y=2$

therefore,

$y=\dfrac{5}{3}x+2$ line passes through

$(-1.2,0)$ and

$(0,2)$ .

Hence,

$y<\dfrac{5}{3}x+2$ includes the region below the line.

Option D satisfies the above conditions.

A graph and the system of inequalities are shown above. Which region of the graph could represent the solution for the system of in equations?

$y > x$

$3y≤-4x+6$

0%
$A$
0%
$B$
0%
$C$
0%
$D$

Explanation

given

$y>x$

and

$3y\leq -4x+6$

first, draw the graph for equations

$y =x$ and

$3y= -4x+6$

$y=x$ is the line which passes through the origin as shown in the above fig

Hence,

$y>x$ includes the above region of the line.

similarly, for

$3y= -4x+6$

substitute y=0 we get,

$-4x+6=0 \implies x=1.5$

substitute x=0 we get,

$3y=6 \implies y=2$

therefore,

$3y= -4x+6$ line passes through (1.5,0) and (0,2) as shown in fig.

Hence,

$3y\leq -4x+6$ includes the region below the line.

the intersection region is the D region as shown in above figure.

If the solution set for the system

$\begin{cases} y>x \\ y\le -\dfrac { 3 }{ 7 } x+5 \end{cases}$ is given by the above figure, then which of the following is NOT a solution to the system?

0%
$(0,3)$
0%
$(1,2)$
0%
$(2,4)$
0%
$(3,3)$

Explanation

As per the given figure,

$(0,3)$ and

$(1,2)$ lie in the desired region, hence these are solutions of the system.

Whereas

$(2,4)$ and

$(3,3)$ lie on the lines but

$(3,3)$ does not satisfy the equation

$y>x$ .

Hence, it is not a solution to the system.

State true or false:

The statement

$0>1$

$\rightarrow$

$\sin x = 2$ .

0%
True
0%
False

Explanation

$\mathrm P$	$\mathrm Q$	$\mathrm P \rightarrow Q$
True	True	True
True	False	False
False	True	True
False	False	True

$0>1$ is False

$\sin x=2$ is False

(since,

$\sin x$ range is

$[-1,1]$ )

From the above truth table,

$\mathrm {False} \rightarrow \mathrm {False}$ is

$\mathrm {True}$

Which equation has the solution shown on the number line?

0%
$-3 > x > -1$
0%
$-5 < x < -2$
0%
$-2 > x > 0$
0%
$-1 > x > -6$

Explanation

Since shaded region on the graph is from

$-5$ to

$-2$

So solution is

$-5<x<-2$

Hence, option B is correct.

Which of the following is the correct graph for

$x\ge 3$ or

$x\le -2$ ?

0%
Line A
0%
Line B
0%
Line C
0%
Line D
0%
Line E

Explanation

It is stated

$x \ge 3$ that means including

$3$ and till infinity (positive)

$= \{3, 4, 5, 6, ....\}$

And, it also stated

$x \le -2$ that means including

$-2$ and till infinity (negative)

$= \{3, 4, 5, 6, ... \}$

So, the line D perfectly defines the two stated i.e.

$x \ge 3$ and

$x \le -2$

$\therefore$ Option D is correct.

2106130_537737_ans_0c2aad2c61b04aab87ab1316046e7ef4.png

The point

$(2,1)$ lies in the _____ region of the solution of linear inequations

$x-2y \leq 0,$ and

$x-y>0$

0%
feasible
0%
non-feasible
0%
unbounded
0%
bounded

Explanation

Given,

$x-2y\leq 0$ and

$x-y>0\implies x>y \implies y<x$

First draw the graph for the equations,

$x-2y= 0$ and

$y=x$

$x-2y= 0$ and

$y=x$ are the lines which pass through the origin as shown in the above fig.

Hence,

$y<x$ includes the below region of the line and

$x-2y\leq 0$ also includes the below region of the line.

Therefore, the blue shaded region is the feasible region which is unbounded.

Given point (2,1). Substituting in the given inequations,

$x-2y\leq 0\implies 2-2*1\leq 0 \implies 0\leq 0$ True

$x-y>0 \implies 2-1>0 \implies 1>0$ True

Therefore, (2,1) belongs to the feasible region which is unbounded.

Which quadrant does the solution lie in?

0%
$I$
0%
$II$
0%
$III$
0%
$IV$

Explanation

Given,

$2x+3y>2$

$x-y<0 \implies y>x$

$y\geq 0$

$x\leq 3$

first, draw the graph for equations

$2x+3y=2$

$x-y=0 \implies x=y$

$y= 0 \implies x-axis$

$x= 3$

$x=y$ is the line which passes through the origin as shown in the above fig

Hence,

$y>x$ includes the above region of the line.

$y= 0 \implies x-axis$ . Hence,

$x\leq 3$ includes the left region of the line.

$x= 3 \implies$ line parallel to x-axis. Hence,

$x\leq 3$ includes the left region of the line.

for

$2x+3y=2$

substituting y=0, we get

$2x=2\implies x=1$

substituting x=0, we get

$3y=2\implies y=\dfrac{2}{3}$

Therefore, line

$2x+3y=2$ passes through (1,0) and (0,2/3) as shown in the above figure

Hence,

$2x+3y>2$ includes the above region of the line.

Therefore, the blue shaded region is the feasible region which is present in I quadrant as shown in the figure.

How many solutions does this set include?

0%
$5$
0%
$10$
0%
$20$
0%
Infinite

Explanation

Given,

$2x+3y>2$

$x-y<0 \implies y>x$

$y\geq 0$

$x\leq 3$

first, draw the graph for equations

$2x+3y=2$

$x-y=0 \implies x=y$

$y= 0 \implies x-axis$

$x= 3$

$x=y$ is the line which passes through the origin as shown in the above fig

Hence,

$y>x$ includes the above region of the line.

$y= 0 \implies x-axis$ . Hence,

$y \geq 0$ includes the region above the x axis.

$x= 3 \implies$ line parallel to x-axis. Hence,

$x\leq 3$ includes the left region of the line.

for

$2x+3y=2$

substituting y=0, we get

$2x=2\implies x=1$

substituting x=0, we get

$3y=2\implies y=\dfrac{2}{3}$

Therefore, line

$2x+3y=2$ passes through (1,0) and (0,2/3) as shown in the above figure

Hence,

$2x+3y>2$ includes the above region of the line.

Therefore, the blue shaded region is the feasible region which is not a closed region or it is not bounded. Hence, there are infinite number of solutions.

Which of the following points lie in the solution set?

0%
$(1,1)$
0%
$(1,2)$
0%
$(2,1)$
0%
$(3,2)$

Explanation

By option verification,

i.e., substituting the options in given linear inequations and verifying

Given,

$2x+3y>2$

$x-y<0$

$y\geq 0$

$x\leq 3$

substituting option A (i.e.,)

$(x,y)=(1,1)$

$2x+3y>2\implies 2\times 1+3\times 1>2 \implies 5>2$ True

$x-y<0 \implies 1-1<0 \implies 0<0$ False

$y\geq 0 \implies 1\geq 0$ True

$x\leq 3 \implies 1\leq 3$ True

substituting option B (i.e.,)

$(x,y)=(1,2)$

$2x+3y>2\implies 2\times 1+3\times 2>2 \implies 8>2$ True

$x-y<0 \implies 1-2<0 \implies -1<0$ True

$y\geq 0 \implies 2\geq 0$ True

$x\leq 3 \implies 1\leq 3$ True

substituting option C (i.e.,)

$(x,y)=(2,1)$

$2x+3y>2\implies 2\times 2+3\times 1>2 \implies 7>2$ True

$x-y<0 \implies 2-1<0 \implies 1<0$ False

$y\geq 0 \implies 1\geq 0$ True

$x\leq 3 \implies 2\leq 3$ True

substituting option D (i.e.,)

$(x,y)=(3,2)$

$2x+3y>2\implies 2\times 3+3\times 2>2 \implies 12>2$ True

$x-y<0 \implies 3-2<0 \implies 1<0$ False

$y\geq 0 \implies 2\geq 0$ True

$x\leq 3 \implies 3\leq 3$ True

Therefore option B satisfies the above linear inequalities.

Consider the linear inequations and solve them graphically:

$3x-y-2 > 0;\,\,\, x+y \leq 4;\,\,\, x >0;\,\,\, y \geq 0$ .

The solution region of these inequations is a convex polygon with _____ sides.

0%
$3$
0%
$4$
0%
$5$
0%
$7$

Explanation

Given,

$3x-y-2>0\implies y<3x-2$

$x+y\leq 4$

$x>0$ and

$y\geq 0$

First, draw the graph for equations

$3x-y-2=0$

$x+y= 4$

$x=0$ and

$y= 0$

$x=0$ is the Y-axis.

Hence,

$x>0$ includes the right side region of the line.

$y=0$ is the X-axis.

Hence,

$y\geq 0$ includes the upper side region of the line.

Similarly, for

$y=3x-2$

Substitute y=0 we get,

$3x=2 \implies x=\dfrac{2}{3}$

Substitute x=0 we get,

$y=-2$

Therefore,

$y=3x-2$ line passes through (2/3,0) and (0,-2).

Hence,

$y<3x-2$ includes the region below the line.

Similarly, for

$x+y= 4$

Substitute y=0 we get,

$x=4$

Substitute x=0 we get,

$y=4$

Therefore,

$x+y= 4$ line passes through (2/3,0) and (0,-2).

Hence,

$x+y\leq 4$ includes the region below the line.

As shown in the above figure, the blue shaded region is the feasible region which is the intersection of all 4 solution sets. Feasible region is the polygon with

$3 \space sides$ as shown in the figure.

Find solution of following inequality also show it graphically.

$x+3<5,x\in R$

Explanation

Given,

$x+3<5\implies x<2$ and

$x\in R$

The real numbers includes all the integers(e.g., -4), fractions(e.g., 3/4) and irrational numbers(e.g.,

$\sqrt{2}$ ).

Option C, represents only integers.

Option B, represents all the numbers

$<2$

CBSE Questions for Class 11 Engineering Maths Linear Inequalities Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Linear Inequalities Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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