MCQ Questions for Class 11 Engineering Maths Linear Inequalities Quiz 3

Find solution of following inequality, also show it graphically.

$x-5\geq-7,x\in R$

Explanation

Given,

$x-5\geq -7\implies x\geq -2$ 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

$\geq -2$

Find solution of following inequality, also show it graphically:

$x+3\leq5,x\in Z$

Explanation

Given,

$x+3\leq 5 \implies x\leq 5-3 \implies x\leq 2$ and

$x \in Z$

ntegers are the numbers which include all the positive and negative numbers including zero. (i.e.,)

$Z=\{...,-2,-1,0,1,2,...\}$

Therefore,

$\{...,-2,-1,1,2\}$ represents

$x\leq 2$ .

Option C represents the set

$\{...,-2,-1,1,2\}$ which are

$x\leq 2$ and

$x\in Z$ .

Formulate the equations for the above problem.
(

$x$ and

$y$ are the number of units of

$A$ and

$B$ manufactured in a day respectively)

0%
$15x+5y \leq 10;\, 24x + 14y \geq 1000$
0%
$15x+5y \leq 600;\, 24x + 14y \geq 1000$
0%
$5x+15y \leq 600;\, 24x + 14y \geq 1000$
0%
$5x+15y \leq 10;\, 24x + 14y \geq 1000$

Explanation

Given, The profit from the product

$A$ is

$Rs. 24$ and

The profit from the product

$B$ is

$Rs. 14$

The maximum number of units of

$B$ manufactured in a day is

$30$

i.e.,

$y\leq 30$ --------- (1)

Let the number of units of

$A$ manufactured in a day be

$x$

Therefore the profit for a day from the product

$A$ is

$24x$

Let the number of units of

$B$ manufactured in a day be

$y$

Therefore the profit for a day from the product

$B$ is

$14y$

The minimum profit for a day is

$Rs.1000$

Therefore, the total profit from products

$A$ and

$B$ should be more than

$Rs.1000$

i.e.,

$24x+14y\geq 1000$ --------- (2)

Given, time taken to manufacture one product of

$A$ is

$15\space min$ and

time taken to manufacture one product of

$B$ is

$5\space min$

Therefore, time taken to manufacture

$x$ products of

$A$ is

$15x\space min$ and

time taken to manufacture

$y$ products of

$B$ is

$5y\space min$

In a day, a worker works for a maximum of

$10\space hrs = 600\space min$

Therefore, the time taken to manufacture products

$A$ and

$B$ should be less than

$600\space min$

I.e.,

$15x+5y\leq 600$ --------- (3)

Find solution of following inequality also show it graphically.

$x<5,x\in Z$ .

Explanation

given

$x<5$ and

$x\in Z$

Integers are the numbers which include all the positive and negative numbers including zero. (i.e.,)

$Z=\{...,-2,-1,0,1,2,...\}$

Therefore,

$\{...,-2,-1,1,2,3,4\}$ represents

$x<5$

Option A includes all the decimal values. Hence it is false.

Option B represents the set

$\{...,-2,-1,1,2,3\}$ which are

$x<5$ and

$x\in Z$ .

Consider the linear inequations and solve them graphically:

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

Which of the following are corner points of the convex polygon region of the solution?

0%
$(0,0)$
0%
$(2,3)$
0%
$(0,4)$
0%
$\left(\dfrac32, \dfrac52\right)$

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.

solving

$3x-y-2=0$ and

$x+y= 4$ we get the intersecting point.

adding the equations

$\implies 3x+x=2+4\implies 4x=6 \implies x=\dfrac{3}{2}$

substituting

$x=\dfrac{3}{2} \implies y=4-\dfrac{3}{2} \implies y=\dfrac{5}{2}$

As shown in the above figure, the blue shaded region is the feasible region with the three corner points

$(\dfrac{3}{2},\dfrac{5}{2}), (\dfrac{2}{3},0), (4,0)$

Find solution of following inequality, also show it graphically:

$x+3\leq5,x\in N$

Explanation

Given,

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

$x \in N$

Natural numbers are counting numbers whose set is

$N=\{1,2,3,...\}$

Therefore,

$\{1,2\}$ represents

$x<5$

Option D graph has

$\{1,2\}$ solution set.

Consider the linear inequations and solve them graphically:

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

Which of the following points belong to the feasible solution region?

0%
$\left(\dfrac12,0\right)$
0%
$\left(\dfrac12,\dfrac52\right)$
0%
$\left(\dfrac32, \dfrac52\right)$
0%
None of the above

Explanation

By option verification,

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

Given,

$3x-y-2>0$

$x+y\leq 4$

$x>0$ and

$y\geq 0$

Substituting option A i.e.,

$(x,y)=(\dfrac{1}{2},0)$

$3*\dfrac{1}{2}-0-2>0\implies -\dfrac{1}{2}>0$ False

Substituting option B i.e.,

$(x,y)=(\dfrac{1}{2},\dfrac{5}{2})$

$3*\dfrac{1}{2}-\dfrac{5}{2}-2>0\implies -3>0$ False

Substituting option C i.e.,

$(x,y)=(\dfrac{3}{2},\dfrac{5}{2})$

$3*\dfrac{3}{2}-\dfrac{5}{2}-2>0\implies 0>0$ False

No option satisfies the given linear inequations, hence option D is correct.

The quantity of

$A$ and

$B$ in one day for which profit will be maximum is:

0%
$25,30$
0%
$30,25$
0%
$25,25$
0%
$30,30$

Explanation

Given, The profit from the product

$A$ is

$Rs. 24$ and

The profit from the product

$B$ is

$Rs. 14$

The maximum number of units of

$B$ manufactured in a day is

$30$

i.e.,

$y\leq 30$ --------- (1)

Let the number of units of

$A$ manufactured in a day be

$x$

Therefore the profit for a day from the product

$A$ is

$24x$

Let the number of units of

$B$ manufactured in a day be

$y$

Therefore the profit for a day from the product

$B$ is

$14y$

The minimum profit for a day is

$Rs.1000$

Therefore, the total profit from products

$A$ and

$B$ should be more than

$Rs.1000$

i.e.,

$24x+14y\geq 1000$ --------- (2)

Given, time taken to manufacture one product of

$A$ is

$15\space min$ and

time taken to manufacture one product of

$B$ is

$5\space min$

Therefore, time taken to manufacture

$x$ products of

$A$ is

$15x\space min$ and

time taken to manufacture

$y$ products of

$B$ is

$5y\space min$

In a day, a worker works for a maximum of

$10\space hrs = 600\space min$

Therefore, the time taken to manufacture products

$A$ and

$B$ should be less than

$600\space min$

I.e.,

$15x+5y\leq 600$ --------- (3)

Now, substituting the options and verifying which will be the maximum and satisfies the 3 constraints.

substituting option A i.e.,

$(x,y)=(25,30)$

$15x+5y\leq 600\implies 15*25+5*30 = 525\leq 600$ True

$24x+14y = 24*25+14*30 = 1020$

substituting option B i.e.,

$(x,y)=(30,25)$

$24x+14y = 24*30+14*25 = 1070$

$15x+5y\leq 600\implies 15*30+5*25 = 575\leq 600$ True

substituting option C i.e.,

$(x,y)=(25,25)$

$24x+14y = 24*25+14*25 = 950$

$15x+5y\leq 600\implies 15*25+5*25 = 500\leq 600$ True

substituting option D i.e.,

$(x,y)=(30,30)$

$24x+14y = 24*30+14*30 = 1140$

$15x+5y\leq 600\implies 15*30+5*30 = 600\leq 600$ True

Therefore,

$(30,30)$ will produce the maximum profit.

A manufacturer produces two products

$A$ and

$B$ . Product

$A$ fetches him a profit of Rs.

$24$ and product

$B$ fetches him a profit of Rs.

$14$ . It takes

$15$ minutes to manufacture one unit of product

$A$ and

$5$ minutes to manufacture one unit of product

$B$ . There is a limit of

$30$ units to the quantity of

$B$ of manufactured due to a constraint on the availability of materials. The minimum profit that the company decides to make is Rs.

$1000$ . The workers work for a maximum of

$10$ hours a day. Find the maximum daily profit.

0%
Rs. $1020$
0%
Rs. $1040$
0%
Rs. $1140$
0%
Rs. $1240$

Explanation

Given, The profit from the product

$A$ is

$Rs. 24$ and

The profit from the product

$B$ is

$Rs. 14$

The maximum number of units of

$B$ manufactured in a day is

$30$

i.e.,

$y\leq 30$ --------- (1)

Let the number of units of

$A$ manufactured in a day be

$x$

Therefore the profit for a day from the product

$A$ is

$24x$

Let the number of units of

$B$ manufactured in a day be

$y$

Therefore the profit for a day from the product

$B$ is

$14y$

The minimum profit for a day is

$Rs.1000$

Therefore, the total profit from products

$A$ and

$B$ should be more than

$Rs.1000$

i.e.,

$24x+14y\geq 1000$ --------- (2)

Given, time taken to manufacture one product of

$A$ is

$15\space min$ and

time taken to manufacture one product of

$B$ is

$5\space min$

Therefore, time taken to manufacture

$x$ products of

$A$ is

$15x\space min$ and

time taken to manufacture

$y$ products of

$B$ is

$5y\space min$

In a day, a worker works for a maximum of

$10\space hrs = 600\space min$

Therefore, the time taken to manufacture products

$A$ and

$B$ should be less than

$600\space min$

I.e.,

$15x+5y\leq 600$ --------- (3)

The total profit from products

$A$ and

$B$ is

$P=24x+14y$

In the above figure, the blue shaded region is the feasible region with three corner points.

$(\dfrac{290}{12},30), (30,30), (\dfrac{340}{9},\dfrac{20}{3})$

$(\dfrac{290}{12},30)$ is the point where

$24x+14y= 1000$ intersects

$y=30$

I.e., substituting

$y=30 \implies 24x+14*30 =1000 \implies x=\dfrac{1000-420}{24}\implies x=\dfrac{290}{12}$

$(30,30)$ is the point where

$15x+5y= 600$ intersects

$y=30$

I.e., substituting

$y=30 \implies 15x+5*30 =600 \implies x=\dfrac{600-150}{15}\implies x=30$

$(\dfrac{340}{9},\dfrac{20}{3})$ is the point where

$24x+14y= 1000$ intersects

$15x+5y= 600$

I.e., solving the two equations, we get

$x=\dfrac{340}{9}$ and

$y=\dfrac{20}{3}$

Now substituting the corner points the profit equation,

substituting

$(\dfrac{290}{12},30) \implies P=24*\dfrac{290}{12}+14*30=1000$

substituting

$(30,30) \implies P=24*30+14*30=1140$

substituting

$(\dfrac{340}{9},\dfrac{20}{3}) \implies P=24*\dfrac{340}{9}+14*\dfrac{20}{3}=1000$

$1140$ is the maximum profit

Solve the following inequality and show it graphically:

$\dfrac{x+4}{x-3}>0,x\in W$

Explanation

Given,

$\dfrac{x+4}{x-3}>0$ and

$x\in W$

whole numbers are positive numbers including zero whose set is

$W=\{0,1,2,3,...\}$

$\implies x\geq 0$

$(x+4)*\dfrac{1}{x-3}>0$

$\implies x+4>0$ and

$\dfrac{1}{x-3}>0$

$\implies x>-4$ and

${x-3}>0$

$\implies x>-4$ and

$x>3$ and

$x\geq 0$

$x>-4$ and

$x\geq 0 \implies x\geq 0$

Therefore,

$\{4,5,6,...\}$ represents the

$x>3$ and

$x\geq 0$

Solve the following inequality and show it graphically:

$|x+3|<4 ,x\in R$

Explanation

Given that

$|x+3|<4$

Which implies

$x+3 < 4$ and

$x+3>-4$

$\Rightarrow x<1$ and

$x>-7$

$\Rightarrow -7<x<1$

Since,

$x\in R$

$x$ can take values

$(-7,1)$ , not including boundary points.

Hence, D is correct.

If the product of

$n$ positive numbers is

$1$ , then their sum is

0%
A positive integer
0%
Divisible by $n$
0%
Equal to $n+\dfrac 1n$
0%
Greater than or equal to $n$

Explanation

Since,

$A.M. \geq G.M.$

let n positive numbers be

$x_1, x_2,...,x_n$

$A.M. =\dfrac{x_1+x_2+...+x_n}{n}$

$G.M. = x_1*x_2*...*x_n$

Therefore,

$\dfrac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1*x_2*...*x_n}$

$\implies x_1+x_2+...+x_n \geq n*\sqrt[n]{x_1*x_2*...*x_n}$

$\implies x_1+x_2+...+x_n \geq n$

Find solution of following inequality, also show it graphically:

$x-5\geq-7,x\in Z$

Explanation

Given,

$x-5\geq -7\implies x\geq -2$ 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 A represents all the numbers

$\geq -2$

Solve the following inequality and show it graphically:

$|x+3|<4 ,x\in Z$

Explanation

Given that

$|x+3|<4$

Which implies

$x+3 < 4$ and

$x+3>-4$

$\Rightarrow x<1$ and

$x>-7$

$\Rightarrow -7<x<1$

Since

$x$ is natural number

$x$ can take values from

$-6$ to

$0$ .

Hence, option

$C$ is correct.

Solve the following inequality and show it graphically:

$-2<x+3<5,x\in N$

Explanation

Given,

$-2<x+3<5$ and

$x \in N$

Natural numbers are counting numbers whose set is

$N=\{1,2,3,...\}$

$\implies x>0$

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

$x+3<5$

$\implies x>-2-3$ and

$x<5-3$

$\implies x>-5$ and

$x<2$

$x>0$ and

$x>-5\implies x>0$

Therefore,

$\{1\}$ represents

$x>0$ and

$x<2$

The shaded region in the figure is the solution set of the inequations.

0%
$5x + 4y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 2$
0%
$5x + 4y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 2$
0%
$5x + 4y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$
0%
$5x + 4y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 2$

Explanation

1. Since, region(shaded) lies above x-axis

$\therefore y\ge 0$
2. Shaded region is on left of

$x=6$

$\therefore x\le 6$
3. Shaded region is below

$y=3$

$\therefore y\le 3$
4. Equation of line passing through (4,0) & (0,5) is

$y-0=\cfrac { 5 }{ -4 } \left( x-4 \right)$

$5x+4y-20=0$

$5x+4y=20$
If

$5x+4y\ge 20$ , Put (0,0) we get

$0\ge 20$ FALSE.

$\therefore$ Region lies away from origin which is true

$\therefore \quad 5x+4y\ge 20$
5. Shaded region lies on right side of y-axis

$\therefore x\ge 0$

$\therefore 5x+4y\ge 20,x\le 6,y\le 3,x\ge 0,y\ge 0$

Solve inequality and show the graph of the solution,

$7x+3 < 5x+9$

Explanation

$7x + 3 < 5x + 9$

$\Rightarrow 7x - 5x < 9 - 3$

$\Rightarrow 2x < 6$

$\Rightarrow x < 3$

Area of the region {

$(x,y):{x}^{2}+{y}^{2}\le 1\le x+y$ } is

0%
$\dfrac{\pi}{4}+\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}-\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}+\dfrac{3}{4}$
0%
$\pi+1$

Explanation

To find area

$x^2+y^2\leqslant |\leqslant x+y$

1) Circle

$x^2+y^2\leqslant |$

$(x-0)^2+(y-0)^2\leqslant |$\Rightarrow $ centere$ (0,0)

$, Radius$ 1$$

$x+y\geqslant |$

Required area is

$A=\int_{1}^{0}\sqrt{1-x^2}-(1-x)$

$A\Rightarrow ^1_0\left[\dfrac{\sin ^{-1}(x)}{2}+\dfrac{x}{2}\sqrt{1-x^2}\right]^1_0\left[x-\dfrac{x^2}{2}\right]$

$\therefore A\Rightarrow \left[\dfrac{\pi}{4}+\dfrac{1}{2}\times 0-0-0\right]-\left[1-0-\dfrac{1}{2}+0\right]$

$\boxed{\therefore A\Rightarrow \dfrac{\pi}{4}-\dfrac{1}{2}}$

Number of integral solutions satisfy inequality

0%
5
0%
6
0%
7
0%
8

Explanation

Case

$(1)\quad x\le -8\Rightarrow$

$-x+3+2x+5 \ge -x-8$

$2x \ge -8$

$x\ge -8\quad \therefore \boxed {x=-8}$

Case

$(2)\quad x\in [-8, -5/2]$

$-x+3+2x+5\ \ge x+8$

$8\ge 8$ true

$\therefore \boxed {x\in [-8, -5/2]}$

Case

$(3)\quad x\in [-5/2, 3]$

$-x+3-2x-5\ \ge x+8$

$-10\ge 4x$

$-5/2 \ge x$

$\therefore \boxed {x =-5/2}$

Case

$(4)\quad x\ge 3$

$x-3-2x-5\ge x+8$

$-16\ge 2x$

$-8\ge x$

$\boxed {No\ solution}$

$\therefore$

Finally

$x\in [-8, -5/2]$

$x\ \in \boxed {-8, -2.5}$

$\therefore$

the integral solution of

$x$ are

$x=-8, -7, -6, -5, -4, -3$

$\therefore$

$B$ Answer.

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:-

0%
$8$
0%
$7$
0%
$14$
0%
$4$

Explanation

Let the four terms be

$\;a-3d\;,a-d\;,a+d\;,and\;a+3d$ with common difference

$2d$ .

$sum=4a=20$

$a=5$

$\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{2}{3}$

$\dfrac{25-9d^2}{25-d^2}=\dfrac{2}{3}$

$d^2=1$

$\Rightarrow d=1$

$Largest \;term=a+3d$

$=5+3(1)$

$=8.$

CBSE Questions for Class 11 Engineering Maths Linear Inequalities Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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