MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Engineering Maths Linear Inequalities Quiz 3 - MCQExams.com
CBSE
Class 11 Engineering Maths
Linear Inequalities
Quiz 3
Find solution of following inequality, also show it graphically.
$$x-5\geq-7,x\in R$$
Report Question
0%
0%
0%
0%
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:
$$x+3\leq5,x\in Z$$
Report Question
0%
0%
0%
0%
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)
Report Question
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.
$$x<5,x\in Z$$.
Report Question
0%
0%
0%
0%
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?
Report Question
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:
$$x+3\leq5,x\in N$$
Report Question
0%
0%
0%
0%
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?
Report Question
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:
Report Question
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.
Report Question
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:
$$\dfrac{x+4}{x-3}>0,x\in W$$
Report Question
0%
0%
0%
0%
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:
$$|x+3|<4 ,x\in R$$
Report Question
0%
0%
0%
0%
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
Report Question
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:
$$x-5\geq-7,x\in Z$$
Report Question
0%
0%
0%
0%
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:
$$|x+3|<4 ,x\in Z$$
Report Question
0%
0%
0%
0%
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:
$$-2<x+3<5,x\in N$$
Report Question
0%
0%
0%
0%
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.
Report Question
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, $$7x+3 < 5x+9$$
Report Question
0%
0%
0%
0%
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
Report Question
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$$
2) $$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}}$$
Number of integral solutions satisfy inequality $$\left| x-3 \right| -\left| 2x+5 \right| \ge \left| x+8 \right| $$ is
Report Question
0%
5
0%
6
0%
7
0%
8
Explanation
Case $$(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.$$ The greatest number is:-
Report Question
0%
$$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.$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Engineering Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
